This deck covers vector-valued functions and the space curves they trace: domain, limits and continuity taken componentwise, and the derivative read as a tangent and velocity vector. It then gives the product rules for dot and cross products, defines smooth curves, integrates with a vector constant, works projectile motion, and shows why a curve of constant magnitude has its position vector orthogonal to its velocity. It targets three classic errors: differentiating a dot or cross product without the product rule, treating the constant of integration as a scalar, and forgetting that the derivative is taken component by component.
Subject: Calculus III · 115 slides · symbolic lesson
Open the interactive version of this deck · Homework for this lesson
Objectives
By the end of this deck you can:
1. Read a vector-valued function and describe the space curve it traces.
2. Find the domain, limits, and points of continuity by working one component at a time.
3. Differentiate a vector function and read the result as a tangent or velocity vector.
4. Apply the product rules for the dot product and the cross product.
5. Integrate a vector function with a vector constant of integration and solve motion problems, including projectiles.
6. Explain why a curve of constant length keeps its position perpendicular to its velocity.
Warm-up
Discussion prompt
Before we open Week 3 - Vector-Valued Functions: without looking back, what was the main idea of Week 2 - Geometry of Space, and what could you do by the end of it that you could not do before?
Hint: One sentence for the idea, one for the skill. If the second one is blank, that is the part to revisit.
Answer:
That deck covers lines and planes in space and the distances between them, then cylinders, the six quadric surfaces, and the cylindrical and spherical coordinate systems. It targets the confusion between a normal vector and a direction vector, the sign pattern that separates a one-sheet surface from a two-sheet one, and the absolute-value error in the distance-to-a-plane formula.
Concept
An ordinary function sends a number to a number. A vector-valued function sends a number to a vector.
The input is usually a time or parameter. The output is an arrow in the plane or in space.
\[ \mathbf{r}(t) = \langle f(t),\, g(t),\, h(t)\rangle = f(t)\,\mathbf{i} + g(t)\,\mathbf{j} + h(t)\,\mathbf{k} \]
vector-valued function — A rule r that assigns to each number t in its domain a vector r(t). In three dimensions r(t) has three scalar component functions f, g, and h.
Counterexample
Discussion prompt
An ordinary function sends a number to a number. A vector-valued function sends a number to a vector.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Answer:
The input is usually a time or parameter. The output is an arrow in the plane or in space.
Picture it
Figure (svg): A curved path with a position vector drawn from the origin to a point moving along the curve
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
Think of the input as a clock. At each instant, the function hands you the position of a moving particle.
Intuition
Think of the input as a clock. At each instant, the function hands you the position of a moving particle.
As the clock runs, the tip of the position vector sweeps out a path. That path is the curve.
Figure (svg): A curved path with a position vector drawn from the origin to a point moving along the curve
Analogy
Discussion prompt
Explain Picture a moving particle by analogy to something with no Calculus III in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
Think of the input as a clock. At each instant, the function hands you the position of a moving particle.
Concept
Everything about a vector function lives in its components. Each component is an ordinary scalar function of the same parameter.
\[ \mathbf{r}(t) = \langle f(t),\, g(t),\, h(t)\rangle \]
This is the key that unlocks the whole topic: nearly every operation is done one component at a time.
Explain it
Discussion prompt
Explain The component functions carry the meaning to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
Everything about a vector function lives in its components. Each component is an ordinary scalar function of the same parameter.
Concept
A line through a point in the direction of a fixed vector:
\[ \mathbf{r}(t) = \langle x_0 + at,\, y_0 + bt,\, z_0 + ct\rangle \]
A circle of radius three in the plane:
\[ \mathbf{r}(t) = \langle 3\cos t,\, 3\sin t\rangle \]
A circular helix, a circle that also climbs:
\[ \mathbf{r}(t) = \langle \cos t,\, \sin t,\, t\rangle \]
Ranking
Put in order
Put the moves of Worked example: plotting points on a helix into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Plug zero into each component: cosine of zero is one, sine of zero is zero, and the third component is the parameter itself.
Worked example
Trace a few points of the helix to see how it climbs.
\[ \mathbf{r}(t) = \langle \cos t,\, \sin t,\, t\rangle \]
Evaluate at t equals zero
Why: Plug zero into each component: cosine of zero is one, sine of zero is zero, and the third component is the parameter itself.
\[ \mathbf{r}(0) = \langle 1,\, 0,\, 0\rangle \]
Evaluate at t equals one quarter turn
Why: At a quarter turn the cosine is zero and the sine is one, while the height equals the parameter.
\[ \mathbf{r}\!\left(\tfrac{\pi}{2}\right) = \left\langle 0,\, 1,\, \tfrac{\pi}{2}\right\rangle \]
Read the pattern
Why: The first two components ride around the unit circle while the third steadily increases, so the curve spirals upward.
Verify the points lie on the surface of a cylinder
Why: The first two components satisfy the unit-circle relation for every t, confirming the curve wraps a cylinder of radius one.
\[ \cos^2 t + \sin^2 t = 1 \]
Picture it
Animation
Shows: Each line of the worked example "plotting points on a helix", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The first two components satisfy the unit-circle relation for every t, confirming the curve wraps a cylinder of radius one.
Concept
As the parameter runs through the domain, the set of all output tips forms a curve.
space curve — The set of all points traced by r(t) as t runs through the domain. The function also records the order and speed of travel, which the bare curve does not.
Two different functions can trace the same curve while running along it at different speeds or in opposite directions.
Definition probe
Sort into buckets
Every line below is part of the definition of vector-valued function or of space curve — one or the other, never both. Put each where it belongs.
Concept
The output vector only exists when all components exist. So the domain is the set of inputs allowed by every component at once.
