# Homework Help: Related Rates

1. Apr 25, 2007

### mathmann

1. The problem statement, all variables and given/known data
A triangular prism has end peices in the shape of inverted isosceles traingles with bases 60cm and heights 40 cm . It is 4m long and water is being pumped into it at a rate of 9 L/s. How fast is the level of water rising when the water is 10cm deep.

2. Relevant equations

dv/dt = dv/dh * dh/dt

3. The attempt at a solution

h = 40 - 10 = 30

dv/dt = 9000 cm(cubed)/s
dv/dh = (.5)bhl
= 360 000

dh/dt = 0.025cm(cubed)/s

Is this correct and is there supposed to be another step to this equation?

2. Apr 25, 2007

### mezarashi

If I have the geometry of your question right, you have a prism lying with the triangle faces on the side. So as the water fills up, it fills up the triangle.

If this is the configuration, you should be alarmed by your answer which is a constant. By looking at this container, you see that as you fill it with water, the sides become narrower and narrower, meaning it will fill faster. Your answer of the change in height over time should then be a function of the height itself.

3. Apr 26, 2007

### mathmann

How do I make it a function?

4. Apr 26, 2007

### HallsofIvy

Mezarashi, because it say an inverted isosceles triangle, I interpret this as meaning that the vertex of the triangle is at the bottom so prism widens toward the top, not narrow.

mathmann, (1/2)bhl is the volume of the entire prism. Draw a picture of the triangle and draw a horizontal line in it representing the water level. It is the volume of the water that you want. Since h is increasing, "b", the length of the base is also increasing. You need to replace b by a function of h.

To do that, look at your picture. You should see that you have two similar triangles- the triangle formed by the end of the prism and the triangle formed by the water. They have exactly the same angles and so are similar. That means that corresponding lenghts are in the same proportions. The top of the prism, divided by the height, 60/40, is equal to the top of the water, b, divided by the height of the water, h: b/h= 60/40. That easily gives you "b" as a function of "h". Replace b in your calculation by that function.

By the way, what course is this? You seem to be consistently posting, in "precalculus", problems that I would consider "calculus".

5. Apr 26, 2007

### mathmann

Thanks Halls of Ivy

For volume of the water I got 30 000 cm(cubed) and for the rate I got 0.3 cm(cubed)/s when the height is 10 cm. Is this now correct?

FYI : It is a intro to calculus class.