Question

a. Show that the curve with parametric equations

x=sint and y=sin(t+sint) for 0≤t≤2π

has two tangent lines at the origin and their equations. Illustrate by graphing the curve and its tangents.

b. Find the slope of the line tangent to the parametric curve

x=2cost and y=2-cos²t for 0≤t≤π

at points (-1, -1). Show the graph the parametric equations and the tangent line.

Answer 1

a.

$$\begin{matrix} x=0 \\ y=0 \end{matrix}, \ 0\leq t\leq \pi$$

$$\begin{matrix} \sin t=0 \\ \sin(t+\sin t)=0 \end{matrix}, \ 0\leq t\leq \pi$$

$$Origin(0, 0): t=0\ or\ t=\pi$$

$$slope=\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{\cos(t+\sin t)\cdot(1+\cos t)}{\cos t}$$

$t_1=0:$

$$slope_1=\dfrac{\cos((0)+\sin (0))\cdot(1+\cos (0))}{\cos (0)}=2$$

The tangent line in point-slope form

$$y-0=2(x-0)$$

The tangent line in slope-intercept form

$$y=2x$$

$t_2=\pi:$

$$slope_2=\dfrac{\cos((\pi)+\sin (0))\cdot(1+\cos (\pi))}{\cos (\pi)}=0$$

The tangent line in point-slope form

$$y-0=0(x-0)$$

The tangent line in slope-intercept form

$$y=0$$

2.

$$\begin{matrix} x=-1 \\ y=-1 \end{matrix}, \ 0\leq t\leq \pi$$

$$\begin{matrix} 2 \cos t=-1 \\ 2-\cos^2t=-1 \end{matrix}, \ \ 0\leq t\leq \pi$$

No solution. Point $(-1, -1)$ does not lie on the given curve.

$$slope=\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}=\dfrac{2\cos t\sin t}{-2\sin t}=-\cos t,$$

$0<t<\pi$

Point $(1, 1.75)$

$$\begin{matrix} x=1 \\ y=1.75 \end{matrix}, \ 0\leq t\leq \pi$$

$$\begin{matrix} 2\cos t=1 \\ 2-\cos^2 t=1.75 \end{matrix}, \ 0\leq t\leq \pi$$

$$t=\pi/3$$ $$slope=\cos(\pi/3)=-0.5$$

The tangent line in point-slope form

$$y-1.75=-0.5(x-1)$$

The tangent line in slope-intercept form

$$y=-0.5x+2.25$$