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mathcalculus

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

  • 8 months ago
  • 8 months ago

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  1. mathcalculus
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    • 8 months ago
  2. mathcalculus
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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

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

    • 8 months ago
  4. mathcalculus
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    yes i know the product rule

    • 8 months ago
  5. Psymon
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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.

    • 8 months ago
  6. phi
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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

    • 8 months ago
  7. mathcalculus
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    so not respect to b?

    • 8 months ago
  8. phi
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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..

    • 8 months ago
  9. mathcalculus
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    I'm confused with finding the derivative.

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

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

    • 8 months ago
  12. Psymon
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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.

    • 8 months ago
  13. mathcalculus
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    right

    • 8 months ago
  14. Psymon
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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.

    • 8 months ago
  15. phi
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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

    • 8 months ago
  16. mathcalculus
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    dA/dt

    • 8 months ago
  17. phi
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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

    • 8 months ago
  18. mathcalculus
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    oh okay so we ignore the 0.5 for now?

    • 8 months ago
  19. phi
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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

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

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

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

    • 8 months ago
  23. phi
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    ?

    • 8 months ago
  24. mathcalculus
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    @phi sorry connection went off

    • 8 months ago
  25. mathcalculus
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    i understand what you're saying up there.

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

    • 8 months ago
  27. mathcalculus
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    @Psymon

    • 8 months ago
  28. mathcalculus
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    @phi

    • 8 months ago
  29. phi
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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

    • 8 months ago
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