What is the normal of y=(x-2)^{2/3},(3,1) ?

The normal of y=(x-2)^{2/3},(3,1) is y=-3/2 x+11/2

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normal of y=(x-2)^{2/3},(3,1)

normal of

y = (x - 2)^{\frac{2}{3}}, (3, 1)

y = - \frac{3}{2} x + \frac{11}{2}

Solution steps

Find the slope of

y = (x - 2)^{\frac{2}{3}} : \frac{d y}{d x} = \frac{2}{3 ( x - 2 ) ^{\frac{1}{3}}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = \frac{2}{3}

Compute the slope of the perpendicular line:

m_{p} = - \frac{3}{2}

Find the line with slope m=

- \frac{3}{2}

and passing through

(3, 1) : y = - \frac{3}{2} x + \frac{11}{2}

y = - \frac{3}{2} x + \frac{11}{2}

What is the normal of y=(x-2)^{2/3},(3,1) ?
The normal of y=(x-2)^{2/3},(3,1) is y=-3/2 x+11/2