What is the slope of the line normal to the tangent line of f(x) = e^(x^2-1)+2x-2 f(x)=ex2−1+2x−2 at x= 1 x=1?

1 Answer
Nallasivam V
May 16, 2017

y=-1/82.34x+22.09y=−182.34x+22.09

Explanation:

Given

y=e^(x^2-1)+2x-2y=ex2−1+2x−2

At x=2x=2

y=2.71828^(2^2-1)+2(2)-2y=2.7182822−1+2(2)−2
y=2.71828^3+4-2y=2.718283+4−2
y=20.085+2=22.085y=20.085+2=22.085

At Point (2, 22.085)(2,22.085) There is a tangent to the curve.
Look at the graph
enter image source here

The slope of the tangent is equal to the slope of the given curve.

The slope of the given curve is-

dy/dx=2xe^(x^2-1)+2dydx=2xex2−1+2

At x=2x=2 the slope of the curve is

dy/dx=2(2)(2.71828^3)+2dydx=2(2)(2.718283)+2
dy/dx=4xx 20.085+2dydx=4×20.085+2
dy/dx=82.34dydx=82.34

The slope of the tangent m_1=82.34m1=82.34

The slope of the normal is m_2=-1xx1/82.34=(-1)/82.34m2=−1×182.34=−182.34

The equation of the Normal is -

y=m_2x+cy=m2x+c
22.085=-1/82.34(2)+c22.085=−182.34(2)+c
22.085+1/82.34=c22.085+182.34=c
22.09=c22.09=c

y=-1/82.34x+22.09y=−182.34x+22.09

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