Question

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

Answer 1

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:

In order to solve this problem, we must first picture a cylinder of height h and radius r (see attached picture).

a) So, in order to find the rate at which the area of the circular surface of the dough is increasing with respect to time, we need to start by using the are formula for a circle:

$A=\pi r^{2}$

So, to find the rate of change of the area, we can now take the derivative of this formula with respect to the radius r:

$dA = \pi(2) r dr$

and divide both sides into dt so we get:

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

and now we can substitute:

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

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

b) In order to solve part b, we can start with the formula for the volume:

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

and solve the equation for h, so we get:

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

So now we can rewrite the equation so we get:

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

and now we can take its derivative so we get:

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

we can rewrite the derivative so we get:

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

we can take the original volume formula and substitute it into our current derivative, so we get:

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

and simplify:

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

so now we can go ahead and substitute the values  provided by the problem:

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

Which simplifies to:

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

c)

Part c was explained as part of part b where we got the expression for the rate of change of the height of the dough with respect to the radius of the dough in terms of the height h and the radius r:

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

Answer 2

The rate of change of the height of the pizza with respect to (w.r.t.) time

can be found given that the volume of the pizza is constant.

• (a) The rate of increase of the surface area with time is 4·π in.²/min

• (b) The rate at which the height of the dough is decreasing is $\underline{4.\overline 3 \ in./min}$

• (c) Rate of change the height of the dough with respect to the radius $\dfrac{dh}{dr}$, is $\underline{-2 \cdot \dfrac{h}{r}}$

Reasons:

The height of the dough when t = k is 13 inches

Radius of the dough = 12 inches

Rate at which the radius of the dough is increasing, $\dfrac{dr}{dt}$ = 2 in.²/min

(a) Required: The rate of increase of the surface area with time

Solution:

The circular surface area, A = π·r²

By chain rule of differentiation, we have;

$\dfrac{dA}{dt} = \mathbf{\dfrac{dA}{dr} \times \dfrac{dr}{dt}}$

$\dfrac{dA}{dt} = \dfrac{d ( \pi \cdot r^2)}{dr} \times \dfrac{dr}{dt} = 2 \cdot \pi \times 2 = 4 \cdot \pi$

The rate of increase of the surface area with time, $\mathbf{\dfrac{dA}{dt}}$ = 4·π in.²/min.

(b) Required: The rate of decrease of the height with respect to time

The volume of the pizza is constant, given by; V = π·r² ·h

Therefore;

$h = \mathbf{ \dfrac{V}{\pi \cdot r^2}}$

$\dfrac{dh}{dt} = \dfrac{d \left( \dfrac{V}{\pi \cdot r^2} \right)}{dr} \times \dfrac{dr}{dt} = \dfrac{-2 \cdot V}{\pi \cdot r^3} = \dfrac{-2 \cdot \pi \cdot r^2 \cdot h}{\pi \cdot r^3} \times \dfrac{dr}{dt} = \mathbf{-2 \cdot \dfrac{h}{r} \times \dfrac{dr}{dt}}$

$\dfrac{dh}{dt} = -2 \cdot \dfrac{h}{r} \times \dfrac{dr}{dt} = -2 \times \dfrac{13}{12} \times 2 = \dfrac{13}{3} = 4. \overline 3$

The rate at which the height of the dough is decreasing, $\mathbf{\dfrac{dh}{dt}}$= $\underline{4.\overline 3 \ in./min}$

(c) Required:]The expression for the rate of change the height of the dough with respect to the radius of the cone.

Solution:

$\dfrac{dh}{dr} = \dfrac{d \left( \dfrac{V}{\pi \cdot r^2} \right)}{dr} = \dfrac{-2 \cdot V}{\pi \cdot r^3} = \dfrac{-2 \cdot \pi \cdot r^2 \cdot h}{\pi \cdot r^3} = -2 \cdot \dfrac{h}{r}$

$\dfrac{dh}{dr} = \mathbf{ -2 \cdot \dfrac{h}{r}}$

The rate of change the height of the dough w.r.t. the radius is $\underline{\dfrac{dh}{dr} = -2 \cdot \dfrac{h}{r}}$

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