Question

Can someone help me with this calculus related rates problem?

Answer 1

A potter forms a piece of clay into a right circular cylinder. As she rolls it, the height h of the cylinder increases and the radius r decreases. Assume that no clay is lost in the process. Suppose the height of the cylinder is increasing by 0.9 centimeters per second. What is the rate at which the radius is changing when the radius is 3 centimeters and the height is 5 centimeters?

Answer 2

\[v = \pi*r^2*h\]

Answer 3

well if you use the information you can find dV/dt

using the volume formula find dV/dh

and you know dh/dt

do dV/dt = dV/dh x dh/dt

does that make sense... as the 1st part

Answer 4

dv/dt=pi*r^2*h
0=pi*h*dr/dt*2r+pi*r^2*dh/dt
o=pi*h*dr/dt*2r+pi*r^2*.9

Answer 5

Am I on the right track?

Answer 6

wow... I would have worked this way

\[\frac{dV}{dh} = \pi r^2\]

so

\[\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt} = \pi r^2 \times 0.9\]

Answer 7

so now you know

\[\frac{dV}{dt} = 0.9 \pi r^2\]

Answer 8

so then doesn't \[\frac{dr}{dt} = \frac{dr}{dV} \times \frac{dV}{dt}\]

Answer 9

so find dV/dr then take the reciprocal to find dr/dV

Answer 10

so its just a suggest... they may be an easier... way... or i have it all muddled up

Answer 11

I am confused what you mean on the last part.

Answer 12

so is that about finding dr/dt...?

Answer 13

Yes

Answer 14

ok... so I split the problem into 2 related rates problems

part 1... identify the change in volume with respect to time dV/dt

then use that and related rates to dr/dt

@freckles may know better

Answer 15

I like your way @campbell_st
it seems to be less work than the other way

Answer 16

so part 2 is

\[\frac{dV}{dr} = 2\pi rh\]

then the recprocal is

\[\frac{dr}{dV} = \frac{1}{2\pi r h}\]

and using this

\[\frac{dr}{dt} = \frac{dr}{dV} \times \frac{dV}{dt} = \frac{1}{2\pi r h} \times 0.9 \pi r^2\]

Answer 17

but anyway, I'm sure someone will come along and correct any mistakes I made...

Answer 18

Thanks for helping me!

Answer 19

It said the answer was wrong when I plugged in the numbers and entered the problem..

Answer 20

(.9*pi*9)/(2*pi*3*5)

Answer 21

well remember r is decreasing

so after simplifying

\[\frac{dr}{dt} = - \frac{0.9 \times 3}{2 \times 5}\]

Answer 22

as the fraction

\[\frac{dr}{dt} = \frac{0.9 \pi r^2}{2\pi rh} = \frac{0.9r}{2h}\]

Answer 23

I got it right. I forgot to put the negative.

Answer 24

Thank you so much!

Answer 25

there you go... nice and simple

Answer 26

Phew... I'm still at high school... just happy to help