# A Curve (C) Has Parametric Equations x=3cost and y=cos2t and find the coordinates on the point on the curve where the gradient is 1?

Aug 3, 2016

$\left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}\frac{9}{4} \\ \frac{1}{8}\end{matrix}\right)$

#### Explanation:

by gradient i think you mean slope ie $\frac{\mathrm{dy}}{\mathrm{dx}}$. [there is a big difference though it comes later in the math journey, when gradient does not mean slope!!]

We can say, from the chain rule, that

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\mathrm{dy}}{\mathrm{dt}} \cdot \frac{\mathrm{dt}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}} = \frac{- 2 \sin 2 t}{- 3 \sin t}$

$= \frac{4 \sin t \cos t}{3 \sin t}$ ....we just used the ID, $\sin 2 A = 2 \sin A \cos A$

$= \frac{4}{3} \cos t$

Now if

${\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]}_{t = {t}_{o}} = 1$

then
$\frac{4}{3} \cos {t}_{o} = 1 \implies \cos {t}_{o} = \frac{3}{4}$

from the Pythagorean ID: ${\sin}^{2} + {\cos}^{2} = 1$

we know that $\sin {t}_{o} = \frac{\sqrt{7}}{4}$

the point on the curve is $\left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}3 \cos t \\ \cos 2 t\end{matrix}\right)$

and from an additional trig ID: $\cos 2 A = 2 {\cos}^{2} A - 1$

we can say that $\cos 2 {t}_{o} = 2 \cdot \frac{9}{16} - 1 = \frac{2}{16}$

so

$\left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}3 \cdot \frac{3}{4} \\ \frac{1}{8}\end{matrix}\right) = \left(\begin{matrix}\frac{9}{4} \\ \frac{1}{8}\end{matrix}\right)$

this can be also done by de-parameterising the equation

from $x = 3 \cos t$, we say that $\cos t = \frac{x}{3}$

we then have $y = \cos 2 t$ and from that double angle identity we say that

$y = 2 {\cos}^{2} t - 1$

$= 2 {\left(\frac{x}{3}\right)}^{2} - 1$

so $y ' = \frac{4 x}{3} \cdot \frac{1}{3} = \frac{4 x}{9}$

$y ' = 1 \implies x = \frac{9}{4}$

and so

$y = 2 {\left(\frac{\frac{9}{4}}{3}\right)}^{2} - 1 = \frac{18}{16} - 1 = \frac{1}{8}$