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 = t² + 1, y = 6t, z = e⁽ᵗ² ⁻ ᵗ⁾, and the specified point is (2, 6, 1).

Answer 1

To find the tangent line's parametric equations at the specified point (2, 6, 1), determine the corresponding t value, differentiate the x, y, z functions to find the tangent line's direction ratios, and compose the parametric equations using these derivatives as the direction of the line.

Step-by-step explanation:

To find the parametric equations for the tangent line to the given curve at the specified point (2, 6, 1), we first need to determine the value of t that corresponds to this point. Since x = t² + 1, we set x to 2 and solve for t. We find t = 1 (since t must be positive, as y = 6t gives us y = 6). Next, we find the derivatives of x, y, and z with respect to t to compute the slope components of the tangent vector: dx/dt = 2t, dy/dt = 6, and dz/dt is found by differentiating e^(t² - t). These derivatives at t = 1 are the direction ratios of the tangent line.

Finally, the parametric equations of the tangent line are:

• x = (2² + 1) + 2(1)(s - 1)
• y = 6(1) + 6(s - 1)
• z = e^((1)² - (1)) + e^0(s - 1)

where s is the parameter along the tangent line.