Question

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point: x = e^(-3t)cos(3t), y = e^(-3t)sin(3t), z = e^(-3t); (1, 0, 1).

Answer 1

To find the parametric equations for the tangent line at the specified point, we need to find the slope of the curve at that point. We can do this by finding the derivative of each of the parametric equations with respect to t.

Step-by-step explanation:

To find the parametric equations for the tangent line at the specified point, we need to find the slope of the curve at that point. We can do this by finding the derivative of each of the parametric equations with respect to t.

The derivative of x with respect to t is dx/dt = -3e^(-3t)sin(3t) - 3e^(-3t)cos(3t)

The derivative of y with respect to t is dy/dt = -3e^(-3t)cos(3t) + 3e^(-3t)sin(3t)

The derivative of z with respect to t is dz/dt = -3e^(-3t)

To find the slope at t = 25, substitute t = 25 into the derivatives.

The slope is given by dx/dt = -3e^(-3*25)sin(3*25) - 3e^(-3*25)cos(3*25)

Now, we have the slope at the specified point. We can use this slope and the point (1, 0, 1) to find the equation of the tangent line using the point-slope form:

y - 0 = m(x - 1)

where m is the slope we just calculated. Thus, the parametric equations for the tangent line are:

x = 1 - t * (-3e^(-3*25)sin(3*25) - 3e^(-3*25)cos(3*25))

y = t

z = 1 - t * -3e^(-3*25)