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 = 1/12t, y = t³ - t, z = t³t. The specified point is (13, 0, 2).

Answer 1

To find the parametric equations for the tangent line to the curve, find the first derivative of each parametric equation, then use the point-slope form of a line to find the equation of the tangent line at the specified point.

Step-by-step explanation:

To find the parametric equations for the tangent line to the curve at the specified point, we need to find the first derivative of each parametric equation. Let's start with x = \frac{1}{12}t:

\frac{dx}{dt} = \frac{1}{12}

Next, let's find the derivative of y = t³ - t:

\frac{dy}{dt} = 3t² - 1

Finally, let's find the derivative of z = t³t:

\frac{dz}{dt} = 4t³

Now that we have the derivatives, we can use the point-slope form of a line to find the equation of the tangent line. The equation of a line is given by:

y - y₁ = m(x - x₁)

where (x₁, y₁) is the point of tangency and m is the slope. Plugging in the values (13, 0) for (x₁, y₁) and the derivatives we found for m, the equation becomes:

y - 0 = \frac{3}{2}t² - \frac{1}{12}(x - 13)

This is the equation of the tangent line to the curve at the specified point.