Question 1

Little Tom has a talent to roll perfect spherical shape snowballs. He enters a snowball building contest at DMCI. Calculus students notice that the rate of which the radius of the sphere shaped snowballs Little Tom is rolling changes 3 cm per second.

a) At what rate is the volume changing when the radius is equal to 16 centimeters?

b) If the snowman Little Tom built had a volume totaling 729 cm³. Given r(0)= 0, approximately how long did it take him to build it?


Solution

a) Answer dV/dt = 3072π cm³ per second.

From the question we can tell that this is going to be a related rates question. We can solve this by:

• Separating the "general" and "particular" information.
"General information" is within the problem and true at all times.
"Particular information" is true only at a particular instant.
e.g. in this problem, the general information is that the radius is growing at a rate of 2 centimeters per second. So, if we let the radius of the sphere be represented by r, we can say that r'(t)= 2 cm/s.

note:
It is important to separate out these two types of information because information about the particular case is used at the very last step of solving the problem so you don’t want to confuse yourself.

• Identify the known rate and the desired rate
In this case we know dr/dt, and we want to find the rate of change of volume, dV/dt.

• Find a relationship using geometry or trigonometry that relates both the known and unknown rate of change
In this case we’re working with spheres, and we’re trying to find its volume so we’re going to be using the equation
V= (4/3)πr³

note:
From using this equation we get both Volume and radius of the sphere represented in the same equation and they are both changing with time so it is best to write it
V(t) = (4/3)π[r(t)]³

• Now differentiate both sides of the equation above with respect to time.
V(t) = (4/3)π[r(t)]³
dV/dt = 4π[r(t)]² dr/dt

• Then just plug in the values as we know r(t) and dr/dt.
dV/dt = 4π[16]²(3)

b) Answer t = ([(3/4π)729]^(1/3)) / 3 seconds or approx 1.8611 seconds.

With the given information we know that V= 729 cm³, and r(0) = 0, r'(t)= 3.

• Antidifferientiate r'(t) so we can an equation for r(t).
∫ r'(t) dt = r(t) = 3t + C

• Since we are given a point on r(t), we can pinpoint our graph to eliminate C.
r(0) = 3(o) + C
C = 0

• Now that we have an equation for r(t), we can use it in the sphere-volume equation.
729 cm³ = (4/3)π[r(t)]³
r(t) = [(3/4π) 729]^(1/3)
3t = [(3/4π) 729]^(1/3)
t = ([(3/4π) 729]^(1/3)) / 3