The slope as a ratio of derivatives against the parameter, second derivatives and concavity, arc length as the integral of a speed, area under a parametric curve, and where the formulas fail.
Subject: Calculus II · 67 slides · symbolic lesson
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Title
Calculus II · Section 7.2
Slopes, areas, lengths and surfaces, all computed through the parameter
Objectives
Section 7.1 described curves by giving both coordinates as functions of a parameter. This lesson does calculus on them without ever solving for y as a function of x.
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, pp. 625-639 — learning objectives 7.2.1 to 7.2.4
In Section 7.1 you learned to describe a curve by saying where a moving point is at each instant: an x coordinate and a y coordinate, both functions of a parameter t. That description handles circles, loops and cusped curves that no single function of x can describe.
This lesson asks the calculus questions about those curves. How steep is the curve at a given instant? Which way does it bend? How much area sits under it, how long is it, and how much surface does it sweep out when you spin it about an axis?
The pleasant surprise is that you never have to eliminate the parameter. Every answer is built from just two derivatives, x prime of t and y prime of t, combined in a different way for each question. The last part of the lesson shows you where each combination can mislead you.
Warm-up
Discussion prompt
Two things from Calculus I. First: if y is a function F of x, and x itself depends on t, what is dy/dt? Second: write the arc length formula for the graph of f from x = a to x = b.
Write both answers before you reveal them. The first is the chain rule in its most basic form: if y depends on x and x depends on t, then the rate of y against t is the rate of y against x times the rate of x against t.
Hold onto that line, because the next few slides simply solve it for the slope. That is the entire proof of Theorem 7.1.
The arc length formula is the one you met in Chapter 2 of this volume, for the graph of a function. By the end of Part 4 you will have a more general formula and you will see this one drop out of it as a special case, which is a good way to check that the new formula is right.
Section
Part 1
Concept
Figure (svg): The line segment from (−1, −10) to (9, 5) traced by x equals 2t plus 3, y equals 3t minus 4. Dots mark t equal to −2, −1, 0, 1, 2 and 3. Between t equal to 0 and t equal to 1 a right triangle shows a horizontal step of 2 and a vertical step of 3.
\[ x(t) = 2t + 3, \quad y(t) = 3t - 4, \quad -2 \le t \le 3 \]
Eliminate the parameter and the slope can be read off; differentiate both coordinates instead and the same number appears as a ratio.
\[ t = \frac{x-3}{2} \;\Longrightarrow\; y = \frac{3x}{2} - \frac{17}{2}, \quad \frac{dy}{dx} = \frac32 \]
\[ x'(t) = 2, \quad y'(t) = 3, \quad \frac{y'(t)}{x'(t)} = \frac32 \]
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, pp. 625-626 — Figure 7.16
Start with the simplest possible parametric curve, a straight line, because you can check every answer by eye. The dots mark whole-number values of t, and between each pair of neighbouring dots the point moves two units right and three units up.
Eliminating the parameter the usual way gives a line with slope three halves. Now compare with the two derivatives: x prime is 2 and y prime is 3, and their ratio is also three halves. The little triangle on the figure is the reason. Per unit of t, the run is 2 and the rise is 3, and slope is rise over run however you measure the step.
The book says this is no coincidence, and the next slide proves it for any curve, not just lines.
Concept
Suppose the parameter could be eliminated, so that y is some differentiable function F of x along the curve. Then y at time t is F evaluated at x at time t.
\[ y(t) = F\left(x(t)\right) \]
\[ y'(t) = F'\left(x(t)\right)\,x'(t) \]
\[ F'\left(x(t)\right) = \frac{y'(t)}{x'(t)} \quad \text{when } x'(t) \ne 0 \]
Derivative of parametric equations (Theorem 7.1) — If x prime and y prime exist and x prime of t is not zero, the slope of the curve at the point for parameter t is y prime of t divided by x prime of t.
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{y'(t)}{x'(t)} \]
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, p. 626 — Theorem 7.1 and its proof
The proof borrows an assumption for a moment: pretend that near the point of interest the curve can be written as the graph of some function F. Then the height at time t is F of the x coordinate at time t, and the chain rule differentiates that composition.
Solving for F prime gives the slope as y prime over x prime. The slope of the graph is exactly the thing called dy by dx, so that is the theorem.
Notice where the hypothesis comes from: you cannot divide by x prime when it is zero. The beauty of the final formula is that it no longer mentions F at all. You compute it straight from the parametrisation, even for curves like a circle that are not the graph of any single function.
Notation
Annotate
On: \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{y'(t)}{x'(t)}, \quad x'(t) \ne 0 \)
Step through the annotations one at a time and keep the moving point in mind. The two derivatives against t are the two components of its velocity: how fast it moves sideways and how fast it moves up.
The slope is not a speed. It is the direction of motion: rise per unit run. Dividing the vertical velocity by the horizontal velocity cancels the time, which is why the answer does not depend on how fast the curve is traced.
The last annotation deserves the most attention. When the horizontal velocity is zero, the point is moving straight up or down, or it has stopped completely. Those are the places where the formula breaks, and Part 6 of the lesson is devoted to them.
Worked example
Find the slope and locate any critical points.
\[ x(t) = t^2 - 3, \quad y(t) = 2t - 1, \quad -3 \le t \le 4 \]
Differentiate both coordinates
Why: Power rule on each.
\[ x'(t) = 2t, \quad y'(t) = 2 \]
Divide
Why: Theorem 7.1: the vertical rate over the horizontal rate.
\[ \frac{dy}{dx} = \frac{2}{2t} = \frac1t \]
Find where the slope fails to exist
Why: The denominator vanishes at one parameter value.
\[ 2t = 0 \;\Longrightarrow\; t = 0 \]
Locate the point
Why: Substitute t equal to 0 into both coordinates.
\[ x(0) = -3, \quad y(0) = -1 \;\Longrightarrow\; (-3,-1) \]
Figure (svg): The parabola x equals t squared minus 3, y equals 2t minus 1 for t from −3 to 4, opening to the right, from (6, −7) up through the vertex (−3, −1) to (13, 7). A dashed vertical line touches the vertex; small arrows show the direction of increasing t.
Check by eliminating the parameter
Why: Solve the y equation for t and treat x as a function of y; its derivative is t, the reciprocal of the slope, and it is zero at the vertex, so the tangent there is vertical.
\[ x = \left(\tfrac{y+1}{2}\right)^2 - 3, \quad \frac{dx}{dy} = \frac{y+1}{2} = t \]
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, p. 627 — Example 7.4a and Figure 7.17
The recipe is always the same three moves: differentiate both coordinates, divide, and look at where the denominator vanishes. Here the slope comes out as one over t, a function of the parameter rather than of x, and that is perfectly normal.
One over t is undefined at t equal to 0, and substituting that value gives the point minus 3, minus 1. The figure shows why: that is the vertex of a parabola opening to the right, where the curve turns around and its tangent line stands vertical.
The check solves the y equation for t and writes x as a function of y. Its derivative is t itself, the reciprocal of the slope, which confirms the formula, and it is zero at the vertex. A tangent with zero slope in the sideways picture is a vertical tangent in the usual picture.
Worked example
\[ x(t) = 2t + 1, \quad y(t) = t^3 - 3t + 4, \quad -2 \le t \le 5 \]
Differentiate both coordinates
Why: The first is linear; the second needs the power rule.
\[ x'(t) = 2, \quad y'(t) = 3t^2 - 3 \]
Divide
Why: The denominator is never zero, so the slope exists everywhere.
\[ \frac{dy}{dx} = \frac{3t^2 - 3}{2} \]
Set the slope to zero
Why: A fraction is zero where its numerator is.
\[ 3t^2 - 3 = 0 \;\Longrightarrow\; t = \pm 1 \]
Locate the two points
Why: Substitute each parameter value.
\[ t = -1: (-1,\,6), \qquad t = 1: (3,\,2) \]
Figure (svg): The curve x equals 2t plus 1, y equals t cubed minus 3t plus 4 for t from −2 to about 2.2. It rises to a relative maximum at (−1, 6), where t is −1, falls to a relative minimum at (3, 2), where t is 1, then rises again. Short horizontal segments mark both turning points.
Check by eliminating the parameter
Why: With t equal to x minus 1 over 2, y becomes a cubic in x; its derivative at x equal to 3, where t is 1, is zero.
\[ \frac{dy}{dx} = \frac32\left(\frac{x-1}{2}\right)^2 - \frac32 = \frac32 - \frac32 = 0 \text{ at } x = 3 \]
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, pp. 627-628 — Example 7.4b and Figure 7.18
Here the horizontal velocity is the constant 2, so the curve is always moving right and the slope exists everywhere. This curve is secretly the graph of a function, since x never turns back.
Critical points now come from the numerator. Three t squared minus 3 is zero at t equal to plus or minus 1, and substituting those values gives the relative maximum at minus 1, 6 and the relative minimum at 3, 2. Be careful to substitute into both coordinate functions: the parameter value is not the x coordinate.
The check writes y as a cubic in x and differentiates at x equal to 3, which is t equal to 1. The slope comes out zero, so both methods agree on the flat spot.
