Equation of the normal to the curve

[val1]

Please can you check my solution for the following question?

Find the equation of the normal to the curve \(\displaystyle y = x^2 + \cos 2x\) at the point where \(\displaystyle x = \pi\) Leave \(\displaystyle \pi\) in your answer.

\(\displaystyle \frac{{dy}}{{dx}} = 2x - 2\sin 2x\)
when \(\displaystyle x = \pi {\rm , }y = \left( {x^2 + 1} \right){\rm and }\frac{{dy}}{{dx}} = 2\pi\)

So the gradient of the normal to the tangent =\(\displaystyle - \frac{1}{{2\pi }}\)

To find the equation of the normal

\(\displaystyle y - \left( {\pi ^2 + 1} \right) = - \frac{1}{{2\pi }}\left( {x - \pi } \right)\)

\(\displaystyle y = - \frac{1}{{2\pi }}x + \frac{1}{2} - \left( {\pi ^2 + 1} \right) = - \frac{1}{{2\pi }}x - \pi ^2 - \frac{1}{2} \\\)

\(\displaystyle 2y = - 2\pi ^2 - \pi x - 1 \\\)

Is this right please?

:?

 

[galactus]

Please can you check my solution for the following question?

Find the equation of the normal to the curve \(\displaystyle y = x^2 + \cos 2x\) at the point where \(\displaystyle x = \pi\) Leave \(\displaystyle \pi\) in your answer.

\(\displaystyle \frac{{dy}}{{dx}} = 2x - 2\sin 2x\)
when \(\displaystyle x = \pi {\rm , }y = \left( {x^2 + 1} \right){\rm and }\frac{{dy}}{{dx}} = 2\pi\)

So the gradient of the normal to the tangent =\(\displaystyle - \frac{1}{{2\pi }}\)

To find the equation of the normal

\(\displaystyle y - \left( {\pi ^2 + 1} \right) = - \frac{1}{{2\pi }}\left( {x - \pi } \right)\)

Check your algebra

\(\displaystyle \L\\y=\frac{-x}{2{\pi}}+{\pi}^{2}+\frac{3}{2}\)

\(\displaystyle y = - \frac{1}{{2\pi }}x + \frac{1}{2} - \left( {\pi ^2 + 1} \right) = - \frac{1}{{2\pi }}x - \pi ^2 - \frac{1}{2} \\\)

\(\displaystyle 2y = - 2\pi ^2 - \pi x - 1 \\\)

Is this right please?

:?