Question

A constant volume of cookie dough is formed into a cylinder with a relatively small height and large radius. When the cookie dough is placed into the oven, the height of the dough decreases as the radius increases, but it retains its cylindrical shape. At time t, the height of the dough is 8 mm, the radius of the dough is 18 mm, and the radius of the dough is increasing at a rate of 2 mm per minute.

Part A: At time t, at what rate is the area of the circular surface of the cookie dough increasing with respect to time? (5 points)

Part B: At time t, at what rate is the height of the dough decreasing with respect to time?

Answer 1

Answer:

a) dA/dt = 125,42 mm/min

b) dh/dt = - 1,78 mm/min

Step-by-step explanation:

The volume of the cylinder is:

V(c) = π*r²*h         ( r and h radius of the base and heigh respectively )

Then the surface area is: Area of the base : π*r²

Plus lateral area :  h*2*π*r

A(c) =  π*r² + 2*π*r*h

With the help of differentials we obtain:

dA/dt = 2*π*r*dr/dt + 2*π*h*dr/dt + 2*π*r*dh/dt        (1)

In this equation and at time t we know:

r = 18 mm        dr/dt = 2 mm/min      h = 8 mm    dh/dt = ??  

dA/dt = increasing rate of the surface area ( with respect to time)

To find dh/dt we know:

V(c) is constant, then

V(c) = π*r²*h     then   h = V(c) / π*r²

dh/dt = - [V(c)*2*π*r*dr/dt ] / π²*r⁴

By subtitution of V(c)

dh/dt = - [ π*r²*h *2*π*r*dr/dt ] / π²*r⁴

dh/dt = - [2*π²h*r³*dr/dt ] /π²*r⁴

dh/dt = - [2*h* dr/dt ] / r

dh/dt =  -  2*8*2 / 18    mm/min

dh/dt = - 1,78 mm/min

By subtitution in equation (1)

dA/dt = 2*π*r*dr/dt + 2*π*h*dr/dt + 2*π*r*dh/dt

dA/dt = 2*π*18*2 + 2*π*8*2 + (- 2*π*18*1,78 )

dA/dt = 226,20 + 100,53 - 201,31

dA/dt = 125,42 mm/min

Answer 2

Following are solutions to the given points:

Formula:

Cylinder volume $v = \pi r^2 h...............(i)$

$\to \frac{dr}{dt} = 2 \frac{mm}{min}\\\\ \to \frac{dh}{dt} = -2 \frac{mm}{min}\\\\ \to h=8\\\\ \to r=18$

Part A:  

Calculating the circular area of the surface:

$\to A=\pi r^2$

So, $\frac{dA}{dt}=2 \pi r \frac{dr}{dt}.........(ii)$

Therefore

$\to \frac{dA}{dt}= 2\pi \times 18 \times 2= 72 \pi \ \frac{mm^2}{min}$

Part B:

$\to \frac{dh}{dt}= - 2 \frac{mm}{min}$

Learn more:

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