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

Question

Quizplus · Accepted Answer

The slope of the tangent line is found by differentiating $y$ with respect to $x$, which gives $\frac{dy}{dx} = 6$. At $t = \frac{\pi}{3}$, $x = \frac{\sqrt{3}}{2}$ and $y = 3\sqrt{3}$. Using the point-slope form, the equation is $y - 3\sqrt{3} = 6(x - \frac{\sqrt{3}}{2})$, simplifying to $y = 6x$.

Find an Equation for the Line Tangent to the Curve x=sin⁡t,y=6sin⁡t,t=π3x = \sin t , y = 6 \sin t , t = \frac { \pi } { 3 }x=sint,y=6sint,t=3π​

Find an Equation for the Line Tangent to the Curve $x = \sin t , y = 6 \sin t , t = \frac { \pi } { 3 }$