Question

At what approximate rate (in cubic meters per minute) is the volume of a sphere changing at the instant when its surface area is 3 square meters and the radius is increasing at the rate of 1/5 meters per minute?

Answer 1

V=(4/3)pi*r^3
A=4pi*r^2=3
and we are given that
dr/dt=1/5
now compute
dV/dt
to find the rate change of volume and watch what formula you get
(remember to use the chain rule because this is the derivative w/respect to t)

Answer 2

am i finding the derivative for surface area or volume?

Answer 3

volume with respect to time
dV/dt
is what the question asks for

Answer 4

oh thats right.

Answer 5

so would it be 4pi(r)^2dr/dt?

Answer 6

the chain rule is going to give
exactly
now what does the first part of that formula look like?

Answer 7

[4pi(r)^2]dr/dt
^^^^^^^
familiar?

Answer 8

yes surface area formula!

Answer 9

so we can just plug in A=3 for that
and dr/dt=1/5
and we're done :)

Answer 10

plus now you know that for a sphere
A=dV/dr
which is a nice thing to know

Answer 11

3=4pi(r)^2(1/5)

Answer 12

you put the 3 where dV/dt should be
dV/dt=4pi(r)^2(dr/dt)=A(dr/dt)
...now plug in for A and dr/dt

Answer 13

3/5

Answer 14

yup

Answer 15

oh wow so it would 3/5 cubic meters per minute

Answer 16

thank you!

Answer 17

anytime!

Answer 18

i have another problem i need help with if you can help me im putting it up :)