One of the practice problems in my Calculus book states Find the equation of the normal to the curve $y=1+cosx$ at the point when $x= \pi/2$.

Can anyone point me in the right direction on this? This is what I got so far (though I don't think I'm supposed to do direct decimal values):




  • $\begingroup$ Your $y'$ is wrong. $\endgroup$ – marty cohen Oct 15 '15 at 21:02

First notice that you have some mistake in your calculations. I suppose that you know that $\sin (\pi/2) =1$ and $\cos (\pi/2)=0$ so you have: $$ y(\pi/2)=1+\cos(\pi/2)=1 $$

the derivative of your function is $y'=-\sin x$ (the derivative of the constant $1$ is $0$), and you have: $$ y'(\pi/2)=-1 $$

so the tangent line at the point $x=\pi/2$ has slope $m=y'(\pi/2)=-1$.

The normal (to the tangent) has slope $m'=-\dfrac{1}{m}=1$ and it is the straight line with this slope passing through the point $(\pi/2,1)$ so its equation is: $$ y-1=1(x-\pi/2) $$


If $y = 1+\cos(x)$, then $y' =-\sin(x) $ (you made a mistake here).

At $x = \pi/2$, $y' =-\sin(\pi/2) =-1 $.

The normal at this point has slope $1$ (since $-\frac1{-1} = 1 $).

So, this gives you the slope of the normal, and you know the point that it passes through. From this, you should be able to get the equation of the line.

  • $\begingroup$ Thanks Marty, but I'm not following? $\endgroup$ – Analytic Lunatic Oct 15 '15 at 21:08
  • $\begingroup$ Where do you stop? $\endgroup$ – marty cohen Oct 15 '15 at 21:09
  • $\begingroup$ I get that derivative of $cos(x)$ is $-sin(x)$, but shouldn't the derivative of $y=1+cos(x)$ then be $y'=1-sin(x)$ -- did you leave out the 1? $\endgroup$ – Analytic Lunatic Oct 15 '15 at 21:10
  • 1
    $\begingroup$ No. Since "1" is a constant, its derivative is zero. $\endgroup$ – marty cohen Oct 15 '15 at 21:12

You are almost done. Use the fact that the slope of the normal line at that point is negative reciprocal of the tangent line at that point. Then use the slope of the normal line and coordinates of the point to derive the equation of the normal line. That's all.

  • $\begingroup$ Can you elaborate H.R.? I'm still not following. $\endgroup$ – Analytic Lunatic Oct 15 '15 at 21:08

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