Answer to Question #258989 in Calculus for lyn

Question

a water tank in the form of an inverted cone is being emptied at the rate of 2 cubic feet per second. the height of the cone is 2.5m and the radius is 1.5m. find the rate of change of the water level when the depth is 2m. ( answer must be in english term)

Answer 1

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

$$\dfrac{r}{h}=\dfrac{1.5\ m}{2.5\ m}=>r=0.6h$$

$$V=V(h)=\dfrac{1}{3}\pi (0.6h)^2h=0.12\pi h^3$$

$$V(h)=0.12\pi h^3$$

Differentiate both sides with respect to $t$

$$\dfrac{dV}{dt}=\dfrac{d}{dt}(0.12\pi h^3)$$

Use the Chain Rule

$$\dfrac{dV}{dt}=0.36\pi h^2\dfrac{dh}{dt}$$

Solve for $\dfrac{dh}{dt}$

$$\dfrac{dh}{dt}=\dfrac{\dfrac{dV}{dt}}{0.36\pi h^2}$$

Given $\dfrac{dV}{dt}=-2ft^3/s, h=2m=6.56167979 ft$

$$\dfrac{dh}{dt}=\dfrac{-2ft^3/s}{0.36\pi (6.56167979 ft)^2}\approx-0.041ft/s$$

The water level decreases at rate of 0.041ft/s when the depth is 2m.