Worked example
\[ x(t) = 5\cos t, \quad y(t) = 5\sin t, \quad 0 \le t \le 2\pi \]
Differentiate both coordinates
Why: Derivatives of cosine and sine.
\[ x'(t) = -5\sin t, \quad y'(t) = 5\cos t \]
Divide and simplify
Why: The fives cancel.
\[ \frac{dy}{dx} = \frac{5\cos t}{-5\sin t} = -\cot t \]
Where the slope is zero
Why: Cosine vanishes at a quarter and three quarters of a turn.
\[ \cos t = 0: \; t = \tfrac{\pi}{2}, \tfrac{3\pi}{2} \;\Longrightarrow\; (0, 5), (0,-5) \]
Where the slope is undefined
Why: Sine vanishes at the start, half way, and the end.
\[ \sin t = 0: \; t = 0, \pi, 2\pi \;\Longrightarrow\; (5, 0), (-5, 0) \]
Figure (svg): The circle x equals 5 cos t, y equals 5 sin t. Horizontal tangent segments touch the top (0, 5) and bottom (0, −5), labelled slope 0; vertical tangent segments touch the sides (5, 0) and (−5, 0), labelled vertical.
Check with implicit differentiation
Why: The circle is x squared plus y squared equals 25; differentiating implicitly gives minus x over y, the same function once x and y are written in t.
\[ \frac{dy}{dx} = -\frac{x}{y} = -\frac{5\cos t}{5\sin t} = -\cot t \]
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, pp. 628-629 — Example 7.4c and Figure 7.19
A circle is the classic curve that fails the vertical line test, so the parametric slope formula really earns its keep here. The fives cancel, and the slope is minus cot t.
Read the two kinds of special point off the two factors. Cosine vanishes at a quarter turn and three quarters of a turn, which are the top and bottom of the circle, where the tangent is horizontal. Sine vanishes at the start, half way round and the end, which are the two sides, where the tangent is vertical.
The check uses implicit differentiation from Calculus I. Differentiating x squared plus y squared equals 25 gives minus x over y, and writing x and y in terms of t turns that into minus cot t. Two completely different methods, one answer.
Worked example
\[ x(t) = t^2 - 4t, \quad y(t) = 2t^3 - 6t, \quad -2 \le t \le 3 \]
Differentiate both coordinates
Why: Power rule on each.
\[ x'(t) = 2t - 4, \quad y'(t) = 6t^2 - 6 \]
Divide and factor
Why: Take 6 out of the top and 2 out of the bottom.
\[ \frac{dy}{dx} = \frac{6t^2 - 6}{2t - 4} = \frac{3(t^2 - 1)}{t - 2} \]
Zero slope: the numerator vanishes
Why: Two parameter values, two points.
\[ t = -1: (5, 4), \qquad t = 1: (-3, -4) \]
Undefined slope: the denominator vanishes
Why: One parameter value.
\[ t = 2: \; (4 - 8,\; 16 - 12) = (-4, 4) \]
Figure (svg): The curve x equals t squared minus 4t, y equals 2t cubed minus 6t for t from −2 to 3. It starts at (12, −4), rises to a horizontal tangent at (5, 4), swings left and down to a horizontal tangent at (−3, −4), turns at a vertical tangent at (−4, 4), and climbs steeply to (−3, 36).
Check that t equal to 2 is a genuine vertical tangent
Why: The numerator is not also zero there, so the curve does not stall; it moves straight up.
\[ x'(2) = 0, \quad y'(2) = 6(4) - 6 = 18 \ne 0 \]
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, p. 629 — Checkpoint 7.4
Try this one yourself before stepping through it. The factoring in the third line is worth doing because it makes both the zeros and the undefined point easy to see.
The numerator vanishes at t equal to plus and minus 1, giving horizontal tangents at 5, 4 and at minus 3, minus 4. The denominator vanishes at t equal to 2, giving the point minus 4, 4.
The final check is a habit to build. A zero denominator only means a vertical tangent if the numerator is not also zero. Here y prime is 18 at t equal to 2, so the point is moving straight up at that instant, and the figure shows the curve turning at the far left and climbing steeply.
Worked example
\[ x(t) = t^2 - 3, \quad y(t) = 2t - 1 \quad \text{at } t = 2 \]
Find the slope at t equal to 2
Why: Example 7.4(a) gave one over t.
\[ \left.\frac{dy}{dx}\right|_{t=2} = \frac12 \]
Find the point
Why: Substitute t equal to 2.
\[ x(2) = 4 - 3 = 1, \quad y(2) = 4 - 1 = 3 \]
Write the point-slope form
Why: Slope one half through the point (1, 3).
\[ y - 3 = \tfrac12(x - 1) \]
Solve for y
Why: Distribute, then add 3.
\[ y = \tfrac12 x - \tfrac12 + 3 = \tfrac12 x + \tfrac52 \]
Figure (svg): The parabola x equals t squared minus 3, y equals 2t minus 1, with the point (1, 3) marked at t equal to 2 and the tangent line y equals one half x plus five halves drawn through it, touching the upper branch.
Check that the line only touches the curve
Why: Substitute the parametric point into the line; the resulting quadratic in t has a double root at 2, which is what tangency means.
\[ 2t - 1 = \tfrac12(t^2 - 3) + \tfrac52 \iff (t - 2)^2 = 0 \]
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, p. 630 — Example 7.5 and Figure 7.20
A tangent line needs two ingredients, a slope and a point, and both come from the same parameter value. The slope is one over t evaluated at 2, which is one half. The point comes from substituting 2 into both coordinates, giving 1, 3.
From there it is Calculus I: point-slope form, then solve for y. The line is one half x plus five halves.
The check is worth understanding. Substituting the curve into the line gives a quadratic in t, and it turns out to be t minus 2, squared. A double root means the line meets the curve at t equal to 2 and touches rather than crosses there, which is exactly what tangent means.
Anomaly
\[ x(t) = t^2 - 4t, \quad y(t) = 2t^3 - 6t, \quad -2 \le t \le 3, \quad \text{tangent at } t = 5? \]
Predict first
The book asks for the tangent line at t = 5, but the curve was only defined for t from −2 to 3. What is the honest answer?
Correct: y = 24x + 100, on the curve extended past t = 3
Why: The formulas work for every t, so extend the curve: the slope is 3(25 − 1)/(5 − 2) = 24 and the point is (5, 220), giving y − 220 = 24(x − 5), or y = 24x + 100. Strictly, t = 5 lies outside the stated interval, a slip in the book; say so, then answer for the extended curve.
\[ \frac{dy}{dx}\Big|_{t=5} = \frac{3(25-1)}{5-2} = 24, \quad (x,y) = (25-20,\; 250-30) = (5, 220) \]
\[ y - 220 = 24(x - 5) \;\Longrightarrow\; y = 24x + 100 \]
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, p. 631 — Checkpoint 7.5
Commit to an answer before revealing. This checkpoint contains a small slip in the book, and noticing it is part of reading mathematics carefully: the curve was defined only for t from minus 2 to 3, and the question asks about t equal to 5.
The formulas themselves work for any t, so the sensible response is to say that you are extending the curve, then compute. The slope is 24 and the point is 5, 220, which gives y equals 24 x plus 100.
If you picked the answer with minus 100, check the arithmetic in point-slope form: 220 minus 24 times 5 is plus 100. If you picked the reciprocal slope, you divided the derivatives upside down, which is the trap two slides from now.
Worked example
Find every horizontal and vertical tangent, and the slope at the origin.
\[ x = \sin 2t, \quad y = 2\sin t, \quad 0 \le t < 2\pi \]
Differentiate and divide
Why: Chain rule on sin 2t; the twos cancel.
\[ \frac{dy}{dx} = \frac{2\cos t}{2\cos 2t} = \frac{\cos t}{\cos 2t} \]
Horizontal tangents: cos t equals zero
Why: Cos 2t is minus 1 there, not zero.
\[ t = \tfrac{\pi}{2}, \tfrac{3\pi}{2} \;\Longrightarrow\; (0, 2), \; (0, -2) \]
Vertical tangents: cos 2t equals zero
Why: Four parameter values; cos t is not zero at any of them.
\[ t = \tfrac{\pi}{4}, \tfrac{3\pi}{4}, \tfrac{5\pi}{4}, \tfrac{7\pi}{4} \;\Longrightarrow\; (\pm 1, \pm\sqrt2) \]
The origin is visited twice
Why: At t equal to 0 and at t equal to pi, with different slopes.
\[ t = 0: \frac{1}{1} = 1, \qquad t = \pi: \frac{-1}{1} = -1 \]
Figure (svg): The closed curve x equals sin 2t, y equals 2 sin t, shaped like a figure eight standing upright and crossing itself at the origin. Two dashed tangent lines of slopes 1 and −1 cross at the origin. Horizontal tangents touch the top (0, 2) and bottom (0, −2); vertical tangents touch the four points with x equal to plus or minus 1.
Check one vertical tangent
Why: At t equal to a quarter of pi the point is (1, root 2), the sideways speed is zero and the upward speed is not.
\[ x'\left(\tfrac{\pi}{4}\right) = 2\cos\tfrac{\pi}{2} = 0, \quad y'\left(\tfrac{\pi}{4}\right) = \sqrt2 \ne 0 \]
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, p. 640 — Exercises 78 and 79
This curve crosses itself, so it is a good test of everything so far. The slope simplifies to cos t over cos 2t. Horizontal tangents come from the numerator and vertical tangents from the denominator, and in each case you should confirm that the other factor is not also zero.
The figure shows the six special points: the top and bottom where the curve is flat, and four points at x equal to plus or minus 1 where it stands vertical.
The most interesting point is the origin. The curve passes through it at t equal to 0 and again at t equal to pi, heading in different directions, so it has two tangent lines there with slopes 1 and minus 1. That is why a parametric slope is always asked for at a value of t, not at a point: a point can be visited twice.
Trap
Example 7.4(a), done from memory:
\[ \frac{dy}{dx} = \frac{x'(t)}{y'(t)} = \frac{2t}{2} = t \]
Wrong. This is the slope of x against y.
