Parametric equations from partial derivatives

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Homework Statement


The surface z=f(x,y)=√(9-2x2-y2) and the plane y=1 intersect in a curve. Find parametric equations for the tangent line at (√(2),1,2).


Homework Equations


Partial derivatives


The Attempt at a Solution


Okay, so I'm just trying to work through an example in my textbook, so technically I have the answer, I just want the in between steps that I can't figure out. I know that you take the partial derivative with respect to x and get

fx(x,y)=0.5(9-2x2-y2)-0.5×(-4x)

and then you plug in the given point (√(2),1,2) and come out with -√(2). But then it goes to say, 'It follows that this line has direction vector <1,0,-√(2)>'.

Where did the 1 and 0 come from? And how do I get the parametric equations from the 'direction vector'?
 

Answers and Replies

  • #2
SammyS
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Homework Statement


The surface z=f(x,y)=√(9-2x2-y2) and the plane y=1 intersect in a curve. Find parametric equations for the tangent line at (√(2),1,2).

Homework Equations


Partial derivatives

The Attempt at a Solution


Okay, so I'm just trying to work through an example in my textbook, so technically I have the answer, I just want the in between steps that I can't figure out. I know that you take the partial derivative with respect to x and get

fx(x,y)=0.5(9-2x2-y2)-0.5×(-4x)

and then you plug in the given point (√(2),1,2) and come out with -√(2). But then it goes to say, 'It follows that this line has direction vector <1,0,-√(2)>'.

Where did the 1 and 0 come from? And how do I get the parametric equations from the 'direction vector'?
Hello nindelic. Welcome to PF !
Find an equation which describes the curve of intersection. This will give z as a function of x along this curve.
 
  • #3
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Hello nindelic. Welcome to PF !
Find an equation which describes the curve of intersection. This will give z as a function of x along this curve.

Thank you, SammyS! And okay, I don't know what that means. A big part of my problem I'd say is understanding math speak. If all I have is numbers then I can do about anything, but throw some words in there and I am completely lost. So what exactly does that mean? Like using different coordinates?

I tried finding the partial derivative with respect to y but that didn't help at all. I tried solving the equation and the partial WRT x for the other variables but that didn't help. I just don't know where the 1 or 0 came from the in the direction vector...
 
  • #4
SammyS
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The initial statement in your Original Post, is:
"The surface z=f(x,y)=√(9-2x2-y2) and the plane y=1 intersect in a curve."​
Can you give the equation describing this curve?
 
  • #5
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The initial statement in your Original Post, is:
"The surface z=f(x,y)=√(9-2x2-y2) and the plane y=1 intersect in a curve."​
Can you give the equation describing this curve?

No, I'm not sure how to find it.
 
  • #6
SammyS
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No, I'm not sure how to find it.

Plug y = 1 into the equation of the surface.
 
  • #7
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Okay so then the equation of the intersecting curve is:

z=f(x,y)=√(9-2x^2-1)?

But how does that help me?
 
  • #8
SammyS
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Okay so then the equation of the intersecting curve is:

z=f(x,y)=√(9-2x^2-1)?

But how does that help me?
Now you can find dz/dx along the curve of intersection?

What can that tell you?
 
  • #9
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Now you can find dz/dx along the curve of intersection?

What can that tell you?

It tells you the rate of change of the intersection with respect to x.

dz/dx= (0.5(8-2x^2)^-0.5)*(-4x)
 
  • #10
SammyS
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It tells you the rate of change of the intersection with respect to x.

dz/dx= (0.5(8-2x^2)^-0.5)*(-4x)

What is this rate when x = 1 ?
 
  • #11
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-2/sqrt(6)
 
  • #12
SammyS
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What is this rate when x = 1 ?

-2/sqrt(6)
DUH! ...

I should have asked, "What is this rate when x = √2 ?".
 

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