Question

Find the parametric equations for the tangent line to the curve with the given parametric equations at the specified point. The parametric equations are x = t, y = e⁽⁻²ᵗ⁾, z = 3t - t³. Find the tangent line at the point (0, 1, 0).

Answer 1

The tangent line to the curve with the given parametric equations at the specified point (0, 1, 0) is x = t, y = 1 - 2t, z = 3t.

Step-by-step explanation:

To find the parametric equations for the tangent line to the curve, we need to find the derivative of each component of the parametric equations. Differentiating x with respect to t gives dx/dt = 1, differentiating y with respect to t gives dy/dt = -2e^(-2t), and differentiating z with respect to t gives dz/dt = 3 - 3t^2.

At t = 0, the point on the curve is (0, 1, 0). Evaluating each component of the derivative at t = 0, we have dx/dt = 1, dy/dt = -2, and dz/dt = 3.

Therefore, the tangent line at the point (0, 1, 0) is given by the parametric equations:

x = 0 + t(1) = t

y = 1 + t(-2) = 1 - 2t

z = 0 + t(3) = 3t