For the Love of God Answer this Straight?

Question

Hey, I really need help on this question. I've been stuck on it for a while, and its driving me crazy.

A trough is 15ft long and 2.4ft across the top. its ends are isosceles triangles with an altitude of 1.7ft and vertex down. Water is being pumped into the trough at a rate of 2.3ft^3/min. How fast is the water level rising when the water is 1.2 ft deep?

thanks in advance!@

Answer 1

This question is most easily answered using calculus. If you're not there yet, sorry - it would help in the future if you'd tell us what class you're in.

In related rate problems, you're given the rate of change of one quantity and asked for the rate of change of another. You need to find a relation between the two quantities and then take the time-derivative of both sides and solve for what you want. Here, you're given the rate of change of the volume and asked the rate of change of the depth.

If you make a sketch of the trough showing the water line somewhere down, you'll notice that the water at the end forms a triangle similar (in the geometric sense) to the trough as a whole. If you think about it, then, you're actually being asked for the rate of change of the altitude of the triangle formed by the water.

The volume of the trough will be (1/2)bhl = (1/2)(2.4)(1.7)(15) = 30.6 cubic feet. For right now, let's just call that V. The volume of water in the trough will be given by the same formula (basically just the area of the triangle times the length of the trough), but both the base and the height of the triangle will be changing over time. However, since the triangle made by the water is similar to that of the trough as a whole, the ratio of the base and the height of the triangle will be the same as that for the trough as a whole. Let B be the base of the water-triangle (how far the surface of the water is from side to side) and H be the height (which is the depth of the water - the thing whose rate of change you want). Then we know that

H/B = 1.7/2.4

So B = (17/24) H

So:

V = (1/2)BH(15)

V = (1/2)(17/24)H * H * 15

V = (255/48)H^2

dV/dt = (255/24)H dH/dt

dH/dt = (24 dV/dt)/(255 H)

= 24 (2.3)/(255 * 1.2) = 0.180 ft/min

Answer 2

Its an interesting problem.

I will basically tell you how to do it but not give you the answer.

First, your volume is constantly increasing at a constant rate.

V(t) = 2.3t

Now what you need to do is find the water level as a function of the time and then take the derivative of that.

So to do that, you need to find the volume as a function of the water level L. (If the trough was 2 ft^3 CUBE the function would be V=2*2*(L) ft cube = 4 ft cube)


(That is easy if you draw a picture.

So you have

V(L). From this you should be able to solve for L in terms of V.
L(V) = ??? (probably something with a cube root)

In my example
V(L) = 4L and V(t) = 2.3ft^3 per min.


Then you can use the chain rule:
Derivative(L(t)) = Derivative(L(V))*Derivative(V(t))
= Derivative(L(v))*2.3
(In my example,
Derivative(L(t)) = Derivative(4L)*(Derivative(2.3t)
= 4*2.3 ft per min.

And that is your answer.

Answer 3

I'm laughing at the doodle answer... So.... how big?

Answer 4

oh god i'm so glad that i don't have to do that stuff any more. i'm so sorry. and i don't remember the answer.

Answer 5

How big is ur doodle

Answer 6

i dont know but i wouldnt worry about it if i were you