The radius of a cone is increasing at a constant rate of 7 meters per minute, and the volume is decreasing at a rate of 236 cubic meters per minute. At the instant when the radius of the cone is 99 meters and the volume is 180 cubic meters, what is the rate of change of the height? The volume of a cone can be found with the equation V=\frac{1}{3}\pi r^2 h.V= 3 1 ​ πr 2 h. Round your answer to three decimal places.

Answer:

The rate of change of the height is 0.021 meters per minute

Step-by-step explanation:

From the formula

[tex]V = \frac{1}{3}\pi r^{2}h[/tex]

Differentiate the equation with respect to time t, such that

[tex]\frac{d}{dt} (V) = \frac{d}{dt} (\frac{1}{3}\pi r^{2}h)[/tex]

[tex]\frac{dV}{dt} = \frac{1}{3}\pi \frac{d}{dt} (r^{2}h)[/tex]

To differentiate the product,

Let r² = u, so that

[tex]\frac{dV}{dt} = \frac{1}{3}\pi \frac{d}{dt} (uh)[/tex]

Then, using product rule

[tex]\frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h\frac{du}{dt}][/tex]

Since [tex]u = r^{2}[/tex]

Then, [tex]\frac{du}{dr} = 2r[/tex]

Using the Chain's rule

[tex]\frac{du}{dt} = \frac{du}{dr} \times \frac{dr}{dt}[/tex]

∴ [tex]\frac{dV}{dt} = \frac{1}{3}\pi [u\frac{dh}{dt} + h(\frac{du}{dr} \times \frac{dr}{dt})][/tex]

Then,

[tex]\frac{dV}{dt} = \frac{1}{3}\pi [r^{2} \frac{dh}{dt} + h(2r) \frac{dr}{dt}][/tex]

Now,

From the question

[tex]\frac{dr}{dt} = 7 m/min[/tex]

[tex]\frac{dV}{dt} = 236 m^{3}/min[/tex]

At the instant when [tex]r = 99 m[/tex]

and [tex]V = 180 m^{3}[/tex]

We will determine the value of h, using

[tex]V = \frac{1}{3}\pi r^{2}h[/tex]

[tex]180 = \frac{1}{3}\pi (99)^{2}h[/tex]

[tex]180 \times 3 = 9801\pi h[/tex]

[tex]h =\frac{540}{9801\pi }[/tex]

[tex]h =\frac{20}{363\pi }[/tex]

Now, Putting the parameters into the equation

[tex]\frac{dV}{dt} = \frac{1}{3}\pi [r^{2} \frac{dh}{dt} + h(2r) \frac{dr}{dt}][/tex]

[tex]236 = \frac{1}{3}\pi [(99)^{2} \frac{dh}{dt} + (\frac{20}{363\pi }) (2(99)) (7)][/tex]

[tex]236 \times 3 = \pi [9801 \frac{dh}{dt} + (\frac{20}{363\pi }) 1386][/tex]

[tex]708 = 9801\pi \frac{dh}{dt} + \frac{27720}{363}[/tex]

[tex]708 = 30790.75 \frac{dh}{dt} + 76.36[/tex]

[tex]708 - 76.36 = 30790.75\frac{dh}{dt}[/tex]

[tex]631.64 = 30790.75\frac{dh}{dt}[/tex]

[tex]\frac{dh}{dt}= \frac{631.64}{30790.75}[/tex]

[tex]\frac{dh}{dt} = 0.021 m/min[/tex]

Hence, the rate of change of the height is 0.021 meters per minute.

Plz help me out ...And pls don’t use my points and don’t say idk sry in the answer’s

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