Worksheet
MA 114
Spring 2013
1. Show that the curve with equation y = cosh 2x − 6 sinh x has a stationary point, and
find its x-coordinate in logarithmic form.
dy
Solution: Find
and where this derivative is 0.
dx
dy
= 2 sinh(2x) − 6 cosh(x)
dx
2 sinh(2x) − 6 cosh(x) = 0
e2x − e−2x
ex + e−x
=3
2
2
e2x − e−2x
=3
ex + e−x
ex − e−x = 3
(ex )2 − 3ex − 1 = 0
√
9+4
e =
√2
3+ 9+4
=
2 √
3+ 9+4
x = ln
2
x
3±
(Because ln must take a positive argument.)
2. Find the exact solution to 3 cosh x = 4 sinh x, giving the answer in terms of a logarithm.
Solution:
3 cosh(x) = 4 sinh(x)
3ex + 3e−x = 4ex − 4e−x
ex − 7e−x = 0
(ex )2 − 7 = 0
√
ex = 7
1
x = ln 7
2
3. Find the equation of the tangent to the curve y = cosh(2x) at x = 1.
dy
Solution:
= 2 sinh(2x). Set x = 1 and m = y 0 (1) = 2 sinh 2, so the equation of the
dx
tangent line is
y − cosh 2 = 2 sinh 2(x − 1).
MA 114
Worksheet
Spring 2013
4. Find the equation of the tangent to the curve y = sinh(x) + cosh(x) at x = 0.
dy
= cosh x + sinh x. Set x = 0 and m = y 0 (0) = cosh 0 + sinh 0 = 1, so the
Solution:
dx
equation of the tangent line is
y − 1 = 1(x − 0),
or y = x + 1.
5. Suppose we know that the hyperbolic cosine of a certain variable is equal to 1. What is
the hyperbolic cosine of twice that variable?
Solution: We are given that cosh a = 1. We must find cosh(2a).
Method 1: There is only one value of a that will make cosh a = 1 and that is a = 0.
Then 2a = 0 and cosh(2a) = 1.
Method 2: Since cosh2 a − sinh2 a = 1 and cosh a = 1, we see that sinh2 a = 0 and thus
sinh a = 0. Then
cosh(2a) = cosh2 a + sinh2 a = 1 + 0 = 1.
6. How is the hyperbolic secant of 10 related to the hyperbolic secant of −10?
Solution: Since cosh x is an even function, it follows that sech x = 1/ cosh x is also even.
Thus, sech(−10) = sech(10).
7. Using the definitions of the hyperbolic functions, show that 2 cosh2 x − cosh 2x = 1.
Solution:
Method 1: Using hyperbolic trig identities:
2 cosh2 x − cosh 2x = 2 cosh2 x − (cosh2 x + sinh2 x)
= cosh2 x − sinh2 x
=1
Method 2
2
e2x + e−2x
ex + e−x
−
2 cosh x − cosh 2x = 2
2
2
2x
−2x
e +2+e
e2x + e−2x
=2
−
4
2
2·2
=
4
=1
2
Page 2 of 7
MA 114
8. Find
Worksheet
Spring 2013
d
coth (e2tx ). Solution:
dt
d
coth e2tx = −csch2 e2tx e2tx (2x)
dt
= −2xe2tx csch2 e2tx
9.
R
cosh(3x) dx
Z
cosh(3x) dx =
10.
R
1
sinh(3x) + C.
3
sinh(x + 1) dx
Z
sinh(x + 1) dx = cosh(x + 1) + C.
11.
R
sinh3 x cosh6 x dx
Solution: Just like regular trig functions, rewrite sinh3 x = sinh x sinh2 x = sinh x(cosh2 x−
1), and then integrate.
Z
3
Z
6
sinh x cosh x dx =
=
12.
R
sinh x(cosh2 x − 1) cosh6 x dx
1
1
cosh9 x − cosh7 x + C
9
7
sinh2 x cosh2 x dx
1
Solution: To really simplify matters, rewrite sinh2 x = (1 + cosh(2x)) and cosh2 x =
2
1
(1 − cosh(2x)).
