Find an equation for the line tangent to the curve at the point defined by the given value of t.
-$$x = t + \cos t , y = 2 - \sin t , t = \frac { \pi } { 6 }$$
A) $y = - \sqrt { 3 } x + \frac { 3 } { 2 }$
B) $y = - \sqrt { 2 } x - \frac { \sqrt { 2 } } { 4 } \pi + 2$
C) $y = - \sqrt { 3 } x + \frac { \sqrt { 3 } } { 6 } \pi + 3$
D) $y = \sqrt { 3 } x - \frac { \sqrt { 3 } } { 6 } \pi$

Question

Quizplus · Accepted Answer

First, we need to find the derivative of the curve at the given point:
$$\frac{dx}{dt} = 1 - \sin t$$
$$\frac{dy}{dt} = -\cos t$$
At $t=\frac{\pi}{6}$, we have:
$$\frac{dx}{dt} = \frac{\sqrt{3}}{2}$$
$$\frac{dy}{dt} = -\frac{\sqrt{3}}{2}$$
So the slope of the tangent line is:
$$m = \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = -1$$
Now we need to find the $y$-intercept of the tangent line. We know it passes through the point $(t+\cos t, 2-\sin t)$, so:
$$y = mx + b$$
$$2 - \sin t = -1(t+\cos t) + b$$
$$b = 3 + \sin t - t - \cos t = 3 + \frac{1}{2} - \frac{\sqrt{3}}{2} - \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{6}\pi + 3$$
Therefore, the equation of the tangent line is:
$$y = -x + \frac{\sqrt{3}}{6}\pi + 3$$
which matches choice C.

Find an Equation for the Line Tangent to the Curve x=t+cos⁡t,y=2−sin⁡t,t=π6x = t + \cos t , y = 2 - \sin t , t = \frac { \pi } { 6 }x=t+cost,y=2−sint,t=6π​

Find an Equation for the Line Tangent to the Curve $x = t + \cos t , y = 2 - \sin t , t = \frac { \pi } { 6 }$