Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x = e^−2t cos 4t, y = e^−2t sin 4t, z = e^−2t; (1, 0, 1)

x=1

y=s

z=1

Step-by-step explanation:

(x, y, z)=(1, 0, 1)

Substitute 0 for y

Confirming if t=0 satisfy the other equation

x = e^−2t cos 4t = e^−2(0)cos(4*0)

= e^(0)cos(0) = 1

z = e^−2t  = e^−2(0)  = 0

Therefore t=0 satisfies the other equation

Finding the tangent vector at t=0

The vector equation of the tangent line is

(1, 0, 1) +s(0,1,0)= (1, s, 1)

The parametric equations are:

x=1

y=s

z=1

me! what did you put for this question?

step-by-step explanation:

r = 78539458390583095803458309458039485093485 or just 3

step-by-step explanation: