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Math Help - Calculus Word Problem Help

  1. #1
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    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?
    Attached Thumbnails Attached Thumbnails Calculus Word Problem Help-photo-3.jpg  
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  2. #2
    Senior Member topher0805's Avatar
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    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?
    Last edited by topher0805; March 4th 2008 at 10:19 PM.
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  3. #3
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    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...
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  4. #4
    Senior Member topher0805's Avatar
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    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?
    Last edited by topher0805; March 4th 2008 at 10:19 PM.
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  5. #5
    Senior Member topher0805's Avatar
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    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.
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  6. #6
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    Yeah, I see that triangle. That was the one I had erased so that the picture was more clean.
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  7. #7
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    since it's mentioned that the picture is a cross-section.... shouldn't we be looking at volume?
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  8. #8
    Senior Member topher0805's Avatar
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    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.

    <br />
V = \frac {1}{2}\frac {H}{4}HL, where L = 10.

    <br />
V = 5\frac {H}{4}H

    So,

    V = \frac {5H^2}{4}
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  9. #9
    Senior Member topher0805's Avatar
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    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?
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  10. #10
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    so...

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

    Or am I way off?
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  11. #11
    Senior Member topher0805's Avatar
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    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.
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  12. #12
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    So.... about .037?
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  13. #13
    Senior Member topher0805's Avatar
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    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?
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  14. #14
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    Thank you so much, you are beyond amazing!

    Yeah, I'm in Calc I
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  15. #15
    Senior Member topher0805's Avatar
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    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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