In words: take the domain of each component, then keep only the inputs they all share.
\[ \text{dom}(\mathbf{r}) = \text{dom}(f) \cap \text{dom}(g) \cap \text{dom}(h) \]
Step zero
Discussion prompt
Worked example: finding a domain — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Condition from the square root
Answer:
Worked example
Find every parameter value this function accepts.
\[ \mathbf{r}(t) = \left\langle \sqrt{t+2},\, \ln(4-t),\, \tfrac{1}{t}\right\rangle \]
Condition from the square root
Why: A real square root needs a nonnegative radicand, so t plus two must be at least zero.
\[ t + 2 \ge 0 \;\Longrightarrow\; t \ge -2 \]
Condition from the logarithm
Why: A logarithm needs a strictly positive argument, so four minus t must be greater than zero.
\[ 4 - t > 0 \;\Longrightarrow\; t < 4 \]
Condition from the reciprocal
Why: Division by zero is undefined, so t cannot equal zero.
\[ t \ne 0 \]
Intersect the three conditions
Why: Keep only inputs that satisfy all three at once, removing the single point t equals zero.
\[ [-2,\,0) \cup (0,\,4) \]
Verify an endpoint and a hole
Why: At t equals negative two the root gives zero (allowed) and the other components are fine; at t equals zero the reciprocal blows up, so excluding it is correct.
Picture it
Animation
Shows: Each line of the worked example "finding a domain", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At t equals negative two the root gives zero (allowed) and the other components are fine; at t equals zero the reciprocal blows up, so excluding it is correct.
Trap
Combining the component domains by taking everything any one component allows.
\[ \text{dom} \overset{?}{=} \text{dom}(f) \cup \text{dom}(g) \cup \text{dom}(h) \]
For the previous function this would wrongly include t equal to five, where the logarithm of a negative number does not exist.
\[ \ln(4-5) = \ln(-1) \;\; \text{undefined} \]
The vector exists only when every component exists at the same input, so intersect the domains.
\[ \text{dom} = \text{dom}(f) \cap \text{dom}(g) \cap \text{dom}(h) \]
One undefined component makes the whole output undefined. The strictest condition wins.
Concept
The limit of a vector function is just the vector of the limits of its components, when each of those limits exists.
\[ \lim_{t\to a} \mathbf{r}(t) = \left\langle \lim_{t\to a} f(t),\; \lim_{t\to a} g(t),\; \lim_{t\to a} h(t)\right\rangle \]
If even one component limit fails to exist, the vector limit fails to exist.
Intuition
A limit asks where the tip of the arrow settles as the clock approaches a value.
Each coordinate settles on its own. Stack the three answers back into one arrow and you have the limiting vector.
Estimation
Predict first
Evaluate the limit as the parameter approaches zero.
Commit before you compute: what does Worked example: a componentwise limit come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify the middle component numerically
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Testing a tiny angle such as t equals one tenth gives sine over angle near 0.9983, consistent with a limit of one.
Worked example
Evaluate the limit as the parameter approaches zero.
\[ \lim_{t\to 0} \left\langle t^2 + 1,\; \frac{\sin t}{t},\; e^{t}\right\rangle \]
First component
Why: This is continuous, so substitute directly: zero squared plus one equals one.
\[ \lim_{t\to 0} (t^2 + 1) = 1 \]
Second component
Why: This is the classic trigonometric limit; the ratio of sine to its angle approaches one.
\[ \lim_{t\to 0} \frac{\sin t}{t} = 1 \]
Third component
Why: The exponential is continuous, so substitute: e to the zero is one.
\[ \lim_{t\to 0} e^{t} = 1 \]
Assemble the vector
Why: Stack the three component limits back into one vector.
\[ \langle 1,\, 1,\, 1\rangle \]
Verify the middle component numerically
Why: Testing a tiny angle such as t equals one tenth gives sine over angle near 0.9983, consistent with a limit of one.
Picture it
Animation
Shows: Each line of the worked example "a componentwise limit", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Testing a tiny angle such as t equals one tenth gives sine over angle near 0.9983, consistent with a limit of one.
Concept
A vector function is continuous at a point exactly when every component is continuous there.
continuous at a point — r is continuous at t equal to a when the limit as t approaches a equals r(a). For a vector function this must hold in each component at the same time.
Fill the middle
Fill in the blanks
From Worked example: where is it continuous — finish the line. Write what belongs on the right of the equals sign before you look.
\mathbf\left\langle \frac{1}{t-1},\; \sqrt{t},\; t^2\right\rangle(t) = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. A rational function is continuous except where its denominator is zero, so avoid t equal to one.
Worked example
Find every parameter where this function is continuous.
\[ \mathbf{r}(t) = \left\langle \frac{1}{t-1},\; \sqrt{t},\; t^2\right\rangle \]
First component
Why: A rational function is continuous except where its denominator is zero, so avoid t equal to one.
\[ t \ne 1 \]
Second component
Why: The square root is continuous only for nonnegative inputs.
\[ t \ge 0 \]
Third component
Why: A polynomial is continuous everywhere, so it adds no restriction.
Intersect the conditions
Why: The function is continuous where all components are, which is the nonnegative numbers with the single hole removed.
\[ [0,\,1) \cup (1,\,\infty) \]
Verify the trouble point
Why: At t equal to one the first component has a zero denominator, so continuity genuinely fails there and excluding it is correct.
Picture it
Animation
Shows: Each line of the worked example "where is it continuous", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At t equal to one the first component has a zero denominator, so continuity genuinely fails there and excluding it is correct.
Pattern
1. Split the vector into its components
Why: Domain, limits, continuity, derivatives, and integrals all reduce to the same operation on each scalar component.
2. Do the ordinary single-variable calculus on each component
Why: Use the tools you already know: limit laws, differentiation rules, and antiderivatives, one component at a time.
3. Reassemble the results into one vector
Why: Stack the component answers back together, remembering that a constant of integration is itself a vector.
Check
Apply the componentwise pattern to this limit.
\[ \lim_{t\to 0} \left\langle t^3,\; \cos t,\; \frac{e^{2t}-1}{t}\right\rangle \]
Check your understanding
What is the limit?
Answer: A
Why: First component: zero cubed is zero. Second: cosine of zero is one. Third: the ratio of e to the 2t minus one over t approaches 2, since the exponent carries a factor of two. So the limit is the vector with components zero, one, two.
Concept
The derivative of a vector function is defined by the same difference quotient as in first-year calculus, now with vector subtraction.
\[ \mathbf{r}'(t) = \lim_{\Delta t \to 0} \frac{\mathbf{r}(t+\Delta t) - \mathbf{r}(t)}{\Delta t} \]
Because subtraction and scaling of vectors happen componentwise, the limit passes into each component.