Rise over run: the change in y goes on top. Test your memory on the line from the start of the lesson, which moves 2 across and 3 up per unit of t, so its slope must be three halves, not two thirds.
\[ \frac{dy}{dx} = \frac{y'(t)}{x'(t)} = \frac{3}{2} \]
Under time pressure the two derivatives get swapped, and the answer is still a perfectly plausible-looking function of t. Nothing in the calculation warns you.
The defence is a test case you trust. The line from the start of the lesson moves 2 across and 3 up per unit of t, so its slope must be three halves. If your remembered formula gives two thirds, it is upside down. Rise always goes on top, and the rise is the change in y.
Fill the middle
\[ x'(t_0) = 0, \quad y'(t_0) \ne 0 \]
Fill in the blanks
At a parameter value where the point stops moving sideways but keeps moving up or down, the tangent line is vertical; where the vertical rate alone is zero, the tangent is horizontal.
Why: A zero horizontal rate with a nonzero vertical rate means the point moves straight up or down at that instant, so the tangent is vertical and the slope is undefined. A zero vertical rate with a nonzero horizontal rate means it moves straight sideways: slope zero, a horizontal tangent.
Fill in both blanks before checking. The question is really about the picture of a moving point: which way is it moving at that instant?
If the sideways speed is zero but the vertical speed is not, the point is moving straight up or down, so its path is vertical there. If the vertical speed is zero but the sideways speed is not, it is moving straight across, so the tangent is horizontal.
Keep the third case in reserve: both speeds zero at once. Then the point has stopped, and the formula cannot say anything. You will meet that case properly in Part 6.
Section
Part 2
Concept
The second derivative is the derivative of the slope with respect to x. The slope is already known, but as a function of t, so call it m of t.
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}\big[m(t)\big] \]
Theorem 7.1 never needed y to be a coordinate: it turns the x-derivative of any function of t into a ratio. Apply it with m in place of y.
\[ \frac{d}{dx}\big[m(t)\big] = \frac{m'(t)}{x'(t)} \]
\[ \frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt} \]
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, p. 631 — equation 7.2
The second derivative measures how the slope changes as you move along the curve in the x direction. The difficulty is that you only know the slope as a function of t.
Here is the key observation. Theorem 7.1 was proved for y, but nothing in the proof used anything special about y. It says that to differentiate any quantity with respect to x, you differentiate it with respect to t and divide by x prime. Apply that to the slope itself.
So the rule is: differentiate the slope in t, then divide by x prime again. The denominator is the same one as before, which makes the formula easy to remember once you see where it comes from.
Notation
Annotate
On: \( \frac{d^2y}{dx^2} = \frac{(d/dt)\left(dy/dx\right)}{dx/dt} \)
Look at the two layers in the numerator. The inner layer is the slope you already found; the outer layer differentiates it with respect to t. This is an ordinary Calculus I derivative, often a quotient rule.
The denominator is where marks are lost. Differentiating the slope in t tells you how fast the slope changes per unit of time, but concavity is about change per unit of x. Dividing by x prime converts one into the other.
The last annotation names the tempting error explicitly. The second derivatives of the coordinates do not appear anywhere in the correct formula. You will see in a moment exactly how badly dividing them goes wrong.
Worked example
\[ x(t) = t^2 - 3, \quad y(t) = 2t - 1, \quad \frac{dy}{dx} = \frac1t \]
Differentiate the slope in t
Why: Power rule on t to the minus one.
\[ \frac{d}{dt}\left(\frac1t\right) = -\frac{1}{t^2} \]
Divide by dx/dt
Why: The horizontal rate is 2t.
\[ \frac{d^2y}{dx^2} = \frac{-1/t^2}{2t} \]
Simplify
Why: Multiply the denominators.
\[ \frac{d^2y}{dx^2} = -\frac{1}{2t^3} \]
Read the sign
Why: Negative for t above zero (the upper branch bends down), positive below (the lower branch bends up).
\[ t > 0: \; -\tfrac{1}{2t^3} < 0, \qquad t < 0: \; -\tfrac{1}{2t^3} > 0 \]
Check on the upper branch as a function
Why: For t positive, y is 2 root of x plus 3, minus 1. Differentiate twice and put back t, which is the root of x plus 3.
\[ y = 2\sqrt{x+3} - 1, \quad y'' = -\tfrac12(x+3)^{-3/2} = -\frac{1}{2t^3} \]
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, p. 631 — Example 7.6
The slope from Example 7.4 was one over t, which differentiates easily to minus one over t squared. Then divide by the horizontal rate, 2t, and simplify to minus one over 2 t cubed.
Reading the sign tells you the shape. For positive t, on the upper half of the parabola, the second derivative is negative, so the curve bends downward, like the top of a square root graph. For negative t it is positive, and the lower half bends upward. Look back at the figure from Example 7.4 to see both.
The check makes the upper half into an honest function of x, the square root curve, and differentiates it twice. Writing the result in terms of t gives exactly the same expression.
Worked example
\[ x(t) = t^2 - 4t, \quad y(t) = 2t^3 - 6t, \quad \frac{dy}{dx} = \frac{3(t^2 - 1)}{t - 2} \]
Differentiate the slope in t
Why: Quotient rule.
\[ \frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{3\left[2t(t-2) - (t^2 - 1)\right]}{(t-2)^2} \]
Simplify the numerator
Why: Expand and collect.
\[ 2t^2 - 4t - t^2 + 1 = t^2 - 4t + 1 \]
Divide by dx/dt
Why: The horizontal rate is 2t minus 4, which is 2 times t minus 2.
\[ \frac{d^2y}{dx^2} = \frac{3(t^2 - 4t + 1)}{(t-2)^2 \cdot 2(t-2)} = \frac{3(t^2 - 4t + 1)}{2(t-2)^3} \]
Classify the flat spots
Why: At t equal to minus 1 the curve bends down, so (5, 4) is a local maximum; at t equal to 1 it bends up, so (minus 3, minus 4) is a local minimum.
\[ t = -1: \; \frac{3(6)}{2(-27)} = -\frac13, \qquad t = 1: \; \frac{3(-2)}{2(-1)} = 3 \]
Check numerically
Why: A difference quotient of the slope at t equal to one half, divided by x prime there, gives 0.33333; the formula gives exactly one third.
\[ \frac{3(\tfrac14 - 2 + 1)}{2(-\tfrac32)^3} = \frac{-2.25}{-6.75} = \frac13 \]
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, p. 631 — Checkpoint 7.6
This time the slope is a quotient, so the quotient rule is needed. Take your time with the numerator: expand, collect, and you get t squared minus 4t plus 1, times 3.
The division by x prime is where the power in the denominator goes up from 2 to 3. Writing x prime as 2 times t minus 2 makes that clean.
The payoff is the second-derivative test from Calculus I, applied to a parametric curve. At the flat spot with t equal to minus 1 the curve bends down, so it is a local maximum; at t equal to 1 it bends up, so it is a local minimum. The numerical check at t equal to one half confirms the formula to five decimal places.
Counterexample
\[ x = t, \quad y = t^2 \]
Discussion prompt
Someone proposes that the second derivative is the second t-derivative of y divided by the second t-derivative of x. Test that rule on the curve above, whose graph you know is the parabola y equals x squared, and say what the test proves.
Write your answer before revealing. The curve x equals t, y equals t squared is just the parabola y equals x squared, whose second derivative you know is 2 everywhere.
The tempting rule divides the second derivative of y by the second derivative of x. But x is t, whose second derivative is zero, so the rule tries to divide by zero and produces nothing at all.
The correct rule gives a slope of 2t, differentiates it to 2, divides by x prime, which is 1, and gets 2, the right answer. Whenever you doubt a formula, test it on a curve whose answer you already know. It costs ten seconds and it is more reliable than memory.
Worked example
\[ x = 3t^2, \quad y = t^3 - t \]
Differentiate both coordinates
Why: Power rule.
\[ x'(t) = 6t, \quad y'(t) = 3t^2 - 1 \]
Form the slope and split it
Why: Splitting the fraction avoids a quotient rule.
\[ \frac{dy}{dx} = \frac{3t^2 - 1}{6t} = \frac{t}{2} - \frac{1}{6t} \]
Differentiate the slope in t
Why: Term by term.
\[ \frac{d}{dt}\left(\frac{t}{2} - \frac{1}{6t}\right) = \frac12 + \frac{1}{6t^2} = \frac{3t^2 + 1}{6t^2} \]
Divide by dx/dt
Why: The horizontal rate is 6t.
\[ \frac{d^2y}{dx^2} = \frac{3t^2 + 1}{6t^2 \cdot 6t} = \frac{3t^2 + 1}{36t^3} \]
Read the sign
Why: The numerator is always positive, so the sign is the sign of t cubed.
\[ t > 0: \text{ concave up}, \qquad t < 0: \text{ concave down} \]
Figure (svg): The curve x equals 3t squared, y equals t cubed minus t for t from −1.6 to 1.6, a loop that crosses itself at (3, 0). The part with t negative is drawn in red and bends downward; the part with t positive is drawn in green and bends upward.
Check at t equal to 1
Why: A central difference of the slope at t equal to 1, divided by x prime of 1, gives 0.11111, and the formula gives four thirty-sixths.
\[ \frac{3 + 1}{36} = \frac19 \approx 0.1111 \]
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, p. 641 — Exercise 98
This curve makes a loop that crosses itself at 3, 0. The question asks where it is concave up and where concave down, as intervals of t.
