A hemispherical bowl of radius 6cm contains water which is flowing into it at a constant rate.When the height of the water is h cm the volume V of water in the bowl is given by V=22/7(6h^2-1/3h^3)cm^3.Find the rate at which the water level is rising when h=3,given that the time taken to fill the bowl is 1minute.

dV/dh=22/7

A hemispherical bowl of radius 6cm contains water which is flowing into it at a constant rate.when the height of the water is h cm, the volume v of the water in the bowl is given by v=π(6h^2-1/3h^3)cm^3.find the rate at which the water level is rising when h=3, given that the time taken to fill the bowl is one minute

To find the rate at which the water level is rising (dh/dt) when h = 3, we need to differentiate the volume function V with respect to height h, and then substitute h = 3 into the derivative.

Given: V = (22/7)(6h^2 - (1/3)h^3) cm^3

To differentiate V with respect to h, we can use the power rule and constant multiple rule:

dV/dh = d/dh [(22/7)(6h^2 - (1/3)h^3)]
= (22/7)(d/dh [6h^2] - d/dh [(1/3)h^3])

Differentiating term by term:
dV/dh = (22/7)(12h - (1/3)(3h^2))
= (22/7)(12h - h^2)

Now we can substitute h = 3 into the derivative:
dh/dt = dV/dt / dV/dh

We are given that it takes 1 minute to fill the bowl, so dV/dt = V/t = (22/7)(6(3)^2 - (1/3)(3)^3) / 1 = (22/7)(54 - 9) = (22/7)(45)

Substituting this into the equation for dh/dt:
dh/dt = (22/7)(45) / (22/7)(12(3) - (3)^2)
= 45 / (12(3) - 3^2)
= 45 / (36 - 9)
= 45 / 27
= 5/3

Therefore, the rate at which the water level is rising when h = 3 is 5/3 cm per minute.

To find the rate at which the water level is rising (dh/dt) when h = 3, we need to differentiate the given volume equation V = (22/7)(6h^2 - (1/3)h^3) with respect to time t.

Using the chain rule, we can express dV/dh as (dV/dt) * (dt/dh). Since the time taken to fill the bowl is 1 minute, dt/dh will be the reciprocal of the rate at which the water level is rising (dh/dt).

So, let's differentiate the volume equation V with respect to h:
dV/dh = (22/7) * [d(6h^2)/dh - d((1/3)h^3)/dh]

Now, let's find the derivative of each term:
d(6h^2)/dh = 12h
d((1/3)h^3)/dh = (1/3) * 3h^2 = h^2

Substituting these derivatives back into the equation:

dV/dh = (22/7) * (12h - h^2)
=> dV/dh = (22/7) * (12h - h^2)

Now, we can solve for dh/dt:

dV/dh = (22/7) * (12h - h^2)
(dt/dh) * (dV/dh) = (dt/dh) * (22/7) * (12h - h^2)

Since dt/dh is the reciprocal of dh/dt, we can write:

1 = (dt/dh) * (22/7) * (12h - h^2)

Rearranging the equation to solve for dt/dh:

(dt/dh) = 7 / [22(12h - h^2)]

Now, we can substitute the given value h = 3 and solve for dt/dh:

(dt/dh) = 7 / [22(12(3) - 3^2)]
= 7 / [22(36 - 9)]
= 7 / [22 * 27]
= 1 / (2 * 3)
= 1/6

Therefore, the rate at which the water level is rising when h = 3 is 1/6.

Why can't you just put the subject in the subject box? Otherwise, the right folks will never look at it.

VV=PI(6h^2-1/3h^3)
dV/dt=PI(12h - h^2)dh/dt

solve for dh/dt when h=3, and

v(h=6)=dv/dt * 60 sec or
dv/dt=PI (6*6^2-1/3 *6^3)/30 in cm^3/sec