[Math] Calculate equation of normal line on any given point

calculus

I know there are alot of topics about this, but almost all of them cover the case when you are given an explicit point like (3,6), but I need the general formula to solve for any given X and I'm a bit lost at the process of calculating it.

Here is what I did so far:

given f(x) = sin( x)

I would like to find the tangent ( and later the normal) for the point P=(x,sin(x))

I know that the normal line is perpendicular to the tangent line of that function, so first I calculate the slope of the tangent line of f(x), like so:

f'(x) = cos(x)

now that I have the slope of the tangent,I could try to find the equation for the tangent line using this:

y= m*x + b , and replacing…

y = cos(x) * x + b

and to get b :

sin(x) = cos(x) * x + b

then:

sin(x) – cos(x) * x = b

So the equation for the tangent line would be this:

y = cos(x) * x + sin(x) – cos(x) * x

As I said earlier ( according to my understanding) the normal line is perpendicular to the slope of the tangent line, so given two slopes:

m1= slope of tangent

m2= slope of normal

m1*m2 = -1 // To be perpendicular

finding m2:

m2 = -1/ cos(x)

SO, if all of the above is correct, the equation for the normal line would be:

y = (-1/cos(x)) * x + sin(x) – cos(x) * x

Did I do it right? And if so, how do I continue?

Thanks

PS: English is not my native language so my apologies if something isn't clear, just say it in the comments and I will try to explain it better

Best Answer

You're very close. You're using the symbol $x$ both for the first coordinate of the point at which you want to find the normal line andas the variable in the equation of the normal line. Let me rewrite:

Given $f(x) = \sin(x)$, compute the normal line at $(a, \sin(a))$. The slope of the tangent is $\cos(a)$, so the slope of the normal is $\frac{-1}{\cos a}, so the equation of the normal is

$$ y = \frac{-1}{\cos a} (x - a) + Q $$ where $Q$ is as yet unknown. But the point $x = a, y = \sin a$ must lie on this line, so we must have $$ \sin a = \frac{-1}{\cos a} (a - a) + Q $$ so $Q = \sin a$. Hence your equation is $$ y = \frac{-1}{\cos (a)} (x - a) + \sin(a). $$

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