integration question

coyoteflare · #1 $Old$ January 12th, 2009, 07:32 AM

$y=\frac{4}{x^2}$

why is this

$y=\frac{4x}{x^3}$

and not

$y=\frac{4x}{0.3333x^3}$

sorry i dont know hwo to do a fraction in a fraction so 0.33333 = one third

Danny · #2 $Old$ January 12th, 2009, 07:44 AM

Quote:

Originally Posted by coyoteflare $View Post$

$y=\frac{4}{x^2}$

why is this

$y=\frac{4x}{x^3}$

and not

$y=\frac{4x}{0.3333x^3}$

sorry i dont know hwo to do a fraction in a fraction so 0.33333 = one third

Are you trying to integrate $\frac{4}{x^2}$ i.e.

$\int \frac{4}{x^2}\, dx$ ?

coyoteflare · #3 $Old$ January 12th, 2009, 07:58 AM

Quote:

Originally Posted by danny arrigo $View Post$

Are you trying to integrate $\frac{4}{x^2}$ i.e.

$\int \frac{4}{x^2}\, dx$ ?

heres the question

A is the point (2,1) on the curve $y=\frac{4}{x^2}$

use calculas to find the gradient of the curve at A

Danny · #4 $Old$ January 12th, 2009, 08:53 AM

Quote:

Originally Posted by coyoteflare $View Post$

heres the question

A is the point (2,1) on the curve $y=\frac{4}{x^2}$

use calculas to find the gradient of the curve at A

Oh, you want the slope of the tangent. In this case you'll need the derivative. So

$\frac{d}{dx} \left( \frac{4}{x^2} \right) = \frac{d}{dx} \left( 4 x^{-2} \right)$

Once you have this, evaluate it at $x = 2$ and you'll have your answer.