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Consider a tank in the shape of an inverted right circular cone that is leaking water. The dimensions of the conical tank are a height of 12 ft and a radius of 10 ft. How fast does the depth of the water change when the water is 10 ft high if the cone leaks at a rate of 8 cubic feet per minute?

Respuesta :

MrRoyal

Answer:

[tex]\frac{dh}{dt} = \frac{216}{1875\pi}[/tex]

Step-by-step explanation:

Given

Represent radius with r and height with h

[tex]h = 12ft[/tex]

[tex]r = 10ft[/tex]

[tex]Rate = \frac{8ft^3}{min}[/tex] when [tex]h = 10ft[/tex]

Express radius in terms of height

[tex]\frac{r}{h} = \frac{10}{12}[/tex]

[tex]\frac{r}{h} = \frac{5}{6}[/tex]

[tex]r = \frac{5}{6}h[/tex]

First, we need to determine the volume of the cone in terms of height.

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

Substitute [tex]r = \frac{5}{6}h[/tex]

[tex]V = \frac{1}{3} * \pi * (\frac{5}{6}h)^2 * h[/tex]

[tex]V = \frac{1}{3} * \pi * \frac{25}{36}h^2 * h[/tex]

[tex]V = \frac{1}{3} * \pi * \frac{25}{36}h^3[/tex]

[tex]V = \frac{25}{108} \pi h^3[/tex]

Differentiate both sides w.r.t time (t)

[tex]\frac{dV}{dt} = 3 * \frac{25}{108} \pi h^{3-1} * \frac{dh}{dt}[/tex]

[tex]\frac{dV}{dt} = \frac{75}{108} \pi h^{2} \frac{dh}{dt}[/tex]

Solve for [tex]\frac{dh}{dt}[/tex]

[tex]\frac{dh}{dt} = \frac{dv}{dt} * \frac{108}{75\pi h^2}[/tex]

At this point, [tex]h = 10[/tex] and [tex]\frac{dv}{dt} = \frac{8ft^3}{min}[/tex]

So, we have:

[tex]\frac{dh}{dt} = 8 * \frac{108}{75\pi 10^2}[/tex]

[tex]\frac{dh}{dt} = \frac{8 * 108}{75\pi 10^2}[/tex]

[tex]\frac{dh}{dt} = \frac{864}{7500\pi}[/tex]

[tex]\frac{dh}{dt} = \frac{216}{1875\pi}[/tex]

Hence:

The rate is [tex]\frac{216}{1875\pi} ft/min[/tex]

Using implicit differentiation, it is found that the depth of water changes at a rate of -0.0764 feet per minute.

The volume of a cone of radius r and height h is given by:

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

Applying implicit differentiation, we can find it's rate of change, thus:

[tex]\frac{dV}{dt} = \frac{2\pi rh}{3}\frac{dr}{dt} + \frac{\pi r^2}{3}\frac{dh}{dt}[/tex]

In this problem:

  • Height of 12 ft and radius of 10 ft, thus [tex]h = 12, r = 10[/tex].
  • Radius does not change, thus [tex]\frac{dr}{dt} = 0[/tex].
  • Water leaks at a rate of 8 cubic feet per minute, thus [tex]\frac{dV}{dt} = -8[/tex]

Then:

[tex]\frac{dV}{dt} = \frac{2\pi rh}{3}\frac{dr}{dt} + \frac{\pi r^2}{3}\frac{dh}{dt}[/tex]

[tex]-8 = \frac{\pi (10)^2}{3}\frac{dh}{dt}[/tex]

[tex]\frac{dh}{dt} = -\frac{24}{100\pi}[/tex]

[tex]\frac{dh}{dt} = -0.0764[/tex]

The depth of water changes at a rate of -0.0764 feet per minute.

A similar problem is given at https://brainly.com/question/9543179