The hyperbolic functions as the even and odd parts of the exponential, the identity that names them, their derivatives and integrals, the inverse functions with their logarithmic formulas, and the catenary.
Subject: Calculus I · 65 slides · symbolic lesson
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Title
Calculus I · Chapter 6 — Applications of Integration
Calculus of the Hyperbolic Functions
Objectives
Five outcomes. A new family built entirely from the exponential, with a parallel to trigonometry that is exact but not blind.
OpenStax Calculus Volume 1, §6.9 Calculus of the Hyperbolic Functions §6.9, pp. 646-656 — the section these objectives are drawn from
Warm-up
Section 6.7 built the exponential securely, and Section 5.4 distinguished even functions from odd ones.
Discussion prompt
The exponential is neither even nor odd. Can it be split into an even part and an odd part?
Hint: Average it with its reflection, and take half the difference.
Answer:
Averaging any function with its reflection gives an even function, and half their difference gives an odd one — and the two add back to the original. Applied to the exponential, those two halves are given names.
\[ \cosh x = \frac{e^{x}+e^{-x}}{2}, \qquad \sinh x = \frac{e^{x}-e^{-x}}{2} \]
The names are borrowed from trigonometry, and the borrowing is not arbitrary: these functions satisfy identities, derivative rules and integral formulas that mirror the circular ones almost exactly. Almost — and this section is largely about where the mirror flips a sign.
Concept
The hyperbolic cosine and sine are the even and odd parts of the exponential. Everything else in this section follows from those two elementary definitions.
the hyperbolic functions — The even and odd parts of the exponential function, together with the four quotients formed from them in the same way as the trigonometric functions.
\[ \cosh x = \frac{e^{x}+e^{-x}}{2}, \quad \sinh x = \frac{e^{x}-e^{-x}}{2} \]
Because the definitions are elementary, every property is provable by a short computation. There is no new theory here — only a new family assembled from a function already secured.
Figure (svg): The hyperbolic sine and cosine as the odd and even parts of the exponential
OpenStax Calculus Volume 1, §6.9 Calculus of the Hyperbolic Functions §6.9, pp. 782-792
Section
Section 1
Concept
The hyperbolic cosine averages the exponential with its reflection and the hyperbolic sine takes half their difference. Their sum is the exponential, and their difference its reflection.
even and odd parts — Any function splits uniquely into an even function and an odd one. For the exponential these parts are the hyperbolic cosine and sine.
\[ \cosh x + \sinh x = e^{x} \]
For large positive inputs both halves approach half the exponential, so the two curves converge — visible in the figure, and the reason the hyperbolic tangent approaches one.
Figure (svg): The hyperbolic sine and cosine as the odd and even parts of the exponential
OpenStax Calculus Volume 1, §6.9 Calculus of the Hyperbolic Functions §6.9, pp. 782-790 — definitions of the hyperbolic functions
Picture it
The exponential split.
Figure (svg): The hyperbolic sine and cosine as the odd and even parts of the exponential
Both curves approach half the exponential on the right, and diverge from each other on the left where the reflected term dominates. Their sum is exactly the exponential everywhere.
Worked example
Example 6.54. Confirming the split.
\[ \text{Show } \cosh \text{ is even, } \sinh \text{ is odd, and their sum is the exponential.} \]
Replace x with its negative in cosh
Why: The two terms swap.
Do the same in sinh
Why: The two terms swap and the sign flips.
Add the definitions
Why: The reflected terms cancel.
\[ e ^{x} \]
Subtract them
Why: The forward terms cancel.
\[ e ^{-x} \]
Note the uniqueness
Why: Any function splits this way once.
Figure (svg): The hyperbolic sine and cosine as the odd and even parts of the exponential
\[ \cosh(-x)=\cosh x, \quad \sinh(-x)=-\sinh x \]
Verify: check the values at zero
Why: At zero the hyperbolic cosine is the average of 1 and 1, which is 1, and the hyperbolic sine is half their difference, which is 0 — matching the circular functions exactly. That agreement at the origin is one of many, and it is part of why the borrowed names are apt. Note also that the decomposition is unique: no other pair of an even and an odd function sums to the exponential.
OpenStax Calculus Volume 1, §6.9 Calculus of the Hyperbolic Functions §6.9, pp. 785-787
Fill the middle
The even part of the exponential.
