Question

1. (10 points) A cylindrical can has a height of 15 mm and an initial radius of 16 mm. The volume of the cylindrical can is decreasing at a rate of 541 cubic mm per minute, with the height being held constant. What is the rate of change, in mm per minute, of the radius when the radius is 3 mm? Remember to include a negative sign if the radius is decreasing.

please write clearly

Answer 1

1. We have, the volume of a cylinder of height = 15(as its constant) and radius r is

$\small V=15\pi r^2$

Differentiating both sides with respect to t we get

$\small {dV\over dt}=30\pi r \; {dr\over dt}$

Now, the volume is decreasing at a rate of 541 cubic mm per minute, so the rate of change of the radius when the radius is = 3mm is given by

$\small -541=30\pi \times 3\times \; {dr\over dt} \Rightarrow {dr\over dt}=-{541\over 90\pi}\text{ mm/minute}=-1.9134\text{ mm/minute}$

this concludes our answer to the first question.

2. We have the function

f(2) 8.6 + 6 3r +7

Then, differentiating with respect to x we have

$\small f'(x)=\frac{\frac{d}{dx}\left(8x+6\right)\left(3x+7\right)-\frac{d}{dx}\left(3x+7\right)\left(8x+6\right)}{\left(3x+7\right)^2}=\frac{8\left(3x+7\right)-3\left(8x+6\right)}{\left(3x+7\right)^2}$

Simplifying the numerator further we get

$\small f'(x)=\frac{38}{\left(3x+7\right)^2}$

Thus, we have

$\small f'(-3)=\frac{38}{\left(3(-3)+7\right)^2}={38\over 2^2}={19\over 2}$

Also, let us calculate f(-3)

$\small f(-3)={8(-3)+6\over 3(-3)+7}={-18\over -2}=9$

Now, a linear approximation to f(x) at x = -3 is given by

$\small f(x)\approx f(-3)+(x-(-3))f'(-3)$

$\small \Rightarrow f(x)\approx 9+{19\over 2}(x+3)$

Thus, using this we can estimate f(-3.1) as

$\small f(-3.1)\approx 9+{19\over 2}(-3.1+3)=9-{19\times 0.1\over 2}=9-{19\over 20}={161\over 20}$

which completes our answer to 2.