Picture it
Figure (svg): A curve with a tangent arrow drawn at one point pointing in the direction of travel
Discussion prompt
Read the picture before the words. What is this showing, and what is the one thing it is built to make obvious? Commit to an answer, then read on.
Hint: Name the parts, then say what changes between them — and if nothing changes, say what is being held still.
Answer:
The difference vector connects a point on the curve to a nearby point. As the two points close in, that little arrow lines up with the curve.
Intuition
The difference vector connects a point on the curve to a nearby point. As the two points close in, that little arrow lines up with the curve.
So the derivative is a tangent vector. If the parameter is time, it is the velocity: the direction of travel, with length equal to the speed.
Figure (svg): A curve with a tangent arrow drawn at one point pointing in the direction of travel
Concept
Differentiate each component separately. That is the whole rule.
\[ \mathbf{r}'(t) = \langle f'(t),\, g'(t),\, h'(t)\rangle \]
Every single-variable rule you know, power, product, chain, applies inside each component.
Worked example
Differentiate this vector function.
\[ \mathbf{r}(t) = \langle t^2,\; \sin t,\; e^{3t}\rangle \]
First component
Why: Power rule: the derivative of t squared is two t.
\[ \frac{d}{dt}\, t^2 = 2t \]
Second component
Why: The derivative of sine is cosine.
\[ \frac{d}{dt}\, \sin t = \cos t \]
Third component
Why: Chain rule: the inside function three t has derivative three, which multiplies the exponential.
\[ \frac{d}{dt}\, e^{3t} = 3e^{3t} \]
Assemble the derivative
Why: Collect the three component derivatives into one vector.
\[ \mathbf{r}'(t) = \langle 2t,\; \cos t,\; 3e^{3t}\rangle \]
Verify at t equals zero
Why: The formula gives velocity zero, one, three at the start; each piece matches the known derivative of its component, confirming the result.
\[ \mathbf{r}'(0) = \langle 0,\, 1,\, 3\rangle \]
Picture it
Animation
Shows: Each line of the worked example "computing r prime of t", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The formula gives velocity zero, one, three at the start; each piece matches the known derivative of its component, confirming the result.
Trap
Trying to differentiate the length of the vector and calling that the derivative of the vector.
\[ \mathbf{r}(t) = \langle \cos t,\, \sin t\rangle,\qquad |\mathbf{r}(t)| = 1 \]
The length is the constant one, so this wrongly concludes the derivative is zero, even though the particle is clearly moving.
\[ \frac{d}{dt}\,|\mathbf{r}(t)| = 0 \;\; \text{(not the derivative of }\mathbf{r}) \]
Differentiate each component; do not collapse the vector to a number first.
\[ \mathbf{r}'(t) = \langle -\sin t,\; \cos t\rangle \]
This velocity is never the zero vector; its length is one, so the particle moves at constant speed around the circle.
\[ |\mathbf{r}'(t)| = \sqrt{\sin^2 t + \cos^2 t} = 1 \]
Concept
A tangent line uses a point on the curve and the tangent direction from the derivative.
\[ \mathbf{L}(s) = \mathbf{r}(t_0) + s\,\mathbf{r}'(t_0) \]
This is just the point-and-direction recipe for a line, with the direction supplied by the velocity.
Fill the middle
Fill in the blanks
From Worked example: a tangent line — finish the line. Write what belongs on the right of the equals sign before you look.
\mathbf\langle 1,\, 1,\, 1\rangle(1) = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. Evaluate the function at one to get the point the line passes through.
Worked example
Find the tangent line at the parameter value one.
\[ \mathbf{r}(t) = \langle t^2,\; t^3,\; t\rangle \]
Find the point
Why: Evaluate the function at one to get the point the line passes through.
\[ \mathbf{r}(1) = \langle 1,\, 1,\, 1\rangle \]
Find the direction
Why: Differentiate componentwise, then evaluate at one to get the tangent direction.
\[ \mathbf{r}'(t) = \langle 2t,\, 3t^2,\, 1\rangle,\qquad \mathbf{r}'(1) = \langle 2,\, 3,\, 1\rangle \]
Write the line
Why: Combine the point and direction with a new parameter.
\[ \mathbf{L}(s) = \langle 1,\, 1,\, 1\rangle + s\,\langle 2,\, 3,\, 1\rangle \]
Verify the line touches the curve
Why: At s equal to zero the line gives the point one, one, one, exactly the point on the curve, so the tangent line is anchored correctly.
Picture it
Animation
Shows: Each line of the worked example "a tangent line", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At s equal to zero the line gives the point one, one, one, exactly the point on the curve, so the tangent line is anchored correctly.
Concept
If the function gives position, the first derivative is velocity and the second derivative is acceleration.
\[ \mathbf{v}(t) = \mathbf{r}'(t),\qquad \mathbf{a}(t) = \mathbf{r}''(t) \]
Velocity and acceleration are vectors. Speed is different.
Intuition
Velocity carries both a direction and a rate. Strip away the direction and keep only how fast, and you have speed.
Speed is a single number, the magnitude of the velocity vector.
\[ \text{speed} = |\mathbf{v}(t)| = |\mathbf{r}'(t)| \]
Step zero
Discussion prompt
Worked example: velocity, speed, acceleration — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Differentiate for velocity
Answer:
Worked example
For this helical motion find velocity, speed, and acceleration.
\[ \mathbf{r}(t) = \langle 3\cos t,\; 3\sin t,\; 4t\rangle \]
Differentiate for velocity
Why: Differentiate each component; the constant multiples ride along.
\[ \mathbf{v}(t) = \langle -3\sin t,\; 3\cos t,\; 4\rangle \]
Take the magnitude for speed
Why: Square the components and add: nine sine squared plus nine cosine squared is nine, plus sixteen is twenty-five.
\[ |\mathbf{v}(t)| = \sqrt{9\sin^2 t + 9\cos^2 t + 16} = \sqrt{25} = 5 \]
Differentiate again for acceleration
Why: Differentiate the velocity componentwise; the constant vertical component has derivative zero.
\[ \mathbf{a}(t) = \langle -3\cos t,\; -3\sin t,\; 0\rangle \]
Verify the speed is constant
Why: The trig identity makes the sine and cosine terms combine to a constant, so the speed is exactly five for every t, a clean sanity check.
