- #1

kreil

Gold Member

- 668

- 67

1. The radius (r) of a sphere is increasing at a constant rate of 0.04 cm/s. (Note: The volume of a sphere with radius r is V=4/3pir³

a) At the time when the radius of the sphere is 10 cm, what is the rate of increase of its volume?

b) At the time when the volume of the sphere is 36pi cubic cm, what is the rate of increase of the area of a cross section through the center of the sphere?

c) At the time when the volume and the radius of the sphere are increasing at the same numerical rate, what is the radius?

I had no problems with parts a or b:

a)V'=4pir²

V'(10)=4pi100

=400pi cubic cm/s

b)36pi=4/3pir³

27pi=pir³

r=3

A=pir²

A'=2pir

A'(3)=2(pi)(3)

=6pi cm/s

c) This is the one I was unsure about...What I did here was set the rate of increase of the radius (0.04) equal to the rate of increase of the volume of the sphere (4pir²) and solve for r, but for some reason it doesn't feel right:

(4)(10^-2)=4pir²

r²=.1/pi

r=(.1/pi)^(1/2)

Did I do this part correctly or not?

2. Let f be the function defined by f(x)=(1+tanx)^(3/2) for -pi/4 <= x < pi/2

a) Write an equation for the line tangent to the graph of f at the point where x=o

b) Using the equation found in part a, approximate f(0.02)

c) Let f^-1 denote the inverse fcn of f. Write an expression that gives

f^-1(x) for all x in the domain of f^-1.

Once again, no problems with a or b:

a) f'(x)=3/2(1+tanx)^1/2 (sec^2(x))

f'(0)=3/2(1+0)^1/2 (sec^2(0))

f'(0)=3/2(1)(1)

f'(0)=3/2

f(0)=1

(y-1)=3/2(x)

b) f(x)=(3/2)x+1

f(0.02)=(3/2)(2)(10^-2)+1

f(0.02)=.03+1

f(0.02)=1.03

c) I THINK all it asks for here is the inverse function of f...but it seems like there are too many conditions for the answer to just be

f^-1(x)=(1+tany)^(3/2) <~ is this correct?

Thank you in advance for all help offered.