Fill in the blanks
\cosh x = \frac-x+e^___}}___
Why: Averaging the exponential with its reflection gives an even function. Taking half their difference instead gives the odd part, the hyperbolic sine.
Worked example
Checkpoint 6.54. Quotients, as in trigonometry.
\[ \text{Define } \tanh, \; \coth, \; \operatorname{sech} \text{ and } \operatorname{csch}. \]
Define the tangent
Why: Sine over cosine.
\[ \tanh = \sinh / \cosh \]
Define the cotangent
Why: Its reciprocal.
\[ \cot h = \cosh / \sinh \]
Define the secant
Why: Reciprocal of the cosine.
\[ \sec h = 1 / \cosh \]
Define the cosecant
Why: Reciprocal of the sine.
\[ \csc h = 1 / \sinh \]
Note the pattern
Why: Exactly the trigonometric one.
Figure (svg): The solution to Worked example the other four functions shown as a ladder of expressions, one row per legal move
\[ \tanh x = \frac{\sinh x}{\cosh x} = \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} \]
Verify: note where the domains differ from the circular case
Why: The hyperbolic cosine is never zero, so the hyperbolic tangent and secant are defined for every real input — unlike the circular tangent, which has infinitely many vertical asymptotes. The hyperbolic sine vanishes only at the origin, so the cotangent and cosecant have a single break there rather than infinitely many. The absence of periodicity removes all the repeated singularities.
OpenStax Calculus Volume 1, §6.9 Calculus of the Hyperbolic Functions §6.9, pp. 787-789
Trap
\[ \cosh(x+2\pi)=\cosh x \]
Carry periodicity over from the circular functions
Why: The student assumes the parallel is complete.
The definitions are built from exponentials, which are strictly monotonic — nothing here repeats.
\[ \cosh \text{ is not periodic; it grows without bound} \]
Check each property against the definitions rather than the analogy
Why: The parallel covers identities and derivatives, not periodicity.
The graph settles it immediately: the hyperbolic cosine rises without bound in both directions, which no periodic function does. The analogy is with the identities, not with the geometry of repetition.
Sorting
Replace the input with its negative.
Sort into buckets
Sort each function.
The parities match the circular functions exactly: cosine and secant even, sine and tangent odd. That correspondence is one of the many that make the borrowed names apt.
Two truths and a lie
All three are about the definitions.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. They are built from exponentials, which are strictly monotonic, so nothing repeats — the hyperbolic cosine grows without bound in both directions. The analogy with trigonometry covers identities and derivatives, not periodicity.
Prediction
Commit before reasoning.
Predict first
Why are the even and odd parts of the exponential worth naming?
Correct: Because they mirror the trigonometric functions usefully.
\[ \cosh^{2}x-\sinh^{2}x=1 \quad \text{against} \quad \cos^{2}+\sin^{2}=1 \]
Why: The decomposition would be a curiosity if the pieces had no structure, but they satisfy an identity like the Pythagorean one, differentiate into each other like sine and cosine, and handle exactly the radicals that circular substitution cannot. They are not simpler than the exponential — each is built from two copies of it — and the names came after the structure was noticed, not before.
Section
Section 2
Concept
The fundamental identity has a minus where the Pythagorean one has a plus, so the point traces a hyperbola rather than a circle — which is where the names come from.
the fundamental hyperbolic identity — The hyperbolic cosine squared minus the hyperbolic sine squared equals one, so the point given by those coordinates lies on the unit hyperbola.
\[ \cosh^{2}x-\sinh^{2}x=1 \]
The parameter is not an angle in either case but twice the area of a sector — which is the deeper correspondence, and the one that makes the parallel exact rather than decorative.
Figure (svg): Why they are called hyperbolic: the point traces a hyperbola
OpenStax Calculus Volume 1, §6.9 Calculus of the Hyperbolic Functions §6.9, pp. 786-794 — hyperbolic identities
Picture it
Two families, two curves.
Figure (svg): Why they are called hyperbolic: the point traces a hyperbola
The identities differ by one sign and so do the curves. Every other difference between the two families traces back to that single change.
Worked example
Example 6.56. A two-line computation.
\[ \text{Prove } \cosh^{2}x-\sinh^{2}x=1. \]
Square the hyperbolic cosine
Why: Expand the binomial.
\[ \frac{e ^{2} x + 2 + e ^{-2} x}{4} \]
Square the hyperbolic sine
Why: Expand.
\[ \frac{e ^{2} x - 2 + e ^{-2} x}{4} \]
Subtract
Why: The outer terms cancel.
\[ \frac{2 + 2}{4} \]
Simplify
Why: Collect.
\[ 1 \]
Note what made it work
Why: The cross terms differ in sign.
