find parametric equations for the tangent line to the curve with the given parametric equations at the…

Explanation:

Step1: Find the value of (t)

Set (x = e^{-4t}\cos(4t)=1), (y = e^{-4t}\sin(4t)=0), (z = e^{-4t}=1). From (z = e^{-4t}=1), we get (t = 0).

Step2: Differentiate (x,y,z) with respect to (t)

Using the product - rule ((uv)^\prime=u^\prime v + uv^\prime), where for (x = e^{-4t}\cos(4t)), (u = e^{-4t}), (u^\prime=-4e^{-4t}), (v=\cos(4t)), (v^\prime=- 4\sin(4t)), so (x^\prime(t)=-4e^{-4t}\cos(4t)-4e^{-4t}\sin(4t)). For (y = e^{-4t}\sin(4t)), (u = e^{-4t}), (u^\prime=-4e^{-4t}), (v=\sin(4t)), (v^\prime = 4\cos(4t)), so (y^\prime(t)=-4e^{-4t}\sin(4t)+4e^{-4t}\cos(4t)). For (z = e^{-4t}), (z^\prime(t)=-4e^{-4t}).

Step3: Evaluate the derivatives at (t = 0)

Substitute (t = 0) into (x^\prime(t)), (y^\prime(t)) and (z^\prime(t)). (x^\prime(0)=-4e^{0}\cos(0)-4e^{0}\sin(0)=-4), (y^\prime(0)=-4e^{0}\sin(0)+4e^{0}\cos(0)=4), (z^\prime(0)=-4e^{0}=-4).

Step4: Write the parametric equations of the tangent line

The parametric equations of a line passing through the point ((x_0,y_0,z_0)=(1,0,1)) with direction - vector (\langle x^\prime(0),y^\prime(0),z^\prime(0)\rangle=\langle - 4,4,-4\rangle) are (x(t)=x_0+x^\prime(0)t=1-4t), (y(t)=y_0 + y^\prime(0)t=4t), (z(t)=z_0+z^\prime(0)t=1-4t).

Answer:

((x(t),y(t),z(t))=(1 - 4t,4t,1 - 4t))