Final answer:
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.