Single Variable Calculus, Chapter 3, 3.8, Section 3.8, Problem 24
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At what rate is the water level rising when the water is 6 inches deep? Suppose that the base are isosceles triangles.
Recall that the volume $V$ = (Area of the base)(length) so,
$ \begin{equation} \begin{aligned} V &= \text{(Area of the triangle)(length)}\\ \\ V &= \frac{bh}{2}(10)\\ \\ V &= 5bh \text{; but } \frac{b}{h} = \frac{3}{1} \Longrightarrow b = 3h\\ \\ V &= 5(3h)h\\ \\ V &= 15h^2 \end{aligned} \end{equation} $
Taking the derivative with respect to time we got,
$ \begin{equation} \begin{aligned} \frac{dV}{dt} &= 15 \cdot \frac{d}{dh} \left( h^2 \right) \frac{dh}{dt}\\ \\ \frac{dV}{dt} &= 15(2h) \frac{dh}{dt}\\ \\ \frac{dV}{dt} &= 30h \frac{dh}{dt}\\ \\ \frac{dh}{dt} &= \frac{\frac{dV}{dt}}{30h} && \text{where} \frac{dV}{dt} = 12 \frac{\text{ft}^3}{\text{min}}, h = 6 \text{inches} \left( \frac{1\text{ft}}{12 \text{inches}}\right) = 0.5 ft\\ \\ \frac{dh}{dt} &= \frac{12}{30(0.5)}\\ \\ \frac{dh}{dt} &= 0.8 \frac{\text{ft}}{\text{min}} \end{aligned} \end{equation} $
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