Figure (svg): Why they are called hyperbolic: the point traces a hyperbola
\[ \cosh^{2}x-\sinh^{2}x=1 \]
Verify: see why the identity has a minus rather than a plus
Why: Adding the squares instead would give twice the hyperbolic cosine of 2x, not a constant — so the minus sign is not a convention but what the algebra produces. It arises because the cross terms in the two expansions have opposite signs and the outer terms are identical, so subtracting cancels the outer terms and doubles the cross terms. That single sign is what makes the traced curve a hyperbola.
OpenStax Calculus Volume 1, §6.9 Calculus of the Hyperbolic Functions §6.9, pp. 789-791
Fill the middle
The fundamental hyperbolic identity.
Fill in the blanks
\cosh^-x___\sinh^___x=1
Why: The minus sign is what expanding the definitions produces, and it is what makes the traced curve a hyperbola rather than a circle.
Worked example
Checkpoint 6.56. The parallel is not blind.
\[ \text{Compare } \cos(2x) \text{ and } \cosh(2x) \text{ in terms of the squares.} \]
Recall the circular version
Why: From trigonometry.
\[ \cos ^{2} - \sin ^{2} \]
Compute the hyperbolic version
Why: From the definitions.
\[ \cosh ^{2} + \sinh ^{2} \]
Compare
Why: The sign is opposite.
Check another
Why: The double-argument sine.
State the rule of thumb
Why: Some signs flip.
Figure (svg): The solution to Worked example which identities differ shown as a ladder of expressions, one row per legal move
\[ \cos 2x=\cos^{2}-\sin^{2}, \quad \cosh 2x=\cosh^{2}+\sinh^{2} \]
Verify: note where the flips occur
Why: The pattern, sometimes called Osborn's rule, is that a sign flips wherever a product of two hyperbolic sines appears — the double-argument cosine hides such a product, and the fundamental identity contains one explicitly. The double-argument sine has only one factor of the sine, so nothing flips. Knowing the rule is less important than knowing that flips happen, since verifying any single identity takes two lines from the definitions.
OpenStax Calculus Volume 1, §6.9 Calculus of the Hyperbolic Functions §6.9, pp. 791-793
Error analysis
A student uses the Pythagorean identity for hyperbolic functions.
Annotate
On: \( \cosh^{2}x+\sinh^{2}x=1 \)
Testing an identity at a single numerical value catches this in seconds, and it is worth doing whenever a property is carried over from one family to the other.
Sorting
Compare each identity with its circular counterpart.
Sort into buckets
Sort each pair.
Three of five flip, so following the parallel blindly is wrong more often than right on these particular items. Verifying from the definitions takes two lines and is the reliable route.
Two truths and a lie
All three are about the identity.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. The sum of the squares is the hyperbolic cosine of twice the input, which is not constant — at an input of 1 it is about 3.76. Only the difference is identically one.
Prediction
Commit before reasoning.
Predict first
For the circular functions the parameter is an angle. What is it for the hyperbolic ones?
Correct: Twice the area of a sector.
\[ \text{both parametrised by twice the sector's area} \]
Why: For the circle, the arc-length interpretation and the sector-area one coincide up to a factor, which is why the angle reading is available there. The hyperbola has no angle, but the sector area works for both — so the area interpretation is the deeper one and it is what makes the parallel exact rather than a formal resemblance. Arc length along a hyperbola is not the parameter and is not even elementary.
Section
Section 3
Concept
The hyperbolic sine and cosine differentiate into each other, but without the minus sign the circular cosine picks up. Everything follows from the definitions in two lines.
hyperbolic derivatives — The derivative of the hyperbolic sine is the hyperbolic cosine and vice versa, both with a plus sign. The remaining derivatives follow by the quotient rule as in trigonometry.
\[ \frac{d}{dx}\sinh x = \cosh x, \qquad \frac{d}{dx}\cosh x = \sinh x \]
The absence of the minus sign is the most consequential difference. It means neither function is a solution of the oscillation equation, and both solve the growth equation instead.
