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Calculus with hyperbolic functions questions
sinh and cosh differentiate into one another with no change of sign to remember, which makes the standard derivatives short. Read backwards, the inverse hyperbolic functions integrate expressions containing x² + a² and x² − a², completing the set begun with the inverse trigonometric functions.
7 original questions · 27 marks · the calculus with hyperbolic functions notes · Hyperbolic functions
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Write down the derivatives of sinh x and cosh x, and note the difference from the trig pattern.
Worked answer
sinh' = cosh and cosh' = sinh. The pair swap with both signs positive. B1 B1 for the two derivatives. Unlike cos, whose derivative is −sin, cosh differentiates with no minus anywhere.Differentiate y = cosh 3x, and find the gradient at x = 0.
Worked answer
Chain rule gives dy/dx = 3 sinh 3x. At x = 0, sinh 0 = 0, so the gradient is 0. M1 A1 for the derivative, A1 for the gradient. The curve has its minimum there, of value cosh 0 = 1.Find the equation of the tangent to y = cosh x at x = ln 2.
Worked answer
Gradient sinh(ln 2) = 3/4 and value cosh(ln 2) = 5/4, so y = 5/4 + (3/4)(x − ln 2). M1 A1 for the gradient, B1 for the value of y, A1 for the tangent. Exact hyperbolic values at logarithmic points come straight from the definitions, with no calculator required.Evaluate ∫ sinh x dx from 0 to ln 3, exactly.
Worked answer
∫sinh = cosh, both signs positive, so the value is [cosh x] from 0 to ln 3 = cosh(ln 3) − 1 = 5/3 − 1 = 2/3. The value cosh(ln 3) = 5/3 comes from (3 + 1/3)/2. M1 for the antiderivative, M1 for cosh(ln 3) = 5/3, A1 for 2/3. Marks are lost here by importing the minus sign from ∫sin x dx = −cos x.Evaluate ∫ 1/√(x² − 9) dx from 3 to 5, giving the answer as a single logarithm.
Worked answer
With a = 3 the integral is arcosh(x/3) from 3 to 5, that is arcosh(5/3) − arcosh 1. In log form this is ln(5/3 + √(25/9 − 1)) = ln(5/3 + 4/3) = ln 3, and arcosh 1 = 0. The answer is ln 3. M1 A1 for the arcosh form, M1 for the logarithmic form, A1 for ln 3.Using the identity cosh 2x = 2 cosh²x − 1, evaluate ∫ cosh²x dx from 0 to 1, exactly.
Worked answer
cosh²x = (1 + cosh 2x)/2, so the integral is [x/2 + sinh 2x/4] from 0 to 1 = 1/2 + sinh 2/4. Since sinh 2 = 2 sinh 1 cosh 1, this is (1 + sinh 1 cosh 1)/2 ≈ 1.407. M1 for rearranging the identity, M1 for integrating, A1 for the antiderivative, A1 for the exact value. The double angle identity carries no sign flip, unlike its cos 2x counterpart.Using the substitution x = 2 sinh u, show that ∫ √(x² + 4) dx from 0 to 2√3 = 2 ln(2 + √3) + 4√3.
Worked answer
dx = 2 cosh u du and x² + 4 = 4 sinh²u + 4 = 4 cosh²u, so √(x² + 4) = 2 cosh u and the integral becomes ∫4 cosh²u du. The limits transform too, with x = 0 giving u = 0 and x = 2√3 giving sinh u = √3, so u = arsinh √3 = ln(√3 + 2). Using cosh²u = (1 + cosh 2u)/2 turns the integral into 2[u + sinh u cosh u]. At the upper limit sinh u = √3 and cosh u = 2, giving 2[ln(2 + √3) + 2√3] = 2 ln(2 + √3) + 4√3, about 9.562. M1 for dx = 2 cosh u du, M1 A1 for reducing the root to 2 cosh u, B1 for the transformed limits, M1 for the double angle identity, A1 for the antiderivative, A1 for the printed result. The two traps are leaving the limits in terms of x and forgetting that cosh u is the positive root throughout.
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