Rate problems

[xtrmk]

The diameter and height of a paper cup in the shape of a cone are both 4in. and water is leaking out at the rate of 0.5in^3/s. Find the rate at which the water level is dropping when the diameter of the surface is 2in. (V=1/3pir^2h)

Can someone help me with this problem..

d = 4 , h = 4
r = -0.5

and:
If the edge of a cube is increased by 1%, find the relative error in the surface area.
How do i proceed without given any volume or surface area

 

[merlin2007]

Re: First one

I'm not sure I understand the second one, but for the first one:

First, we need to make the radius and height the same variable. Assuming that the cone is a right cone, the ratio of the height to the radius is 2, by similar triangles.

Then
V = (1/3)(Pi)(r^2)(2r)=(2/3)(Pi)(r^3)
V' = 2(Pi)(r^2)(r')

From the above equation, we know the rate at which the volume is changing, and we know the current radius. Plugging in, we get r' = .5/(2 Pi) = 1/ (4 Pi) inches / second.

I misread the problem, I thought it was asking for rate of radius change.

 

[galactus]

The diameter of the top of the cone is 4" when the height is 4", therefore, the height when the diameter is 2" is 2".

Use similar triangles:

\(\displaystyle \L\\\frac{r}{h}=\frac{2}{4}\Rightarrow{r=\frac{1}{2}h}\)

\(\displaystyle \L\\V=\frac{1}{3}{\pi}r^{2}h\)

\(\displaystyle \L\\V=\frac{1}{3}{\pi}(\frac{1}{2}h)^{2}h\)

\(\displaystyle \L\\V=\frac{1}{3}{\pi}\frac{1}{4}h^{3}\)

\(\displaystyle \L\\V=\frac{1}{12}{\pi}h^{3}\)

\(\displaystyle \L\\\frac{dV}{dt}=\frac{1}{4}{\pi}h^{2}\frac{dh}{dt}\)

\(\displaystyle \L\\\underbrace{\frac{-1}{2}}_{\text{rate out}}=\frac{1}{4}{\pi}\underbrace{(2)^{2}}_{\text{height at\\dia. 2}}\frac{dh}{dt}\)

Find dh/dt.