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Math 137 - Calculus I for the Life Sciences
Tuesday, February 26
∗∗ Related Rates, Trig Derivatives, and Exponential Derivatives
1. A street light is mounted on top of a 20-foot pole. A man and his top hat are together 7
feet tall. This man is waking away from the pole at 5 feet per second. How fast is the tip
of his shadow moving when he is 50 feet from the pole?
Let x (t) give the man’s distance from the pole at time t, and let s(t) be the length of his
shadow. Let y(t) = x (t) + s(t) be the distance from the pole to the tip of the shadow. Now,
7
the similar triangles in the diagram imply x20
+s = s . From this, we see that 20s = 7( x + s ),
7 dx
7
so s = 13
x. Taking the derivative with respect to t, we find ds
dt = 13 dt . Finally, we note
that, since the man is moving at 5 feet per second (He is a brisk walker.), dx
dt = 5. Then we
have
dy
dx ds
7 dx
7
=
+
= 5+
= 5 + (5) ≈ 7.69,
dt
dt
dt
13 dt
13
therefore the tip of his shadow is moving at approximately 7.69 feet per second.
2. Let a be a constant. Find the first four derivatives of f ( x ) = e ax . What is the 100th
derivative?
f 0 ( x ) = ae ax
f 00 ( x ) = a( ae ax ) = a2 e ax
f 000 ( x ) = a2 ( ae ax ) = a3 e ax
f (4) ( x ) = a3 ( ae ax ) = a4 e ax
f (100) = a100 e ax
d
csc x = − csc x cot x by rewriting csc x in terms of sin x, and using the fact
3. Verify that dx
d
that dx sin x = cos x.
We write csc x =
1
sin x ,
so
d
d 1
csc x =
= −(sin x )−2 (cos x ) = −
dx
dx sin x
4. Find the derivative of each of the following.
1
sin x
cos x = − csc x cot x.
sin x
(a) f ( x ) = sin( x ) cos( x )
By the product rule, f 0 ( x ) = cos2 x − sin2 x.
(b) f ( x ) = tan(cot( x ) + x2 + π )
By the chain rule, f 0 ( x ) = sec2 (cot x + x2 + π )(− csc2 x + 2x ).
(c) f ( x ) = sec3 ( x3 )
By the chain rule, f 0 ( x ) = 3 sec2 ( x3 )(sec( x3 ) tan( x3 )(3x )) = 9x sec3 ( x3 ) tan( x3 ).
(d) f ( x ) = ecsc x
By the chain rule, f 0 ( x ) = ecsc x (− csc x cot x ).
2
(e) f ( x ) = cos(100)( x )
2
By the chain rule and equation (4.9) on page 179, f 0 ( x ) = ln(cos(100)) cos(100)( x ) (2x ).
(f) f ( x ) = 2g( x) , where g is a differentiable function of x
By the chain rule and equation (4.9), f 0 ( x ) = (ln 2)2g( x) g0 ( x ).
5. (# 70 on page 173) This was not on the worksheet; however, I already had a solution typed
up, so I left it in for those who’d like to see.
Suppose that we pump water into an inverted right circular conical tank at the rate of 5
cubic feet per minute (i.e., the tank stands with its point facing downward). The tank has
a height of 6 ft and the radius on top is 3 ft. What is the rate at which the water level is
rising when the water is 2 ft deep? (Note that the volume of a right circular cone of radius
r and height h is V = 31 πr2 h.)
First, we note that the volume of water in the tank is a function of time, as are the depth
and radius of the surface of the water. That is, if t is measured in minutes, V (t) =
1
2
3 π (r ( t )) h ( t ). We also see similar triangles formed by the axis of the cone, and the lines
perpendicular to the base (labelled with lengths 3 and r). Then the ratios of their sides are
equal, so 36 = hr , hence r = 2h and we can rewrite the volume as
1
1
h(t) 2
1
2
V (t) = π (r (t)) h(t) = π
h(t) =
π (h(t))3 .
3
3
2
12
By the chain rule, V 0 (t) = 14 π (h(t))2 h0 (t). Since the volume of water in the tank is increasing by 5 cubic feet per minute, we have 5 = V 0 (t) = 14 π (h(t))2 h0 (t). Now, when the
water is 2 ft deep, we have h(t) = 2, so 5 = 14 π (2)2 h0 (t) = πh0 (t), and h0 (t) = π5 , that is,
the water level is rising at a rate of π5 feet per minute.