Math 1210
Quiz 5
February 14th, 2014
Answer the following three (3) questions. The value of every question is
indicated at the beginning of it. You may use scratch paper, but you can
only turn in this sheet. Please write your answer in the space provided. You
have 25 minutes.
Name:
UID:
1. (10 points) Sand is pouring from a pipe at a rate of 16 cubic feet per second. If the
falling sand forms a conical pile on the ground whose altitude is always 14 of the diameter
of the base, how fast is the altitude increasing when the pile is 4 feet high? Refer to
figure 1 and use the fact that the volume is given by V = 13 πr2 h.
Solution: At some given time t, the pile of sand will be a vertical cone: let’s denote its
height by h and the radius of its base by r. The volume of the cone will then be given
by
1
V (h, r) = πr2 h
3
1
Since the height is always 4 of the diameter of the base we have
1
r
h = (2r) =
4
2
so r = 2h and hence we may write the volume of the cone as a function of h only, namely:
1
4
V (h) = π(2h)2 h = πh3
3
3
We know that the volume is increasing constantly at a rate of 16 cubic feet per second, so
that V 0 (t) = 16 (that is, the derivative of V with respect to t). Computing the derivative
of V with respect to t we thus have the following (recall that h = h(t) is the height of
the cone at some time t, so it is itself a function of t)
16 = V 0 (t) =
4
· 3πh2 h0 = 4πh2 h0
3
and solving for h0 we conclude
h0 (t) =
4
π[h(t)]2
so when h(t) = 4 we will have
4 1
h (t) =
=
π[h(t)]2 h(t)=4 4π
0
2. (10 points) The angle between the two equal sides of an isosceles triangle measures
θ = π3 rad and the two equal sides are a = 3cm long. How much does the length L of the
third side vary if θ increases 0.01rad? Recall that L = 2a tan 2θ .
Solution: The length of the third side varies with θ and we are being told that L(θ) =
2a tan 2θ . If we vary θ a small amount dθ, we know we can approximate the resulting
value of L as follows:
L(θ + dθ) ' L(θ) + L0 (θ)dθ
so the variation of L is given by L0 (θ)dθ.
Note that
L0 (θ) = 2a ·
so if we start with an angle θ =
approximately given by
L0 (θ)dθ '
π
3
3
cos2
π
6
a
1
1
=
·
θ
2 cos2 2
cos2 2θ
and we vary it dθ = 0.01, the variation in L is
3
· 0.01 = √ 2 · 0.01 = 0.06
2
2
3. (10 points) Find the maximum and the minimum values of the function
f (x) = sin x − cos x
over the interval [0, π]. Recall that sin π4 = cos π4 =
√
2
2
The extreme points of f will be among the critical points of f over the interval [0, π].
(i) Values at endpoints: f (0) = sin 0 − cos 0 = −1 and f (π) = sin π − cos π = 1.
(ii) The function f has no singular points.
(iii) Values at stationary points (namely, points c such that f 0 (c) = 0). In our case
f 0 (x) = cos x + sin x
so f 0 (x) = 0 if cos x + sin x = 0, so tan x = −1 and the only x in [0, π] for which
this happens is x = 3π
. The value of f at this point is
4
√ !
√
√
3π
3π
3π
2
2
= sin
− cos
=
= 2
f
− −
4
4
4
2
2
So the minimum value of f is f (0) = −1 and the maximum value is f
Page 2
3π
4
=
√
2.