related rates. Derivative question

[Cuddles]

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...?

 

[soroban]

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?

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 . . .

 

[Cuddles]

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

 

[Cuddles]

ok, I got 1/3 but the answer sheet says its 9/2

 

[skeeter]

y = (-2/27)x + 1/3

set y = 0, not x.