Picture it
Animation
Shows: Each line of the worked example "velocity, speed, acceleration", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The trig identity makes the sine and cosine terms combine to a constant, so the speed is exactly five for every t, a clean sanity check.
Check
Read the question carefully: it asks for speed, a number.
\[ \mathbf{r}(t) = \langle \sin t,\; \cos t,\; t\rangle \]
Check your understanding
What is the speed at t equals zero?
Answer: A
Why: Velocity is cosine, negative sine, one, which at zero is the vector 1, 0, 1. Speed is its magnitude, the square root of one plus zero plus one, which is the square root of 2.
Concept
Sums and scalar multiples behave exactly as expected.
\[ (\mathbf{u} + \mathbf{v})' = \mathbf{u}' + \mathbf{v}', \qquad (c\,\mathbf{u})' = c\,\mathbf{u}' \]
A scalar function times a vector function needs a product rule, just like in single-variable calculus.
\[ \big(k(t)\,\mathbf{u}(t)\big)' = k'(t)\,\mathbf{u}(t) + k(t)\,\mathbf{u}'(t) \]
Concept
The dot product of two vector functions is a scalar function, and it obeys a product rule.
\[ \frac{d}{dt}\big(\mathbf{u}\cdot\mathbf{v}\big) = \mathbf{u}'\cdot\mathbf{v} + \mathbf{u}\cdot\mathbf{v}' \]
The order of the factors in each dot product does not matter, because the dot product is commutative.
Worked example
Differentiate the dot product of these two functions.
\[ \mathbf{u}(t) = \langle t,\, t^2,\, 1\rangle,\qquad \mathbf{v}(t) = \langle 1,\, t,\, t^3\rangle \]
Differentiate each factor
Why: Differentiate componentwise to get the two derivative vectors.
\[ \mathbf{u}'(t) = \langle 1,\, 2t,\, 0\rangle,\qquad \mathbf{v}'(t) = \langle 0,\, 1,\, 3t^2\rangle \]
Apply the product rule
Why: Add the derivative of the first dotted with the second to the first dotted with the derivative of the second.
\[ \mathbf{u}'\cdot\mathbf{v} + \mathbf{u}\cdot\mathbf{v}' = (1 + 2t^2) + (t^2 + 3t^2) \]
Simplify
Why: Combine like terms to get the derivative of the scalar dot product.
\[ 1 + 6t^2 \]
Verify by expanding first
Why: Compute the dot product directly as t plus two t cubed, then differentiate to get one plus six t squared, matching the product-rule answer exactly.
\[ \mathbf{u}\cdot\mathbf{v} = t + 2t^3 \;\Longrightarrow\; \frac{d}{dt}(t + 2t^3) = 1 + 6t^2 \]
Picture it
Animation
Shows: Each line of the worked example "differentiating a dot product", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Compute the dot product directly as t plus two t cubed, then differentiate to get one plus six t squared, matching the product-rule answer exactly.
Trap
Guessing that the derivative of a dot product is just the dot product of the derivatives.
\[ \frac{d}{dt}(\mathbf{u}\cdot\mathbf{v}) \overset{?}{=} \mathbf{u}'\cdot\mathbf{v}' \]
For the previous example this gives the dot product of the derivative vectors, which is wrong.
\[ \mathbf{u}'\cdot\mathbf{v}' = (1)(0) + (2t)(1) + (0)(3t^2) = 2t \]
Use the product rule: keep both cross terms.
\[ \frac{d}{dt}(\mathbf{u}\cdot\mathbf{v}) = \mathbf{u}'\cdot\mathbf{v} + \mathbf{u}\cdot\mathbf{v}' \]
That gives one plus six t squared, which matches expanding first. The naive answer of two t is off by the missing term.
\[ \mathbf{u}'\cdot\mathbf{v} + \mathbf{u}\cdot\mathbf{v}' = 1 + 6t^2 \]
Concept
The cross product of two vector functions is another vector function, and it also has a product rule.
\[ \frac{d}{dt}\big(\mathbf{u}\times\mathbf{v}\big) = \mathbf{u}'\times\mathbf{v} + \mathbf{u}\times\mathbf{v}' \]
Here the order matters. The cross product is not commutative, so keep the first factor first in every term.
Estimation
Predict first
Differentiate the cross product of these two functions.
Commit before you compute: what does Worked example: differentiating a cross product come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify by crossing first
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Directly, the cross product is zero, zero, t squared, whose derivative is zero, zero, two t, matching the product-rule result.
Worked example
Differentiate the cross product of these two functions.
\[ \mathbf{u}(t) = \langle t,\, 0,\, 0\rangle,\qquad \mathbf{v}(t) = \langle 0,\, t,\, 0\rangle \]
Differentiate each factor
Why: Componentwise differentiation gives two constant derivative vectors.
\[ \mathbf{u}'(t) = \langle 1,\, 0,\, 0\rangle,\qquad \mathbf{v}'(t) = \langle 0,\, 1,\, 0\rangle \]
First term of the product rule
Why: Cross the derivative of the first with the second, keeping the order.
\[ \mathbf{u}'\times\mathbf{v} = \langle 1,0,0\rangle \times \langle 0,t,0\rangle = \langle 0,\, 0,\, t\rangle \]
Second term of the product rule
Why: Cross the first with the derivative of the second, again keeping the order.
\[ \mathbf{u}\times\mathbf{v}' = \langle t,0,0\rangle \times \langle 0,1,0\rangle = \langle 0,\, 0,\, t\rangle \]
Add the terms
Why: Sum the two contributions to get the derivative of the cross product.
\[ \mathbf{u}'\times\mathbf{v} + \mathbf{u}\times\mathbf{v}' = \langle 0,\, 0,\, 2t\rangle \]
Verify by crossing first
Why: Directly, the cross product is zero, zero, t squared, whose derivative is zero, zero, two t, matching the product-rule result.
\[ \mathbf{u}\times\mathbf{v} = \langle 0,0,t^2\rangle \;\Longrightarrow\; \frac{d}{dt}\langle 0,0,t^2\rangle = \langle 0,0,2t\rangle \]
Picture it
Animation
Shows: Each line of the worked example "differentiating a cross product", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Directly, the cross product is zero, zero, t squared, whose derivative is zero, zero, two t, matching the product-rule result.
