Question

A snowball melts so that its surface area decreases at a rate of 1cm^2/min. Find a rate at which the diameter decreases when then diameter is 10 cm.

Answer 1

right or not ????

cause if diameter is 10 that means radius is 5,
therefore area of the snowball we have as \[4\pi r ^{2}\]\[=4*25*\pi\]\[=100\pi\],
therefore given it decreases by 1cm^2 /min it becomes,
\[99\pi\]

ie,
\[4\pi r ^{2}=99\pi\]\[4\ r ^{2}=99\]\[r ^{2}=99/4\]\[r =3\sqrt{11}/2\]
therefore \[diameter =3\sqrt{11}\],
Hence it decreases by \[10 - 3\sqrt{11}\]

Answer 2

At some point you should have calculated the derivative with respect to time

Answer 3

?????? @hero please explain what u mean ??

Answer 4

You mean, you want me to provide the solution

Answer 5

didnt get what u meant by 'derivative with respect to time' or perhaps havent studied it (cause am in 8th grade) !!

Answer 6

Calculus in eighth grade? I'm sure you could pull it off :P

Answer 7

this is not calculus dont even know what is calculus this is just surface area and volumes really

Answer 8

perhaps u could help me know a little more

Answer 9

I probably could but it is very late here where I am and I'm not exactly in the lecturing mood.

Answer 10

no problem still will google it or something but sorry for the asker for not providing the answer, perhaps u could help in this regard

Answer 11

Actually, for this problem, the first step would be to re-write SA = pi r^2 in terms of d first.

Answer 12

ok asker has o understand I am over and out, first time heard the word calculus and lokks like it is not for me as yet but still will search on its basics for now thnx !!

Answer 13

Good luck with that

Answer 14

\[A = 4\pi(\frac{d}{2})^{2} = \pi d^{2}\]
\[\frac{dA}{dd} = 2\pi d\]
\[\frac{dA}{dt} = 1\]
\[\rightarrow \frac{dd}{dt} = \frac{dA}{dt}*\frac{dd}{dA} = 1*\frac{1}{2\pi d}\]
when d=10
\[\frac{dd}{dt} = \frac{1}{20\pi}\]

Answer 15

Thanks everyone. we went over it today in class and 1/20pi is correct. I just wanted a head start