Question

Show that the curve with parametric equations

x = sin t and y = sin(t + sin t) for 0 ≤ t ≤ 2π

Answer 1

$$x_t'=\cos t, y'_t=\cos(t+\sin t)(1+\cos t)$$ $$y_x'=\dfrac{y_t'}{x_t'}=\dfrac{\cos(t+\sin t)(1+\cos t)}{\cos t}$$

At the origin we have: $x=0$ and $y=0.$

$$x=0=>\sin t=0=>t=\pi n, n\in \Z$$

For $0\leq t \leq 2\pi$ we have $t_1=0, t_2=\pi , t_3=2\pi.$

Check for $y$

$$t_1=0, y(0)=\sin (0+\sin(0)) =0, True$$ $$t_2=\pi, y(\pi)=\sin (\pi+\sin(\pi)) =0, True$$ $$t_3=2\pi, y(2\pi)=\sin (2\pi+\sin(2\pi)) =0, True$$

So $y$ is also satisfied. Put these values of $t$ into $y_x'$ to find the gradient of the tangent line:

$$t_1=0,$$ $$y_x'|_{t=0}=\dfrac{\cos(0+\sin (0))(1+\cos (0))}{\cos (0)}=2$$ $$slope_1=m_1=2$$

The equation of the tangent line is

$$y-0=2(x-0)$$ $$y=2x$$ $$t_2=\pi,$$ $$y_x'|_{t=\pi}=\dfrac{\cos(\pi+\sin (\pi))(1+\cos (\pi))}{\cos (\pi)}=0$$ $$slope_2=m_2=0$$

The equation of the tangent line is

$$y-0=0(x-0)$$ $$y=0$$ $$t_3=2\pi,$$ $$y_x'|_{t=2\pi}=\dfrac{\cos(2\pi+\sin (2\pi))(1+\cos (2\pi))}{\cos (2\pi)}=0$$ $$slope_3=m_3=2=m_1$$

The equation of the tangent line is

$$y=2x$$

Therefore our two tangents at the origin will be:

$$y_1=2x, y_2=0$$