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- anonymous

approximately by what percentage does the volume of a sphere increase if it's surface area is increased 2%?

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- anonymous

- chestercat

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- anonymous

\[V =\frac{ 4 }{ 3 }\pi r^3\]
\[surface area = 4\pi r^2\]

- anonymous

answer is a 3% increase ( need to know how though)

- anonymous

\[+0.02 = 8 \pi r \frac{ dr }{ dt }\]
after plugging in and taking derivative of the surface area forumla:

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- anonymous

find out what percentage r changes if you increase the area by 2% and then find out what percentage volume changes when you change r by (whatever you find from the first part)

- anonymous

i dont think you need calculus to solve this

- anonymous

how do i find the % by which r changes

- anonymous

brb, im gonna leave this question open

- anonymous

first choose a number for r, find the surface area with this r, then change the surface area to 102% of that area and solve for r, the percent difference will be [(r2/r1)-1]*100

- anonymous

back

- anonymous

ive got a couple of ideas, ill try yours and mine

- anonymous

i chose r =1... i only got 1% for the answer hmmm

- anonymous

i think calculus is necessary

- anonymous

I got 3%

- anonymous

what did you choose for radius

- anonymous

r1 = 1
A1 = 4pi
A2 = 4pi*1.02
r2 = sqrt(1.02)
percent difference between r1 and r2 = ((sqrt(1.02)/1)-1)*100

- anonymous

which is about 1 percent, but that's what percent the radius changes, so now we know that
if the area changes 2 percent, the radius changes 1 percent, so if we find out how much the
volume changes when we change the radius by 1 percent, we will have found out how much
the volume changes when we change the area by 2 percent

- anonymous

sorry ugly formatting

- anonymous

now we find the volume using r1 and then using r2 and find the percent difference.

- anonymous

o, now i get the answer

- anonymous

thanks!

- anonymous

np :)

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