calculus: related rates of change

[Mel Mitch]

Hello everyone,
Can someone who understand the below question explain it to me......:?:

Problem

Oil is dripping onto a surface at the rate of 1/10 (pi) cm^3/s and forms a circular film which may be considered to have a uniform depth of 0.1 cm.

Find the rate at which the radius of the circular film is increasing when the radius is 5 cm.

Mel

 

[BigGlenntheHeavy]

\(\displaystyle OK, \ we \ have \ a \ cylinder \ with \ a \ constant \ height, \ but \ an \ increasing \ radius.\)

\(\displaystyle V_{cyl} \ = \ \pi r^{2}h, \ \frac{dV}{dt} \ = \ \pi[2rh\frac{dr}{dt}+r^{2}\frac{dh}{dt}].\)

\(\displaystyle Now, \ \frac{dV}{dt} = \frac{1}{10\pi} \ and \ \frac{dh}{dt} \ = \ 0 \ as \ h \ is \ a \ constant.\)

\(\displaystyle Hence, \ when \ r \ = \ 5: \ \frac{1}{10\pi} \ = \ \pi[2(5)(.1)\frac{dr}{dt}+(25)(0)]\)

\(\displaystyle \frac{1}{10\pi} \ = \ (\pi)\frac{dr}{dt}, \ therefore \ \frac{dr}{dt} \ = \ \frac{1}{10\pi^{2}} \ cm \ per \ sec. \ increase.\)

 

[Mel Mitch]

thanks again bigG...the pi is not below its above and the other will divide it......so the answer it 0.1cm per sec...thanks for the help