Date: 10/10/2016
Instructor: Jay Taylor
Math 125 - Calculus 1 (Fall 2016)
Exam 3-A
Solutions
1. T F
If f is a function which is differentiable at a then
f (x) ≈ f (a) + f 0 (a)(x − a)
when x is near a.
T F
If f (x) is a function satisfying f 0 (x) = 0 on an open interval a < x < b then
f (x) is constant on a 6 x 6 b.
T F
cosh2 (x) + sinh2 (x) = 1.
T F
If f is continous on a 6 x 6 b and differentiable on a < x < b, then there
exists a number c, with a < c < b, such that
f (b) − f (a) = f 0 (c)(b − a).
T F
tanh(x) is an odd function.
2.
1
f (x) =
2
0
cosh(3x)
tan(x) + 1
−1/2
(tan(x) + 1)3 sinh(3x) − cosh(3x) sec2 (x)
(tan(x) + 1)2
3. A possible graph is as follows.
4. Taking the derivative
f 0 (x) = −e −x sin(x) + e −x cos(x) = e −x (cos(x) − sin(x)).
This is defined everywhere so critical points occur when f 0 (x) = 0, i.e., when
cos(x) − sin(x) = 0 ⇒ cos(x) = sin(x).
From the hint we have there are two critical points at π/4 and 5π/4. Taking the second
derivative we have
f 00 (x) = −e −x (cos(x) − sin(x)) + e −x (− sin(x) − cos(x))
= e −x (− cos(x) + sin(x) − sin(x) − cos(x))
= −2 cos(x)e −x .
Possible points of inflection occur when f 00 (x) = 0, so when cos(x) = 0. Thus π/2
and 3π/2 are possible inflection points. As cos(x) changes sign over these points so
does f 00 (x) so these are inflection points. Checking the second derivative of the critical
points we see that
cos(π/4) > 0 ⇒ f 00 (π/4) < 0 ⇒ π/4 is a local max,
cos(5π/4) < 0 ⇒ f 00 (5π/4) > 0 ⇒ 5π/4 is a local min.
5. (a) Implicitly differentiating we have
dy
d
= e −xy (−xy ) = −e −xy
dx
dx
dy
y +x
dx
.
Rearranging we thus have
dy
dy
−y e −xy
−y 2
dy
+ xe −xy
= −y e −xy ⇒
=
=
.
dx
dx
dx
1 + xe −xy
1 + xy
(b) A point on the curve has a horizontal tangent line if dy /dx = 0. This happens
only when y = 0 but no point on the curve satisfies y = 0 because otherwise we
would have 0 = e −x0 = e 0 = 1, which is nonsense.
6. (a) The side of the cone is the radius of the original circle, which is 3. Applying
Pythagoras’ Theorem to the right angled triangle inside the cone we have 9 =
h2 + r 2 so r 2 = 9 − h2 . Hence, we have the volume of the cone as a function of h
is given by
π
π
V (h) = (9 − h2 )h = (9h − h3 ).
3
3
Differentiating we have
π
V 0 (h) = (9 − 3h2 ).
3
This is defined everywhere so critical points occur when
V 0 (h) = 0 ⇒ 9 − 3h2 = 0 ⇒ h2 = 3 ⇒ h =
√
3.
Note h must be positive because it is a distance. Differentiating again we have
V 00 (h) = −
2π
h.
3
√
√
Therefore V 00 ( 3) < 0 so h = 3 is a local maximum. This must be a global
maximum because it is the only critical point with h > 0 and V (0) = V (3) = 0.
(b) We already know the height that maximises the volume of the cone, we just need
√
to find x in terms of h. Firstly, when h = 3 we have
r 2 = 9 − h2 ⇒ r =
√
9−3=
√
6.
Comparing the circumferences we have
√
√
6π − x = 2π 6 ⇒ x = 6π − 2π 6.