Figure (svg): The derivatives, proved directly from the exponential definitions
OpenStax Calculus Volume 1, §6.9 Calculus of the Hyperbolic Functions §6.9, pp. 790-798 — derivatives and integrals
Picture it
Each from the definitions.
Figure (svg): The derivatives, proved directly from the exponential definitions
The minus in the exponent supplies every sign in the answers, and no separate rules need to be remembered. Each derivative is a two-line computation.
Worked example
Example 6.57. Straight from the definitions.
\[ \text{Find the derivatives of } \sinh \text{ and } \cosh. \]
Differentiate the hyperbolic sine
Why: Term by term.
\[ \frac{e ^{x} + e ^{-x}}{2} \]
Recognise the result
Why: The even part.
Differentiate the hyperbolic cosine
Why: Term by term.
\[ \frac{e ^{x} - e ^{-x}}{2} \]
Recognise it
Why: The odd part.
Note the missing minus
Why: Unlike the circular cosine.
Figure (svg): The derivatives, proved directly from the exponential definitions
\[ \sinh' = \cosh, \qquad \cosh' = \sinh \]
Verify: see what the missing minus changes
Why: Differentiating twice returns the original function rather than its negative, so both satisfy the equation stating the second derivative equals the function — the growth equation, whose solutions are exponentials. The circular pair satisfy the equation with a minus, the oscillation equation, whose solutions repeat. That one sign is the difference between growth and oscillation, and it is why the hyperbolic functions describe hanging chains rather than vibrations.
OpenStax Calculus Volume 1, §6.9 Calculus of the Hyperbolic Functions §6.9, pp. 793-795
Fill the middle
No sign change here.
Fill in the blanks
\frac+___\cosh x = ___\sinh x
Why: Differentiating the definition gives the odd part with a plus sign. The circular cosine's minus does not carry over, and this is the section's most consequential difference.
Worked example
Checkpoint 6.57. Where they earn their place.
\[ \text{Which substitution suits } \int\frac{dx}{\sqrt{x^{2}+4}}? \]
Examine the radicand
Why: A sum of squares.
Recall the identity
Why: With the minus sign.
\[ \cosh ^{2} - \sinh ^{2} = 1 \]
Rearrange it
Why: Add.
\[ 1 + \sinh ^{2} = \cosh ^{2} \]
Choose the substitution
Why: To match the radicand.
\[ x = 2 \sinh t \]
Note the result
Why: The root simplifies.
\[ \sqrt{4 \cosh ^{2} t} = 2 \cosh t \]
Figure (svg): Where hyperbolic functions earn their place in integration
\[ x = 2\sinh t \;\Longrightarrow\; \sqrt{x^{2}+4}=2\cosh t \]
Verify: compare with the circular alternative
Why: A circular substitution would need one plus the square of a tangent, which does give a secant squared and also works — so both routes exist for this integrand. What the hyperbolic version avoids is the awkward integral of the secant, which the circular route produces and which needs a memorised trick. That is the practical reason hyperbolic substitutions are preferred for sums of squares.
OpenStax Calculus Volume 1, §6.9 Calculus of the Hyperbolic Functions §6.9, pp. 795-797
Trap
\[ \frac{d}{dx}\cosh x = -\sinh x \]
Copy the circular cosine's derivative
Why: The student follows the analogy.
\[ \frac{d}{dx}\frac{e^{x}+e^{-x}}{2} = \frac{e^{x}-e^{-x}}{2} = +\sinh x \]
Differentiating the definition gives a plus sign, and the two-line check settles it immediately.
\[ \frac{d}{dx}\cosh x = \sinh x \]
Differentiate the definition rather than recalling the analogue
Why: The parallel does not extend to this sign.
This is the single most consequential sign in the section, since it is what makes both functions solve the growth equation rather than the oscillation one.
Matching
The identity decides.
Match the pairs
Why: A difference of squares matches the circular identity and a sum matches the rearranged hyperbolic one. Reading which shape the radicand has determines the substitution immediately.
Sorting
The sign in the second derivative decides.
Sort into buckets
Sort each function.
The split is exactly along the missing minus sign. It is why hyperbolic functions describe hanging chains and circular ones describe vibrations, and the two equations are the standard models for those two behaviours.
Prediction
Commit before reasoning.
Predict first
The hyperbolic cosine's derivative has no minus sign. What is the consequence?