Splitting the slope into two simple terms before differentiating is a small trick that avoids a quotient rule. After dividing by 6t the concavity is 3t squared plus 1 over 36 t cubed.
The numerator is always positive, so the sign is entirely decided by t cubed, which has the sign of t. On the figure, the red half with negative t bends down, and the green half with positive t bends up. The check at t equal to 1 compares the formula with a numerical difference quotient; both give one ninth.
Error analysis
\[ x = \tfrac12 t^2, \quad y = \tfrac13 t^3, \quad t = 2 \]
Annotate
On: \( \frac{d^2y}{dx^2} = \frac{y''(t)}{x''(t)} = \frac{2t}{1} = 4 \quad \text{at } t = 2 \)
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, p. 641 — Exercise 96
Read the line and find the flaw before revealing the notes. The computation itself is carried out correctly; the mistake is in the first equals sign.
The correct route: the slope is t, differentiating it in t gives 1, and dividing by x prime, which is t, gives one over t, so one half at t equal to 2. The wrong line gives 4, which is eight times too big.
The third note shows an independent check by eliminating the parameter. That is the most convincing kind of evidence: a completely different method that agrees with the right answer and disagrees with the wrong one.
Comparison
Comparison matrix
| curve | at | dy/dx | d²y/dx² |
|---|---|---|---|
| x = 2t + 3, y = 3t − 4 | any t | 3/2 | 0 |
| x = t² − 3, y = 2t − 1 | t = 1 | 1 | −1/2 |
| x = t²/2, y = t³/3 | t = 2 | 2 | 1/2 |
| x = 5 cos t, y = 5 sin t | t = π/4 | −1 | −2√2/5 ≈ −0.566 |
Fill in each blank before checking. Every entry uses the same two formulas, so this is practice at running the procedure quickly.
The straight line has constant slope, so its second derivative is zero. For the parabola at t equal to 1, the slope one over t is 1 and the concavity minus one over 2 t cubed is minus one half. For the curve from the error-analysis slide at t equal to 2, the slope is t, so 2.
The circle row is worth checking a second way. At a quarter of pi the point is at 45 degrees, where the slope is minus 1, and the circle bends downward there, which matches the negative concavity.
Section
Part 3
Intuition
Under a curve, area is the integral of height times a sliver of width. On a parametric curve the height is y of t, and the width of a sliver is how far x moves while t moves by dt.
\[ A = \int y\,dx, \qquad x = x(t), \quad dx = x'(t)\,dt \]
\[ A = \int_{t=a}^{t=b} y(t)\,x'(t)\,dt \]
It is the substitution rule. Two things follow: the limits must be values of t, and if x runs backwards as t increases, x prime is negative and so is the answer.
Nothing about area has changed: it is still height times width, added up. The only question is how to express a sliver of width when x is itself a function of t.
If x depends on t, a small change dt moves x by x prime of t times dt. Substituting that into the integral of y dx gives the parametric area formula. So this is the substitution rule you already know, with the substitution handed to you.
Two consequences are worth saying now, because they cause most of the mistakes later. After a substitution, the limits must be values of the new variable, t. And if x decreases as t increases, then x prime is negative, and the integral comes out negative.
Concept
Figure (svg): One arch of the cycloid x equals t minus sin t, y equals 1 minus cos t, over eight rectangles built on equal steps of t. The rectangles are narrow near the ends and wide in the middle, because x moves slowly near the cusps and quickly at the top.
Cut the parameter interval into equal steps. Each rectangle's height is y at a sample parameter, and its width is how far x moved during that step.
\[ A_n = \sum_{i=1}^{n} y(\bar t_i)\left[x(t_i) - x(t_{i-1})\right] \]
\[ A_n = \sum_{i=1}^{n} y(\bar t_i)\,\frac{x(t_i) - x(t_{i-1})}{\Delta t}\,\Delta t \]
\[ A = \lim_{n\to\infty} A_n = \int_a^b y(t)\,x'(t)\,dt \]
Area under a parametric curve (Theorem 7.2) — For a curve that does not cross itself, with x differentiable, the area under it is the integral of y of t times x prime of t with respect to t.
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, pp. 632-633 — Figure 7.22 and Theorem 7.2
The book derives the formula from rectangles, and the figure shows the one surprising feature. The rectangles are built on equal steps of t, but they are not equally wide. Near the ends of the arch the point moves slowly sideways and the rectangles are thin; at the top it moves fastest and they are wide.
Each width is a difference of x values. Multiplying and dividing by the step in t turns that difference into a difference quotient, which becomes x prime in the limit. That is where the factor x prime comes from.
Theorem 7.2 includes the hypothesis that the curve does not cross itself. If it did, parts of the region would be counted with opposite signs, and the integral would no longer be an area.
Estimation
\[ x = t - \sin t, \quad y = 1 - \cos t, \quad 0 \le t \le 2\pi \]
Predict first
A wheel of radius 1 rolls once along the floor, and a point on its rim draws one arch of a cycloid. The wheel's own area is π. How many times the wheel's area is the area under the arch?
Correct: 3 times
Why: The area under one arch is exactly 3π, three wheels' worth. The arch spans 2π and rises to height 2, so its bounding box has area 4π; the arch fills three quarters of it. The next slide proves it.
Make a guess before revealing. The arch is 2 pi wide and 2 tall, so its bounding box has area 4 pi, and the arch clearly fills most but not all of the box.
The answer is exactly three times the wheel's area. This is one of the classical results of seventeenth-century mathematics, found by Roberval before calculus was fully developed.
Keep your guess in mind as you work through the next slide, and notice how little effort the integral needs to settle a question that is hard to answer by looking.
Worked example
\[ A = \int_0^{2\pi} y(t)\,x'(t)\,dt, \quad y = 1 - \cos t, \quad x' = 1 - \cos t \]
Substitute
Why: Height and horizontal rate happen to be the same function.
\[ A = \int_0^{2\pi} (1 - \cos t)(1 - \cos t)\,dt \]
Expand the square
Why: Three terms.
\[ A = \int_0^{2\pi} \left(1 - 2\cos t + \cos^2 t\right)dt \]
Reduce the power
Why: Cos squared is one half of 1 plus cos 2t.
\[ A = \int_0^{2\pi} \left(\tfrac32 - 2\cos t + \tfrac12\cos 2t\right)dt \]
Integrate
Why: One term at a time.
\[ A = \left[\tfrac32 t - 2\sin t + \tfrac14\sin 2t\right]_0^{2\pi} \]
Evaluate
Why: Every sine term is zero at both ends.
\[ A = \tfrac32(2\pi) - 0 + 0 = 3\pi \]
Figure (svg): One arch of the cycloid shaded underneath, with the wheel of radius 1 that generates it drawn rolling along the x-axis, centred at (π/2, 1). A dot on the rim marks the point that traces the curve. The label says the shaded area is 3π, three times the wheel's area π.
Check with rectangles
Why: The unequal-width Riemann sums with 8, 16 and 64 rectangles give 9.345, 9.405 and 9.4235, closing in on 3 pi.
\[ A_{64} = 9.4235 \;\to\; 3\pi = 9.4248 \]
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, p. 633 — Example 7.7
The height and the horizontal rate are the same function here, 1 minus cos t, so the integrand is its square. Expand, then reduce cos squared with the double-angle identity, the standard move from Section 3.2.
After integrating, every sine term vanishes at both 0 and 2 pi, leaving three halves times 2 pi, which is 3 pi. The limits were already t values, and x runs from 0 to 2 pi as t does, so no sign trouble arises.
The check connects back to the rectangles. The unequal-width sums with 8, 16 and 64 rectangles give 9.345, 9.405 and 9.4235, closing in on 3 pi, which is 9.4248. The picture shows the region holding three copies of the wheel.
Step zero
\[ x(t) = 3\cos t + \cos 3t, \quad y(t) = 3\sin t - \sin 3t, \quad 0 \le t \le \pi \]
Discussion prompt
Checkpoint 7.7 asks for the area under this curve. Before writing any integral, find where the curve starts and ends, and which way x moves as t increases. What does that predict about the sign of the integral of y x′ dt?
Write your answer before revealing. The question is not asking you to integrate; it is asking you to predict what the integral will look like.
At t equal to 0 the curve is at 4, 0, and at t equal to pi it is at minus 4, 0. So as t increases, the point sweeps from right to left. That means x prime is negative, and the integral of y times x prime will be negative too.
Doing this check first turns a confusing sign at the end into a confirmed prediction. It takes two substitutions and it saves you from hunting for an error that does not exist.
Worked example
Differentiate x
Why: Chain rule on cos 3t.
\[ x'(t) = -3\sin t - 3\sin 3t \]
Set up the integral
Why: Limits are the parameter values 0 and pi.
\[ \int_0^{\pi}(3\sin t - \sin 3t)(-3\sin t - 3\sin 3t)\,dt \]
Factor out minus 3 and expand
Why: The product of the brackets has three terms.
\[ -3\int_0^{\pi}\left(3\sin^2 t + 2\sin t\sin 3t - \sin^2 3t\right)dt \]
Use three standard integrals
Why: Each squared sine averages one half; sin t times sin 3t integrates to zero over this interval.
\[ \int_0^{\pi}\sin^2 t\,dt = \int_0^{\pi}\sin^2 3t\,dt = \frac{\pi}{2}, \quad \int_0^{\pi}\sin t\sin 3t\,dt = 0 \]
Combine
Why: Three halves of pi minus one half of pi.
\[ -3\left(\tfrac{3\pi}{2} + 0 - \tfrac{\pi}{2}\right) = -3\pi \]
Interpret the sign
Why: The curve was traced right to left, as predicted, so the area is the absolute value.
\[ A = \left|-3\pi\right| = 3\pi \]
Figure (svg): The hypocycloid x equals 3 cos t plus cos 3t, y equals 3 sin t minus sin 3t, a four-cusped star. Its upper half, for t from 0 to pi, is drawn solid and shaded underneath; arrows show it is traced from (4, 0) at t equal to 0 leftward to (−4, 0) at t equal to pi. The lower half is dashed.
