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?

:?