Question

find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x = e^(-4t) cos(4t), y = e^(-4t) sin(4t), z = e^(-4t), (1, 0, 1) (x(t), y(t), z(t)) = (⟨-1/√3, 1/√3, -1/√3⟩)

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 \(\langle x^\prime(0),y^\prime(0),z^\prime(0)
angle=\langle - 4,4,-4
angle\) 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:

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