Q: Finding the Rate of Gasoline Level Change in a Tanker Truck

[pine_apple]

please help me with this assignment question.

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.

 

[HallsofIvy]

please help me with this assignment question.

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.

A good feeling that this is wrong? I always have a bad feeling that I am wrong!

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 [itex]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
[tex]\int_{1.25}^{1.25-h}(1.25-h-\sqrt{1.565- x^2} dx[/itex]
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.

 

[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?

 

[courtrigrad]

[tex]V = 15\int_{1.25}^{1.25-h}(1.25-h-\sqrt{1.565- x^2} dx[/tex]So taking the derivative of that would be [tex]15(1.25-h-\sqrt{1.565- x^2} dx[/tex]

 

[pine_apple]

courtrigrad, thank you very much for your reply.
but i was wondering where "dx" came from; does "d" represents the diameter?

 

[courtrigrad]

my fault, it should just be: [tex]15(1.25-h-\sqrt{1.565- x^2}) [/tex]because if we have [tex] F(x) = \int_{a}^{x} f(t) \; dt [/tex] then [tex] F'(x) = f(x) [/tex] for every [tex] x [/tex] in [tex] [a,b] [/tex].There is no [tex] dx [/tex]

 

[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?

 

[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.

 

[pine_apple]

Aero, thank you so much ! :D

 

[pine_apple]

HallsofIvy, Coutrigrad, and Aero, thank you :)