Here's the question you clicked on:
iop360
approximately by what percentage does the volume of a sphere increase if it's surface area is increased 2%?
\[V =\frac{ 4 }{ 3 }\pi r^3\] \[surface area = 4\pi r^2\]
answer is a 3% increase ( need to know how though)
\[+0.02 = 8 \pi r \frac{ dr }{ dt }\] after plugging in and taking derivative of the surface area forumla:
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)
i dont think you need calculus to solve this
how do i find the % by which r changes
brb, im gonna leave this question open
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
ive got a couple of ideas, ill try yours and mine
i chose r =1... i only got 1% for the answer hmmm
i think calculus is necessary
what did you choose for radius
r1 = 1 A1 = 4pi A2 = 4pi*1.02 r2 = sqrt(1.02) percent difference between r1 and r2 = ((sqrt(1.02)/1)-1)*100
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
now we find the volume using r1 and then using r2 and find the percent difference.