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Math Help - Calculus Applications

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    Calculus Applications

    The cross-section of a water trough is an equilateral triangle with a horizontal top edge. If the trough is 5m long and 25 cm deep, and water is flowing in at a rate of 0.25m3/min, how fast is the water level rising when the water is 10 cm deep at the deepest point?
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    Quote Originally Posted by mathamatics112 View Post
    the cross-section of a water trough is an equilateral triangle with a horizontal top edge. If the trough is 5m long and 25 cm deep, and water is flowing in at a rate of 0.25m3/min, how fast is the water level rising when the water is 10 cm deep at the deepest point?
    Calculus Applications-capture.jpg

    First we need to be careful with the units convert everything into meters

    using the pythagorean theorem we can find the area of the front of the tank interms of only h.

    h^2 +\left( \frac{l}{2}\right)^2=l^2

    solving for l we get

    l=\frac{2}{\sqrt{3}}h

    Since we have two triangles to make the front of the trough we get

    A=2 \left( \frac{1}{2}\right)\left( \frac{l}{2}\right)h=\frac{h^2}{\sqrt{3}}

    So the volume is

    V=\frac{5m}{\sqrt{3}}h^2

    now taking the derivative with respect to time we get

    \frac{dV}{dt}=\frac{10 m}{\sqrt{3}}h\frac{dh}{dt}

    Now plugging in all of our data we get

     <br />
\frac{0.25m^3}{min}=\frac{10m}{\sqrt{3}}(.1m)\frac  {dh}{dt}<br />

     <br />
\frac{0.25m^3}{min}=\frac{m^2}{\sqrt{3}}\frac{dh}{  dt}<br />

    \left( \frac{0.25m^3}{min}\right) \left( \frac{\sqrt{3}}{m^2}\right)=\frac{dh}{dt}

    \frac{dh}{dt}=\frac{\sqrt{3}m}{4min}
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    Thank you so much
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