anonymous
  • anonymous
write equation of normal line to y= 1/x at x=2
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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ChillOut
  • ChillOut
First we gotta find a tangent line to 1/x. Do you know how to do it?
anonymous
  • anonymous
im not sure. this is what i did but im not sure if its right |dw:1370051592641:dw|
ChillOut
  • ChillOut
Erm.. The first derivative tells the slope of the function. In this case, you want to find the slope of the tangent line. Remember that the equation of a straight line is y - y0 = m(x-x0). If you take the derivative of 1/x, you find the slope of its tangent line (m).

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ChillOut
  • ChillOut
Remember that:\[m=\frac{d}{dx}\ and \ m_{\tan}*m_{normal} =-1.\ \ \ So\ \frac{-1}{x²}*m_{normal} = -1\]
ChillOut
  • ChillOut
Do you understand what I am saying?
anonymous
  • anonymous
no not exactly
anonymous
  • anonymous
wouldnt you just do f(x+h)-f(x) over h to find the slope of the tan line?
ChillOut
  • ChillOut
That's the definition of derivative.\[\lim_{h \rightarrow 0}\frac{f(x+h) - f(x)}{h} = \frac{d}{dx} f(x)=m\] The slope of the tan line is -1/x², which is -1/4 at point 2. Now let's find y0, which is equal to the original graph's y coordinate (y=1/x -> 1/2 = 1/2) Now m(tan)*m(normal)= -1, that is,\[\frac{-1}{4}m_{normal}=-1,m_{normal}=4\] This is the value of the normal line's slope. Plug in the values for y0 and x0 in the straight line equation y-y0 = m(x-x0) -> y-1/2 = 4*(x-2)
anonymous
  • anonymous
so thats the equation?
ChillOut
  • ChillOut
Yes. It's the equation of the normal line. Do you want me to draw so you can understand what I did?
anonymous
  • anonymous
yes please
ChillOut
  • ChillOut
|dw:1370054499981:dw|
ChillOut
  • ChillOut
The line should read "normal line of slope m(normal) = -1/m(tan)
anonymous
  • anonymous
ok great i think i get it. thank you for your help!
ChillOut
  • ChillOut
You're welcome :)

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