Finding Parametric Equations for Tangent Line of Surface Intersection

sapiental
Messages
110
Reaction score
0

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
 
Physics news on Phys.org
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.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Back
Top