Step 1
Write the expression to find the parametric equations for a line through the point \((x_{0},\ y_{0},\ z_{0})\ \text{are parallel to the vector}\ v =\ <\ a,\ b,\ c\ >.\)

\(x=x_{0}\ +\ at,\ y=y_{0}\ +\ bt,\ z=z_{0}\ +\ ct\) Step 2 Write the parametric equations of the curveas follows. \(x=e^{-8t}\ \cos(8t),\ y=e^{-8t}\ \sin(8t),\ z=e^{-8t}\) Step 3 Write the vector equation from the parametric equations of the curve as follows. \(r(t)=\langle\ e^{-8t}\ \cos(8t),\ e^{-8t}\ \sin(8t),\ e^{-8t}\rangle\) Step 4 The tangent vector of the curve is the derivative of the vector function r(t). To find the derivative of the vector function, differentiate each component of the vector function. \(r'(t)=\Bigg\langle\ \frac{d}{dt}(e^{-8t}\ \cos(8t)),\ \frac{d}{dt}(e^{-8t}\ \sin(8t)),\ \frac{d}{dt}(e^{8t}) \Bigg\rangle\)

\(=\ \langle\ e^{-8t}(-8\ \sin(8t)\ +\ \cos(8t)(-8e^{-8t}),\ e^{-8t}(8\ \cos(8t))\ +\ \sin(8t)(-8e^{-8t},\ -8e^{-8t}) \rangle\)

\(=\ \langle\ -8e^{-8t}\ \sin(8t)\ -\ 8e^{-8t}\ \cos(8t),\ 8e^{-8t}\ \cos\ -\ 8e^{-8t}\ \sin(8t),\ -8e^{-8t}) \rangle\) Step 5 Given that the point \((1,\ 0,\ 1)\). That is, \(1=e^{-8t}\ \cos(8t)\ \Rightarrow\ e^{-8t}=1\ or\ \cos(8t)=1\ \Rightarrow\ t=0\)

\(0=e^{-8t}\ \sin(8t)\ \Rightarrow\ e^{-8t}\ \neq\ 0\ so\ \sin(8t)=0\ \Rightarrow\ t=0\)

\(1=e^{-8t}\ \Rightarrow\ t=0\) As the specified point \((1,\0,\1)\ \text{corresponds to}\ t = 0\), consider the value of scalar parameter t as 0 and substitute in the parametric equations of the curve to obtain the point, which is on the required line. Substitute 0 for t in equation (2), \(x=e^{-8(0)}\ \cos(8(0)),\ y=e^{-8(0)}\ \sin(8(0)),\ z=e^{-8(0)}\)

\(x=e^{0}\ \cos(0),\ y=e^{0}\ \sin(0),\ z=e^{-8(0)}\)

\(x = 1,\ y = 0,\ z = 1\) The point on the required line is \((1,\ 0,\ 1).\) As the point on the required line is same as the specified point \((1,\ 0,\ 1),\ \text{the tangent vector is}\ r'(t)\ at\ t = 0.\) Step 6 Substitute 0 for t in \(r'(t)\)

\(r'(t)=\ \Big\langle\ -8e^{-8(0)}\ \sin(8(0))\ -\ 8e^{-8(0)}\ \cos(8(0)),\ 8e^{-8(0)}\ \cos(8(0))\ -\ 8e^{-8(0)}\ \sin(8(0)),\ -8^{-8(0)} \Big\rangle\)

\(=\ \langle\ -8(1)(0)\ -\ 8(1)(1),\ 8(1)(1)\ -\ 8(1)(0),\ -8(1)\ \rangle\)

\(=\ \langle\ -8,\ 8,\ 8\ -\ 8\ \rangle\) The required line passes through the point \((1,\ 0,\ 1)\ \text{and parallel to the vextor}\ \ \langle\ -8,\ 8,\ 8\ -\ 8\ \rangle\) Step 7 Substitute 1 for \(x_{0},\ 0\ for\ y_{0},\ 1\ for\ z_{0},\ –8\) for a, 8 for b, and –8 for c in equation (1), \(x=(1)\ +\ (-8)t,\ y=0\ +\ 8t,\ z=1\ +\ (-8)t\)

\(x=1\ -\ 8t,\ y=8t,\ z=1\ -\ 8t\) Thus the parametric equations for the tarhent line to the curve with the parametric equations \(x=e^{-8t},\ y=e^{-8t},\ z=e^{-8t}\ \text{at the point}\ (1,\ 0,\ 1)\ are\ \underline{x=1\ -\ 8t,\ y=8t,\ z=1\ -\ 8t}\)

