Solutions
Calculus I - Related Rates Extra - Fall 2011 [Shaw]
1. Spherical balloon. A spherical balloon is being inflated with a large helium tank at a constant rate. You
estimate that as the balloon is being inflated, when the radius of the balloon is 4 inches, the radius is increasing
at about 1 inch per second. What is the rate at which the tank expels air, in cubic inches per second? (Fact:
the volume of a sphere with radius r is 34 πr3 .)
In order to calculate this rate, and assuming that there are no leaks so that all the air from the tank is truly
going into the balloon, we really need to know the rate of increase of the volume of the sphere. Thus, we use
the given equation and take the derivative of both sides.
dV
4
dr
dr
= 3 · πr2
= 4πr2
dt
3
dt
dt
Now we interpret the remainder of the problem: when r = 4,
dr
dt
= 1, so we get
dV
dt
= 4π · 42 · 1 = 64π.
2. Endzone run. An observer sitting on the 40-yard line of a football field watches a player catch the ball around
midfield and run straight toward the endzone. The observer is 30 yards from midfield, and the player runs at a
constant speed of 9 yards per second. At what rate is the player’s distance from the observer increasing when
the player has run 25 yards?
We set x to be the distance the player travels down the field, and y to be the distance from the runner to the
observer. A simplified labeled picture is below.
dy
The given information in the problem states that dx
dt is constant at 9 yps. We want to know what dt is at the
time when x = 25.
The equation relating all of these values is the Pythagorean theorem, namely that x2 + 302 = y 2 . We solve this
equation for y and then differentiate.
p
y = x2 + 900
− 1
x dx
2x dx
dy
1 2
dx
= √ dt
=
x + 900 2 · 2x
= √ dt
dt
2
dt
2 x2 + 900
x2 + 900
Substituting x = 25 and
dx
dt
= 9, we get:
dy
25 · 9
45
=√
= √ ≈ 5.76 yps. [Any correct form would have been fine.]
2
dt
61
25 + 900
1