# Consider the line perpendicular to the surface z = x^{2} + y^{2} at the point where x = 4 and y =...

## Question:

Consider the line perpendicular to the surface {eq}\,z = x^{2} + y^{2}\, {/eq} at the point where {eq}\,x = 4\, {/eq} and {eq}\,y = 1 {/eq}.

Find a vector parametric equation for this line in terms of the parameter {eq}\,t {/eq}.

## Finding Vector Parametric Equation:

The vector parametric equation is defined as {eq}x =x_{0}+at, \ y=y_{0}+bt, \ z=z_{0}+ct {/eq}. This equation has point and the direction vector of the line. The direction vector is found by evaluating the gradient at the point using the given surface.

Writing the vector parametric equation:

{eq}\begin{align*} \text{Point on the surface}: x &= 4 \\ y &= 1 \\ z &= x^{2} + y^{2} \\ z &= (4)^{2} + (1)^{2} \\ z &= 17 \\ (x_{0}, \ y_{0}, \ z_{0}) &=(4, \ 1, \ 17) \\ \\ \text{Direction vector}: \\ f(x, y, z) &=x^{2} + y^{2}-z \\ f_{x}(x, y, z) &=\frac{\partial }{\partial x}\left(x^2+y^2-z\right)=2x \\ f_{x}(4, 1, 17) &=2(4) =8 \\ \\ f_{y}(x, y, z) &=\frac{\partial \:}{\partial \:y}\left(x^2+y^2-z\right)=2y \\ f_{y}(4, 1, 17) &=2(1)=2 \\ \\ f_{z}(4, 1, 17) &=\frac{\partial \:}{\partial \:z}\left(x^2+y^2-z\right)=-1 \\ \nabla f(4, 1, 17) &=\left \langle 8, 2, -1 \right \rangle \\ \\ \text{Vector parametric equation}: \\ x &=x_{0}+at, \quad y=y_{0}+bt, \quad z=z_{0}+ct \\ x &=4+8t, \quad y=1+2t, \quad z=17-1t \end{align*} {/eq}

Thus, the vector parametric equations are {eq}x =4+8t, \ y=1+2t, \ z=17-t {/eq}.