Question

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.

x = 1 + 6t, y = t3 − t, z = t3 + t; (7, 0, 2), y = t3 − t, z = t3 + t; (7, 0, 2)

Answer 1

The parametric equations for the curve are x = 1 + 6t, y = t^3 − t, z = t^3 + t. To find the tangent line at the specified point (7, 0, 2), substitute the values into the parametric equations and calculate the derivatives. The equation of the tangent line is y - 15100 = 312.33(x - 151).

Step-by-step explanation:

The parametric equations for the curve are given by:

x = 1 + 6t

y = t^3 − t

z = t^3 + t

To find the tangent line at the specified point (7, 0, 2), we need to substitute the values into the parametric equations and also calculate the derivatives of x, y, and z with respect to t.

At t = 25 s, the x-coordinate is 1 + 6(25) = 151, the y-coordinate is (25^3) − 25 = 15100, and the z-coordinate is (25^3) + 25 = 15625.

The derivatives of x, y, and z are: dx/dt = 6, dy/dt = 3t^2 − 1, dz/dt = 3t^2 + 1.

Substituting t = 25 into the derivatives, we get:

dx/dt = 6

dy/dt = 3(25^2) − 1 = 1874

dz/dt = 3(25^2) + 1 = 1876

Therefore, the slope of the tangent line is given by dy/dt / dx/dt = 1874 / 6 = 312.33.

Using the point-slope form of the equation of a line, y-y1 = m(x-x1), we can write the equation of the tangent line as:

y - 15100 = 312.33(x - 151)