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What are the critical points for the plane curve defined by the equations x(t)=cot t,y(t)=sin t, and 0 < 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 .

What are the critical points for the plane curve defined by the equations $x(t)=\cot t, y(t)=\sin t$ , and \( 0

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Write a linear function: word problems

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

x(t)=\cot t, y(t)=\sin t

, and

0<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

.

Question Prompt: Question prompt: What are the critical points for the plane curve defined by the equations $x(t)=\cot t$ , $y(t)=\sin t$ , for 0 < t < \pi?
Derivatives of $x(t)$ and $y(t)$ : Determine the derivatives of $x(t)$ and $y(t)$ with respect to $t$ . The critical points occur where the derivative of $x(t)$ or $y(t)$ is undefined or $0$ .
Derivative of $x(t)$ : Find the derivative of $x(t) = \cot t$ . The derivative of $\cot t$ is $-\csc^2 t$ . $\frac{dx}{dt} = -\csc^2 t$
Derivative of $y(t)$ : Find the derivative of $y(t) = \sin t$ . The derivative of $\sin t$ is $\cos t$ . $\frac{dy}{dt} = \cos t$
Identify Undefined or Zero Values: Identify the values of $t$ where $\frac{dx}{dt}$ and $\frac{dy}{dt}$ are undefined or zero. $\newline$ For $\frac{dx}{dt} = -\csc^2 t$ to be undefined, $\sin t$ must be zero since $\csc t = \frac{1}{\sin t}$ . $\newline$ For $\frac{dy}{dt} = \cos t$ to be zero, $\cos t$ must be zero.
Solve for $\sin t = 0$ : Solve for $t$ where $\sin t = 0$ within the interval $(0, \pi)$ . $\newline$ $\sin t = 0$ at $t = 0$ and $t = \pi$ , but these are not within the open interval $(0, \pi)$ , so they are not considered.
Solve for $\cos t = 0$ : Solve for $t$ where $\cos t = 0$ within the interval $(0, \pi)$ . $\newline$ $\cos t = 0$ at $t = \frac{\pi}{2}$ .
Conclude Critical Points: Conclude the critical points. $\newline$ The only critical point within the interval $(0, \pi)$ is at $t = \frac{\pi}{2}$ .

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