Finding Parametric Equations for Tangent Line of Surface Intersection

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SUMMARY

The discussion focuses on finding the parametric equations for the tangent line to the curve of intersection of the surfaces defined by the equations z² = x² + y² and x² + 2y² + z² = 66 at the point (3, 4, 5). The gradients of the functions f(x,y,z) = x² + y² - z² and g(x,y,z) = x² + 2y² + z² are calculated, yielding grad f(3,4,5) = <6, 8, -10> and grad g(3,4,5) = <6, 16, 10>. The cross product of these gradients is computed to find the tangent vector, which was initially miscalculated. The correct tangent vector is determined to be <240, -120, 48> after reevaluating the determinant.

PREREQUISITES
  • Understanding of vector calculus, specifically gradients and cross products.
  • Familiarity with parametric equations and their applications in 3D space.
  • Knowledge of surface equations and their intersections.
  • Ability to compute determinants and evaluate vector operations.
NEXT STEPS
  • Study the properties of gradients in vector calculus.
  • Learn how to compute cross products of vectors in three-dimensional space.
  • Explore the concept of parametric equations in the context of curves and surfaces.
  • Review examples of tangent lines to curves of intersection for better understanding.
USEFUL FOR

Students studying multivariable calculus, mathematicians working on surface intersections, and educators teaching vector calculus concepts.

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Homework Statement




5. Find parametric equations for the tangent line to the curve of intersection of the surfaces
z^2 = x^2 + y^2 and x^2 + 2y^2 + z^2 = 66 at the point (3, 4, 5).


The Attempt at a Solution



f(x,y,z) = x^2 + y^2 - z^2
g(x,y,z) = x^2 + 2y^2 + z^2

Partial derivz:

f'x = 2x
f'y = 2y
f'z = -2z

g'x = 2x
g'y = 4y
g'z = 2z

grad f = <2x,2y,-2z>
grad g = <2x,4y,2z>


grad f (3,4,5) = < 6,8,-10>
grad g (3,4,5) = <6,16,10>

then v = [(grad f) X (grad g )(cross product)

= |i j k | = 180i + 220j - 48k
|6 8 -10|
|6 16 10|


this approach is wrong because the solutions manual gives me

-10i + 4j - 2k for that vector


please help.

Thanks
 
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There is something going wrong with your determinant evaluation. I get 240i-120j+48k. Also remember there is not a single tangent vector - you can always multiply by a constant and still have a tangent vector.
 

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