# Consider the set of parametric equations x=sin(t), y=sin(2t). What is the equation of the tangent line at the origin with positive slope?

Nov 11, 2017

$y = 2 x$

#### Explanation:

Slope of tangent at a point $\left({x}_{0} , f \left({x}_{0}\right)\right)$ on the curve $y = f \left(x\right)$ is given by the value of the derivative $\frac{\mathrm{df}}{\mathrm{dx}}$ at $\left({x}_{0} , f \left({x}_{0}\right)\right)$.

Here we are given a parametric equation of the type $\left(x \left(t\right) , y \left(t\right)\right)$. Slope of such parametric equation is given by $\frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}}$

As $x = \sin t$ and $y = \sin 2 t$, $\frac{\mathrm{dy}}{\mathrm{dt}} = 2 \cos 2 t$ and $\frac{\mathrm{dx}}{\mathrm{dt}} = \cos t$

Now at origin i.e. $\left(0 , 0\right)$, we have $t = 0$, as it makes both $x$ and $y$ equal to $0$, and hence slope of tangent is given by

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{\mathrm{dt}}}{\frac{\mathrm{dx}}{\mathrm{dt}}} = \frac{2 \cos 2 t}{\cos} t$

and hence value of slope at $t = 0$ is $2$ (observe that slope is positive) and equation of tangent is $y = 2 x$

graph{(y-2xsqrt(1-x^2))(y-2x)=0 [-2.5, 2.5, -1.25, 1.25]}