Rates of change 3

[whiteti]

An inverted cone has a volume V cm^3 and a height h cm, and its base has a radius r cm. The volume of the cone is increasing at a rate of 20picm^3/min while its height is increasing at a rate of 3cm/min. At what rate is the base changing if its volume is 400picm^3 and its height is 12cm?
hint: V=1/3pir^2h

Ive got

d/dt((1/3pir^2)h)
= (1/3pir^2)'*h+(1/3pir^2)*(h)'

I'm stuck

 

[whiteti]

anyone?

 

[Deleted member 4993]

\(\displaystyle \frac{d}{dt}\left [\frac{1}{3}\pi * r^2\right ] \)

=\(\displaystyle \left [\frac{2}{3}\pi * r \right ] \frac{dr}{dt} \)

 

[HallsofIvy]

These are basic derivative formulas: (cf(x))'= c f(x)' for c a constant and (x^n)'= nx^(n-1).

\(\displaystyle \left(\frac{1}{3}\pi r^2\right)'= \frac{1}{3}\pi (r^2)'= \frac{1}{3}\pi(2r^1)= \frac{2}{3}\pi r\)

By the way, you "bumped" this thread after only 20 minutes. That can get you banned. People do not just sit around waiting for you to post a question. Most of us can spend a few minutes a day. When people are trying to help you don't complain that they are not helping you fast enough!

 

[soroban]

Hello, whiteti!

\(\displaystyle \text{An inverted cone has volume }V\text{ cm}^3,\text{ height }h\text{ cm, and its base has a radius }r\text{ cm. }\)
\(\displaystyle \text{The volume of the cone is increasing at a rate of }20\pi\text{ cm}^3\text{/min while its height}\)
\(\displaystyle \text{is increasing at a rate of }3\text{ cm/min.}\;\text{ At what rate is the }radius \text{ changing when}\)
\(\displaystyle \text{its volume is }400\pi\text{ cm}^3\text{ and its height is 12 cm? }\;\text{ Hint: }\,V\:=\:\frac{\pi}{3}r^2h\)

We have: .\(\displaystyle V \:=\:\frac{\pi}{3}r^2h\)


Differentiate with respect to time \(\displaystyle (t):\)

. . \(\displaystyle \dfrac{dV}{dt} \;=\;\dfrac{\pi}{3}\left(r^2\dfrac{dh}{dt} + 2rh\dfrac{dr}{dt}\right)\) .[1]

We are given: .\(\displaystyle \begin{Bmatrix}\dfrac{dV}{dt} &=& 20\pi \\ \dfrac{dh}{dt} &=& 3 \\ V &=& 400\pi \\ h &=& 12 \end{Bmatrix}\)


When \(\displaystyle V = 400\pi,\:h = 12\text{, we have: }\:400\pi \:=\:\frac{\pi}{3}r^2(12) \)

We have: .\(\displaystyle r^2 \,=\,100 \quad\Rightarrow\quad r \,=\,10\)


Substitute into [1]: .

. . \(\displaystyle 400\pi \:=\:\dfrac{\pi}{3}\left(10^2(3) + 2(10)(12)\dfrac{dr}{dt}\right) \)


Solve for \(\displaystyle \dfrac{dr}{dt}\)