Question

A conical vessel whose height is 4 metres and of base radius 2 metres is being filled with water at the rate of 0.75 cubic metres per minute. Find the rate at which the level of the water is rising when the depth of water is 1.5 metres.

Answer 1

$5445222222222113131$

cjccjccjccjcv

Answer 2

The rate at which the level of the water is rising when the depth of water is 1.5 metres is $\frac{4}{3}$π metres per minute.

Step-by-step explanation:

According to the given information, it is given that, a conical vessel whose height is 4 metres and of base radius 2 metres is being filled with water at the rate of 0.75 cubic metres per minute.

This clearly refers to the fact that the conical vessel is being filled with water at the rate that is $\frac{dV}{dt}$ equal to 0.75 cubic metres per minute. Also, it is given that, we need to find the rate at which the level of the water is rising when the new depth of water is 1.5 metres.

We know that, the volume of a circular cone is equal to V = $\frac{1}{3}$πr²h.

Now, putting the values given in the given problem, we get,

V= $\frac{1}{3}$π($\frac{h}{2)}$²h.

= $\frac{1}{12}$πh³.

Let this be equation (1).

Now, differentiating equation (1) with respect to t, we get,

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

=$\frac{1}{4}$π$h^{2} \frac{dh}{dt}$

Let this be equation (2).

Now, the new depth of water is 1.5 metres and let $\frac{dV}{dt}$ have the value 0.75, then, putting these values in equation (2), we get,

0.75 = $\frac{1}{4}$π$(1.5)^{2} \frac{dh}{dt}$

= $\frac{1}{4}$π$3 \frac{dh}{dt}$

Or, we get,

$\frac{dh}{dt} = \frac{4}{3}$π metres per minute.

Thus, the rate at which the level of the water is rising when the depth of water is 1.5 metres is $\frac{4}{3}$π metres per minute.

Learn more here

https://brainly.in/question/26807303

Learn more

https://brainly.in/question/6596510