Finding the tangent line of the curve of intersection

  • Thread starter Lucree
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  • #1
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Given the paraboloid z = 6 - x - x2 -2y2 and the plane x = 1, find curve of intersection and the parametric equations of the tangent line to this curve at point (1,2,-4).


So I plugged x=1 into the paraboloid equation and got z = 4-2y2.

Then I take the derivative of the curve of intersection:
dz/dy = -4y

Then I plug in 2 for t since y=t=2 at the given point and get the slope -8.

The parametric equation of a line is r = r0 +tv
I have r0 = (1,2,-4) and I have a slope -8. I'm not sure how to get a vector.
 
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Answers and Replies

  • #2
Simon Bridge
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Consider in general:

dy/dx=m ... means that every 1 unit you go in x, you go m units in y right?
The vector that does that must be (1,m)t right?
 
  • #3
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yes, so if -8 is dz/dy then for every unit in y I go down 8 in z. so the vector would be <0, 1, -8>? then the parametric equation is

x = 1
y = 2 + t
z = -4 -8t
 
  • #4
Simon Bridge
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There you go :)
 

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