Check with the triple-angle identities
Why: Cos 3t is 4 cos cubed minus 3 cos, so x is 4 cos cubed t and y is 4 sin cubed t: an astroid with a equal to 4, whose full area is 3 pi a squared over 8.
\[ x = 4\cos^3 t, \; y = 4\sin^3 t, \quad \tfrac12\cdot\frac{3\pi(4)^2}{8} = 3\pi \]
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, p. 633 — Checkpoint 7.7
The set-up follows the formula exactly, with the parameter values as limits. The expansion is the only fiddly step: multiply out the brackets carefully and the middle terms combine into 2 sin t sin 3t.
Three standard integrals finish it. A squared sine over a whole number of half periods averages one half, so each gives pi over 2, and sin t times sin 3t integrates to zero over this interval. The result is minus 3 pi, negative exactly as predicted, so the area is 3 pi.
The check reveals what this curve really is. The triple-angle identities turn the coordinates into 4 cos cubed t and 4 sin cubed t, an astroid, and the known area of an astroid confirms the answer. Seeing a familiar curve behind unfamiliar formulas is always worth a moment.
Prediction
Figure (svg): The ellipse x equals 3 cos t, y equals 2 sin t, traced counterclockwise. The upper half is shaded green and labelled as traversed right to left, contributing minus its area; the lower half is shaded red and labelled as traversed left to right while y is negative, which also contributes minus its area.
\[ x = a\cos t, \quad y = b\sin t, \quad 0 \le t < 2\pi \]
Predict first
Integrate y(t) x′(t) over one full counterclockwise turn. What comes out?
Correct: −πab
Why: The integrand is b sin t times −a sin t, which is −ab sin² t, and sin² t integrates to π over a full turn. So the integral is −πab: the enclosed area, with a minus sign because a counterclockwise loop runs right to left along its top. It is not zero: the two halves do not cancel, they reinforce.
\[ \int_0^{2\pi} b\sin t\,(-a\sin t)\,dt = -ab\int_0^{2\pi}\sin^2 t\,dt = -\pi ab \]
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, p. 641 — Exercise 102
Commit to an answer first. It is tempting to guess zero, reasoning that the top half and the bottom half should cancel. Look at the figure to see why they do not.
Along the top, y is positive and the point moves left, so x prime is negative: the product is negative. Along the bottom, y is negative and the point moves right, so x prime is positive: the product is negative again. Both halves contribute the same sign.
So a full counterclockwise loop gives minus the enclosed area, minus pi a b. This is a genuinely useful fact: for a closed curve traced once, the integral measures the area inside it, with the sign recording the direction of travel.
Trap
Checkpoint 7.7, with the limits read off the picture:
\[ A = \int_{-4}^{4} y(t)\,x'(t)\,dt \]
Wrong. The integrand is in t; the limits are in x.
After a substitution the limits must be values of t. Here x equal to 4 is t equal to 0 and x equal to minus 4 is t equal to pi. Written left to right in x, the integral runs from pi down to 0.
\[ A = \int_{\pi}^{0} y(t)\,x'(t)\,dt = 3\pi \]
This mistake comes from reading the limits off the picture. The picture shows the region stretching from x equal to minus 4 to x equal to 4, but the integral is written in t, so those numbers mean nothing to it.
The fix is to translate each end of the region into a parameter value. The right end, x equal to 4, is t equal to 0, and the left end is t equal to pi. If you insist on reading the region left to right, the integral runs from pi down to 0, and the minus sign from the reversed limits cancels the minus sign from x prime, giving 3 pi directly.
Section
Part 4
Picture it
Figure (svg): Three copies of the upper semicircle of radius 3, each with an inscribed polygon of equal parameter steps: 2 chords with total length 8.485, 4 chords with total 9.184, and 8 chords with total 9.364. The exact arc length is 3 pi, about 9.425.
Join the points at equal parameter steps with straight chords and add up their lengths. Each chord is the distance formula applied to one step.
\[ s \approx \sum_{k=1}^{n}\sqrt{\left(x(t_k) - x(t_{k-1})\right)^2 + \left(y(t_k) - y(t_{k-1})\right)^2} \]
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, pp. 633-634 — Figure 7.23 and equation 7.4
To measure a curved length, approximate it by something you can measure: straight chords. The three panels use 2, 4 and 8 chords on the same semicircle.
Each chord is shorter than the arc it replaces, because a straight line is the shortest path between two points. So every polygon underestimates, and the totals climb: 8.485, then 9.184, then 9.364, toward the true length 3 pi, about 9.425.
The formula on the slide is just the distance formula applied to each chord and added up. The next slide turns that sum into an integral.
Concept
The Mean Value Theorem turns each difference into a derivative times the step, at some point inside the step.
\[ x(t_k) - x(t_{k-1}) = x'(\hat t_k)\,\Delta t, \quad y(t_k) - y(t_{k-1}) = y'(\tilde t_k)\,\Delta t \]
\[ s \approx \sum_{k=1}^{n}\sqrt{\left(x'(\hat t_k)\right)^2 + \left(y'(\tilde t_k)\right)^2}\;\Delta t \]
That is a Riemann sum. As the steps shrink, both sample points are squeezed into the same tiny interval, so they converge together.
Arc length of a parametric curve (Theorem 7.3) — If x and y are differentiable functions of t, with continuous derivatives, the length of the curve from the first parameter value to the second is the integral of the square root of x prime squared plus y prime squared.
\[ s = \int_{t_1}^{t_2}\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt \]
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, pp. 634-635 — Theorem 7.3
The Mean Value Theorem says that over each small step, the change in x equals the derivative of x at some point inside the step, times the step. The same holds for y, possibly at a different point.
Pull the step out of the square root and the sum becomes a Riemann sum for the speed. As the steps shrink, the two sample points are trapped in the same tiny interval, so they come together, and the limit is the integral in Theorem 7.3.
The hypothesis of continuous derivatives is what makes that limit behave. For every curve you will meet in this course it holds automatically.
Picture it
Figure (svg): The curve x equals 3t squared, y equals 2t cubed, from the origin up to the right. At the point (3, 2), where t is 1, a small right triangle is drawn along the tangent: a horizontal leg labelled dx equals x prime of t dt, a vertical leg labelled dy equals y prime of t dt, and the hypotenuse labelled ds.
Over a tiny parameter step the curve is indistinguishable from its tangent line. The step moves the point across by x prime dt and up by y prime dt, and Pythagoras gives the length moved.
\[ ds = \sqrt{(dx)^2 + (dy)^2} = \sqrt{x'(t)^2 + y'(t)^2}\;dt \]
This picture is the fastest way to remember the formula. Zoom in on the curve far enough and it looks like a straight segment along its tangent line.
Over a small parameter step dt the point moves across by x prime times dt and up by y prime times dt. Those are the legs of the small right triangle on the figure, and the hypotenuse is the length ds that the point actually travels. Pythagoras does the rest.
Factor dt out of the square root and what remains is the speed of the point. Arc length is speed times time, added up.
Notation
Annotate
On: \( s = \int_{t_1}^{t_2}\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt \)
Read the integrand as physics. The two derivatives are the components of the velocity; the square root combines them into the speed; multiplying by dt gives a small distance.
Two consequences follow straight from this reading. A speed is never negative, so an arc length can never be negative, unlike the area integral. And the integral measures distance travelled during the time interval, which equals the length of the curve only if no part of it is covered twice.
Keep that last point in mind; a trap later in this part turns on it.
Concept
Suppose the parameter can be eliminated, so that y is F of x. The chain rule writes y prime in terms of x prime.
\[ y'(t) = F'(x)\,x'(t) \]
\[ \sqrt{x'(t)^2 + \left(F'(x)\,x'(t)\right)^2} = x'(t)\sqrt{1 + \left(F'(x)\right)^2} \quad (x' > 0) \]
\[ s = \int_{t_1}^{t_2}\sqrt{1 + \left(\frac{dy}{dx}\right)^2}\;x'(t)\,dt \]
\[ s = \int_a^b \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\,dx, \quad a = x(t_1), \; b = x(t_2) \]
The last step is the substitution x equals x of t run backwards: x prime dt becomes dx and the limits become x-values. This is the formula from the warm-up.
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, p. 635 — the side derivation after Theorem 7.3
A new formula should agree with the old one wherever both apply. Here the check is a short piece of algebra. If y is a function F of x, the chain rule writes y prime as F prime times x prime.
Substitute that into the square root and factor x prime squared out. Taking x prime to be positive, which just means the curve is traced left to right, it comes out of the root unchanged.
Finally, x prime dt is dx, and the limits become the x values at the two ends. What remains is the arc length formula from the warm-up. So the parametric formula contains the old one and extends it to curves that are not graphs.
