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A sculptor uses a constant volume of modeling clay to form a cylinder with a large height and a relatively small radius. The clay is molded in such a way that the height of the clay increases as the radius decreases, but it retains its cylindrical shape. At time t=c, the height of the clay is 8 inches, the radius of the clay is 3 inches, and the radius of the clay is decreasing at a rate of 12 inch per minute. (a) at time t=c, at what rate is the area of the circular cross section of the clay decreasing with respect to time? show the computations that lead to your answer. Indicate units of measure.

Respuesta :

The rate [tex]\frac{dr}{dh} = - \frac{r}{2h}[/tex]  iat which the area of the circular cross section of the clay decreasing with respect to time.

Step-by-step explanation of the given problem:

The formula for calculating a cylinder's volume is: V = πr²h,

Where V is the volume, r is the cylinder's radius, and h is the cylinder's height.

The rate of change of the radius of the clay with respect to the height of the clay is to be expressed in terms of height h and radius r.

That is, dr / dh is expressed in terms of height h and radius r. To do so, we must first differentiate the volume from the height, i.e. find dV / dh:

V = πr²h

dV/dh = d/dh(π[tex]r^{2}[/tex]h)

dV/dh = π[tex]r^{2}[/tex] + 2πrh (dr/dh)

The volume is constant, hence dV/dh = 0.

We get,

0 = π[tex]r^{2}[/tex]h

dV/dh = π[tex]r^{2}[/tex] + 2πrh (dr/dh)

-π[tex]r^{2}[/tex] = 2πrh (dr/dh)

dr/dh = - π[tex]r^{2}[/tex]/2πrh

dr/dh = - r/2h

To learn more about Cylinder's volume, visit: https://brainly.com/question/28946667

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