Question

a conical tank is 10 feet across and 12 feet deep If water is flowing into the tank at a rate of 10 cubic feet per minute find the rate of change of the depth of the water when the water is 8 feet deep

Answer 1

Answer:

0.286 ft per minute ( approx )

Step-by-step explanation:

Let r represents the radius and h represents the height of the cone,

If the conical tank is 10 feet across and 12 ft deep,

Diameter = 10 ft, h = 10 ft

⇒ 2r = 10 ⇒ r = 5 feet,

Now, radius of a cone and its height is always in same proportion,

$\frac{r}{h}=\frac{5}{12}$

$\implies r =\frac{5}{12}h-----(1)$

Now, volume of a cone,

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

From equation (1),

$V = \frac{1}{3}\pi (\frac{5}{12} h)^2 h$

$V = \frac{1}{3}\pi (\frac{25}{144} h^2)h$

$V=\frac{25}{432}\pi h^3$

Differentiating with respect to t ( time ),

$\frac{dV}{dt}=\frac{25}{432}\pi (3h^2)\frac{dh}{dt}$

$\frac{dV}{dt} = \frac{25}{144}\pi h^2\frac{dh}{dt}$

Here, $\frac{dV}{dt}=10\text{ cubic ft per min}, h = 8\text{ ft }$

$10 =\frac{25}{144}\pi (8)^2 \frac{dh}{dt}$

$10 = \frac{25\times 64\pi }{144} \frac{dh}{dt}$

$10 =\frac{25\times 4\pi }{9}\frac{dh}{dt}$

$\implies \frac{dh}{dt}=\frac{10\times 9}{100\pi }=0.286\text{ ft per minute}$

Hence, the rate of change of the depth of the water would be 0.286 ft per minute