Worked example
\[ x(t) = 3\cos t, \quad y(t) = 3\sin t, \quad 0 \le t \le \pi \]
Differentiate both coordinates
Why: Derivatives of cosine and sine.
\[ x'(t) = -3\sin t, \quad y'(t) = 3\cos t \]
Square and add
Why: Each square carries a factor 9.
\[ x'^2 + y'^2 = 9\sin^2 t + 9\cos^2 t \]
Use the Pythagorean identity
Why: Sine squared plus cosine squared is 1.
\[ 9\left(\sin^2 t + \cos^2 t\right) = 9, \quad \sqrt9 = 3 \]
Integrate the constant speed
Why: Speed 3 for a time of pi.
\[ s = \int_0^{\pi} 3\,dt = 3t\Big|_0^{\pi} = 3\pi \]
Figure (svg): Two panels. Left: the upper semicircle of radius 3 traced by x equals 3 cos t, y equals 3 sin t for t from 0 to pi. Right: the speed, the square root of x prime squared plus y prime squared, plotted against t; it is the constant 3, and the shaded rectangle under it from 0 to pi has area 3 pi.
Check against geometry
Why: Half a circumference is pi times the radius, and the radius is 3.
\[ \tfrac12(2\pi r) = \pi(3) = 3\pi \approx 9.425 \]
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, pp. 635-636 — Example 7.8 and Figure 7.24
The Pythagorean identity collapses the speed to the constant 3. That is typical of circles: a point going round a circle with this parametrisation moves at constant speed, the radius.
The figure makes the result visible. On the right, the speed is the flat line at height 3 for a time of pi, and the area under it, 3 pi, is the distance the point travels along the semicircle on the left.
The check is the one the book itself makes: half the circumference of a circle of radius 3 is 3 pi. Calculus has just reproduced a formula from geometry, which is always a reassuring sign.
Worked example
\[ x(t) = 3t^2, \quad y(t) = 2t^3, \quad 1 \le t \le 3 \]
Differentiate both coordinates
Why: Power rule.
\[ x'(t) = 6t, \quad y'(t) = 6t^2 \]
Square, add and factor
Why: Take out 36 t squared, a perfect square.
\[ \sqrt{36t^2 + 36t^4} = \sqrt{36t^2(1 + t^2)} = 6t\sqrt{1 + t^2} \]
Substitute u equal to 1 plus t squared
Why: Then du is 2t dt, so 6t dt is 3 du; the limits become 2 and 10.
\[ s = \int_1^3 6t\sqrt{1+t^2}\,dt = 3\int_2^{10} u^{1/2}\,du \]
Integrate
Why: Power rule.
\[ 3\cdot\tfrac23 u^{3/2}\Big|_2^{10} = 2\left(10^{3/2} - 2^{3/2}\right) \]
Evaluate
Why: Ten root ten minus two root two, doubled.
\[ s = 2\left(10\sqrt{10} - 2\sqrt2\right) \approx 57.589 \]
Check against the chord
Why: The straight line from (3, 2) to (27, 54) must be shorter than the curve; it is, but only just, because this stretch of curve is nearly straight.
\[ \sqrt{24^2 + 52^2} = \sqrt{3280} \approx 57.271 < 57.589 \]
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, p. 636 — Checkpoint 7.8
Most arc length integrals cannot be done in closed form, because the square root does not simplify. This one is designed so that it does: 36 t squared is a perfect square and comes out of the root, leaving 6t times the root of 1 plus t squared.
The substitution u equal to 1 plus t squared is then exactly right, since its derivative 2t is sitting outside the root. Remember to change the limits from 1 and 3 to 2 and 10.
The check compares the curve with the straight chord between its endpoints. The curve must be at least as long as the chord, and it is, 57.589 against 57.271. The small gap tells you this part of the curve is nearly straight, which you can see in the speed triangle figure.
Concept
\[ x = t - \sin t, \quad y = 1 - \cos t, \quad 0 \le t \le 2\pi \]
\[ x'^2 + y'^2 = (1 - \cos t)^2 + \sin^2 t = 2 - 2\cos t \]
\[ 2 - 2\cos t = 4\sin^2\tfrac{t}{2} \;\Longrightarrow\; \text{speed} = 2\sin\tfrac{t}{2} \]
\[ s = \int_0^{2\pi} 2\sin\tfrac{t}{2}\,dt = \left[-4\cos\tfrac{t}{2}\right]_0^{2\pi} = 4 + 4 = 8 \]
Figure (svg): Two panels. Left: one arch of the cycloid x equals t minus sin t, y equals 1 minus cos t. Right: its speed, 2 sin of t over 2, a single hump from 0 at t equal to 0, up to 2 at t equal to pi, back to 0 at 2 pi; the shaded area under the hump is 8.
The half-angle identity takes the square root cleanly, and sin of t over 2 is not negative anywhere on the interval, so no absolute value is lost. A numerical integral of the speed agrees: 8.0000.
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, p. 641 — Exercise 116
The speed of the cycloid simplifies with the identity 1 minus cos t equals 2 sin squared of t over 2. Taking the square root then gives 2 sin of t over 2, and since that is never negative on this interval, no absolute value is lost.
The figure puts the arch and its speed side by side. The speed is zero at both ends, at the cusps, and greatest at the top, where the point on the rim is moving at twice the speed of the wheel's centre.
The area under the speed curve is 8. So the arch is exactly 8 units long, four times its height, another result that looks like a coincidence until you compute it.
Intuition
Trace the unit circle once at speed 1, or once at speed 2 in half the time. The speeds differ, the times differ, and the product does not.
\[ x = \cos t, \; y = \sin t, \; 0 \le t \le 2\pi: \quad \int_0^{2\pi} 1\,dt = 2\pi \]
\[ x = \cos 2t, \; y = \sin 2t, \; 0 \le t \le \pi: \quad \int_0^{\pi} 2\,dt = 2\pi \]
Figure (svg): Two panels. Left: the segment from (0, 1) to (1, 0) on the line x plus y equals 1, with arrows running both ways along it, labelled length the square root of 2, traced six times. Right: the speed, root 2 times the absolute value of sin 2t, for t from 0 to 3 pi: six identical humps, total shaded area 6 root 2.
Changing the parametrisation is a substitution in the integral, so the length of a curve traced once does not depend on how it is traced. Tracing it twice is another matter.
Two different parametrisations of the unit circle: one goes round at speed 1 in time 2 pi, the other at speed 2 in time pi. Speed times time is 2 pi in both cases, and so is the length.
In general, changing the parameter is a substitution in the integral, and a substitution does not change the value. That is why arc length is a property of the curve and not of how you trace it, provided you trace it exactly once.
The figure shows what happens when that proviso fails. The point runs back and forth along one short segment, and the speed integral adds up every trip. That is the next trap.
Tweak it
Parameter explorer
A point on the rim of a wheel of radius r traces x = r(t − sin t), y = r(1 − cos t). The graph is its speed against t for one turn. Slide r: how do the height of the hump and the area under it change?
\[ s = \int_0^{2\pi} 2{r}\sin\tfrac{t}{2}\,dt = 8({r}) \]
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, p. 641 — Exercise 114, with r equal to 4, gives 32
Start with the radius at 1 and you see the speed hump from the previous slides, peaking at 2 with area 8 underneath. Now drag the radius up to 2.
The hump doubles in height but keeps the same width, because the wheel still makes one turn as t goes from 0 to 2 pi. So the area under it doubles too, and the arch length becomes 16. In general, it is 8 times the radius.
That is what you would expect from scaling: making the whole picture twice as large doubles every length. Exercise 114 uses a wheel of radius 4, so its arch is 32 long.
Trap
Exercise 115, reported as the curve's length:
\[ x = \sin^2 t, \; y = \cos^2 t, \; 0 \le t \le 3\pi \]
\[ \int_0^{3\pi}\sqrt2\,|\sin 2t|\,dt = 6\sqrt2 \]
Wrong as a length. Right as a distance travelled.
Since x plus y is always 1, the point never leaves the segment from (0, 1) to (1, 0), of length root 2. It runs along it and back again, six trips in all. Theorem 7.3 assumes the curve is traced once; the integral adds every pass.
\[ \text{length} = \sqrt2 \]
\[ \text{distance} = 6\sqrt2 \approx 8.485 \]
The integral is computed correctly. The error is in what it is claimed to measure. Because sine squared plus cosine squared is 1, the point always satisfies x plus y equals 1, and both coordinates stay between 0 and 1, so it never leaves one short segment.
As t runs from 0 to 3 pi, the point goes along the segment and back again six times. The integral of the speed adds all six trips: 6 root 2 is the distance travelled, not the length of the curve. Whenever the coordinates are periodic, ask how many times the curve is covered before quoting an arc length.
Worked example
A pitch leaves the hand at the origin. Find the distance the ball has travelled, and its speed, a third of a second later.
\[ x(t) = 140t, \quad y(t) = -16t^2 + 2t \quad (\text{feet, seconds}) \]
Write the distance as a function of time
Why: Use v as the variable of integration so that t is free to be the upper limit.
\[ s(t) = \int_0^t \sqrt{140^2 + (-32v + 2)^2}\,dv \]
Differentiate with the Fundamental Theorem
Why: The derivative of an integral with a variable upper limit is the integrand at that limit.
\[ s'(t) = \sqrt{140^2 + (-32t + 2)^2} \]
Expand under the root
Why: 140 squared is 19600; the rest is the square of minus 32t plus 2.
\[ s'(t) = \sqrt{1024t^2 - 128t + 19604} = 2\sqrt{256t^2 - 32t + 4901} \]
Evaluate the speed at one third
Why: Substitute t equal to one third.
\[ s'\left(\tfrac13\right) = 2\sqrt{\tfrac{256}{9} - \tfrac{32}{3} + 4901} \approx 140.27 \text{ ft/s} \]
Evaluate the distance at one third
Why: The book's closed form, from the table integral of the root of a squared plus u squared, agrees with a numerical integral.
\[ s\left(\tfrac13\right) \approx 46.69 \text{ ft} \]
Figure (svg): The path of the pitched ball, x equals 140t and y equals minus 16t squared plus 2t, from the pitcher's hand at the origin for about 0.45 seconds. The vertical scale is stretched about ten times. A dot at t equal to one third marks the ball at x about 46.7 feet, 1.1 feet below release height; a dashed vertical line marks home plate at 60.5 feet.
