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

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Velocity as a rate of change

Full solution

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

\newline

x(t)=\cot t

,

y(t)=\sin t

, and

\newline

0 < t < \pi

? Write your answer as a list of values of

\newline

t

, separated by commas. For example, if you found

\newline

t=1

or

\newline

t=2

, you would enter

1,2

.

\newline

Provide your answer below:

Find Derivative of $x(t)$ : To find the critical points of the plane curve defined by the parametric equations $x(t) = \cot(t)$ and $y(t) = \sin(t)$ , we need to find the values of $t$ where the derivatives $\frac{dx}{dt}$ and $\frac{dy}{dt}$ are both zero or undefined, since these correspond to horizontal and vertical tangents respectively.
Find Derivative of $y(t)$ : First, let's find the derivative of $x(t)$ with respect to $t$ , which is $\frac{dx}{dt}$ . The derivative of $\cot(t)$ is $-\csc^2(t)$ .
Identify Critical Points: Next, we find the derivative of $y(t)$ with respect to $t$ , which is $\frac{dy}{dt}$ . The derivative of $\sin(t)$ is $\cos(t)$ .
Check for Undefined Values: Now, we look for the values of $t$ where $\frac{dx}{dt}$ and $\frac{dy}{dt}$ are zero or undefined. The derivative $\frac{dx}{dt} = -\csc^2(t)$ is undefined when $\sin(t) = 0$ , which occurs at $t = 0$ and $t = \pi$ within the given interval $(0, \pi)$ . However, $t = 0$ and $t = \pi$ are not included in the interval $(0, \pi)$ , so we do not consider these points.
Check for Zero Values: The derivative $\frac{dy}{dt} = \cos(t)$ is zero when $t = \frac{\pi}{2}$ , since $\cos(\frac{\pi}{2}) = 0$ . This is within the interval $(0, \pi)$ .
Final Critical Point: Therefore, the only critical point for the plane curve on the interval $(0, \pi)$ is at $t = \frac{\pi}{2}$ .

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