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 = 16t, y = t³ - t, z = t³t. The specified point is (7, 0, 2).

Answer 1

The parametric equations for the tangent line to the given curve at the specified point are x = t, y = 9.125t - 63.875, z = t³t.

Step-by-step explanation:

The parametric equations for the given curve are:
x = 16t, y = t³ - t, z = t³t.

To find the tangent line to the curve at the specified point (7, 0, 2), we need to find the derivatives of x, y, and z with respect to t.

Taking the derivatives, we have:

dx/dt = 16, dy/dt = 3t² - 1, dz/dt = 3t².

The slope of the tangent line is given by dy/dx, which is (dy/dt) / (dx/dt).

So, the slope of the tangent line at t = 7 is:

slope = (dy/dt) / (dx/dt) = [(3(7)² - 1) / 16] = 146/16 = 9.125.

Now, we can use the point-slope form of a line to find the equation of the tangent line:

y - y₁ = m(x - x₁), where (x₁, y₁) is the specified point and m is the slope.

Plugging in the values, we have:

y - 0 = 9.125(x - 7)

y = 9.125x - 63.875.

Therefore, the parametric equations for the tangent line to the curve at the specified point are x = t, y = 9.125t - 63.875, z = t³t.