# Related Rates - Set up

1. Jan 3, 2008

### rocomath

Related Rates - Is my set-up correct?

A water trough is 10ft long and its ends are isosceles triangles that are 3ft across the top and have height 1ft. If $$\frac{dV}{dt}=12\frac{\mbox{ft}^{3}}{\mbox{min}}$$, how fast is the water level rising when the water is 6 inches deep?

$$V=\frac{1}{2}(b_1 + b_2)h \rightarrow \frac{1}{2}[b_1 + (b_1 + 2r)]h$$

... By similar triangles $$\frac{3}{1}=\frac{r}{h} \rightarrow r = 3h$$

$$V=\frac{1}{2}(20+6h)h \rightarrow (10+3h)h$$

$$\frac{dV}{dt}=(10+6h)\frac{dh}{dt}$$

Is this correct so far?

Last edited: Jan 3, 2008
2. Jan 3, 2008

### dynamicsolo

I'm not sure I understand your choice of a volume formula. The trough is a triangular prism with its long axis (10') horizontal and its triangular cross-sections (base: 3', height: 1') oriented apex downward. The critical issue here is that your formula only has two dimensions of length and seems to be the formula for the area of a trapezoid...

3. Jan 3, 2008

### rocomath

Well I looked at the SM of the following problem b/c I wasn't sure what a trough was, and that is the same picture they used :-\

4. Jan 3, 2008

### HallsofIvy

Staff Emeritus
Did you look up the definition of "trough" in a dictionary? I suspect in the "following problem", they specified that the ends of the trough were trapezoids since you are trying to work with two bases. In this problem, it is specified that the ends are isosceles triangles. That's the difficulty with just "mimicing" another problem solution.

The volume of water is the end area times the length of the trough. The end of the whole trough is an isosceles triangle with height 1 and base 3: its area is (1/2)bh= (1/2)(3)(1)= 3/2 square feet. The water in the trough is a triangle "similar" to the entire triangle. You can get its base, and then area, by using the fact that "corresponding" sides are in the same proportions.

5. Jan 3, 2008

### rocomath

I would still need to take into consideration that it has 2 bases, right?

$$V=A(b_1 + b_2)$$

A = the isosceles triangles, which has units squared which multiplied with the bases will give a cubic.

Last edited: Jan 3, 2008
6. Jan 3, 2008

### dynamicsolo

The third dimension of the trough, which is a triangular prism, is simply the length of the trough. So the volume should be V = (1/2)·b·h·L . For your related rates problem, L will be the constant 10 feet. The trough is being filled from the apex of the triangle upward, so the current "height" of the volume of water will be h and the base will be given by the relation you already found. The filled volume will itself always be a triangular prism.

(It's somewhat amusing that, in an age where the great majority of people live in cities, there are still problems based on water troughs, which hardly anyone outside of rural areas have seen much.)

Last edited: Jan 3, 2008
7. Jan 4, 2008

### HallsofIvy

Staff Emeritus
What two "bases" are you talking about? The ends are triangles and the "3 feet" is the base of those triangles. The only other number given is the 10 foot length. The area of the end triangles is (1/2)(3)(1)= 3/2 square feet. The volume of the entire trought is that area times the length: (3/2)(10)= 15 cubic feet.

NOW, you need to do the actual problem. If the water in the trough has height 6"= 1/2 foot, what is its volume? The length of the trough is still 10 feet, the height is now 1/2 foot and you can get the length of the base by using "similar triangles".

8. Nov 19, 2008

### Calculus!

what we have:
dv/dt = 12
h = .5 ft.
l = 10 ft

what we want:
dh/dt

v = .5bhl
v = .5bh*10
v = .5(3h)10h
v = 15h^2

dv/dt = 30h*dh/dt
dh/dt = dv/dt / 30h
dh/dt = 12/30(10)
dh/dt = 12/300
dh/dt = .04 ft/min.

The height is changing at a rate of .04 ft/min.