1. rate/water level word problem

a trough has a triangular cross section. The trough is 6 ft across the top, 6 ft deep, and 16 ft long. Water is being pumped into the trough at the rate of 4ft^3 per minute. Find the rate at which the height of the water is increasing at the instant that the height is 4 ft.

2. Originally Posted by shepherdm1270
a trough has a triangular cross section. The trough is 6 ft across the top, 6 ft deep, and 16 ft long. Water is being pumped into the trough at the rate of 4ft^3 per minute. Find the rate at which the height of the water is increasing at the instant that the height is 4 ft.

We need to figure out the volume of the trough

$\displaystyle V=A_{base}\cdot l$

To find the area of the base we need to use similar triangles
3
*******
* b *
6*** *
* *
x *
*
so $\displaystyle \frac{x}{6}=\frac{b}{3} \iff b=\frac{1}{2}x$

The area is $\displaystyle 2 \frac{1}{2}x \cdot b=x \cdot \frac{1}{2}x=\frac{x^2}{2}$

So the Volume is $\displaystyle V=\frac{x^2}{2} \cdot 16=8x^2$

$\displaystyle \frac{dV}{dt}=16x\frac{dx}{dt} \iff \frac{1}{16x}\frac{dV}{dt}=\frac{dx}{dt}$

We want to find $\displaystyle \frac{dx}{dt}$ and we know $\displaystyle \frac{dV}{dt}=4 \mbox{ when } h=4$

$\displaystyle \frac{dx}{dt}=\frac{1}{16(4)} \cdot 4=\frac{1}{16}$

3. Originally Posted by shepherdm1270
a trough has a triangular cross section. The trough is 6 ft across the top, 6 ft deep, and 16 ft long. Water is being pumped into the trough at the rate of 4ft^3 per minute. Find the rate at which the height of the water is increasing at the instant that the height is 4 ft.
A similar problem that might be of interest is discussed at this thread.

4. ok, so what is the rate though. 1/16 whats? ft^3?

5. Originally Posted by shepherdm1270
ok, so what is the rate though. 1/16 whats? ft^3?

$\displaystyle \frac{dx}{dt}=\frac{1}{16 \mbox{ft}(4 \mbox{ft})} \cdot \frac{4 ( \mbox{ft})^3}{min}=\frac{1}{16}\frac{\mbox{ft}}{\m box{min}}$

6. An acknowledgement - Thanks for that!, say - of the given solution, followed by asking
Originally Posted by shepherdm1270
ok, so what is the rate though. 1/16 whats? ft^3?
would not be inappropriate here.

Also: TheEmptySet did all the work! Seriously - if you've taken the effort to actually understand his/her solution (as opposed to just copying it out) surely you can do that one small thing and work out the unit yourself.

I ask: How can dx/dt possibly have a unit of volume?

7. May I show you a short cut to these types of related rates problems?.

By making the observation that the rate of change of volume is equal to the cross-sectional area at that instant times the rate of change of the height.

$\displaystyle \frac{dV}{dt}=A(t)\cdot\frac{dh}{dt}$

$\displaystyle \frac{dh}{dt}=\frac{\frac{dV}{dt}}{A(t)}$

We know that dV/dt=4.

The area of the water surface, A(t), when it is 4 feet deep is just a rectangle with area 16*4=64.

So, we have $\displaystyle \frac{dh}{dt}=\frac{4}{64}=\frac{1}{16}$

8. Originally Posted by galactus
May I show you a short cut to these types of related rates problems?.

By making the observation that the rate of change of volume is equal to the cross-sectional area at that instant times the rate of change of the height.

$\displaystyle \frac{dV}{dt}=A(t)\cdot\frac{dh}{dt}$

$\displaystyle \frac{dh}{dt}=\frac{\frac{dV}{dt}}{A(t)}$

We know that dV/dt=4.

The area of the water surface, A(t), when it is 4 feet deep is just a rectangle with area 16*4=64.

So, we have $\displaystyle \frac{dh}{dt}=\frac{4}{64}=\frac{1}{16}$
But what's the unit, galactus!

9. Oh...I forgot.

It's $\displaystyle \frac{1}{16} \;\ N\cdot{m^{2}}/kg^{2}$

Oops.....that's gravity. I always get that and $\displaystyle ft/min$ mixed up.

10. see - this all looks like gibberish to me. My mind can not conceptualize math terms whatsoever. I have a 4.0, and calculus is killin me right now... i've never had something i couldn't figure out, but calculus baffles me. And it's extremely frustrating. So thanks for all your help. I only hope I can recreate it on the final. *crosses fingers*