May 2, 2011
18.01 Problem Set 12 Solutions
Part II: 15 points
1a) (5 points) Remember that the hyperbolic sine and cosine are the functions defined by
sinh x = (ex − e−x )/2,
cosh x = (ex + e−x )/2.
Prove the formulas
cosh2 x − sinh2 x = 1,
d sinh x
= cosh x.
dx
cosh2 x − sinh2 x = (ex + e−x )2 /4 − (ex − e−x )2 /4
= (e2x + 2 + e−2x )/4 − (e2x − 2 + e−2x )/4 = 4/4 = 1.
d
d sinh x
=
((ex − e−x )/2) = (ex + e−x )/2 = cosh(x).
dx
dx
b) Sketch the graph of sinh x, and explain why the inverse function sinh−1 (x) is defined
for all values of x.
Because ex is always strictly positive, the formula for cosh shows that cosh(x) > 0. Therefore
sinh(x) is strictly increasing (since its derivative is positive). It’s also easy to see from the formulas
that
lim sinh(x) = +∞,
lim sinh(x) = −∞.
x→+∞
x→−∞
Therefore sinh takes every possible value (by the Intermediate Value Theorem), exactly once each (since
it’s strictly increasing). That means that the equation y = sinh(x) has a unique solution x = sinh−1 (y)
for each y.
c) Show that the derivative of the inverse hyperbolic sine is given by the formula
p
d sinh−1 x
= 1/ 1 + x2 .
dx
The defining equation of sinh−1 is
sinh(sinh−1 (x)) = x.
Differentiating this equation by the chain rule gives
cosh(sinh−1 (x))
d sinh−1 x
= 1.
dx
q
Since cosh − sinh = 1, we have cosh(y) = 1 + sinh2 (y). So the equation for the derivative is
2
2
q
1 + [sinh(sinh−1 (x))]2
which is
d sinh−1 x
= 1,
dx
p
d sinh−1 x
1 + x2
= 1.
dx
Now solve for the derivative.
d) Show that
Z
x
0
√
dt
= sinh−1 x.
1 + t2
Because sinh(0) = 0, sinh−1 (0) = 0; so this is (c) plus the Fundamental Theorem of Calculus.
2. (5 points) Evaluate
Z
√
dt
1 + t2
using the trigonometric substitution t = tan θ. Your final answer will involve logarithms
and square roots.
We get dt = sec2 θ dθ, so the integral is
Z
Z
Z
sec2 θ dθ
sec2 θ dθ
√
√
=
=
sec θ dθ.
1 + tan2 θ
sec2 θ
This is one of the integrals you’re supposed to memorize; it’s
ln(sec θ + tan θ).
Plugging in sec θ =
√
1 + tan2 θ and t = tan θ, the conclusion is
Z
p
dt
√
= ln(t + 1 + t2 ).
1 + t2
3. (5 points) Solve the equation
2y = ex − e−x
for x.
p
Problems 1(d) and 2 found two antiderivatives for 1/ 1 + y 2 :
sinh−1 y
and
ln(y +
p
1 + y 2 ).
Each antiderivative is equal to 0 when x = 0. Since two antiderivatives must differ by a constant, the
conclusion is
p
sinh−1 y = ln(y + 1 + y 2 ).
But by definition, x = sinh−1 (y) is the unique solution of the equation
y = sinh(x) = (ex − e−x )/2.
p
So we’ve solved the equation: x = ln(y + 1 + y 2 ).
You can also solve the equation just by algebra:
y 2 + 1 = (e2x − 2 + e−2x )/4 + 4/4 = (e2x + 2 + e−2x )/4 = [(ex + e−x )/2]2 ,
so
p
y 2 + 1 = (ex + e−x )/2.
Therefore
y+
and so
p
y 2 + 1 = (ex − e−x )/2 + (ex + e−x )/2 = ex ,
ln(y +
p
y 2 + 1) = x.