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The water level will be dropping at the rate of [tex]0.64 cm/sec[/tex] when the height of the water reaches [tex]3cm[/tex]
We are give a cone with dimensions as seen in the diagram in the attached image.
The volume changes at the rate
[tex]\frac{dV}{dt}=-2cm^2/sec[/tex]
the negative sign is used because the volume of water is reducing.
what we need to know is the rate the height is changing when [tex]h=3cm[/tex]. That is, we need to evaluate
[tex]\frac{dh}{dt}|_{h=3cm}[/tex]
The relationship between the volume and the height is given by
[tex]V=\frac{1}{3}\pi r^2h[/tex]
but both [tex]h[/tex] and [tex]r[/tex] are changing when [tex]V[/tex] changes. So we need to eliminate [tex]r[/tex] somehow. From the diagram, we can use the similar triangle property to accomplish this
[tex]\frac{r}{2}=\frac{h}{6}\\\\\implies r=\frac{h}{3}[/tex]
substitute into the volume formula to eliminate [tex]r[/tex]
[tex]V=\frac{\pi h^3}{27}[/tex]
differentiate both sides with respect to time, [tex]t[/tex], since both the volume and height are functions of time. They will both be differentiated implicitly
[tex]\frac{dV}{dt}=\frac{\pi h^2}{9}\frac{dh}{dt}[/tex]
to find [tex]\frac{dh}{dt}|_{h=3cm}[/tex], substitute the values for [tex]\frac{dV}{dt}[/tex] and [tex]h[/tex]
[tex]-2=\frac{\pi 3^2}{9}\frac{dh}{dt}\\\\\frac{dh}{dt}=-\frac{2}{\pi}\\\\\approx -0.64 cm/sec[/tex]
So the rate at which the height is reducing when it reaches [tex]3cm[/tex] is [tex]0.64cm/sec[/tex]
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