What are the critical points for the plane curve defined by the equations x(t)=4sin(2t),y(t)=-t, and 0

Question

What are the critical points for the plane curve defined by the equations x(t)=4sin(2t),y(t)=-t, and 0 <= t < 2pi ? 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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Find x'(t): To find the critical points of the plane curve, we need to determine where the derivatives of x(t) and y(t) are zero or undefined. Let's start by finding the derivative of x(t) with respect to t. The derivative of x(t) = 4sin(2t) with respect to t is x'(t) = 4 * cos(2t) * 2 = 8cos(2t).Find critical points for x(t): Now, we need to find the values of t where x'(t) = 8cos(2t) is equal to zero within the interval 0 <= t < 2pi. The equation 8cos(2t) = 0 simplifies to cos(2t) = 0. The cosine function is zero at π/2 and 3π/2 within one period of the cosine function. Since we have 2t inside the cosine function, we divide these values by 2 to find the corresponding t values. So, t = π/4 and t = 3π/4 are the points where x'(t) is zero.Find y'(t): Next, we find the derivative of y(t) with respect to t. The derivative of y(t) = -t with respect to t is y'(t) = -1. Since y'(t) is a constant and never equals zero, there are no critical points for y(t) based on its derivative.Combine critical points: Now, we combine the critical points found from the derivatives of x(t) and y(t). Since y'(t) does not contribute any critical points, we only consider the critical points from x(t). Therefore, the critical points for the plane curve are at t = π/4 and t = 3π/4.

What are the critical points for the plane curve defined by the equations $x(t)=4 \sin (2 t), y(t)=-t$ , and 0 \leq t<2 \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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