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Math Help - Could I get some help finding how fast a triangle's area is changing?

  1. #1
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    Could I get some help finding how fast a triangle's area is changing?

    The question is:

    The height h of an equilateral triangle is increasing at a rate of 3cm/min. How fast is the area changing when h is 5 cm? Give the answer to 2 decimal places.

    How can I figure this question out?

    Thank You!
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  2. #2
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    earboth's Avatar
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    Quote Originally Posted by Meeklo Braca View Post
    The question is:

    The height h of an equilateral triangle is increasing at a rate of 3cm/min. How fast is the area changing when h is 5 cm? Give the answer to 2 decimal places.

    How can I figure this question out?

    Thank You!
    1. The area of a triangle is calculated by

    a = \dfrac12 \cdot s \cdot h

    2. In an equilateral triangle the sides are equal and the height is calculated by:

    h^2+\left(\frac12 s\right)^2 = s^2~\implies~h^2 = \dfrac34 s^2~\implies~s = \dfrac23 \sqrt{3} \cdot h

    3. Plug in this term instaed of s into the equation of the area:

    a(h)=\dfrac13 \sqrt{3} \cdot h^2

    4. Calculate the first derivation of a to get the speed of change:

    a'(h) = \dfrac23 \sqrt{3}  \cdot h

    Now calculate a'(5):

    a'(5) = \dfrac23 \sqrt{3}  \cdot 5 ~\approx 5.77 \ square\ units
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  3. #3
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    Isnt the derivative equation

    da/dt=(1/3)sqrt3h * dh/dt

    Isnt your equation missing the dh/dt with a final answer being 8.66?
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  4. #4
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    Quote Originally Posted by Meeklo Braca View Post
    Isnt the derivative equation

    da/dt=(1/3)sqrt3h * dh/dt

    Isnt your equation missing the dh/dt with a final answer being 8.66?
    earboth has given you \frac{da}{dh}. To get the final answer you must multiply this by \frac{dh}{dt} = 3 cm/min.
    Last edited by mr fantastic; March 10th 2009 at 06:02 AM. Reason: Fixed some latex .... Huh ... it was *my* post I fixed. There's one for the books lol!
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