Concept
If the parameter is itself a function of another variable, the chain rule threads through.
\[ \frac{d}{ds}\,\mathbf{r}\big(k(s)\big) = k'(s)\,\mathbf{r}'\big(k(s)\big) \]
The scalar derivative of the inner function multiplies the whole velocity vector.
Concept
A curve is smooth when it has a continuously turning tangent that never stalls.
smooth curve — On an interval, r is smooth when its derivative is continuous and is never the zero vector there. A zero velocity can allow a sharp corner or cusp.
Where the velocity becomes the zero vector, the curve can reverse or form a cusp, so smoothness can fail.
Matching
Match the pairs
Match each term to the definition this lesson gave it — not the one you would guess from the word.
Why: These are the working definitions of vector-valued function, space curve, continuous at a point, smooth curve as Week 3 - Vector-Valued Functions uses them. Pairing them correctly is the test of whether you could state each one with the slide switched off.
Ranking
Put in order
Put the moves of Worked example: finding where smoothness fails into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. Componentwise differentiation gives the velocity vector.
Worked example
Decide where this plane curve fails to be smooth.
\[ \mathbf{r}(t) = \langle t^3,\; t^2\rangle \]
Differentiate
Why: Componentwise differentiation gives the velocity vector.
\[ \mathbf{r}'(t) = \langle 3t^2,\; 2t\rangle \]
Set the velocity to the zero vector
Why: Smoothness can fail only where every component of the velocity is zero at once.
\[ 3t^2 = 0 \;\text{and}\; 2t = 0 \;\Longrightarrow\; t = 0 \]
Interpret
Why: At the parameter zero the velocity vanishes, and the curve has a cusp there, so it is smooth for every nonzero parameter but not at zero.
Verify the cusp
Why: Near zero the horizontal component changes far slower than the vertical, producing the sharp point, which confirms the velocity truly vanishes only at t equal to zero.
Picture it
Animation
Shows: Each line of the worked example "finding where smoothness fails", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Near zero the horizontal component changes far slower than the vertical, producing the sharp point, which confirms the velocity truly vanishes only at t equal to zero.
Trap
Declaring a curve smooth just because its components are polynomials.
\[ \mathbf{r}(t) = \langle t^3,\, t^2\rangle \;\;\text{smooth for all } t\;? \]
Polynomials are differentiable, but that is not the test. This curve has a cusp at the origin.
Smoothness requires the velocity never be the zero vector on the interval.
\[ \mathbf{r}'(0) = \langle 0,\, 0\rangle \]
Since the velocity vanishes at zero, the curve is smooth only away from that point, not everywhere.
Break the constraint
Discussion prompt
The rule this trap just fixed:
Smoothness requires the velocity never be the zero vector on the interval.
Now break it on purpose. Build a case that violates it and follow the consequences until something visibly fails. Where does the failure first show up — and would you have noticed it if you had not been looking?
Hint: The dangerous rules are the ones whose violation still produces an answer. If yours fails loudly, try to find one that fails quietly.
Check
Differentiate componentwise, and watch the chain rule.
\[ \mathbf{r}(t) = \langle e^{2t},\; t^2 + 3,\; \sin(3t)\rangle \]
Check your understanding
What is r prime of t?
Answer: A
Why: First component uses the chain rule: the derivative of e to the 2t is 2 e to the 2t. Second: the derivative of t squared plus three is 2t. Third: the chain rule on sine of 3t gives 3 cosine 3t.
Concept
To integrate a vector function, integrate each component. Integration undoes componentwise differentiation.
\[ \int \mathbf{r}(t)\,dt = \left\langle \int f(t)\,dt,\; \int g(t)\,dt,\; \int h(t)\,dt\right\rangle \]
Concept
Each component antiderivative brings its own arbitrary constant. Collected together, those constants form a single arbitrary vector.
\[ \int \mathbf{r}(t)\,dt = \mathbf{R}(t) + \mathbf{C}, \qquad \mathbf{C} = \langle C_1,\, C_2,\, C_3\rangle \]
This is the single most common slip in the whole topic: the constant must be a vector, not one number added everywhere.
Worked example
Integrate this vector function.
\[ \int \langle 2t,\; \cos t,\; e^{t}\rangle \, dt \]
Integrate the first component
Why: The antiderivative of two t is t squared.
\[ \int 2t\,dt = t^2 \]
Integrate the second component
Why: The antiderivative of cosine is sine.
\[ \int \cos t\,dt = \sin t \]
Integrate the third component
Why: The exponential is its own antiderivative.
\[ \int e^{t}\,dt = e^{t} \]
Assemble with a vector constant
Why: Stack the antiderivatives and add one arbitrary vector constant, not a scalar.
\[ \langle t^2,\; \sin t,\; e^{t}\rangle + \mathbf{C} \]
Verify by differentiating back
Why: Differentiating the result componentwise returns two t, cosine, e to the t, the original integrand, so the antiderivative is correct.
\[ \frac{d}{dt}\langle t^2,\, \sin t,\, e^{t}\rangle = \langle 2t,\, \cos t,\, e^{t}\rangle \]
Picture it
Animation
Shows: Each line of the worked example "an indefinite integral", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Differentiating the result componentwise returns two t, cosine, e to the t, the original integrand, so the antiderivative is correct.
Trap
Adding one number to the whole vector after integrating.
\[ \int \langle 2t,\, \cos t,\, e^{t}\rangle\,dt \overset{?}{=} \langle t^2,\, \sin t,\, e^{t}\rangle + C \]
A single scalar cannot be added to a vector, and it cannot capture three independent constants of integration.
Use a vector constant so each component gets its own freedom.
\[ \langle t^2,\, \sin t,\, e^{t}\rangle + \langle C_1,\, C_2,\, C_3\rangle \]
Only then can an initial condition pin down all three constants at once.
Hypothesis
Predict first
Worked example: a definite integral is about to be worked. State your hypothesis first: which rule or definition decides this one, and what is the first move it forces? Then watch whether the example agrees with you.
Correct: Antidifferentiate componentwise
Why: Integrate each component; no constant is needed for a definite integral.
A hypothesis you wrote down is falsifiable; a vague sense of how it will go is not. If the example opens somewhere else, that gap is the thing worth chasing.
