A constant volume of pizza dough is formed into a cylinder with a relatively small height and large radius. The dough is spun an

Question

3 months ago

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A constant volume of pizza dough is formed into a cylinder with a relatively small height and large radius. The dough is spun an

d tossed into the air in such a way that the height of the dough decreases as the radius increases, but it retains its cylindrical shape. At time t=k, the height of the dough is 13 inch, the radius of the dough is 12 inches, and the radius of the dough is increasing at a rate of 2 inches per minute.
(a) At time t=k, at what rate is the area of the circular surface of the dough increasing with respect to time? Show the computations that lead to your answer. Indicate units of measure.

(b) At time t=k, at what rate is the height of the dough decreasing with respect to time? Show the computations that lead to your answer. Indicate units of measure. (The volume V of a cylinder with radius r and height h is given by V=πr2h.)

(c) Write an expression for the rate of change of the height of the dough with respect to the radius of the dough in terms of height h and radius r.

Mathematics

1 answer:

tester [12.3K] · Answer 1 · 2026-02-24T05:46:38+03:00

Answer:

a) $\frac{dA}{dt} = 48 \pi\frac{in^{2}}{min}$

b) $\frac{dh}{dt} = - \frac{13}{3} \frac{in}{min}$

c) $\frac{dh}{dt} = - 2\frac{h}{r} \frac {dr}{dt}$

Step-by-step explanation:

To tackle this issue, we need to visualize a cylinder with height h and radius r (refer to the attached image).

a) To determine the rate of change in the area of the dough's circular surface over time, we should begin with the area formula for a circle:

$A=\pi r^{2}$

Next, to find the rate at which the area changes, we differentiate this formula with respect to the radius r:

$dA = \pi(2) r dr$

We divide both sides by dt, resulting in:

$\frac{dA}{dr} = 2\pi r \frac{dr}{dt}$

Now we can perform substitution:

$\frac{dA}{dr} = 2\pi(12in)(2\frac{in}{min})$

$\frac{dA}{dt} = 48\pi\frac{in^{2}}{min}$

b) For part b, we initiate with the formula for volume:

$V=\pi r^{2} h$

We can rearrange the equation to isolate h, yielding:

$h=\frac{V}{\pi r^{2}}$

Now we can restate the equation as:

$h=\frac{V}{\pi}r^{-2}$

Now, we will differentiate it to find:

$dh=\frac{V}{\pi} (-2) r^{-3} dr$

We express the derivative in another form so we have:

$\frac{dh}{dt}=-2\frac{V}{\pi r^{3}}\frac{dr}{dt}$

We take our original volume equation and substitute it into the current derivative, giving us:

$\frac{dh}{dt}= -2\frac{\pi r^{2} h}{\pi r^{3}} \frac{dr}{dt}$

Then we can simplify:

$\frac{dh}{dt} =-2\frac{h}{r} \frac{dr}{dt}$

Now we can replace the values provided by the question:

$\frac{dh}{dt} =-2\frac{13in}{12in} (2\frac{in}{min})$

Which simplifies to:

$\frac{dh}{dt} = - \frac{13}{3} \frac{in}{min}$

c)

Part c has already been covered in part b, where we derived the expression for how the height of the dough changes with respect to the radius in terms of height h and radius r:

$\frac{dh}{dt} =-2\frac{h}{r} \frac{dr}{dt}$