I am trying to find the equation of the line normal to the curve. the equation is y=2sin1/2x were x=(3pie)/2.

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mlehuzzah | Student, Graduate | (Level 1) Associate Educator

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I'm guessing you mean `y=2 "sin" ((1/2)x)` and not `y=2 "sin" ((1)/(2x))`

First, at `x=(3 pi)/2`, `y= 2 "sin" ((3 pi)/(4)) = 2* (sqrt(2))/(2)=sqrt(2)`

So we want the line to go through the point `((3 pi)/(2), sqrt(2)`

Next, we want to know the slope of the normal line. The normal line is perpendicular to the tangent line. So we want to find the slope of the tangent line, and then take the negative reciprocal. The slope of the tangent line is just the derivative, with `(3 pi)/(2)` plugged in for x. Thus:

`y=2 "sin" ((1/2)x)`

`y'= 2 "cos" ((1/2)x) * (1/2) = "cos" ((1/2)x) `

Plug in `x=(3 pi)/(2)` and we have:

`y'((3 pi)/(2))="cos" ((3 pi)/(4)) = -(sqrt(2))/(2)`

The negative reciprocal of `-(sqrt(2))/(2)` is `(2)/(sqrt(2))` which is just `sqrt(2)`

Thus we want a line that goes through the point `((3 pi)/(2), sqrt(2)` and has slope `sqrt(2)`

In point-slope form, the line is:

`y - sqrt(2) = sqrt(2) (x - (3 pi)/(2))`



 The black is the original graph; the blue is the tangent line; the red is the normal line

(they don't look quite perpendicular because the scale on the x and the y is not the same; that is, 1 unit in the x direction is larger than 1 unit in the y direction)