Question

The surface area of a sphere is decreasing at the constant rate of 3π sq. cm/sec . At what rate is the volume of the sphere decreasing at the instant its radius is 2 cm ?

Answer 1

getting stuck on this site not loading
what is the surface area of a sphere?

Answer 2

im searching on it

Answer 3

oh doh it is the anti derivative of the volume
making it
\[S=4\pi r^2\]

Answer 4

so basically i just substitute now?

Answer 5

yeah since
\[V'=4\pi r^2r'\]

Answer 6

dont i have to solve for dr/dt?

Answer 7

well let me think
maybe not since the derivative of the volume is the surface are but maybe

Answer 8

but you are told the rate of change of the surface area right? that is what is fixed
so maybe it is a good idea to write the volume in terms of the surface area and skip the radius all together

Answer 9

\[\large V=\frac{1}{6\sqrt{\pi}}S^{\frac{3}{2}}\] i think

Answer 10

so it ends up like this?, 4pi(2)^(2) and what is r^I

Answer 11

then
\[V'=\frac{1}{4\sqrt{\pi}}\sqrt{S}S'\]

Answer 12

the answer must be in cubic cm/sec

Answer 13

you are given \(S'=-3\pi\) or something (hard for me to read) so it is easier so skip the radius in this

Answer 14

ok so what would be the final answer?

Answer 15

yes of course because it is a volume
volume per second

Answer 16

is the rate \(-3\pi\) ?

Answer 17

\[S'=-3\pi\]
\[r=2\]
\[S=16\pi\] put them in here \[V'=\frac{1}{4\sqrt{\pi}}\sqrt{S}S'\]

Answer 18

in fact now we see we didn't need any of this
the derivative of the volume IS the surface area
you get \(-3\pi\)

Answer 19

i just typed it in my calculator and did the procedure like you just did thanks, im going to be posting some questions