Respuesta :
Answer:
Equations of tangent lines are
y= 2 x
y = 0
Step-by-step explanation:
x = sin t -- (1)
y = sin(t + sin(t)) -- (2)
Differentiating both equations w.r.to t to find slopes.
[tex]\frac{dx}{dt}=\frac{d(sin(t))}{dt}\\\\\frac{dx}{dt}=cos(t)--(3)[/tex]
[tex]\frac{dy}{dt}=\frac{d}{dt}(sin(t+sin(t))\\\\\frac{dy}{dt}=cos(t+sin(t))\frac{d}{dt}(t+sin(t))\\\\\frac{dy}{dt}=cos(t+sin(t)(1+cos(t))\\\\\frac{dy}{dt}=(1+cos(t))cos(t+sin(t))--(4)[/tex]
Dividing (2) by (1) to find slope
[tex]\frac{dy}{dx}=\frac{(1+cos(t))cos(t+sin(t))}{cos(t)}\\[/tex]
at tangent point x=y=0
From (1)
sin (t) = 0
⇒ t = 0, π
At t = 0
[tex]\frac{dy}{dx}\Big|_{t=0}=\frac{(1+cos(t))cos(t+sin(t))}{cos(t)}\\\\\\\frac{dy}{dx}\Big|_{t=0}=\frac{(1+cos(0))cos(0+sin(0))}{cos(0)}\\\\\\\frac{dy}{dx}\Big|_{t=0}=\frac{(1+1)cos(0+0)}{1}\\\\\\\frac{dy}{dx}\Big|_{t=0}=2\\[/tex]
At t= π
[tex]\frac{dy}{dx}\Big|_{t=\pi}=\frac{(1+cos(t))cos(t+sin(t))}{cos(t)}\\\\\\\frac{dy}{dx}\Big|_{t=\pi}=\frac{(1+cos(\pi))cos(\pi+sin(\pi))}{cos(\pi)}\\\\\\\frac{dy}{dx}\Big|_{t=\pi}=\frac{(1-1)cos(\pi+0)}{-1}\\\\\\\frac{dy}{dx}\Big|_{t=\pi}=0\\[/tex]
Equation of tangent
[tex](y-y_o)=m_t(x-x_o)\\[/tex]
[tex]Tangent\,\,point=(x_o,y_o)=(0,0)\\\\For\,\,t=0\\\\(y-0)=(2)(x-0)\\\\y=2x\\\\for\,\,t=\pi\\\\(y-0)=(0)(x-0)\\\\y=0[/tex]
