Parametric Equations and Curves — Question 8

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Question 8

Problem

A moving point satisfies x=2+3cos⁡tx=2+3\cos t, y=−1+3sin⁡ty=-1+3\sin t. Without eliminating tt, determine its center, radius, starting point, and initial direction.

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Original worksheet page 1: question and worked solution for 3-1-008
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Question 8 – Solution

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Solution

  1. Compare the equations x=2+3cos⁡t,y=−1+3sin⁡tx=2+3\cos t, \qquad y=-1+3\sin t with the standard circle parametrization x=h+Rcos⁡t,y=k+Rsin⁡t.x=h+R\cos t, \qquad y=k+R\sin t.

  2. It follows immediately that (h,k)=(2,−1)andR=3.(h,k)=(2,-1) \qquad\text{and}\qquad R=3. Thus the curve is a circle centered at (2,−1)\boxed{(2,-1)} with radius 3\boxed{3}.

  3. Evaluate the coordinates at t=0t=0: x(0)=2+3cos⁡0=5,y(0)=−1+3sin⁡0=−1.x(0)=2+3\cos0=5, \qquad y(0)=-1+3\sin0=-1. Therefore, the starting point is (5,−1)\boxed{(5,-1)}.

  4. Differentiate to find the velocity: 𝒓′(t)=(−3sin⁡t,3cos⁡t).\mathbf r'(t)=(-3\sin t,3\cos t). At t=0t=0, 𝒓′(0)=(0,3),\mathbf r'(0)=(0,3), which points straight upward from the rightmost point of the circle.

  5. Upward motion from the rightmost point is counterclockwise, so the initial direction is .

Original worksheet page 2: question and worked solution for 3-1-008

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