Tangent Line MultiVar Calc Problem

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SUMMARY

The discussion focuses on finding the slope of the tangent line to the curve formed by the intersection of the vertical plane defined by the equation x - y + 1 = 0 and the surface z = x² + y² at the point (1, 2, 5). Participants utilize normal vectors to derive the tangent vector, with Normal Vector 1 as i - j and Normal Vector 2 as 2i + 4j - k. The cross product of these vectors yields the tangent direction, which is then normalized. The slope is defined as the ratio of the z component to the projection on the xy-plane, expressed mathematically as slope = z/(sqrt(x² + y²)).

PREREQUISITES
  • Understanding of vector calculus, specifically normal and tangent vectors.
  • Familiarity with the equations of planes and surfaces in three-dimensional space.
  • Knowledge of parameterization of curves and differentiation techniques.
  • Proficiency in calculating cross products and unit vectors.
NEXT STEPS
  • Study the method of parameterizing curves in three-dimensional space.
  • Learn about the application of the cross product in vector calculus.
  • Explore the concept of gradients and their relation to tangent planes.
  • Investigate the implications of slope in multivariable calculus, particularly in optimization problems.
USEFUL FOR

Students and educators in multivariable calculus, particularly those focusing on vector calculus and geometric interpretations of surfaces and planes.

harrietstowe
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Homework Statement



Find the slope of the tangent line to the curve of intersection of the vertical plane x-y+1= 0 and the surface z=x^2+ y^2 at the point (1, 2, 5) .

Homework Equations





The Attempt at a Solution


Normal Vector 1 is i-j
Normal Vector 2 is 2i+4j-k
Cross them to get i+j+6k
then turn that vector into a unit vector and after that I am unsure what to do
 
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first you need to decide what exactly you mean by slope...

Something like dz/du, where du is position change projected on the xy-plane, and dz is the change in vertical position. This lines up with the usual notion of slope.

If you are confident in you tangent vector (which i haven't checked) call it t:
t = xi+yj+zk

then the slope as defined above will be given by the ratio of z component to the projection on the xy plane
slope = z/(sqrt(x^2 + y^2))

Another easy way to find the tangent vector direction would be to parameterise the line in terms of t=x, giving r(t). Then differentiate in terms of t.
 

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