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Math Help - Rates of Change involving a triangle.

  1. #1
    Junior Member
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    May 2009
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    Rates of Change involving a triangle.

    The altitude of a triangle is increasing at the rate of 1cm/min while the area of the triangle is increasing at the rate of 2cm^2/min. At what rate is the base of the triangle changing when the altitude is 10cm and the area is 100cm^2?

    I can't seem to get a single answer as I have got a few possible answers from -1.6cm/min, 2cm/min and 0.08cm/min. It depends on how you split up the db/dt notation to include the height and area variables.

    I have trouble deciding which way to split the notation up. i.e. I get db/dt=db/dA*dA/dh*dh/dt which yeilds an answer of 2cm/min or db/dt=db/dh*dh/dA*dA/dt which yeilds an answer of 0.08cm/min or
    dA/dt = [bdh/dt + db/dth] which yeilds an answer of -1.6cm/min.

    dA/db=5 and b=20 substituting in the known h and A values. db/dA then goes on to equal 2/5.

    If anyone could see what they think it would be great. I think the answer may be 0.08cm/min as this seems to link the most variables. What does anyone think?
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  2. #2
    Senior Member
    Joined
    Dec 2008
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    First, we take all the information given:

    \begin{aligned}<br />
\frac{dh}{dt} &= 1\,\frac{\mbox{\small cm}}{\mbox{\small min}} \\<br />
\frac{dA}{dt} &= 2\,\frac{\mbox{\small cm}^2}{\mbox{\small min}} \\<br />
h &= 10\,\mbox{\small cm} \\<br />
A &= 100\,\mbox{\small cm}^2.<br />
\end{aligned}

    Then we remember the formula for the area of a triangle:

    A=\frac{1}{2}bh.

    Knowing the information above, we differentiate both sides using the Product Rule:

    \begin{aligned}<br />
\frac{dA}{dt}&=\frac{d}{dt}\left(\frac{1}{2}bh\rig  ht)\\<br />
2&=\frac{1}{2}\frac{db}{dt}h+\frac{1}{2}b\frac{dh}  {dt}\\<br />
2&=5\frac{db}{dt}+\frac{1}{2}\left(\frac{2A}{h}\ri  ght)\\<br />
2&=5\frac{db}{dt}+10\\<br />
-\frac{8}{5}\,\frac{\mbox{\small cm}}{\mbox{\small min}}&=\frac{db}{dt}.<br />
\end{aligned}
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