Question $\newline$ $24$ Calculate the derivative $\frac{d y}{d x}$ for the following parametrically defined plane curve and locate any critical points on its graph. $\newline$ $x(t)=2 \sin (2 t), \quad y(t)=2 \cos (t), \quad 0 \leq t \leq 2 \pi .$ $\newline$ Select the correct answer below: $\newline$ $\frac{d y}{d x}=-\frac{\sin (2 t)}{\cos (t)}$ , critical points at $(0, \pm 2),( \pm 2, \pm \sqrt{2})$ $\newline$ $\frac{d y}{d x}=-\frac{\sin (t)}{\cos (2 t)}$ , critical points at $(0, \pm \sqrt{2}),( \pm 2, \pm \sqrt{2})$ $\newline$ $\frac{d y}{d x}=-\frac{\sin (t)}{2 \cos (2 t)}$ , critical points at $(0, \pm 2),( \pm 2, \pm \sqrt{2})$ $\newline$ $\frac{d y}{d x}=-\frac{\sin (t)}{\cos (t)}$ , critical points at $(0, \pm 2),( \pm \sqrt{2}, \pm 2)$ $\newline$ FEEDBACK $\newline$ MORE INSTRUCTION

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Q. Question

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24

Calculate the derivative

\frac{d y}{d x}

for the following parametrically defined plane curve and locate any critical points on its graph.

\newline

x(t)=2 \sin (2 t), \quad y(t)=2 \cos (t), \quad 0 \leq t \leq 2 \pi .

\newline

Select the correct answer below:

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\frac{d y}{d x}=-\frac{\sin (2 t)}{\cos (t)}

, critical points at

(0, \pm 2),( \pm 2, \pm \sqrt{2})

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\frac{d y}{d x}=-\frac{\sin (t)}{\cos (2 t)}

, critical points at

(0, \pm \sqrt{2}),( \pm 2, \pm \sqrt{2})

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\frac{d y}{d x}=-\frac{\sin (t)}{2 \cos (2 t)}

, critical points at

(0, \pm 2),( \pm 2, \pm \sqrt{2})

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\frac{d y}{d x}=-\frac{\sin (t)}{\cos (t)}

, critical points at

(0, \pm 2),( \pm \sqrt{2}, \pm 2)

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FEEDBACK

\newline

MORE INSTRUCTION

Find Derivative: To find the derivative $(\frac{dy}{dx})$ for the parametrically defined plane curve, we need to find the derivatives $\frac{dx}{dt}$ and $\frac{dy}{dt}$ first and then divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ .
Find $\frac{dx}{dt}$ : Let's find $\frac{dx}{dt}$ . Given $x(t) = 2\sin(2t)$ , the derivative with respect to $t$ is $\frac{dx}{dt} = \frac{d}{dt} [2\sin(2t)]$ . Using the chain rule, we get $\frac{dx}{dt} = 2 \cdot \cos(2t) \cdot 2 = 4\cos(2t)$ .
Find $\frac{dy}{dt}$ : Now, let's find $\frac{dy}{dt}$ . Given $y(t) = 2\cos(t)$ , the derivative with respect to $t$ is $\frac{dy}{dt} = \frac{d}{dt} [2\cos(t)]$ . Using the derivative of cosine, we get $\frac{dy}{dt} = -2\sin(t)$ .
Calculate $(dy)/(dx)$ : To find $(dy)/(dx)$ , we divide $dy/dt$ by $dx/dt$ : $(dy)/(dx) = (dy/dt) / (dx/dt) = (-2\sin(t)) / (4\cos(2t))$ . Simplifying this, we get $(dy)/(dx) = -(\sin(t)) / (2\cos(2t))$ .
Identify Critical Points: Next, we need to find the critical points on the graph of the parametric curve. Critical points occur where the derivative $(\frac{dy}{dx})$ is undefined or zero.
Find t values: $(dy)/(dx)$ is undefined when the denominator is zero. So we need to find the values of $t$ where $2\cos(2t) = 0$ . This occurs when $\cos(2t) = 0$ , which happens at $t = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$ within the interval $[0, 2\pi]$ .
Locate Critical Points: $(\frac{dy}{dx})$ is zero when the numerator is zero. So we need to find the values of $t$ where $\sin(t) = 0$ . This occurs when $t = 0$ , $\pi$ , $2\pi$ within the interval $[0, 2\pi]$ .
Summary of Critical Points: Now we need to find the corresponding $x$ and $y$ values for these $t$ values to locate the critical points. $\newline$ For $t = 0, \pi, 2\pi$ , $x(t) = 2\sin(2t) = 0$ and $y(t) = 2\cos(t) = \pm2$ . $\newline$ For $t = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$ , $x(t) = 2\sin(2t) = \pm2$ and $y(t) = 2\cos(t) = \pm\sqrt{2}$ .
Summary of Critical Points: Now we need to find the corresponding $x$ and $y$ values for these $t$ values to locate the critical points. $\newline$ For $t = 0, \pi, 2\pi$ , $x(t) = 2\sin(2t) = 0$ and $y(t) = 2\cos(t) = \pm2$ . $\newline$ For $t = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$ , $x(t) = 2\sin(2t) = \pm2$ and $y(t) = 2\cos(t) = \pm\sqrt{2}$ .Therefore, the critical points on the graph are at $(0, \pm2)$ and $y$ $0$ .