Question

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Answer 1

Answer:

0.64ft/min

Explanations

The rate of change of the water in the conical flask is expressed as;

$\frac{dV}{dt}=\frac{dV}{dh}\cdot\frac{dh}{dt}$

Given the volume of the shaded region expressed as:

$\begin{gathered} V=\frac{1}{27}\pi h^3 \\ \frac{dV}{dh}=\frac{3}{27}\pi h^2=\frac{1}{9}\pi h^2 \end{gathered}$

Substitute the given values and expression into the rate of change as shown;

$\begin{gathered} 11=\frac{1}{9}\pi h^2\cdot\frac{dh}{dt} \\ 99=\pi(7)^2\cdot\frac{dh}{dt} \end{gathered}$

Simplify and determine dh/dt, the rate at which the height is increasing

$\begin{gathered} \frac{99}{49\pi}=\frac{dh}{dt} \\ \frac{dh}{dt}=\frac{99}{49(3.14)|} \\ \frac{dh}{dt}=\frac{99}{153.86} \\ \frac{dh}{dt}\approx0.64ft\text{/min} \end{gathered}$

Therefore the height of the water in the tank is increasing at a rate of 0.64ft/min