Question

A conical tank (as in Example 2) has height 3 m and radius 2 m at the base. Water flows in at a rate of 2 m3/min. How fast is the water level rising when the level is 1 m?

Answer 1

Answer:

$\dfrac{dy}{dt}=1.432\ m/min$

Explanation:

given,

height of cone = 3 m

radius of cone = 2 m

flow rate = 2 m³/min

water level rate when height at 1 m = ?

volume of cone

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

now,

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

where radius of cone at y depth

$r' = \dfrac{r}{h}y$

so,

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

differentiating both side

$\dfrac{dV}{dt}= \pi \dfrac{r^2}{h^2}y^2\dfrac{dy}{dt}$

now, h = 1 m

$2 = \pi \dfrac{2^2}{3^2}\times 1^2\dfrac{dy}{dt}$

$\dfrac{dy}{dt}=1.432\ m/min$

Answer 2

The rate at which the water level rises when the level is 1 m is 0.174m/s

The formula of the conical tank is expressed using the formula:

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

Get the rate at which the water is flowing in the tank

$\frac{dV}{dt} =\frac{dV}{dr} \frac{dr}{dt} + \frac{dV}{dh} \frac{dh}{dt} \\\frac{dV}{dt} =\frac{dV}{dr} \frac{dr}{dt} + \frac{dV}{dh} \frac{dh}{dt} \\\frac{dV}{dt} =\frac{2}{3} \pi rh \frac{dr}{dt} + \frac{1}{3} \pi r^2 \frac{dh}{dt}$

Given the following parameters

r = 2m

h = 3m

$\frac{dV}{dt}=2m^3/s$

$\frac{dh}{dt} = 1m^2/s$

Substituting the given parameters into the formula

$2=\frac{2}{3} (3.14) (2)(3) \frac{dr}{dt} + \frac{1}{3}(3.14) 2^2 (1)\\2=12.56 \frac{dr}{dt} + 4.186\\12.56 \frac{dr}{dt} = 2 - 4.186\\12.56 \frac{dr}{dt} = -2.186\\\frac{dr}{dt} =\frac{2.186}{12.56}\\\frac{dr}{dt} = 0.174m/s$

Hence the rate at which the water level rises when the level is 1 m is 0.174m/s

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