Check the numbers
Why: The ball is barely off a straight line, so s should just exceed the horizontal distance 46.67, and it does. The book prints 140.34 feet per second, having put minus 16 in place of minus 32 under the root; the correct speed is 140.27, about 95.6 miles per hour.
\[ 46.69 > x\left(\tfrac13\right) = 46.67, \quad 140.27 \cdot \tfrac{3600}{5280} \approx 95.6 \]
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, pp. 636-638 — the baseball, after Checkpoint 7.8
This is the application the book promised at the start of the section. The ball's path is a parabola, and the distance travelled after time t is an arc length with a variable upper limit. The variable of integration is renamed v so that t is free to be the upper limit.
The speed does not need the messy antiderivative at all. The Fundamental Theorem of Calculus says the derivative of an integral with respect to its upper limit is just the integrand there. So the speed is the square root expression, evaluated at one third: about 140.27 feet per second.
The book prints 140.34, because it replaced minus 32 by minus 16 inside the root; the correct factoring of the expression gives 140.27, which is about 95.6 miles per hour, a major-league fastball. The distance, 46.69 feet, only just exceeds the horizontal distance, as it must for a path this close to straight.
Real world
Discussion prompt
A phone logs your position once per second as you run: a list of points (x, y) at t = 0, 1, 2, and so on. How does the app turn that list into a distance, which formula from this lesson is it approximating, and will its answer tend to be a little too high or a little too low?
Write your answer before revealing. You have all the ingredients from the polygon slide at the start of this part.
The app never sees a formula for your path, only a list of positions. The natural estimate joins them with straight chords and adds their lengths, which is equation 7.4 exactly. Since each chord cuts the corner of whatever curve you actually ran, the total is slightly too small on a winding route.
Logging more often means shorter chords and a better estimate, just as 8 chords beat 2 on the semicircle. In practice GPS noise pushes the other way, since random jitter adds length, which is why apps smooth the data before measuring it.
Matching
Match the pairs
Why: Every formula in the section is built from the same two derivatives. Their ratio gives a slope; the length of the velocity vector gives a speed, and speed times dt is a sliver of arc; height times the sideways step x′ dt is a sliver of area; circumference 2πy times a sliver of arc is a sliver of surface.
Match each expression to its meaning before checking. All four are built from the same two derivatives, and seeing that is the main point of the exercise.
The ratio of the derivatives is the slope. The square root of the sum of their squares is the speed, the length of the velocity vector. Height times the sideways step is a strip of area, carrying a sign from the direction of travel. The circumference 2 pi y times a sliver of arc is a band of surface.
If you can rebuild each formula from its meaning like this, you will never need to memorise them separately.
Section
Part 5
Concept
Figure (svg): The upper semicircle of radius 2 drawn solid, its mirror image below dashed, and several thin elliptical hoops showing it revolved about the x-axis into a sphere. One narrow band between two hoops is highlighted; its radius is the height y of the curve and its slant width is ds.
Revolve a short piece of the curve about the x-axis. It sweeps out a thin band whose circumference is 2 pi times the height y, and whose width is the arc length ds.
\[ dS = 2\pi y\,ds = 2\pi y(t)\sqrt{x'(t)^2 + y'(t)^2}\,dt \]
\[ S = 2\pi\int_a^b y(t)\sqrt{\left(x'(t)\right)^2 + \left(y'(t)\right)^2}\,dt \]
It is the Cartesian formula with ds written in the parameter. The curve must stay on or above the axis, so that y is a genuine radius.
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, p. 638 — equation 7.6 and Figure 7.25
The picture shows a semicircle spun about the x-axis into a sphere, with a few hoops drawn to show the rotation. Concentrate on the single highlighted band.
That band comes from a tiny piece of the curve, of length ds, at height y. Spinning it around makes a thin ribbon whose circumference is 2 pi times y and whose width is ds. Multiply and add up the ribbons, and you have the surface area formula.
Writing ds in terms of the parameter gives equation 7.6. The condition that y is not negative matters because y is playing the role of a radius here; if the curve dipped below the axis, the formula would subtract surface instead of adding it.
Worked example
Revolve the upper semicircle of radius r about the x-axis.
\[ x(t) = r\cos t, \quad y(t) = r\sin t, \quad 0 \le t \le \pi \]
Find the speed
Why: The same computation as Example 7.8, with r in place of 3.
\[ \sqrt{r^2\sin^2 t + r^2\cos^2 t} = \sqrt{r^2} = r \]
Substitute into equation 7.6
Why: The radius of each band is y, which is r sin t.
\[ S = 2\pi\int_0^{\pi} r\sin t \cdot r\,dt = 2\pi r^2\int_0^{\pi}\sin t\,dt \]
Integrate
Why: The antiderivative of sine is minus cosine.
\[ 2\pi r^2\left[-\cos t\right]_0^{\pi} = 2\pi r^2(1 + 1) \]
Simplify
Why: The familiar formula.
\[ S = 4\pi r^2 \]
Check with a numerical integral
Why: For r equal to 1, Simpson's rule on 2 pi sin t over 0 to pi gives 12.5664, which is 4 pi.
\[ 2\pi\int_0^{\pi}\sin t\,dt \approx 12.5664 = 4\pi \]
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, pp. 638-639 — Example 7.9 and Figure 7.26
The speed of the semicircle is simply r, as you found for radius 3 in Example 7.8. So the integrand is the radius of each band, r sin t, times the speed r.
The remaining integral of sin t from 0 to pi is 2, and the result is 4 pi r squared, the formula for the surface of a sphere that you have probably known since school without ever seeing where it came from.
The numerical check takes r equal to 1 and confirms that the integral really is 4 pi, about 12.566.
Worked example
\[ x(t) = t^3, \quad y(t) = t^2, \quad 0 \le t \le 1, \quad \text{about the x-axis} \]
Find the speed
Why: Take t squared out of the root; t is not negative here.
\[ \sqrt{9t^4 + 4t^2} = t\sqrt{9t^2 + 4} \]
Set up equation 7.6
Why: Radius t squared times speed.
\[ S = 2\pi\int_0^1 t^2 \cdot t\sqrt{9t^2 + 4}\,dt = 2\pi\int_0^1 t^3\sqrt{9t^2+4}\,dt \]
Substitute u equal to 9t squared plus 4
Why: Then t squared is u minus 4 over 9, t dt is du over 18, and the limits become 4 and 13.
\[ S = 2\pi\int_4^{13}\frac{u-4}{9}\,\sqrt u\,\frac{du}{18} = \frac{\pi}{81}\int_4^{13}\left(u^{3/2} - 4u^{1/2}\right)du \]
Integrate
Why: Power rule on each term.
\[ \frac{\pi}{81}\left[\tfrac25 u^{5/2} - \tfrac83 u^{3/2}\right]_4^{13} \]
Evaluate at both limits
Why: At 13 the bracket is 494 root 13 over 15; at 4 it is minus 128 over 15.
\[ \frac{\pi}{81}\cdot\frac{494\sqrt{13} + 128}{15} = \frac{\pi\left(494\sqrt{13} + 128\right)}{1215} \]
Figure (svg): The curve x equals t cubed, y equals t squared for t from 0 to 1, rising from a sharp point at the origin to (1, 1), with its mirror image dashed below and elliptical hoops at x equal to 0.25, 0.5, 0.75 and 1 showing the horn-shaped surface made by revolving it about the x-axis.
Check numerically
Why: Simpson's rule on the original integral gives 4.93642, matching the exact value. The book's answer key prints 64 in place of 128, which would give 4.771; that is a misprint.
\[ \frac{\pi(494\sqrt{13} + 128)}{1215} \approx 4.9364 \]
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, p. 639 — Checkpoint 7.9
The speed simplifies by taking t squared out of the root, which is legitimate because t is not negative on this interval. Multiplying by the radius t squared gives t cubed times the root of 9 t squared plus 4.
The substitution is the delicate part. Setting u equal to 9 t squared plus 4 handles the root, but you still need to express the leftover t squared in terms of u: it is u minus 4 over 9. Then the integral becomes two simple powers of u.
The final numbers deserve care. At 13 the bracket is 494 root 13 over 15, and at 4 it is minus 128 over 15, so subtracting gives plus 128. The book's answer key prints 64, which is a misprint: a numerical integral of the original expression gives 4.936, matching 128 and not 64.
Section
Part 6
Concept
Figure (svg): Two arches of the cycloid x equals t minus sin t, y equals 1 minus cos t, meeting at a sharp point on the x-axis at (2π, 0). Arrows along the curve just before and just after the cusp point almost straight down and almost straight up. The label says x prime and y prime are both 0 at t equal to 2 pi.
\[ x'(t) = 1 - \cos t, \quad y'(t) = \sin t, \quad x'(2\pi) = y'(2\pi) = 0 \]
At t equal to 2 pi Theorem 7.1 gives zero over zero, which says nothing. Look instead at the slope just before and just after.
\[ \frac{dy}{dx} = \frac{\sin t}{1 - \cos t} = \frac{2\sin\tfrac t2\cos\tfrac t2}{2\sin^2\tfrac t2} = \cot\tfrac{t}{2} \]
\[ t \to 2\pi^-: \cot\tfrac t2 \to -\infty, \qquad t \to 2\pi^+: \cot\tfrac t2 \to +\infty \]
The point comes straight down, stops, and leaves straight up. There is no tangent line: the curve has a cusp.
