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galactus · May 10, 2009, 01:54 PM

Here is a related rates y'all may like to tackle.

"a circular tank with height 2 feet, radius on the top of R and radius on the bottom of r where R>r.

If R=6 ft and r=4.5ft, and it's filling at a rate of 15 ft^3/min, how fast is the height of the water changing when the water's radius is 5?

Of course, this forms a frustum of a cone. But, there is a trick that makes this quite easy to find dh/dt.

galactus · May 11, 2009, 02:42 PM

Since it is a frustum, extend the sides on down to make a cone, then treat it as such.

If we extend the sides down to the apex, it is at (0,-6).

That means the cone is 8 inches high overall.

By using similar triangles: $\frac{r}{h}=\frac{6}{8}\Rightarrow r=\frac{3h}{4}$

When r=5, then h=20/3.

$V=\frac{\pi}{3}(\frac{3h}{4})^{2}h=\frac{3{\pi}h^{ 3}}{16}$

$\frac{dV}{dt}=\frac{9{\pi}h^{2}}{16}\cdot\frac{dh} {dt}$

We have to find dh/dt, when h=20/3.

$15=\frac{9{\pi}(\frac{20}{3})^{2}}{16}\cdot\frac{d h}{dt}$

Solving for dh/dt, we get $\frac{dh}{dt}=\frac{3}{5\pi}$

Now, the easy way I mentioned was by noting that $\frac{dh}{dt}=\frac{\frac{dV}{dt}}{A(t)}$

Where A(t) is the area of the water surface at some time t. In this case, we want to know the surface of the water area when r=5.

That is easy enough, $25{\pi}$

Now, all we do is $\frac{15}{25\pi}=\frac{3}{5\pi}$ . Same as above only much quicker.