# Parametric equations from partial derivatives

## 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'?

SammyS
Staff Emeritus
Homework Helper
Gold Member

## 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.

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...

SammyS
Staff Emeritus
Homework Helper
Gold Member
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?

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.

SammyS
Staff Emeritus
Homework Helper
Gold Member
No, I'm not sure how to find it.

Plug y = 1 into the equation of the surface.

Okay so then the equation of the intersecting curve is:

z=f(x,y)=√(9-2x^2-1)?

But how does that help me?

SammyS
Staff Emeritus
Homework Helper
Gold Member
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?

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)

SammyS
Staff Emeritus
Homework Helper
Gold Member
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 ?

-2/sqrt(6)

SammyS
Staff Emeritus