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What are the critical points for the plane curve defined by the equations x(t)=-sin(3t),y(t)=5t, 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)=sin(3t),y(t)=5t x(t)=-\sin (3 t), y(t)=5 t , and 0 \leq t<\pi ? Write your answer as a list of values of t t , separated by commas. For example, if you found t=1 t=1 or t=2 t=2 , you would enter 11,22 .

Full solution

Q. What are the critical points for the plane curve defined by the equations x(t)=sin(3t),y(t)=5t x(t)=-\sin (3 t), y(t)=5 t , and 0t<π 0 \leq t<\pi ? Write your answer as a list of values of t t , separated by commas. For example, if you found t=1 t=1 or t=2 t=2 , you would enter 11,22 .
  1. Find x(t)x'(t): To find the critical points of the plane curve, we need to find the values of tt where the derivatives of x(t)x(t) and y(t)y(t) are both zero or do not exist. Let's start by finding the derivative of x(t)x(t) with respect to tt.
    x(t)=sin(3t)x(t) = -\sin(3t)
    dxdt=cos(3t)3\frac{dx}{dt} = -\cos(3t) \cdot 3
  2. Find y(t)y'(t): Now, let's find the derivative of y(t)y(t) with respect to tt.y(t)=5ty(t) = 5tdydt=5\frac{dy}{dt} = 5
  3. No critical points for y(t)y(t): Since dydt=5\frac{dy}{dt} = 5 is never zero and does not depend on tt, there are no critical points for y(t)y(t) based on its derivative. Therefore, we only need to consider the critical points for x(t)x(t) based on dxdt\frac{dx}{dt}.
  4. Set x(t)x'(t) to zero: We set the derivative of x(t)x(t) equal to 00 to find the critical points for x(t)x(t).0=cos(3t)×30 = -\cos(3t) \times 3This implies that cos(3t)\cos(3t) must be equal to 00.
  5. Solve for cos(3t)=0\cos(3t)=0: The cosine function is zero at odd multiples of π2\frac{\pi}{2}. Therefore, we need to find the values of tt such that 3t3t is an odd multiple of π2\frac{\pi}{2} within the given interval 0 \leq t < \pi.
  6. Find values of t: Let's solve for t:
    3t=(2n+1)π23t = \frac{(2n+1)\pi}{2}, where n is an integer.
    t=(2n+1)π6t = \frac{(2n+1)\pi}{6}
  7. Solve for tt: We need to find the values of nn such that tt is in the interval 0 \leq t < \pi. Let's start with n=0n=0 and increase nn to find all possible values of tt.\newlineFor n=0n=0: t=π6t = \frac{\pi}{6}\newlineFor n=1n=1: nn00\newlineFor nn11: nn22
  8. Find possible values of \newlinett: If we increase \newlinenn to \newline33, we get \newlinet=7π6t = \frac{7\pi}{6}, which is outside the interval \newline0 \leq t < \pi. Therefore, we only consider \newlinen=0n=0, \newlinen=1n=1, and \newlinen=2n=2.
  9. Consider n=0,1,2n=0,1,2: The critical points for the plane curve defined by the equations x(t)=sin(3t)x(t)=-\sin(3t) and y(t)=5ty(t)=5t are at t=π6t=\frac{\pi}{6}, t=π2t=\frac{\pi}{2}, and t=5π6t=\frac{5\pi}{6}.

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