2
Z
2
2
Z
1
1
(1 + cosh(2x)) (1 − cosh(2x)) dx
2
2
Z
1
=−
(1 − cosh2 (2x)) dx
4
Z 1
1
=−
1 − (1 + cosh(4x)) dx
4
2
Z
1
=
(−1 + cosh(4x)) dx
8
1
1
−x + sinh(4x) + C
=
8
4
x
1
=− +
sinh(4x) + C
8 32
sinh x cosh x dx =
Page 3 of 7
MA 114
Worksheet
Spring 2013
d
13. Define the gudermannian function by gd(x) = tan−1 (sinh x). Show that
gd(x) =
dx
sech x.
Solution: We differentiate.
d
d
(gd(x)) =
tan−1 (sinh x)
dx
dx
cosh x
=
1 + sinh2 x
cosh x
=
cosh2 x
1
=
cosh x
= sech x
π
. Prove that gd(x) = f (x) by showing that they have the
2
same derivative and that f (0) = gd(0).
14. Let f (x) = 2 tan−1 (ex ) −
Solution: We differentiate.
π
d 2ex
2 tan−1 (ex ) −
=
dx
2
1 + e2x
2ex e−x
=
1 + e2x e−x
2
= x
e + e−x
1
=
cosh x
= sech x
d
=
(gd(x))
dx
Thus, since the functions have the same derivative, they must differ only by a constant,
π
i.e., gd(x) + C = 2 tan−1 (ex ) − . By comparing the two functions at x = 0 we can find
2
Page 4 of 7
MA 114
Worksheet
Spring 2013
the value of this constant.
π
gd(0) + C = 2 tan−1 (e0 ) −
2
π
tan−1 (sinh 0) + C = 2 tan−1 (1) −
π π 2
−1
tan (0) + C = 2
−
4
2
π π
0+C = −
2
2
C=0
Thus, the two functions are equal.
15. Show that
i. cosh(x) = sec(gd(x)).
sec(gd(x)) = sec tan−1 (sinh x)
q
1 + sinh2 (x)
=
1
q
= cosh2 (x)
= cosh x
ii. tanh(x) = sin(gd(x)).
sin(gd(x)) = sin tan−1 (sinh x)
sinh x
=q
1 + sinh2 (x)
sinh x
cosh(x)
= tanh x
=
iii. coth(x) = csc(gd(x)).
csc(gd(x)) = csc tan−1 (sinh x)
q
1 + sinh2 (x)
=
q sinh x
cosh2 (x)
=
sinh x
cosh x
=
sinh x
= coth x
Page 5 of 7
MA 114
Worksheet
Spring 2013
iv. sech(x) = cos(gd(x)).
cos(gd(x)) = cos tan−1 (sinh x)
1
=q
1 + sinh2 (x)
1
=q
cosh2 (x)
1
cosh x
= sech x
=
v. csch(x) = cot(gd(x)).
cot(gd(x)) = cot tan−1 (sinh x)
1
=
sinh(x)
= csch x
16. Find
i.
d
ln(cosh x))5
dx
4 1
d
ln(cosh x))5 = 5 ln(cosh x)
sinh x
dx
cosh x
4
= 5 tanh(x) ln(cosh x)
ii.
d
sinh(ln x)
dx
1
1
sinh(ln x) =
x−
2
x
d
1
1
sinh(ln x) =
1+ 2
dx
2
x
iii.
iv.
d tanh x
e
dx
d
tanh(ex )
dx
d tanh x
e
= sech2 x · etanh x
dx
d
tanh(ex ) = ex sech2 (ex )
dx
Page 6 of 7
MA 114
v.
Worksheet
Spring 2013
d
sinh(cosh3 x)
dx
d
sinh(cosh3 x) = cosh cosh3 x 3 cosh2 x sinh x
dx
17. Show that
(cosh(x) + sinh(x))n = cosh(nx) + sinh(nx)
for any integer n.
Solution:
ex + e−x ex − e−x
(cosh(x) + sinh(x)) =
+
2
2
x n
2e
=
2
nx
=e
n
n
Likewise:
enx + e−nx enx − e−nx
+
cosh(nx) + sinh(nx) =
2
2
nx 2e
=
2
nx
=e
Thus, clearly (cosh(x) + sinh(x))n = cosh(nx) + sinh(nx) for any integer n.
18. At what point on the curve y = cosh x does the tangent have slope 1?
d
Solution: We need to find the value of x so that
cosh x = 1.
dx
d
cosh x = sinh x
dx
sinh x = 1
x
e − ex
=1
2
ex − ex = 2
(ex )2 − 2ex − 1 = 0
√
√
2± 8
e =
=1± 2
2
x
Thus, x = ln 1 +
√ 2 .
Page 7 of 7