Correct: They satisfy the growth equation rather than the oscillation one.
\[ y''=y \quad \text{against} \quad y''=-y \]
Why: Differentiating twice returns the function itself rather than its negative, which is the equation whose solutions are exponentials — so hyperbolic functions grow without bound instead of oscillating. That single sign separates the physics they describe: hanging chains, cable shapes and certain heat problems on one side, vibrations and waves on the other.
Section
Section 4
Concept
Inverting a hyperbolic function means solving a quadratic in the exponential, so the inverses have closed forms built from logarithms and square roots.
inverse hyperbolic functions — The inverses of the hyperbolic functions, each expressible in elementary terms because inverting reduces to a quadratic in the exponential.
\[ \operatorname{arcsinh} x = \ln\left(x+\sqrt{x^{2}+1}\right) \]
No such formulas exist for the inverse trigonometric functions, and the reason is structural: the hyperbolic functions are built from elementary algebra applied to the exponential, and the circular ones are not.
Figure (svg): The inverse hyperbolic functions have closed forms in logarithms
OpenStax Calculus Volume 1, §6.9 Calculus of the Hyperbolic Functions §6.9, pp. 794-802 — inverse hyperbolic functions
Picture it
Each with its domain.
Figure (svg): The inverse hyperbolic functions have closed forms in logarithms
The domains differ because the functions do: the hyperbolic sine is one-to-one everywhere, the cosine only on a half-line, and the tangent has a bounded range.
Worked example
Example 6.59. A quadratic in the exponential.
\[ \text{Derive } \operatorname{arcsinh} x = \ln\left(x+\sqrt{x^{2}+1}\right). \]
Set the function equal to x
Why: And write the definition.
\[ x = \frac{e ^{y} - e ^{-y}}{2} \]
Multiply through
Why: By twice the exponential.
\[ 2 x e ^{y} = e ^{2} y - 1 \]
Recognise a quadratic
Why: In the exponential.
\[ (e ^{y}) ^{2} - 2 x e ^{y} - 1 = 0 \]
Apply the quadratic formula
Why: And discard the negative root.
\[ e ^{y} = x + \sqrt{x ^{2} + 1} \]
Take logarithms
Why: To solve for y.
Figure (svg): The inverse hyperbolic functions have closed forms in logarithms
\[ \operatorname{arcsinh} x = \ln\left(x+\sqrt{x^{2}+1}\right) \]
Verify: justify discarding the other root and check a value
Why: The other root subtracts the square root, which exceeds the magnitude of x, giving a negative value — and an exponential is never negative, so that root is inadmissible. Checking at x equal to 0 gives the logarithm of 1, which is 0, matching the hyperbolic sine's value there. Note that the whole derivation used only the quadratic formula and a logarithm, both elementary, which is exactly why no such formula exists for the inverse sine.
OpenStax Calculus Volume 1, §6.9 Calculus of the Hyperbolic Functions §6.9, pp. 797-799
Fill the middle
The inverse hyperbolic sine.
Fill in the blanks
\operatorname1 x = \ln\left(x+\sqrt___+___}\right)
Why: The constant comes from the quadratic formula's discriminant. The plus sign under the root, rather than a minus, is what makes the formula valid for every real input.
Worked example
Checkpoint 6.59. Algebraic, like the circular ones.
\[ \text{Find } \frac{d}{dx}\operatorname{arcsinh} x. \]
Differentiate the logarithmic formula
Why: The chain rule.
Differentiate the argument
Why: One plus the root's derivative.
\[ 1 + x / \sqrt{x ^{2} + 1} \]
Combine over a common denominator
Why: The numerator matches the argument.
Simplify
Why: Everything cancels but the root.
\[ 1 / \sqrt{x ^{2} + 1} \]
Compare with the circular case
Why: The inverse sine.
Figure (svg): The solution to Worked example the derivatives of the inverses shown as a ladder of expressions, one row per legal move
\[ \frac{d}{dx}\operatorname{arcsinh} x = \frac{1}{\sqrt{x^{2}+1}} \]
Verify: notice which Section 5.7 gap this fills
Why: Section 5.7 handled the integral of one over the root of a constant minus a square, giving an inverse sine — and could say nothing about a constant PLUS a square. This derivative fills exactly that gap: the integrand with the plus sign has the inverse hyperbolic sine as its antiderivative. So this family completes the table of radical integrands that chapter began, which is its main computational contribution.
