# Rational Functions?

• May 17th 2009, 03:31 PM
reuger_
Rational Functions?
Suppose p varies directly as the cube of x, and inversely with y. If p=5/8 when x=4 and y=10, find p when x=2 and y=16

How do you set up the steps?
• May 17th 2009, 03:43 PM
Jen
When P varies directly with x, it means, $p=k_1x$ for some constant k.

If P varies directly with the cube of x then, $P = k_1 x^3$

When P varies inversely with y it means, $P = \frac{k_2}{y}$

Now if you use the initial conditions provided you should be able to solve for the constants $k_1 \mbox{ and } k_2$
• May 17th 2009, 03:51 PM
reuger_
thanks
Quote:

Originally Posted by Jen
When P varies directly with x, it means, $p=k_1x$ for some constant k.

If P varies directly with the cube of x then, $P = k_1 x^3$

When P varies inversely with y it means, $P = \frac{k_2}{y}$

Now if you use the initial conditions provided you should be able to solve for the constants $k_1 \mbox{ and } k_2$

Thanks for the quick response, I appreciate it.
• May 17th 2009, 04:02 PM
Jen
No Problem. :D