Worked example
Evaluate this definite integral over the interval from zero to one.
\[ \int_{0}^{1} \langle 2t,\; 3t^2,\; 1\rangle \, dt \]
Antidifferentiate componentwise
Why: Integrate each component; no constant is needed for a definite integral.
\[ \left[\,\langle t^2,\; t^3,\; t\rangle\,\right]_{0}^{1} \]
Evaluate at the bounds
Why: Plug in one and subtract the value at zero, componentwise.
\[ \langle 1,\, 1,\, 1\rangle - \langle 0,\, 0,\, 0\rangle = \langle 1,\, 1,\, 1\rangle \]
Verify each component separately
Why: The three scalar integrals give one each by the power rule and the integral of a constant, confirming the vector answer.
Picture it
Animation
Shows: Each line of the worked example "a definite integral", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The three scalar integrals give one each by the power rule and the integral of a constant, confirming the vector answer.
Concept
Given a derivative and a starting value, integrate to recover the function, then use the starting value to find the constant vector.
Substitute the initial parameter, set the result equal to the known vector, and solve for the constant, one component at a time.
Step zero
Discussion prompt
Worked example: integrating with an initial condition — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Integrate the derivative
Answer:
Worked example
Recover the position from its velocity and starting point.
\[ \mathbf{r}'(t) = \langle 2t,\; 1,\; e^{t}\rangle, \qquad \mathbf{r}(0) = \langle 1,\, 0,\, 2\rangle \]
Integrate the derivative
Why: Antidifferentiate each component and add a vector constant.
\[ \mathbf{r}(t) = \langle t^2,\; t,\; e^{t}\rangle + \mathbf{C} \]
Apply the initial condition
Why: Substitute t equal to zero; the polynomial pieces vanish and the exponential gives one.
\[ \mathbf{r}(0) = \langle 0,\, 0,\, 1\rangle + \mathbf{C} = \langle 1,\, 0,\, 2\rangle \]
Solve for the constant vector
Why: Subtract componentwise to isolate the vector constant.
\[ \mathbf{C} = \langle 1,\, 0,\, 1\rangle \]
Write the solution
Why: Put the constant back into the antiderivative.
\[ \mathbf{r}(t) = \langle t^2 + 1,\; t,\; e^{t} + 1\rangle \]
Verify the initial condition
Why: At t equal to zero the solution gives one, zero, two, exactly the given starting vector, so the constant was found correctly.
Picture it
Animation
Shows: Each line of the worked example "integrating with an initial condition", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At t equal to zero the solution gives one, zero, two, exactly the given starting vector, so the constant was found correctly.
Check
Integrate componentwise, then fix the vector constant with the starting value.
\[ \mathbf{r}'(t) = \langle 6t,\; \cos t,\; 2\rangle, \qquad \mathbf{r}(0) = \langle 2,\, 1,\, 0\rangle \]
Check your understanding
What is r of t?
Answer: A
Why: Integrating gives 3t squared, sine t, 2t plus a vector constant. At zero that is 0, 0, 0 plus C equals 2, 1, 0, so C is 2, 1, 0, giving 3t squared plus 2, sine t plus 1, 2t.
Concept
Near the ground the only acceleration is gravity, pointing straight down.
\[ \mathbf{a}(t) = \langle 0,\; -g\rangle \]
Integrate acceleration once with the initial velocity to get velocity, then again with the initial position to get position.
Intuition
The horizontal motion coasts at constant velocity because nothing pushes sideways.
All the curving comes from the vertical component, where gravity steadily pulls the speed down and then back up in reverse.
Missing information
Discussion prompt
A projectile launches from a height of six feet with initial velocity fifty feet per second across and forty feet per second up. Use gravity of thirty-two feet per second squared.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
Antidifferentiate componentwise and add a vector constant.
Worked example
A projectile launches from a height of six feet with initial velocity fifty feet per second across and forty feet per second up. Use gravity of thirty-two feet per second squared.
\[ \mathbf{a}(t) = \langle 0,\; -32\rangle,\quad \mathbf{v}(0) = \langle 50,\, 40\rangle,\quad \mathbf{s}(0) = \langle 0,\, 6\rangle \]
Integrate acceleration for velocity
Why: Antidifferentiate componentwise and add a vector constant.
\[ \mathbf{v}(t) = \langle 0,\; -32t\rangle + \mathbf{C}_1 \]
Use the initial velocity
Why: Set t equal to zero and match the given initial velocity to find the first constant.
\[ \mathbf{C}_1 = \langle 50,\, 40\rangle \;\Longrightarrow\; \mathbf{v}(t) = \langle 50,\; 40 - 32t\rangle \]
Integrate velocity for position
Why: Antidifferentiate again componentwise and add a new vector constant.
\[ \mathbf{s}(t) = \langle 50t,\; 40t - 16t^2\rangle + \mathbf{C}_2 \]
Use the initial position
Why: Set t equal to zero and match the given starting point to find the second constant.
\[ \mathbf{C}_2 = \langle 0,\, 6\rangle \;\Longrightarrow\; \mathbf{s}(t) = \langle 50t,\; 40t - 16t^2 + 6\rangle \]
Verify both initial conditions
Why: At t equal to zero position is zero, six and velocity is fifty, forty, both matching the givens, so the integration constants are correct.
Picture it
Animation
Shows: Each line of the worked example "projectile position", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At t equal to zero position is zero, six and velocity is fifty, forty, both matching the givens, so the integration constants are correct.
Fill the middle
Fill in the blanks
From Worked example: maximum height — finish the line. Write what belongs on the right of the equals sign before you look.
\mathbf\langle 50,\; 40 - 32t\rangle(t) = ___
Why: Producing the right-hand side unprompted is the difference between recognising this line and being able to use it. At the top of the arc the upward speed is momentarily zero.
Worked example
Using the projectile from the previous example, find the greatest height reached.
\[ \mathbf{v}(t) = \langle 50,\; 40 - 32t\rangle \]
Set the vertical velocity to zero
Why: At the top of the arc the upward speed is momentarily zero.
\[ 40 - 32t = 0 \;\Longrightarrow\; t = 1.25 \]
Evaluate the height there
Why: Substitute the time into the vertical position component.
\[ 40(1.25) - 16(1.25)^2 + 6 = 50 - 25 + 6 = 31 \]
Verify the answer is a maximum
Why: The vertical velocity is positive before this time and negative after, so the height thirty-one feet is indeed the peak.