OpenStax Calculus Volume 1, §6.9 Calculus of the Hyperbolic Functions §6.9, pp. 799-801
Error analysis
A student inverts the hyperbolic sine.
Annotate
On: \( e^{y}=x-\sqrt{x^{2}+1} \;\Longrightarrow\; y=\ln\left(x-\sqrt{x^{2}+1}\right) \)
Whenever a quadratic arises in an inversion, one root must usually be discarded on domain grounds. Checking which root is admissible is part of the derivation rather than an afterthought.
Ranking
Inverting a hyperbolic function.
Put in order
Why: Step c is the insight and step d is where the domain reasoning enters. The whole derivation uses only the quadratic formula and a logarithm, which is why an elementary closed form exists here and not for the inverse trigonometric functions.
Sorting
Circular or hyperbolic inverse?
Sort into buckets
Sort each integrand by its antiderivative's family.
Section 5.7 handled the first column and could say nothing about the second. Together the two families cover every radical of this shape, which is the practical reason the hyperbolic functions appear in integral tables.
Prediction
Commit before reasoning.
Predict first
Why do the inverse hyperbolic functions have elementary formulas when the inverse trigonometric ones do not?
Correct: Because inverting reduces to a quadratic in the exponential.
\[ (e^{y})^{2}-2xe^{y}-1=0: \; \text{a quadratic} \]
Why: The hyperbolic functions are elementary algebraic combinations of the exponential, so setting one equal to a value gives a polynomial equation in the exponential — which the quadratic formula solves, after which a logarithm finishes. Sine and cosine are not algebraic combinations of any elementary function, so no analogous route exists, and their inverses are genuinely new functions rather than rearrangements of old ones.
Section
Section 5
Concept
A flexible chain hanging under its own weight takes the shape of a hyperbolic cosine, not a parabola — a question Galileo got wrong and calculus settled.
catenary — The curve of a uniform flexible chain hanging under gravity. It is a scaled hyperbolic cosine, and it differs from the parabola that Galileo conjectured.
\[ y = a\cosh\frac{x}{a} \]
The parabola is the catenary's second-order approximation, so the two agree closely near the lowest point — which is why the error stood for decades before the calculus existed to expose it.
Figure (svg): The catenary: a hanging chain, and why it is not a parabola
OpenStax Calculus Volume 1, §6.9 Calculus of the Hyperbolic Functions §6.9, pp. 798-806 — applications: the catenary
Picture it
Two curves, close at the bottom.
Figure (svg): The catenary: a hanging chain, and why it is not a parabola
Near the lowest point the curves are nearly indistinguishable and they separate toward the ends. The parabola is the catenary's quadratic approximation, which is exactly why the confusion arose.
Worked example
Example 6.60. Comparing the two curves.
\[ \text{Compare } \cosh x \text{ with } 1+\tfrac{x^{2}}{2} \text{ near zero and far from it.} \]
Expand the hyperbolic cosine
Why: From the exponential's series.
\[ 1 + x ^{2} / 2 + x ^{4} / 24 +... \]
Identify the parabola
Why: The first two terms.
Compare at x = 0.5
Why: Both curves.
\[ 1.1276\text{ against } 1.1250 \]
Compare at x = 2
Why: Further out.
\[ 3.762\text{ against } 3.000 \]
State the conclusion
Why: They diverge.
Figure (svg): The catenary: a hanging chain, and why it is not a parabola
\[ \cosh x = 1+\frac{x^{2}}{2}+\frac{x^{4}}{24}+\cdots \]
Verify: understand why the error survived so long
Why: Near the lowest point the two curves agree to within a quarter of a percent, which is well inside what seventeenth-century measurement could distinguish — so Galileo's conjecture was not careless but simply beyond the evidence available. Settling it required both the calculus and the hyperbolic functions, and the answer came from the Bernoullis and Leibniz in 1691. The fourth-order term is the whole difference, and it is invisible near the bottom.
OpenStax Calculus Volume 1, §6.9 Calculus of the Hyperbolic Functions §6.9, pp. 801-803
Fill the middle
A chain hanging under its own weight.
Fill in the blanks
y = a\,\cosh\frac______
Why: The parameter a scales both the curve and its argument, so the shape is the same for every chain and only its scale changes. The parabola is its quadratic approximation.
