Question

Water is poured into a conical paper cup at the rate of 3/2 in3/sec (similar to Example 4 in Section 3.7). If the cup is 6 inches tall and the top has a radius of 3 inches, how fast is the water level rising when the water is 4 inches deep?

Answer 1

Answer:

The water level rising when the water is 4 inches deep is $\frac{3}{8\times \pi} inch/s$.

Step-by-step explanation:

Rate of water pouring out in the cone = R=$\frac{3}{2} inch^3/s$

Height of the cup = h = 6 inches

Radius of the cup = r = 3 inches

$\frac{r}{h}=\frac{3 inch}{6 inch}=\frac{1}{2}$

r = h/2

Volume of the cone = $V=\frac{1}{3}\pi r^2h$

$V=\frac{1}{3}\pi r^2h$

$\frac{dV}{dt}=\frac{d(\frac{1}{3}\pi r^2h)}{dt}$

$\frac{dV}{dt}=\frac{d(\frac{1}{3}\pi (\frac{h}{2})^2h)}{dt}$

$\frac{dV}{dt}=\frac{1}{3\times 4}\pi \times \frac{d(h^3)}{dt}$

$\frac{dV}{dt}=\frac{1\pi }{12}\times 3h^2\times \frac{dh}{dt}$

$\frac{3}{2} inch^3/s=\frac{1\pi }{12}\times 3h^2\times \frac{dh}{dt}$

h = 4 inches

$\frac{3}{2} inch^3/s=\frac{1\pi }{12}\times 3\times (4inches )^2\times \frac{dh}{dt}$

$\frac{3}{2} inch^3/s=\pi\times 4\times \frac{dh}{dt} inches^2$

$\frac{dh}{dt}=\frac{3}{8\times \pi} inch/s$

The water level rising when the water is 4 inches deep is $\frac{3}{8\times \pi} inch/s$.