Volume problem

[Guest]

Hi guys,

Here's the Prob,
If dr/dt = - 4 and dh/dt = 10, then when r = 1 and h = 1, the volume of the cylinder is:

Ok this one is confusing the heck out of me,
So I know V = pi r^2h, and dont know where to go from there.

im just not getting these related rates probs can you guys please help?
Thanks,
Chris

 

[Gene]

Start with the product rule and take dV/dt. You'll get an equation in dh/dt and dr/dt which you know. Substitute them.
But if the question is

volume of the cylinder is:

there are no related rates and you just stick r = h = 1 in the volume equation.

 

[soroban]

Hello, Chris!

I'll reword the problem . . . and insert my own units.

We have a circular cylinder.
The radius is decreasing at 4 m/sec.
The height is increasing at 10 m/sec.
When the radius is 1m and the height is 1m,
what is the rate of change of the volume?

We know: .\(\displaystyle V\:=\:\pi r^2h\)

Differentiate with respect to time. .(product rule!)

. . \(\displaystyle \frac{dV}{dt}\;=\;\pi r^2\left(\frac{dh}{dt}\right)\,+\,\pi\cdot2r\left(\frac{dr}{dt}\right)\cdot h\;=\;\pi r^2\left(\frac{dh}{dt}\right)\,+\,2\pi rh\left(\frac{dr}{dt}\right)\)


We are given: .\(\displaystyle \frac{dr}{dt}\,=\,-4,\;\frac{dh}{dt}\,=\,10,\;r\,=\,1,\;h\,=\,1\)

. . Plug them in: .\(\displaystyle \frac{dV}{dt}\;=\;\pi(1^2)(10)\,+\,2\pi(1)(1)(-4)\;=\;2\pi\)


At that instant, the volume is increasing at \(\displaystyle 2\pi\) m\(\displaystyle ^3\)/sec.

 

[Guest]

ok thx gene,

How does this look?

Volume of cylinder = pi.r^2.h dh/dt = 10, dr/dt = -4

dV/dt = pi[r^2.dh/dt + 2hr.dr/dt]

= pi[r^2(10) +2hr((-4)]

At the time when r = 1 and h = 1 this gives

dV/dt = pi[10 - 8]

= pi[2]

= 6.28318

 

[Guest]

sorry soroban didn't see your reply, thank you!

 

[Guest]

Now let's say for arguments sake that I wanted to take this a step further and determine if the volume is inc, dec, etc... how would I go about doing that?

 

[Daniel_Feldman]

Reread soroban's reply.


1. Calculate dV/dt (change in volume/change in time)

2. If dV/dt at a certain time t is positive, then the volume is increasing at time t. If dV/dt at time t is negative, the volume is decreasing at t.


You can use this to help you with functions where you need to apply calculus.


If you have a function f(x):

f(x) is increasing where its derivative, f'(x), is greater than 0. f(x) is decreasing where f'(x)<0.