# [chain rule] Related Rates

1. Dec 2, 2006

### pine_apple

Q: gasoline is pumped from the tank of a tanker truck at a rate of 20L/s. if the tank is a cylinder 2.5 m in diameter and 15 m long, at what rate is the level of gasoline falling when the gasoline in the tank is 0.5m deep? express in exact answer in cm/min [1L=1000 cm^3]

i did:

dv/dt = - 1200000cm^3/min

dv/dt = dv/dh*dh/dt

Vcylinder = pi r^2 l {l=h=length of the cylinder}
r = 125h/50 cm = 5h/2
V= pi(5/2h)^2l
dv/dh = pi 25/2 hl

-1200000 = pi 25/2 (50)(1500) dh/dt
dh/dt = -96/75 pi

i've got a good feeling that this is wrong. please help.

2. Dec 2, 2006

### HallsofIvy

Staff Emeritus
A good feeling that this is wrong? I always have a bad feeling that I am wrong!:rofl:

You say "r = 125h/50 cm = 5h/2" Why would this be true? I see that the "125" is 1.25 m converted to cm but where is the 50 from? Is it the "0.5 m deep". That shouldn't be in the general formula but only applied after differentiating. And why would muliplying h by that give the radius? I take it you are assuming that the cylinder is lying on its side like it would be on a truck. In that case, as the level of gasoline falls the gasoline does not form cylinders of decreasing radius, it forms a decreasing part of the original cylinder. Set up a coordinate system with the center of the circle, of radius 1.25 m, at the origin. Then it can be written $x^2+ y^2= 1.5625. At a given depth h, the top of the gasoline is at distance 1.25- h from the center. The area, against the ends of the cylinder, of the gasoline is given by $$\int_{1.25}^{1.25-h}(1.25-h-\sqrt{1.565- x^2} dx$ and so the volume of gasoline is [tex]15\int_{1.25}^{1.25-h}(1.25-h-\sqrt{1.565- x^2} dx[/itex] Don't do that integral! Just use the "fundamental theorem of calculus" to differentiate it and set h= 0.5. 3. Dec 2, 2006 ### pine_apple HallsofIvy, thank you very much for your reply. I had realized that the gasoline doesn't form cylinders of decreasing radius too, but I couldn't figure out how to express it. We have not learned the "fundamental theorem of calculus" yet. Is it possible to calculate it in another way without using the theorem? 4. Dec 2, 2006 ### courtrigrad [tex]V = 15\int_{1.25}^{1.25-h}(1.25-h-\sqrt{1.565- x^2} dx$$

So taking the derivative of that would be $$15(1.25-h-\sqrt{1.565- x^2} dx$$

Last edited: Dec 2, 2006
5. Dec 2, 2006

### pine_apple

but i was wondering where "dx" came from; does "d" represents the diameter?

Last edited: Dec 2, 2006
6. Dec 2, 2006

my fault, it should just be: $$15(1.25-h-\sqrt{1.565- x^2})$$

because if we have $$F(x) = \int_{a}^{x} f(t) \; dt$$ then $$F'(x) = f(x)$$ for every $$x$$ in $$[a,b]$$.

There is no $$dx$$

Last edited: Dec 2, 2006
7. Dec 2, 2006

### pine_apple

Thank you.
But if you multiply those, how would it give you the volume? Doesn't the product become the area?
I don't understand why h-(1.565-x^2)^1/2 is being subtracted from 1.25-h.
Can someone please tell me why?

8. Dec 5, 2006

### Aero

Consider the top surface of the liquid. Find its area as a function of the height
(remember that the cylinder is tipped on its side).

dh/dt * area of top surface = dv/dt

Plug in h = 50cm and solve for dh/dt.

9. Dec 7, 2006

### pine_apple

Aero, thank you so much ! :D

10. Dec 7, 2006

### pine_apple

HallsofIvy, Coutrigrad, and Aero, thank you :)