Solve the problem.
-Find all points on the curve $y = \sin x , 0 \leq x \leq 2 \pi$, where the tangent line is parallel to the line $y = \frac { 1 } { 2 } x$.
A) $\left( \frac { \pi } { 3 } , \frac { \sqrt { 3 } } { 2 } \right) , \left( \frac { 5 \pi } { 3 } , - \frac { \sqrt { 3 } } { 2 } \right)$
B) $\left( \frac { \pi } { 6 } , \frac { 1 } { 2 } \right) , \left( \frac { 11 \pi } { 6 } , - \frac { 1 } { 2 } \right)$
C) $\left( \frac { \pi } { 3 } , \frac { \sqrt { 3 } } { 2 } \right) , \left( \frac { 2 \pi } { 3 } , \frac { \sqrt { 3 } } { 2 } \right)$
D) $\left( \frac { \pi } { 3 } , \frac { 1 } { 2 } \right) , \left( \frac { 2 \pi } { 3 } , \frac { 1 } { 2 } \right)$

Question

Quizplus · Accepted Answer

The slope of the tangent line to the curve $y = \sin x$ is given by the derivative $y' = \cos x$. For the tangent to be parallel to $y = \frac{1}{2}x$, we need $y' = \cos x = \frac{1}{2}$. Solving $\cos x = \frac{1}{2}$ within $0 \leq x \leq 2\pi$ gives $x = \frac{\pi}{3}$ and $x = \frac{5\pi}{3}$. Substituting these $x$ values into $y = \sin x$ gives the points $\left( \frac{\pi}{3}, \frac{\sqrt{3}}{2} \right)$ and $\left( \frac{5\pi}{3}, -\frac{\sqrt{3}}{2} \right)$.

Solve the Problem y=sin⁡x,0≤x≤2πy = \sin x , 0 \leq x \leq 2 \piy=sinx,0≤x≤2π

Solve the Problem $y = \sin x , 0 \leq x \leq 2 \pi$