Picture it
Animation
Shows: Each line of the worked example "maximum height", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The vertical velocity is positive before this time and negative after, so the height thirty-one feet is indeed the peak.
Concept
Here is a beautiful and useful fact. If a vector function keeps a constant length, its position is always perpendicular to its velocity.
\[ |\mathbf{r}(t)| = \text{const} \;\Longrightarrow\; \mathbf{r}(t)\cdot\mathbf{r}'(t) = 0 \]
Intuition
Constant length means the tip stays on a sphere centered at the origin. To stay on the sphere the tip can only move along the surface.
Motion along the surface is perpendicular to the radius. The radius is the position vector, so velocity is perpendicular to position.
Missing information
Discussion prompt
Prove that a constant length forces position and velocity to be perpendicular.
What do you need to know — or decide — before the first line can be written? List everything the problem has to hand you.
Hint: Anything you would have to invent to get started is a thing the problem must supply.
Answer:
The right side is constant, so its derivative is zero; differentiate the dot product with the product rule.
Worked example
Prove that a constant length forces position and velocity to be perpendicular.
\[ |\mathbf{r}(t)|^2 = \mathbf{r}(t)\cdot\mathbf{r}(t) = \text{const} \]
Differentiate both sides
Why: The right side is constant, so its derivative is zero; differentiate the dot product with the product rule.
\[ \frac{d}{dt}\big(\mathbf{r}\cdot\mathbf{r}\big) = \mathbf{r}'\cdot\mathbf{r} + \mathbf{r}\cdot\mathbf{r}' = 2\,\mathbf{r}\cdot\mathbf{r}' \]
Set the derivative to zero
Why: Because the length is constant, the whole expression equals zero.
\[ 2\,\mathbf{r}\cdot\mathbf{r}' = 0 \;\Longrightarrow\; \mathbf{r}\cdot\mathbf{r}' = 0 \]
Conclude
Why: A zero dot product means the two vectors are orthogonal.
Verify on the unit circle
Why: For position cosine t, sine t the velocity is negative sine t, cosine t, and their dot product is negative cosine sine plus sine cosine, which is zero, confirming the theorem.
\[ \langle \cos t, \sin t\rangle \cdot \langle -\sin t, \cos t\rangle = -\cos t\sin t + \sin t\cos t = 0 \]
Picture it
Animation
Shows: Each line of the worked example "proving the orthogonality property", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: For position cosine t, sine t the velocity is negative sine t, cosine t, and their dot product is negative cosine sine plus sine cosine, which is zero, confirming the theorem.
Trap
Believing the orthogonality holds for every curve, no matter what.
\[ \mathbf{r}(t)\cdot\mathbf{r}'(t) \overset{?}{=} 0 \;\; \text{always} \]
Take a straight ray outward; its length grows, and position and velocity point the same way.
\[ \mathbf{r}(t) = \langle t,\, t\rangle,\quad \mathbf{r}'(t) = \langle 1,\, 1\rangle,\quad \mathbf{r}\cdot\mathbf{r}' = 2t \]
The property holds only when the length is constant. That is the hypothesis.
\[ |\mathbf{r}(t)| = \text{const} \;\Longrightarrow\; \mathbf{r}\cdot\mathbf{r}' = 0 \]
For the growing ray the dot product is two t, zero only at the origin, so position and velocity are not perpendicular in general.
Two truths and a lie
Sort into buckets
Some of these hold up and some are the exact mistakes this lesson is built to prevent. Sort them.
Elimination
Eliminate the wrong options
What must the dot product of r and its derivative equal?
3 of these 4 are wrong. Strike them one at a time, and say what rules each one out before you strike the next. The survivor is the answer.
Survives elimination: A
Why: Differentiating r dotted with r gives twice r dotted with r prime, and this equals the derivative of a constant magnitude squared, which is zero. So r dotted with r prime is zero.
Check
Suppose a vector function has constant magnitude for all values of the parameter.
Check your understanding
What must the dot product of r and its derivative equal?
Answer: A
Why: Differentiating r dotted with r gives twice r dotted with r prime, and this equals the derivative of a constant magnitude squared, which is zero. So r dotted with r prime is zero.
Pattern
1. Identify what you are given
Why: Usually acceleration or velocity plus one or two initial conditions.
2. Integrate up, adding a vector constant each time
Why: Acceleration to velocity, velocity to position; every integration introduces its own vector constant.
3. Pin down each constant with an initial condition
Why: Substitute the starting parameter and match componentwise to solve for the constant vectors.
4. Answer the actual question
Why: Speed is the magnitude of velocity; peak height sets the vertical velocity to zero; distance travelled integrates the speed.
Real world
Discussion prompt
Outside this lesson: where does Week 3 - Vector-Valued Functions actually turn up? Name one concrete situation — a job, a piece of software someone ships, a decision somebody has to make — and say which part of Recipe for motion problems is doing the work in it.
Hint: Vague is the failure mode here. "Engineering" is not a situation; "deciding whether this build is fast enough to ship" is.
Answer:
That deck covers vector-valued functions and the space curves they trace: domain, limits and continuity taken componentwise, and the derivative read as a tangent and velocity vector. It then gives the product rules for dot and cross products, defines smooth curves, integrates with a vector constant, works projectile motion, and shows why a curve of constant magnitude has its position vector orthogonal to its velocity. It targets three classic errors: differentiating a dot or cross product without the product rule, treating the constant of integration as a scalar, and forgetting that the derivative is taken component by component.
Concept
The distance a particle travels over a time interval is the accumulated speed, the integral of the magnitude of velocity.
\[ \text{distance} = \int_{a}^{b} |\mathbf{r}'(t)| \, dt \]
Integrate the magnitude, a scalar, not the velocity vector itself.
Explain it
Discussion prompt
Explain Distance travelled is the integral of speed to a student a year behind you. No notation, no jargon they have not met — and it still has to be true.
Hint: If your explanation needs a symbol they have never seen, you are describing the notation rather than the idea.
Answer:
The distance a particle travels over a time interval is the accumulated speed, the integral of the magnitude of velocity.
Estimation
Predict first
Find the distance travelled from time zero to time two along the earlier helix.
