# AP Calculus

Gold Member
I have a couple questions about some problems on a take-home quiz that I got (no calculator allowed) Bear with me, this post is long...

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.

Related Introductory Physics Homework Help News on Phys.org
HallsofIvy
Homework Helper
kreil said:
I have a couple questions about some problems on a take-home quiz that I got (no calculator allowed) Bear with me, this post is long...

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
Actually you DO have a problem with parts a) and b). What you have found here is dV/dr. The "rate of increase of volume" is dV/dt. Use the chain rule:
dV/dt= (dV/dr)(dr/dt) and you are given dr/dt.

b)36pi=4/3pir³
27pi=pir³
r=3

A=pir²
A'=2pir
A'(3)=2(pi)(3)
=6pi cm/s
Same thing. A'= dA/dr is NOT the "rate of increase of the area". That is
dA/dt= (dA/dr)(dr/dt).

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?
Once again, the "rate of change of the volume" is NOT dV/dr: it is
dV/dt= (dV/dr)(dr/dt). If (dV/dr)(dr/dt)= dr/dt, what do that tell you about dV/dr??

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
Yes, those are correct.

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.
Look carefully at your answer! You have f-1(x) on one side (the independent variable is x) but there is no x on the other side of the equation! In fact, all you did is replace "x" by "y"!
What you were vaguely remembering, I suspect, is this: rewrite the equation as y= (1+ tan(x))3/2 (just replace "f(x)" by y).
Now, SWAP x and y: x= (1+ tan(y))3/2. That's what really gives the "inverse" function: f goes "from x to y" and the inverse function reverses that. In order write this as a function, however, you now have to solve for y.

Gold Member
My teacher just informed me today that we haven't learned how to do #1 yet...I guess that explains why I totally blew it.

As to the inverse question...how do you propose that I solve for y in
x=(1+tany)^3/2 ?

Expanding the expression seems to get me nowhere.

Start by raising each side to the power of 2/3.

--J

Gold Member
Does this work?

x=(1+tany)^3/2

1.raise each side to the 2/3:

x^2/3 -1=tany

2. Multiply each side by cot(y):

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

Thanks for the hint

Keep in mind that tan(y)*cot(y) = 1, not y. Also, arctan(tan(y)) = y.

--J

Gold Member
Ok, so....

x^2/3 -1=tany

arctan(tan(y))=arctan(x^2/3 -1)

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

Check:

f^-1(f(x))=arctan(((1+tanx)^(3/2))^2/3)-1)=arctan(1+tanx-1)=arctan(tanx)=x
f(f^-1(x))=(1+tan(arctan(x^2/3 -1)))^(3/2)=(1+x^3/2-1)^2/3=x

Thanks Justin!

Gold Member
Different problem, same take home quiz...

Consider the curve defined by: -8x^2 + 5xy + y^3 = -149
a)find dy/dx
b)write an equation for the line tangent to the curve at the point (4,-1)
c)There is a number k so that the point (4.2, k) is on the curve. Using the tangent line found in part b, approximate the value of k.
d)write an equation that can be solved to find the actual value of k so that the point (4.2, k) is on the curve.
e)solve the equation found in part b for the value of k

still no calculator.

I solved parts a,b, and c. If it matters, the approximation I found for k in part c was -0.4. On part d, I substituted the point in the equation:

-8(4.2^2) + 5(4.2)(k) + k^3 = -149
-141.12 + 21k + k^3 = -149
k^3 + 21k + 7.88 = 0

And so my problem is part e. I cannot graph it since I am not allowed to use a calculator,and I am pretty sure this will not factor. It is not a quadratic so I cannot use quadratic formula. Of course, it is entirely possible my equation from part d is wrong...

What does everyone think? All hints/suggestions/thoughts welcome.