Question

water is draining from the bottom of a cone shaped funnel at the rate of 0.03 ft³/s. the height of the funnel is two feet and the radius is one foot. sketch. at what rate is the height of the water in the funnel changing when the water height is ½ ft? first find dh/dt. h = 2r. r = h/2. v cone = πh³/12. dv/dt = 3πh²/12 dh/dt. -0.03 = π/16 dh/dt. dh/dt and dr/dt. dv/dt

Explanation:

Step1: Relate radius and height

Since the cone has height \(h = 2\) and radius \(r = 1\), by similar - triangles, \(r=\frac{h}{2}\).

Step2: Volume formula of a cone

The volume of a cone is \(V=\frac{1}{3}\pi r^{2}h\). Substitute \(r = \frac{h}{2}\) into the volume formula: \(V=\frac{1}{3}\pi(\frac{h}{2})^{2}h=\frac{\pi h^{3}}{12}\).

Step3: Differentiate with respect to time \(t\)

Using the chain - rule \(\frac{dV}{dt}=\frac{dV}{dh}\cdot\frac{dh}{dt}\). Differentiate \(V=\frac{\pi h^{3}}{12}\) with respect to \(h\): \(\frac{dV}{dh}=\frac{\pi h^{2}}{4}\). So \(\frac{dV}{dt}=\frac{\pi h^{2}}{4}\cdot\frac{dh}{dt}\).

Step4: Substitute known values

We know that \(\frac{dV}{dt}=- 0.03\) (negative because the volume is decreasing). When \(h = 1\) (assuming the problem is asking for the rate when the water height is \(h = 1\) foot, since it's not clearly specified in the hand - written content, but following the proportion \(r=\frac{h}{2}\)).
Substitute \(\frac{dV}{dt}=-0.03\) and \(h = 1\) into \(\frac{dV}{dt}=\frac{\pi h^{2}}{4}\cdot\frac{dh}{dt}\):
\(-0.03=\frac{\pi(1)^{2}}{4}\cdot\frac{dh}{dt}\).

Step5: Solve for \(\frac{dh}{dt}\)

\(\frac{dh}{dt}=\frac{-0.03\times4}{\pi}=-\frac{0.12}{\pi}\approx - 0.038\) ft/s.

Answer:

The rate at which the height of the water is changing is \(\frac{dh}{dt}=-\frac{0.12}{\pi}\approx - 0.038\) ft/s.