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.

Answer and Explanation:

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}.


Learn more about this topic:

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Evaluating Parametric Equations: Process & Examples

from Precalculus: High School

Chapter 24 / Lesson 3
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