# What are the equations of the tangent and the normal to the curve y=cos(2x)e^-x at the point where x=0?

## Solutions: ${y}_{T} = - x + 1$ ${y}_{N} = x + 1$ PLEASE EXPLAIN STEP BY STEP. Thanks.

May 30, 2018

In order to find the tangent, we need to differentiate $y$ to find $\frac{\mathrm{dy}}{\mathrm{dx}}$. In this case we have a product of functions and so we can consider there to be two different functions, $\cos \left(2 x\right)$ and ${e}^{-} x$, and they each take their turns being differentiated using the product rule.

$f \left(x\right) = \cos \left(2 x\right)$
$f ' \left(x\right) = - 2 \sin \left(2 x\right)$
and
$g \left(x\right) = {e}^{-} x$
$g ' \left(x\right) = - {e}^{-} x$

so $\frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right) g \left(x\right) + f \left(x\right) g ' \left(x\right)$

and substituting in what we've just found...

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 {e}^{- x} \sin \left(2 x\right) - {e}^{- x} \cos \left(2 x\right)$

where $x = 0$, $y = \cos \left(2 \cdot 0\right) \cdot {e}^{0}$ so $y = 1$
and $\frac{\mathrm{dy}}{\mathrm{dx}} = - 2 \cdot {e}^{0} \cdot \sin \left(0\right) - {e}^{0} \cdot \cos \left(2 \cdot 0\right)$ so $\frac{\mathrm{dy}}{\mathrm{dx}} = - 1$.

Since the derivative of the function gives the gradient at $x$, we know that $m = - 1$ where $x = 0$.

Having found the value of $y$ at $x = 0$, we can simply plug these numbers into the general equation for a line, $y - b = m \left(x - a\right)$ for a set of points $\left(a , b\right)$ and a gradient $m$.

So with our points being $\left(0 , 1\right)$, we can jump straight in and say $y - 1 = - 1 \left(x - 0\right)$ and simplifying that gives $y = - x + 1$.

Now that we have the equation of the tangent line, we can go ahead and say that the normal will be at a right angle to this tangent. Since the product of the gradients of two tangents (when you multiply the values of the tangents of each line together) will always be equal to $- 1$ for lines which meet at right angles, we know that since the gradient of our tangent is $- 1$, then the gradient of the normal will be the number which, when multiplied by $- 1$ equals $- 1$, so it's obviously just $1$. We could make an equation with the gradient of the normal being $x$, and the gradient of the tangent being $- 1$,

so $- 1 \cdot x = - 1$

therefore $x = 1$ and the gradient of the normal is $- 1$

Now that we know the gradient of the normal, we can say that the equation of the tangent line must be
$y = x + 1$.

Notice that we still add the $+ 1$ so that the intercept is the same; keep in mind we only want to flip the gradient so that it meets at a right angle, and not move it up or down.

Hope this helps :-)