• Mar 16th 2009, 11:56 AM
mathrox
hi there, a friend told me this site helped her to understand well so i figured i'd give it a try! i have a few things i'm confused about on my calculus II homework. any tips or pointers would be super

f(x)= (x)/(x-3). find the derivative of the inverse of f(x) at x=2
a) -3
b) -1/3
c) 1/3
d) 1
e) 3

i found the inverse to be x= y/y-3 but i'm stuck..

An ellipse with equation x^2 + 9y^2= 9 is the base of a solid in which each cross section perpendicular to the x axis is a square. find the volume of the solid.
a) 4
b) 8
c) 16
d) 32
e) 64

i drew the graph of the ellipse and now i'm stuck

anyone have any ideas for either of these problems? thank you
• Mar 16th 2009, 12:25 PM
Reckoner
Quote:

Originally Posted by mathrox
f(x)= (x)/(x-3). find the derivative of the inverse of f(x) at x=2
a) -3
b) -1/3
c) 1/3
d) 1
e) 3

i found the inverse to be x= y/y-3 but i'm stuck..

$x=\frac y{y-3}$

$\Rightarrow x(y-3)=y\Rightarrow (x-1)y=3x$

$\Rightarrow y=\frac{3x}{x-1}$

So $f^{-1}(x)=\frac{3x}{x-1}.$

Quote:

An ellipse with equation x^2 + 9y^2= 9 is the base of a solid in which each cross section perpendicular to the x axis is a square. find the volume of the solid.
a) 4
b) 8
c) 16
d) 32
e) 64

i drew the graph of the ellipse and now i'm stuck
The volume of a solid with cross sections, perpendicular to the $x$-axis, of known area $A(x)$ on $[a,b]$ is

$V=\int_a^bA(x)\,dx.$

If we solve for $y$ in your equation, we get $y=\pm\sqrt{1-\frac{x^2}9}.$ So, the area of the cross section at $x$ is

$A(x)=\left[2\sqrt{1-\frac{x^2}9}\right]^2$ (squaring because the cross section is a square).

The total volume is then

$V=4\int_{-3}^3\left(1-\frac{x^2}9\right)\,dx.$

I use $-3$ and $3$ for the limits because the major axis of the ellipse is of length $6$ and it is centered at the origin.
• Mar 16th 2009, 12:36 PM
mathrox
thank you!! i can do it from there now(Happy)