Commit before you compute: what does Worked example: distance along a helix come out to? A rough magnitude and the right form is enough — the point is to have something concrete to be wrong about.
Correct: Verify with a sanity check
Why: A prediction you can defend turns the computation into a check rather than a leap of faith — and an answer that contradicts it is caught on the spot. Constant speed five for two time units gives distance ten, exactly the integral, which confirms the result.
Worked example
Find the distance travelled from time zero to time two along the earlier helix.
\[ \mathbf{r}(t) = \langle 3\cos t,\; 3\sin t,\; 4t\rangle \]
Recall the speed
Why: This helix has constant speed five, found earlier from the magnitude of the velocity.
\[ |\mathbf{r}'(t)| = 5 \]
Integrate the speed
Why: Integrate the constant speed over the interval from zero to two.
\[ \int_{0}^{2} 5 \, dt = 5(2) = 10 \]
Verify with a sanity check
Why: Constant speed five for two time units gives distance ten, exactly the integral, which confirms the result.
Picture it
Animation
Shows: Each line of the worked example "distance along a helix", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: Constant speed five for two time units gives distance ten, exactly the integral, which confirms the result.
Concept
The parameter increasing gives the curve a direction, an orientation, that the bare set of points does not carry.
Replacing the parameter with its negative, for instance, traces the same points in the opposite order, which matters for line integrals later in the course.
Analogy
Discussion prompt
Explain Orientation: the direction of travel by analogy to something with no Calculus III in it at all — a queue, a recipe, a map, a bank balance, whatever fits. Then say where your analogy breaks.
Hint: An analogy that never breaks is not an analogy, it is the same idea wearing a hat. Find the seam — that is the part that is actually new.
Answer:
The parameter increasing gives the curve a direction, an orientation, that the bare set of points does not carry.
Step zero
Discussion prompt
Worked example: recognizing a line — before any calculation: what is the plan? Name the moves in order, in plain English, without doing the arithmetic.
Hint: It starts with: Read off a point
Answer:
Worked example
Describe the curve traced by this vector function.
\[ \mathbf{r}(t) = \langle 1 + 2t,\; 3 - t,\; t\rangle \]
Read off a point
Why: Setting the parameter to zero gives a point the curve passes through.
\[ \mathbf{r}(0) = \langle 1,\, 3,\, 0\rangle \]
Read off a direction
Why: The coefficients of the parameter form the constant direction vector, so the derivative is constant.
\[ \mathbf{r}'(t) = \langle 2,\, -1,\, 1\rangle \]
Identify the curve
Why: A constant velocity means straight-line motion through the point in the direction of that vector.
Verify a second point lies on it
Why: At t equal to one the function gives three, two, one, which equals the base point plus the direction vector, confirming it is the line.
\[ \mathbf{r}(1) = \langle 3,\, 2,\, 1\rangle = \langle 1,3,0\rangle + \langle 2,-1,1\rangle \]
Picture it
Animation
Shows: Each line of the worked example "recognizing a line", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: At t equal to one the function gives three, two, one, which equals the base point plus the direction vector, confirming it is the line.
Concept
When a scalar function multiplies a vector function, use the product rule, keeping the scalar and vector pieces in place.
\[ \big(k(t)\,\mathbf{u}(t)\big)' = k'(t)\,\mathbf{u}(t) + k(t)\,\mathbf{u}'(t) \]
Counterexample
Discussion prompt
When a scalar function multiplies a vector function, use the product rule, keeping the scalar and vector pieces in place.
That is stated as though it always holds. Do one of two things: produce a case where it fails, or say precisely what rules such a case out. "It just does" is not on the menu.
Hint: Hunt at the extremes first — zero, one, negative, empty, equal. If every extreme survives, the reason they survive is the proof.
Ranking
Put in order
Put the moves of Worked example: scalar times vector derivative into the order they have to happen.
Why: These are the moves of the worked example in the order it makes them, and each one is set up by the one before it. The scalar derivative is two t and the vector derivative is componentwise.
Worked example
Differentiate the scalar function times the vector function.
\[ k(t) = t^2,\qquad \mathbf{u}(t) = \langle \cos t,\; t,\; 1\rangle \]
Differentiate each piece
Why: The scalar derivative is two t and the vector derivative is componentwise.
\[ k'(t) = 2t,\qquad \mathbf{u}'(t) = \langle -\sin t,\; 1,\; 0\rangle \]
Apply the product rule
Why: Add the scalar derivative times the vector to the scalar times the vector derivative.
\[ 2t\,\langle \cos t, t, 1\rangle + t^2\,\langle -\sin t, 1, 0\rangle \]
Combine componentwise
Why: Add matching components to get one vector.
\[ \langle 2t\cos t - t^2\sin t,\; 3t^2,\; 2t\rangle \]
Verify by multiplying first
Why: The product is t squared cosine t, t cubed, t squared, whose componentwise derivative matches the answer exactly.
\[ \frac{d}{dt}\langle t^2\cos t,\, t^3,\, t^2\rangle = \langle 2t\cos t - t^2\sin t,\; 3t^2,\; 2t\rangle \]
Picture it
Animation
Shows: Each line of the worked example "scalar times vector derivative", appearing one at a time.
The same working the example does, in the order a tutor would write it.
Takeaway: The product is t squared cosine t, t cubed, t squared, whose componentwise derivative matches the answer exactly.
Connect it up
Draw it
One page, no notation unless you need it: draw how these connect — The master pattern: work componentwise · Recipe for motion problems · A vector-valued function · Picture a moving particle · The component functions carry the meaning. Put an arrow wherever one of them is what makes another possible, and label the arrow with why.
Recap
A vector-valued function packages several scalar functions into one moving arrow, and its tip traces a space curve.
Domain, limits, continuity, derivatives, and integrals are all handled one component at a time.
The derivative is a tangent and velocity vector; its magnitude is the speed, a single number.
| Operation | How to do it |
|---|---|
| Derivative of a dot product | Product rule: u prime dot v plus u dot v prime |
| Derivative of a cross product | Product rule, keeping the order of factors |
| Integral of a vector function | Integrate componentwise, add a vector constant |
| Constant length | Position is orthogonal to velocity |
Next up: unit tangent and normal vectors, arc length, and curvature, which build directly on the derivative and speed you mastered here.
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