Perpendicular line to the Surface

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SUMMARY

The discussion focuses on finding the vector parametric equation of a line perpendicular to the surface defined by the equation z = x^2 + y^2 at the point (-1, 2). The gradient of the function, ∇F(x,y,z) = <2x, 2y, -1>, is calculated at the specified point, yielding ∇F(-1, 2, 0) = <-2, 4, -1>. The correct vector parametric equation is L(t) = <-1, 2, z> + t<-2, 4, -1>, where z must be evaluated correctly at the point (-1, 2) to avoid miscalculations.

PREREQUISITES
  • Understanding of vector calculus and gradients
  • Familiarity with parametric equations
  • Knowledge of surface equations and their properties
  • Basic skills in evaluating functions of multiple variables
NEXT STEPS
  • Study the concept of gradients in multivariable calculus
  • Learn how to derive parametric equations from surface equations
  • Explore the implications of miscalculating function values at specific points
  • Practice problems involving lines perpendicular to surfaces in three-dimensional space
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Homework Statement


Consider the line perpendicular to the surface z=x^2+y^2 at the point where x = −1 and y = 2 Find a vector parametric equation for this line in terms of the parameter t.

The Attempt at a Solution


I wasn't quite sure how to go about with this problem so I just went along with the following ideas. I first took the gradient of the function at that point:

0=x^2+y^2-z
∇F(x,y,z)= &lt;2x,2y,-1&gt;
∇F(-1,2,0)= &lt;-2,4,-1&gt;

Then I constructed the vector parametric equation of the line at that point:

L(t) = P + t∇F
L(t) = &lt;-1,2,0&gt; + t&lt;-2,4,-1&gt;

Afterwards, I submitted this equation, only finding that it was incorrect; can someone explain to me what went wrong here?
 
Physics news on Phys.org
When ##x=-1## and ##y=2##, ##z## isn't zero.
 
Wow haha that was a horrible miscalculation on my part. Thanks for pointing that out! :redface:
 

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