the radius of a sphere is increasing at a constant of
2 cm/sec. at the instant when the volume of the sphere is increasing at 32pi cm^3. sec, the surface area of the sphere is?

Question

the radius of a sphere is increasing at a constant of 
2 cm/sec. at the instant when the volume of the sphere is increasing at 32pi cm^3. sec, the surface area of the sphere is?

liizzyliizz · Answer

I am not sure where to start in this problem o.o help?

anonymous · Answer

first we need an equation that relates surface area, radius, and volume

amistre64 · Answer

surface are of sphere = 4pi r  right?

liizzyliizz · Answer

yes it is :)

amistre64 · Answer

or is it 4pi r^2

liizzyliizz · Answer

r^2

anonymous · Answer

should be squared since it is an area right?

anonymous · Answer

i will take your word for the 4

liizzyliizz · Answer

the second one is correct :P

amistre64 · Answer

so rate of change of SA is:

8 pi r * rate of change of radius

amistre64 · Answer

$$S=4\pi r^2$$
$$S' = 8\pir*\frac{dr}{dt}$$

anonymous · Answer

preview isn't working for me either, for the last few days

amistre64 · Answer

when the rate of change of the volume is:

so Volume = 4/3 pi r^3

V' = 4 pi r^2 * dr/dt

anonymous · Answer

Answer is 16 cm^2

amistre64 · Answer

im thinking it says:

when V' = 32 and the dr/dt is 3 that we can find the radius and in turn find the surface area; but surface area IS V' :)

anonymous · Answer

dr/dt = 2

anonymous · Answer

yeah i jumped the gun. time for a piece of paper

anonymous · Answer

dV/dt = S.(dr/dt)

anonymous · Answer

hence, S = 0.5(dV/dt) = 0.5 *32 = 16

amistre64 · Answer

yeah, trying to parse one long string of information can get challenging at this age:

the radius of a sphere is increasing at a constant of 2 cm/sec.

 at the instant when the volume of the sphere is increasing at 32pi cm^3. sec, 
               the surface area of the sphere is? 

Instant rate is the derivative; V' which is also the Sa :)

32 = 4 pi r^2 * r'

8/(pi r') = r^2

r = 2 sqrt(2/pi r')

r = $2\sqrt{\frac{2}{2pi}}$

r = $2\sqrt{\frac{1}{pi}}$ = 2(pi)^(-1/2)

that looks fun lol

anonymous · Answer

god i kept thinking it was 3

amistre64 · Answer

i get 16 too :)

liizzyliizz · Answer

o.o God help me during the AP exam. lol

anonymous · Answer

we have 
$$V'=4\pi r^2 r'$$
$$V'=8\pi r^2$$ and we know that 
$$V' = 32 \pi$$ so 
$$8 \pi r^2 = 32\pi$$
$$r^24, r=2$$

anonymous · Answer

that is 
$$r^2=4\implies r = 2$$

amistre64 · Answer

32pi?  ugh

anonymous · Answer

replace r by 2 in the formula for surface area and get the answer

anonymous · Answer

yeah pi should not be in answer

anonymous · Answer

$$S=4\pi \times 2^2 = 16\pi$$ what mainaknag said

anonymous · Answer

problem was not that it was hard, that we both (amistre and myself) got off to a false start because usually the problem says "how fast" not "what is"

anonymous · Answer

:) I meissed pi in the dV/dt

amistre64 · Answer

i heard the starting pistol and ran the other way :)

anonymous · Answer

in fact mainaknag solution was snappy two step one

anonymous · Answer

I think working extra hard for act exams, missing a lot of information in the question

liizzyliizz · Answer

:O