I answered 32 pi on my test and it was incorrect. Can you help me with this problem below?

Question

At what rate, in stere per minute, is the volume of a sphere increasing if the radius of the sphere is two meters in length and is increasing at a rate of three meters per minute.

Answer 1

A stere is a very old measure of volume, equal to one cubic metre.
The increase in volume per minute is 4/3 * pi * (5^3 - 2^3)
= 4/3 * pi * 117 = 156 * pi = 490

Answer 2

James L. is correct. Because the rate is changing as the radius increases, the radius at a given time is critical. The only thing I would add is that the answer should be in stere per minute. (A stere is another term for a cubic meter)

48 st/min

Answer 3

Actually the 36 pi is the volume of the sphere if its radius were 3m. The increase in volume is
Final Volume - Initial Volume
=[(4/3)*pi*(5^3)] - [(4/3)*pi*(2^3)]
=[(500/3)*pi] - [(32/3)*pi]
=468/3*pi =156pi stere/min
But this is the increase in volume for the first minute only.
Increase in Volume at the nth minute is
[(4/3)*pi*{(2+3n)^3}] - [(4/3)*pi*{(2+3(n-1))^3}]

Answer 4

The 4/3 pi cancels between volume 1 and volume 2. All
you have to know is that volume increases with the cube
of the radius. An increasing rate of radius growth of 3mpm
translates to 3^3 m^3pm or 27 m^3pm. Read as
27 cubic meters per minute.

Answer 5

V=4/3(pi)r^3

Therefore dV/dr = 4(pi)r^2

Therefore, the rate of increase is;

4 (pi) (3)^2= 36pi steres/min
The initial radius of 2 meters has nothing to do with the problem.

Answer 6

V = 4*pi*r^3/3

Differentiate with respect to t:

dV/dt = 4*pi*r^2*dr/dt

Therefore, the initial radius most certainly DOES matter. The larger the radius, the faster the volume increases.

Plug in r=2 and dr/dt=3:

dV/dt = 4*pi*4*3 = 48 m^3/min.

Answer 7

the volume of a ball is 4/3*pi*r^3.
at time zero the volume is 4/3*pi*2^3
at time 1 minute the volume is 4/3*pi*5^3
the difference is 4/3*pi*117
another way is to find the rate of change by computing the tangent function of the volume 4/3*pi*(R+3dR)^3 with R as the variable, but that does not seem to be the required answer

thinking about your problem i think i have a better solution and it goes like this:
V - volume of the ball
R - radius of the ball
t - time
V=4/3*pi*R^3
R=2+3t
V=4/3*pi*(2+3t)^3 = 4/3*pi*(8+12t+18t^2+27t^3)
dV/dt = 4/3*pi*(12+36t+81t^2)
if t=0, dV/dt=4/3*pi*(36+81) = 4/3*pi*117 = 156*pi
if t=1, dV/dt =4/3*pi*(12+36+81) = 4/3*pi*129 = 172*pi

Answer 8

Its radius is unimportant to the answer

It is increasing at 3m per minute

So it will be increasing its volume by 4/3 * (pi) * 3^3

= 36 * Pi

Answer 9

at 0 min the vol is 4/3 pi 2^3
after 1 min it is 4/3 pi5^3
increase=4/3pi{5^3-2^3]
dV/dt=4pir^2dr/dt
=4pi*4*3=48 pi/min