# Math Help - Calculus Word Problem Help

1. ## Calculus Word Problem Help

I don't know if there is a limit to the amount you can post... I don't want to annoy anyone, but I'm having a hard time with these word problems...

A water trough is 10 feet long and has a cross-section that has the shape indicated in the attachment. If the trough is being filled with water at the rate of 1.25 ft cubed/minute, how fast is the water level rising when the water is 1.5 feet high?

2. First find an equation for the volume of the trough. To do this you need to consider two separate areas, the rectangular one and the triangular one.
Volume of a rectangular box is LWH. Volume of a triangular box is 1/2BHL.

Can you do it from there?

3. That's actually all that I have so far (I erased it from the picture to make it more clean).

I'm not sure how to link that into an equation that relates the water fill rate...

4. Assign a common variable between the two equations. Let's say that H is the height of water measured on the far left wall of your diagram. So the equation for the volume of the rectangular portion is 10*3*H, or 30H.

Now we need to do the same thing for the triangular portion. To do this we need to use similar triangles and Pythagoras. In your diagram there is a triangle of dimensions 0.5*2*x do you see this triangle?

5. Just for reference for our discussion.

Our knowledge of triangles tells us that as H decreases, and subsequently b and x decrease, the ratio of any of these two sides stays the same.

6. Yeah, I see that triangle. That was the one I had erased so that the picture was more clean.

7. since it's mentioned that the picture is a cross-section.... shouldn't we be looking at volume?

8. Ok, so we have the ratio that H/b will always be 4.

We can rearrange that formula to get b = H/4.

So we can then use that to get a new equation for the volume of the triangular portion.

$
V = \frac {1}{2}\frac {H}{4}HL$
, where L = 10.

$
V = 5\frac {H}{4}H$

So,

$V = \frac {5H^2}{4}$

9. So we now have an equation for the total volume of the trough.

$V = 30H + \frac {5H^2}{4}$

What we are looking for is the "rate of change of the water level with respect to time". Or, $\frac {dH}{dt}$

We have been given $\frac {dV}{dt}$, and it is equal to 1.25 cubit feet/minute.

Can you solve it from here?

10. so...

$dv/dt = 30*dH/dt + 4(10H)/4^2*dH/dt$?

Or am I way off?

11. Exactly. Take the derivative of both sides with respect to time.

$\frac {dV}{dt} = 30\frac {dH}{dt} + \frac {5H}{2}\frac {dH}{dt}$

$\frac {dV}{dt}$ is 1.25, and we are looking for $\frac {dH}{dt}$ when H is equal to 1.5 so just plug in those values and you have your answer.

13. Lol, I'm too tired to check right now but I'm sure it's right.

Just factor out $\frac {dH}{dt}$, then divide 1.25 by (30+7.5/2) to get your answer.

Yeah, I think it's right.

BTW, what class are you in right now? Calculus I?

14. Thank you so much, you are beyond amazing!

Yeah, I'm in Calc I

15. Dude, I'm in Calculus I also so I know how frustrating this stuff can be. I just spent the past few hours learning the stuff I just showed you so I don't mind. Gives me more practice.

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