\(x=x_{0}\ +\ at,\ y=y_{0}\ +\ bt,\ z=z_{0}\ +\ ct\) Step 2 Write the parametric equations of the curveas follows. \(x=e^{-8t}\ \cos(8t),\ y=e^{-8t}\ \sin(8t),\ z=e^{-8t}\) Step 3 Write the vector equation from the parametric equations of the curve as follows. \(r(t)=\langle\ e^{-8t}\ \cos(8t),\ e^{-8t}\ \sin(8t),\ e^{-8t}\rangle\) Step 4 The tangent vector of the curve is the derivative of the vector function r(t). To find the derivative of the vector function, differentiate each component of the vector function. \(r'(t)=\Bigg\langle\ \frac{d}{dt}(e^{-8t}\ \cos(8t)),\ \frac{d}{dt}(e^{-8t}\ \sin(8t)),\ \frac{d}{dt}(e^{8t}) \Bigg\rangle\)

\(=\ \langle\ e^{-8t}(-8\ \sin(8t)\ +\ \cos(8t)(-8e^{-8t}),\ e^{-8t}(8\ \cos(8t))\ +\ \sin(8t)(-8e^{-8t},\ -8e^{-8t}) \rangle\)

\(=\ \langle\ -8e^{-8t}\ \sin(8t)\ -\ 8e^{-8t}\ \cos(8t),\ 8e^{-8t}\ \cos\ -\ 8e^{-8t}\ \sin(8t),\ -8e^{-8t}) \rangle\) Step 5 Given that the point \((1,\ 0,\ 1)\). That is, \(1=e^{-8t}\ \cos(8t)\ \Rightarrow\ e^{-8t}=1\ or\ \cos(8t)=1\ \Rightarrow\ t=0\)

\(0=e^{-8t}\ \sin(8t)\ \Rightarrow\ e^{-8t}\ \neq\ 0\ so\ \sin(8t)=0\ \Rightarrow\ t=0\)

\(1=e^{-8t}\ \Rightarrow\ t=0\) As the specified point \((1,\0,\1)\ \text{corresponds to}\ t = 0\), consider the value of scalar parameter t as 0 and substitute in the parametric equations of the curve to obtain the point, which is on the required line. Substitute 0 for t in equation (2), \(x=e^{-8(0)}\ \cos(8(0)),\ y=e^{-8(0)}\ \sin(8(0)),\ z=e^{-8(0)}\)

\(x=e^{0}\ \cos(0),\ y=e^{0}\ \sin(0),\ z=e^{-8(0)}\)

\(x = 1,\ y = 0,\ z = 1\) The point on the required line is \((1,\ 0,\ 1).\) As the point on the required line is same as the specified point \((1,\ 0,\ 1),\ \text{the tangent vector is}\ r'(t)\ at\ t = 0.\) Step 6 Substitute 0 for t in \(r'(t)\)

\(r'(t)=\ \Big\langle\ -8e^{-8(0)}\ \sin(8(0))\ -\ 8e^{-8(0)}\ \cos(8(0)),\ 8e^{-8(0)}\ \cos(8(0))\ -\ 8e^{-8(0)}\ \sin(8(0)),\ -8^{-8(0)} \Big\rangle\)

\(=\ \langle\ -8(1)(0)\ -\ 8(1)(1),\ 8(1)(1)\ -\ 8(1)(0),\ -8(1)\ \rangle\)

\(=\ \langle\ -8,\ 8,\ 8\ -\ 8\ \rangle\) The required line passes through the point \((1,\ 0,\ 1)\ \text{and parallel to the vextor}\ \ \langle\ -8,\ 8,\ 8\ -\ 8\ \rangle\) Step 7 Substitute 1 for \(x_{0},\ 0\ for\ y_{0},\ 1\ for\ z_{0},\ –8\) for a, 8 for b, and –8 for c in equation (1), \(x=(1)\ +\ (-8)t,\ y=0\ +\ 8t,\ z=1\ +\ (-8)t\)

\(x=1\ -\ 8t,\ y=8t,\ z=1\ -\ 8t\) Thus the parametric equations for the tarhent line to the curve with the parametric equations \(x=e^{-8t},\ y=e^{-8t},\ z=e^{-8t}\ \text{at the point}\ (1,\ 0,\ 1)\ are\ \underline{x=1\ -\ 8t,\ y=8t,\ z=1\ -\ 8t}\)