Worked example
Checkpoint 6.60. An arc length that works out.
\[ \text{Find the length of } y=\cosh x \text{ on } [-a,a]. \]
Differentiate
Why: The hyperbolic sine.
\[ y' = \sinh x \]
Square and add one
Why: The arc length integrand.
\[ 1 + \sinh ^{2} x \]
Apply the identity
Why: It simplifies exactly.
\[ \cosh ^{2} x \]
Take the root
Why: The hyperbolic cosine is positive.
Integrate
Why: Its antiderivative is the sine.
\[ 2 \sinh a \]
Figure (svg): The solution to Worked example the catenary's arc length shown as a ladder of expressions, one row per legal move
\[ L = \int_{-a}^{a}\cosh x\,dx = 2\sinh a \]
Verify: notice how rare this is
Why: Section 6.4 showed that almost no curve has an elementary arc length — even a sine's is an elliptic integral. The catenary's works out exactly because the identity turns one plus the derivative squared into a perfect square, which is a property of this family specifically. That the curve describing a real hanging chain also has a computable length is a genuine piece of luck, and it is why catenary calculations appear in engineering handbooks.
OpenStax Calculus Volume 1, §6.9 Calculus of the Hyperbolic Functions §6.9, pp. 803-805
Trap
\[ \text{model a suspension cable as } y=1+\tfrac{x^{2}}{2} \]
Use the parabola because it is easier
Why: The student takes the approximation for the curve.
For a shallow cable the error is small, but for a deep sag it reaches tens of percent and the tension calculation fails with it.
\[ y = a\cosh\frac{x}{a} \]
Use the parabola only where the sag is shallow, and check
Why: It is a second-order approximation, valid near the bottom.
A suspension bridge's main cable is genuinely close to parabolic, because the deck's weight dominates the cable's own — a different physical situation, not an approximation to this one.
Two truths and a lie
All three are about the catenary.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one — Galileo's conjecture, settled in 1691. The two curves agree to within a quarter of a percent near the bottom and differ by 25 percent by an argument of 2, so the error was invisible to seventeenth-century measurement.
Sorting
The physical situation decides.
Sort into buckets
Sort each cable.
The bridge case is not an approximation but a genuinely different problem: a load spread uniformly along the horizontal gives a parabola exactly, while a load along the cable's own length gives a catenary. Which curve applies is a question about the loading, not about the accuracy wanted.
Prediction
Commit before reasoning.
Predict first
Why did Galileo's parabola conjecture stand for decades?
Correct: Because the parabola is the second-order approximation.
\[ \cosh x - \left(1+\tfrac{x^{2}}{2}\right) = \tfrac{x^{4}}{24}+\cdots \]
Why: The two curves' series agree in their first two terms and differ only at fourth order, so near the lowest point they match to within a fraction of a percent — well inside seventeenth-century measurement. Distinguishing them required either very deep sags or the calculus, and the calculus arrived first. That a wrong model can be locally excellent is a general lesson, and the same one Section 6.8's logistic comparison made.
Comparison
Fill the blanks. Three of these differ by a sign.
Comparison matrix
| Circular | Hyperbolic | |
|---|---|---|
| Identity | cos^2 + sin^2 = 1 | cosh^2 - sinh^2 = 1 |
| Curve traced | a circle | a hyperbola |
| Derivative of the cosine | minus the sine | plus the sine |
| Second derivative | minus the function: oscillation | the function itself: growth |
The last row is what the parallel is really about. One sign separates the mathematics of vibration from the mathematics of hanging cables, and every other difference follows from it.
Pattern
Given a hyperbolic function to work with.
Step one is the section's discipline. Three of the five compared properties differ in sign from their circular counterparts, so following the analogy blindly is wrong more often than right.
Stewart, Calculus: Early Transcendentals 8e, §3.11 Hyperbolic Functions §3.11, pp. 259-265
Check
The identity.
Check your understanding
Which identity do the hyperbolic functions satisfy?
Answer: A
Why: Expanding the definitions cancels the outer terms and leaves one.
Check
Derivatives.
Check your understanding
What is the derivative of cosh x?
Answer: A
Why: Differentiating the definition gives the odd part with a plus sign.
Check
The catenary.
Check your understanding
What shape does a chain hanging under its own weight take?
Answer: A
Why: Galileo conjectured a parabola; the Bernoullis and Leibniz settled it in 1691.
