A curve is defined by parametric equations x = t^2 + 1, y = t^3 - 3t. At what values of parameter t is the tangent line horizontal?

Answer and explanation

Correct answer: A. t = 1 and t = -1

  1. Step 1: Horizontal tangents occur when dy/dt = 0 while dx/dt != 0.
  2. Step 2: Compute dy/dt: d/dt(t^3 - 3t) = 3t^2 - 3.
  3. Step 3: Set dy/dt = 0 => 3(t^2 - 1) = 0 => t = 1 or t = -1.
  4. Step 4: Compute dx/dt: d/dt(t^2 + 1) = 2t. At t = +-1, dx/dt = +-2 != 0.
  5. Step 5: Thus horizontal tangents occur at t = 1 and t = -1.
Answer
t = 1 and t = -1.
Why not the others: 't = 0 only' makes dx/dt = 0 (vertical tangent); 't = sqrt(3) and t = -sqrt(3)' are the roots of y(t) = 0; 't = 3 and t = -3' misfactors 3t^2 - 3.

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