1. ## bonus problem help

1. 1.[Bonus problem, show your work] Water is pumped into a inverted cone-shaped tank with the height 3 m and diameter 2m at the top. If the water level (h) is rising at a rate of 20cm/min in the tank, find out the rate of increasing for the water volume (V) in the tank when the water level is 1.5 m tall.

2. Given is

$\frac{dh}{dt}=20 \times 10^{-2} \, \text{m/min}$.

$\left. \frac{dV}{dt}\right|_{h=1.5}$

Volume of a cone w/ radius r and height h is

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

You want to put V in terms of h alone then differentiate w/ respect to t. Using similarity of triangles you get the equation

$\frac{r}{h}=\frac{1}{3}$.

Solve for r, plug it in the volume equation, then differentiate w/ respect to t.

3. Originally Posted by Black
Given is

$\frac{dh}{dt}=20 \times 10^{-2} \, \text{m/min}$.

$\left. \frac{dV}{dt}\right|_{h=1.5}$

Volume of a cone w/ radius r and height h is

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

You want to put V in terms of h alone and differentiate w/ respect to t. Using similarity of triangles you get the equation

$\frac{r}{h}=\frac{1}{3}$.

Solve for r, plug it in the volume equation, then differentiate w/ respect to t.
Ok thanks.. But I don't know how to do it when you say differentiate w/ respect to t

4. Plugging r in the volume equation gives us

$V=\frac{1}{27}\pi h^3$.

Differentiating wrt t gives us

$\frac{dV}{dt}=\frac{1}{9}\pi h^2 \frac{dh}{dt}$.

5. ok so what do I do with the (1/9)(pi)(h^2)

6. Plug in h = 1.5 m and dh/dt= .2 m/min into the equation to get your answer.

7. ok thanks

8. so i should get (9pi/180) ??

9. Originally Posted by tbenne3
so i should get (9pi/180) ??
Aren't bonus problems meant to be your own work ....?

Thread closed. (pm me if I'm incorrect in the above assumption ....)