calculus help

dolphin6

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Joined
Nov 26, 2005
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11
find the equation of the line tangent to the curve at the point\
xy^2=12;(3,-2)
 
Solve to get
y²=12/x
Get the derivitive
2y*dy = -12/x² *dx
dy/dx = -12/(2yx²)
{Substitute the given values.}
dy/dx = -12/(2(-2)(3²))
dy/dx is the slope m in y-y<sub>1</sub> = m(x-x<sub>1</sub>)
where (x<sub>1</sub>,y<sub>1</sub>) is the point.
 
Yup. Thats the point of interest. It might be a good idea to check the first equation to make sure that is a "good" point. They have been known to have a warped sense of humor.
 
He's already done almost all of it for you. Where are you stuck?

Eliz.
 
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