Question

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Answer 1

0.64ft/min

Explanations

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

`(dV)/(dt)=(dV)/(dh)\cdot(dh)/(dt)`

Given the volume of the shaded region expressed as:

`\begin{gathered} V=(1)/(27)\pi h^3 \\ (dV)/(dh)=(3)/(27)\pi h^2=(1)/(9)\pi h^2 \end{gathered}`

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

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

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

`\begin{gathered} (99)/(49\pi)=(dh)/(dt) \\ (dh)/(dt)=(99)/(49(3.14)|) \\ (dh)/(dt)=(99)/(153.86) \\ (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