Thread: Related Rate question regarding application of the deriviative..

1. Related Rate question regarding application of the deriviative..

I have tried this problem, but it's not working out.

Sand poured on the ground at a rate of 3 meters cubed per minute forms a conical pile whose height is 1/3 the diameter of the base. How fast is the altitude increasing when the radius of the base is 2 m.

First I replace r in the conical V=1/3 PIE r^2h
I take h = 1/3 d
h=1/3(2r)
So I get r = 3h/2
So then I put it in V=1/3 PIE (3h/2)^2h
When I do the derivative I get
dV/dt=3PIEh^2 dh/dt
I plug in the numbers
3=3PIE(2)^2 dh/dt
and get .0795 m/min

What am I doing wrong in this problem, if someone can help thanks.
Joanne

2. You have to apply the chain rule when differentiating.

$\displaystyle V(t)=\frac{9}{12}\pi h^3$

$\displaystyle \frac{dV}{dt}=\frac{9}{12}\pi(3h^2)\frac{dh}{dt}$

Your h should be $\displaystyle h=\frac{1}{3}(2r)=\frac{4}{3}$ when r=2
So just plug in your values. and solve for $\displaystyle \frac{dh}{dt}$

3. I am a bit lost here. Why is it 9/12 PIE h^3?? I have tried to figure it out and it's not working.
grrrrrrrrrr on not getting it
Jo

UPDATE here is what I got and finally got the answer at plugging at it for a while.
h=1/3(2r)
From this I got h=4/3 and r =3h/2
Then V=1/3Pi(3h/2)^2h
V=1/3 Pi (9h/4)^3
dV/dt=1/3 Pi(27/4)h^2 dh/dt
3=1/3 Pi(27/4)(4/3)^3
dh/dt = 0.239 m/min

I was shocked myself when I got it.
I could not figure yours out above, the way you did it.

4. The volume of a cone is:

$\displaystyle V=\frac{1}{3}\pi r^2h$

and you're given the relationship between the height and diameter.

$\displaystyle h=\frac{1}{3}(2r)=\frac{2}{3}r$

so

$\displaystyle r=\frac{3}{2}h$

The formula for volume becomes:

$\displaystyle V=\frac{1}{3}\pi\left(\frac{9}{4}h^2\right)h=\frac {9}{12}\pi h^3$

$\displaystyle V=\frac{9}{12}\pi h^3$

$\displaystyle \frac{dV}{dt}=\frac{9}{12}\pi (3h^2)\frac{dh}{dt}=\frac{27}{12}\pi h^2\frac{dh}{dt}$

So you want to find $\displaystyle \frac{dh}{dt}$ where you're given $\displaystyle \frac{dV}{dt}=3$ and when$\displaystyle r=2$ you know that $\displaystyle h \frac{4}{3}$.

$\displaystyle 3=\frac{27}{12}\pi\left(\frac{16}{9}\right)\frac{d h}{dt}=\frac{432}{108}\pi\frac{dh}{dt}$

I'm calculating $\displaystyle \frac{dh}{dt}=\frac{324}{432\pi}\approx .239$