What are the critical points for the plane curve defined by the equations $x(t)=-\sin (3 t), y(t)=5 t$ , and $0 \leq t<\pi$ ? Write your answer as a list of values of $t$ , separated by commas. For example, if you found $t=1$ or $t=2$ , you would enter $1$ , $2$ .

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Solve trigonometric equations II

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Q. What are the critical points for the plane curve defined by the equations

x(t)=-\sin (3 t), y(t)=5 t

, and

0 \leq t<\pi

? Write your answer as a list of values of

t

, separated by commas. For example, if you found

t=1

t=2

, you would enter

1

2

Find $x'(t)$ : To find the critical points of the plane curve, we need to find the values of $t$ where the derivatives of $x(t)$ and $y(t)$ are both zero or do not exist. Let's start by finding the derivative of $x(t)$ with respect to $t$ .
$x(t) = -\sin(3t)$
$\frac{dx}{dt} = -\cos(3t) \cdot 3$
Find $y'(t)$ : Now, let's find the derivative of $y(t)$ with respect to $t$ . $y(t) = 5t$ $\frac{dy}{dt} = 5$
No critical points for $y(t)$ : Since $\frac{dy}{dt} = 5$ is never zero and does not depend on $t$ , there are no critical points for $y(t)$ based on its derivative. Therefore, we only need to consider the critical points for $x(t)$ based on $\frac{dx}{dt}$ .
Set $x'(t)$ to zero: We set the derivative of $x(t)$ equal to $0$ to find the critical points for $x(t)$ . $0 = -\cos(3t) \times 3$ This implies that $\cos(3t)$ must be equal to $0$ .
Solve for $\cos(3t)=0$ : The cosine function is zero at odd multiples of $\frac{\pi}{2}$ . Therefore, we need to find the values of $t$ such that $3t$ is an odd multiple of $\frac{\pi}{2}$ within the given interval $0 \leq t < \pi$ .
Find values of t: Let's solve for t:
$3t = \frac{(2n+1)\pi}{2}$ , where n is an integer.
$t = \frac{(2n+1)\pi}{6}$
Solve for $t$ : We need to find the values of $n$ such that $t$ is in the interval $0 \leq t < \pi$ . Let's start with $n=0$ and increase $n$ to find all possible values of $t$ . $\newline$ For $n=0$ : $t = \frac{\pi}{6}$ $\newline$ For $n=1$ : $n$ $0$ $\newline$ For $n$ $1$ : $n$ $2$
Find possible values of $\newline$ $t$ : If we increase $\newline$ $n$ to $\newline$ $3$ , we get $\newline$ $t = \frac{7\pi}{6}$ , which is outside the interval $\newline$ $0 \leq t < \pi$ . Therefore, we only consider $\newline$ $n=0$ , $\newline$ $n=1$ , and $\newline$ $n=2$ .
Consider $n=0,1,2$ : The critical points for the plane curve defined by the equations $x(t)=-\sin(3t)$ and $y(t)=5t$ are at $t=\frac{\pi}{6}$ , $t=\frac{\pi}{2}$ , and $t=\frac{5\pi}{6}$ .