# Related Rates Problem

1. Apr 24, 2006

### tangents

Hey guys, I'm having some trouble figuring out this problem and was wondering if someone would be kind enough to look over my work so I know that I am doing it correctly =).
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There is an open right cylinder cone that has a heigh of 10cm and diameter of 10cm. the depth h is changing at -3/10cm/hr. V=1/3(pi)r^2(h).

a) find volume of water at h=5

V= 1/3(pi)(5^2)(5)= 125/3(pi)cm^3

b) find dv/dt when h=5cm

dv/dt= 1/3(pi)(5^2)(-3/10)=-5/2cm^3/hr

c) show that rate of change of volume of water is directly proportional to the surface area of water. what is the constant of proportionality

dv/dt=1/3(pi)r^2(dh/dt)
surface area=(pi)r^2
1/3(pi)r^2(dh/dt)=(pi)r^2
constant is (pi)r^2
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My main problem is with parts b and c as I'm not sure if I am getting the concept of related rates correctly =/

2. Apr 24, 2006

### mathmike

for one you must understand that diamater of the cone is changing with the height.

3. Apr 24, 2006

### tangents

yes I figured that since the height decreases, so does the radius except the problem didn't give me dr/dt or any real mention so I'm not sure how to apply it to the problem if I need to.

4. Apr 24, 2006

### mathmike

well since the diameter is 10 when the height is 10 you can define one of them in terms of the other one, then you can take the derivitave wrt that variable. do you know how to do this?

5. Apr 25, 2006

### HallsofIvy

Staff Emeritus
Draw a picture. The side of a cone is a straight line so you can use "similar triangles" to find the function relating r to h (or just find the equation of the line through (0,0) and (10, 5)).