related rates. Derivative question

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Cuddles

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Nov 6, 2007
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I have a figure in which line l is tangent to y=1/x^2 at point P, whose coordinates are (w, 1/w^2) and w is >0. Point Q has coordinates (w,0). Line 1 crosses the x-axis at point R, with coordinates (k, 0).

A. Find K when w=3

Ok, so P=(3, 1/9) and Q is (3,0). I need to find the derivative and keep getting answers that don't make any sense. I tried:

y=1/x^2 therefore y=x^-2

y'=-2x^-3

but that doesn't seem right...?
 
Hello, Cuddles!

I have a figure in which line \(\displaystyle L_1\) is tangent to \(\displaystyle y\:=\:\frac{1}{x^2}\) at point \(\displaystyle P\),
. . whose coordinates are \(\displaystyle \left(w,\:\frac{1}{w^2}\right)\) and \(\displaystyle w\,>\,0\).
Point \(\displaystyle Q\) has coordinates \(\displaystyle (w,\,0)\).
\(\displaystyle L_1\) crosses the x-axis at point \(\displaystyle R\), with coordinates \(\displaystyle (k,\,0)\)

(A) Find \(\displaystyle k\) when \(\displaystyle w\,=\,3\).

Ok, so \(\displaystyle P\,=\,\left(3,\,\frac{1}{9}\right)\) and \(\displaystyle Q\,=\,(3,0)\)

I need to find the derivative and keep getting answers that don't make any sense.

I tried: \(\displaystyle \:y\,=\,x^{-2}\;\;\Rightarrow\;\;y' \,=\,-2x^{-3}\)

but that doesn't seem right ... ? . why not?
Click to expand...

At \(\displaystyle P\left(3,\,\frac{1}{9}\right)\), the slope is: \(\displaystyle \:y' \:=\:\frac{-2}{3^2} \:=\:-\frac{2}{27}\)

The equation of \(\displaystyle L_1\) is: \(\displaystyle \:y\,-\,\frac{1}{9} \;=\;-\frac{2}{27}(x\,-\,3)\;\;\Rightarrow\;\;\L y \:=\:-\frac{2}{27}x\,+\,\frac{1}{3}\)

Now find the x-intercept . . .

 
OH!!! Thanks soroban! I forgot that derivative is slope! xD Woooow. Thank you.
 
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