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, pp. 631-632 — the cycloid, Figure 7.21
The figure shows two arches of the cycloid meeting at a sharp point on the floor. At that instant both derivatives are zero, so Theorem 7.1 produces zero over zero.
Zero over zero is not an answer, it is an invitation to look closer. Simplifying the slope with the double-angle identities gives cot of t over 2, and that runs off to minus infinity just before 2 pi and to plus infinity just after.
Physically, the point on the rim comes straight down to the floor, stops for an instant as it touches, and leaves straight up. There is no single tangent line at such a point. Calculus is right to be silent.
Sorting
Sort into buckets
Each item gives x′ and y′ at one parameter value. Sort by what Theorem 7.1 tells you there.
Sort each case by looking at which derivative is zero. The signs of the nonzero derivatives affect which way the point is moving, but not which bucket the case goes into.
When only the horizontal rate is zero the tangent is vertical, and when only the vertical rate is zero it is horizontal. When neither is zero there is an ordinary slope.
The case with both zero is different in kind. The point has stopped, and what happens next depends on how it starts moving again. The cycloid makes a cusp; the next slide shows a curve that passes straight through.
Prediction
\[ x = t^3, \quad y = t^3, \quad x'(0) = y'(0) = 0 \]
Predict first
At t = 0 both derivatives vanish, just as at the cycloid's cusp. What does this curve actually do at the origin?
Correct: It passes straight through: it is the line y = x
Why: Here y equals x for every t, so the curve is the straight line y = x with slope 1 everywhere; the point just slows to a stop at the origin and speeds up again. Compare x = t³, y = t² from Checkpoint 7.9, which has a real cusp. Zero over zero means the parametrisation stalls, not that the curve is broken.
\[ \frac{dy}{dx} = \frac{3t^2}{3t^2} = 1 \quad (t \ne 0) \]
Commit to an answer. It is natural to expect a cusp, since the derivatives vanish just as they did on the cycloid.
But this curve is the line y equals x, traced by a point that slows to a halt at the origin and then speeds up again. The line has slope 1 everywhere, including at the origin. The trouble was in the parametrisation, which stalls, not in the curve.
Compare the curve from Checkpoint 7.9, x equals t cubed and y equals t squared. Its derivatives also vanish at 0, and it does have a real cusp. Both derivatives being zero tells you only that you must investigate.
Intuition
| formula | assumes | if it fails |
|---|---|---|
| slope y′/x′ | x′(t) ≠ 0 | vertical tangent (y′ ≠ 0) or look closer (y′ = 0) |
| concavity | x′(t) ≠ 0, divided twice | undefined at the same parameter values |
| area ∫ y x′ dt | curve does not cross itself; t-limits | sign flips when x runs right to left |
| arc length ∫ speed dt | curve traced once | counts distance travelled, every pass |
| surface 2π ∫ y ds | y ≥ 0 on the interval | negative y is not a radius |
None of these failures is exotic: vertical tangents, backward tracing and retracing all appeared in the book's own examples. Each one is detected by looking at the derivatives before trusting a number.
Every formula in this section comes with a hypothesis, and this table collects them. Read across each row: what the formula needs, and what goes wrong when that need is not met.
Notice that each failure has a clear geometric meaning. A vertical tangent, a region traced right to left, a curve covered twice, a curve dipping below the axis. None of them is a mistake in the calculus; each is information about the curve.
The practical habit is to look at the two derivatives, and at where the curve starts and ends, before trusting any number.
Section
Part 7
Pattern
Figure (svg): A flow diagram. Start from x of t and y of t and differentiate both. Then branch by the question asked: a slope divides y prime by x prime; concavity differentiates that slope in t and divides by x prime again; length integrates the speed; area integrates y times x prime with t-limits; surface area integrates 2 pi y times the speed. A warning at the bottom says to check where x prime or both derivatives vanish.
This is the workflow for the whole section. It always begins the same way, by differentiating both coordinates, and then branches according to the question.
The warnings on the right of the diagram are the ones from the previous slide. Slopes break where x prime vanishes; concavity divides by x prime twice; area carries the direction as a sign; length and surface count every pass along the curve.
When a problem feels unfamiliar, come back to this diagram and ask which branch you are on.
Ranking
Put in order
Order the steps for finding the tangent line to a parametric curve at a given parameter value.
Why: Derivatives first, then the hypothesis check before the division, then the slope and the point, then the line. The point and slope could be found in either order, but the check must come before dividing.
Put the steps in order before checking. The main thing to get right is the hypothesis check, which must come before the division it protects.
The point and the slope could be computed in either order, since each needs only the given parameter value. The line itself always comes last, in point-slope form.
Check
Check your understanding
For x = eᵗ, y = (t − 1)², what is the slope of the tangent line at the point (1, 1)?
Answer: A
Why: The point (1, 1) comes from t = 0, since e⁰ = 1 and (0 − 1)² = 1. There x′ = e⁰ = 1 and y′ = 2(0 − 1) = −2, so the slope is −2 / 1 = −2.
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, p. 640 — Exercise 77
The question gives a point, not a parameter value, so the first job is to find which t produces that point. Since e to the t is 1 only when t is 0, the parameter value is 0.
Then the recipe runs as usual: x prime is 1, y prime is minus 2, and the slope is minus 2. The most tempting wrong answer comes from substituting 1, the x coordinate, as if it were the parameter.
Check
Check your understanding
For x = t², y = t³ with t > 0, what is d²y/dx²?
Answer: B
Why: The slope is 3t²/(2t) = 3t/2. Its t-derivative is 3/2, and dividing by x′ = 2t gives 3/(4t). Check: the curve is y = x^(3/2), whose second derivative (3/4)x^(−1/2) is 3/(4t).
Run the correct procedure: find the slope, three halves t, differentiate it in t to get three halves, then divide by x prime, which is 2t.
Each wrong option corresponds to one specific slip: forgetting the second division, dividing second derivatives, or inverting the answer. The check by eliminating the parameter, writing the curve as y equals x to the three halves, confirms three over 4t.
Check
Check your understanding
What is the length of x = cos 2t, y = sin 2t for 0 ≤ t ≤ π/2?
Answer: B
Why: The speed is √(4 sin² 2t + 4 cos² 2t) = 2, constant, for a time of π/2, so the length is 2 · π/2 = π. Geometrically, 2t runs from 0 to π: half of the unit circle, whose length is π.
OpenStax Calculus Volume 2, §7.2 Calculus of Parametric Curves §7.2, p. 641 — Exercise 110
The speed is the constant 2, because the point goes round the unit circle twice as fast as usual. Over a time of pi over 2 it travels pi.
The geometric check is quicker than the integral: as t goes from 0 to pi over 2, the angle 2t goes from 0 to pi, which is half of the unit circle, and half the circumference of a unit circle is pi.
Explain it to yourself
Discussion prompt
The area integral for Checkpoint 7.7 came out as minus 3 pi, but no arc length integral can ever come out negative. Explain the difference in two sentences, using the integrands.
Write your two sentences before revealing. The answer is visible in the two integrands side by side.
The area integrand contains x prime itself, and x prime is negative whenever the point is moving left. So the area integral keeps track of the direction of travel as a sign. The length integrand is a square root of squares, a speed, and a speed is never negative.
The same contrast explains the retracing trap: the length integral cannot notice that the point is going back over old ground, because it has thrown the direction away.
Exit ticket
\[ x = t^2, \quad y = t^3 - 3t \]
Discussion prompt
Find the slope and the concavity at t = 2, and the parameter value where the curve has a vertical tangent.
This brings together the first two parts of the lesson on one curve. Work each part on paper before revealing.
The slope at t equal to 2 is nine quarters. For the concavity, split the slope into three halves t minus three over 2t, differentiate to three halves plus three over 2 t squared, which is fifteen eighths at t equal to 2, and divide by x prime, which is 4. The result is fifteen thirty-seconds, so the curve bends upward there.
The vertical tangent is at t equal to 0, where x prime vanishes and y prime is minus 3, so the point is moving straight down through the origin.
Recap
| quantity | formula in t | watch for |
|---|---|---|
| slope | dy/dx = y′(t) / x′(t) | x′ = 0: vertical tangent or cusp |
| concavity | d²y/dx² = (dy/dx)′ / x′(t) | never y″ / x″ |
| area | A = ∫ y(t) x′(t) dt | t-limits; the sign records direction |
| arc length | s = ∫ √(x′² + y′²) dt | a retraced curve is counted every pass |
| surface area | S = 2π ∫ y √(x′² + y′²) dt | y must not be negative |
Next, Section 7.3 describes points by a distance and an angle instead of two coordinates, and Section 7.4 redoes this lesson's area and length in that language.
Stewart, Calculus: Early Transcendentals 8e, §10.2 Calculus with Parametric Curves §10.2, pp. 649-657 — the same material in Stewart
Five quantities, one ingredient. Every formula in this section is built from the two derivatives x prime and y prime, combined as a ratio for the slope, a ratio of a ratio for concavity, a product with y for area, and a speed for length and surface area.
The third column is the part to remember hardest. Vertical tangents and cusps where x prime vanishes, the correct second-derivative formula, the sign that records direction, and the fact that the length integral counts every pass.
Section 7.3 introduces polar coordinates, which are a special kind of parametrisation, and Section 7.4 reuses exactly these ideas to find areas and lengths of polar curves.
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