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mathcalculus

  • 2 years ago

hello, i have a related rate problem with area. (attached below)

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  1. mathcalculus
    • 2 years ago
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  2. mathcalculus
    • 2 years ago
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    i first started the problem like this: facts: altitude: increasing 2.000 A=4.000 altitude: increasing 11.500 A=82.000 so Area= (1/2) * base* height At what rate is the base of the triangle changing when the altitude is 11.500 centimeters and the area is 82.000 square centimeters? 82.000=(1/2)*base* 11.500 82.000=5.75b 14.2609=b ?? help

  3. mathcalculus
    • 2 years ago
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    the derivative of 1/2bh?

  4. mathcalculus
    • 2 years ago
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    yes i know the product rule

  5. Psymon
    • 2 years ago
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    Yep. Thats what these related rate problems require. If you have a change in area then you definitely need an area equation. And since the area is changing, you know its the derivative of the area. And yeah, the derivative of 1/2bh.

  6. phi
    • 2 years ago
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    if it helps, the area A, base B and height H are all functions of time (they are changing) A(t) = 0.5 * H(t) * B(t) take the derivative with respect to time

  7. mathcalculus
    • 2 years ago
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    so not respect to b?

  8. phi
    • 2 years ago
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    the change in Area with respect to time is also known as the rate you want dA/dt , and so on..

  9. mathcalculus
    • 2 years ago
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    I'm confused with finding the derivative.

  10. mathcalculus
    • 2 years ago
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    alli know is this formula: a=pi^2 r but im guessing it's not relevant for this problem.

  11. mathcalculus
    • 2 years ago
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    and also a=1/2*b*h

  12. Psymon
    • 2 years ago
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    This is a triangle, though, so we don't even have pi. But yes, you would use 1/2bh since its the area of a triangle, which is what we have.

  13. mathcalculus
    • 2 years ago
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    right

  14. Psymon
    • 2 years ago
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    Now its just that every time you take the derivative of a variable, you also write dx/dt, dy/dt, etc along with it. These dx/dt things represent the changes given in your problem. So if you decided to make the altitude h, when you take the derivative of it, you would have dh/dt. This dh/dt is what gets replaced with 2.000cm/min value given. But if I'm confusing ya I'm sure phi can help.

  15. phi
    • 2 years ago
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    to take the derivative of A with respect to time you write a little d in front, and divide by dt

  16. mathcalculus
    • 2 years ago
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    dA/dt

  17. phi
    • 2 years ago
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    yes that is the left side you have dA/dt = 0.5 * d/dt ( B * H) use the product rule on B*H

  18. mathcalculus
    • 2 years ago
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    oh okay so we ignore the 0.5 for now?

  19. phi
    • 2 years ago
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    You don't ignore constants, but you can keep them on the outside of the derivative just like d/dt (2x) = 2 *dx/dt

  20. mathcalculus
    • 2 years ago
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    so here: dA/dt=1/2(dh/dt)+(db/dt)

  21. phi
    • 2 years ago
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    except that is not the product rule http://www.1728.org/chainrul.htm

  22. phi
    • 2 years ago
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    the product rule of d/dt (x * y) = x * dy/dt + y * dx/dt

  23. phi
    • 2 years ago
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    ?

  24. mathcalculus
    • 2 years ago
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    @phi sorry connection went off

  25. mathcalculus
    • 2 years ago
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    i understand what you're saying up there.

  26. mathcalculus
    • 2 years ago
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    is this right? dA/dt= 1/2 (b*dh/dt)+(h*db/dt)

  27. mathcalculus
    • 2 years ago
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    @Psymon

  28. mathcalculus
    • 2 years ago
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    @phi

  29. phi
    • 2 years ago
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    is this right? dA/dt= 1/2 (b*dh/dt)+(h*db/dt) yes, but the 1/2 multiplies both terms \[ \frac{dA}{dt}= \frac{1}{2}\left( b\frac{dh}{dt} + h\frac{db}{dt} \right)\\ 2\frac{dA}{dt}= b\frac{dh}{dt} + h\frac{db}{dt} \] now fill in the various quantities and solve for db/dt

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