Real world
An engineer sizing the towers for a cable car must know the cable's shape, its length, and the tension at the anchor points, given the span and the sag at the middle.
Discussion prompt
Explain which curve applies, why the arc length is computable, and where the tension comes from.
Hint: The cable's own weight is what loads it.
Answer:
With no deck to carry, the cable is loaded by its own weight distributed along its length — which is the catenary's defining condition, so the shape is a hyperbolic cosine and not a parabola. The scale parameter is fixed by the span and sag together.
\[ y = a\cosh\frac{x}{a}, \qquad L = 2a\sinh\frac{s}{2a} \]
The arc length is computable exactly, which Section 6.4 showed is rare — almost no curve has an elementary arc length, and the catenary's works out only because the identity turns one plus the derivative squared into a perfect square. The engineer therefore gets a formula rather than a numerical integration, which is why these appear in handbooks.
The tension follows from the shape. The horizontal component is constant along the cable and equals the weight per unit length times the scale parameter; the total tension at any point is that times the hyperbolic cosine there, so it is greatest at the anchors where the cable is steepest. Sizing the towers means evaluating that at the ends.
Note the practical consequence of the parabola error: for a deep sag it underestimates both the length and the anchor tension. Ordering cable by the parabolic length would leave the installation short, and sizing anchors by it would under-specify them — the failure is in the unsafe direction, which is the worst kind.
Commit first
Answer, then rate your confidence honestly.
Predict first
What is the most consequential difference between the hyperbolic and circular functions?
Correct: The missing minus sign in the cosine's derivative.
\[ y''=y \quad \text{against} \quad y''=-y \]
Why: Differentiating twice returns the function itself rather than its negative, so hyperbolic functions grow without bound while circular ones repeat. That single sign is what makes one family describe hanging chains and cable shapes and the other vibrations and waves. The identity's sign is the same difference seen algebraically, and both trace to the minus in the exponent of the definition.
Explain it
They assumed every trigonometric identity has a hyperbolic twin with the same signs.
Discussion prompt
In four sentences or fewer, show them the risk.
Hint: Ask them to test one numerically.
Answer:
Ask them to test the Pythagorean version at an input of 1: the hyperbolic cosine squared plus the hyperbolic sine squared is about 3.76, not 1. The hyperbolic identity has a MINUS sign, and three of the five standard comparisons flip in the same way.
The safe route is to expand the definitions in exponentials, which settles any identity in two lines. The analogy is a good way to guess which identity might exist, and a bad way to decide what its signs are.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the definitions, remember they are the exponential's even and odd parts. For identities, expand from the definitions rather than trusting the analogy. For derivatives, the hyperbolic cosine's has no minus. For inverses, set the function equal to the input and solve a quadratic in the exponential. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top, write both definitions and sketch the two curves with the half-exponential dashed between them. Below, prove the fundamental identity in four lines and mark where the minus sign comes from. Beside it, draw a circle and a hyperbola with their identities beneath. In the middle of the page, write the five circular-hyperbolic comparisons in two columns and circle the three whose signs differ, boxing the cosine's derivative as the most consequential. Below that, write the three radicand shapes with the substitution each calls for. In the lower half, derive the inverse hyperbolic sine's formula in five lines, noting which root is discarded and why. At the bottom, sketch a catenary with its parabolic approximation dashed, and write the arc length result with one sentence on why it is unusual.
If your two columns show the same signs throughout, check them numerically — three of the five differ, and finding that out by testing rather than by being told is what makes the discipline stick.
Recap
Five things, and the discipline behind all of them is to check rather than to assume.
| If you see | Then |
|---|---|
| An identity in doubt | Expand the definitions in exponentials |
| The hyperbolic cosine differentiated | No minus sign |
| A radicand that is a sum of squares | Substitute the hyperbolic sine |
| An inverse hyperbolic function | A logarithm and a square root exist |
| A hanging chain | A hyperbolic cosine, not a parabola |
| A uniformly loaded suspension cable | A parabola, genuinely |
| Any borrowed trigonometric property | Verify it: three of five flip signs |
That completes Calculus I. From a limit in Chapter 2 to the derivative, its applications, the integral and the theorem uniting them, and finally to the families of functions those tools were built to handle — the whole subject now rests on foundations laid within it.
OpenStax Calculus Volume 1, §6.9 Calculus of the Hyperbolic Functions §6.9, pp. 646-656 — everything on these slides traces back here
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