- #1

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## Homework Statement

Find the projection of the intersection between the two surfaces

S

_{1}: z = 4-x^2 - y^2 and

S

_{2}: 4x^2y = 1 (x>0)

in the xy-plane

**2. The attempt at a solution**

4-x^2 - y^2 = 4x^2y -1

Is this all I need to do?

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- Thread starter kasse
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- #1

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Find the projection of the intersection between the two surfaces

S

S

in the xy-plane

4-x^2 - y^2 = 4x^2y -1

Is this all I need to do?

- #2

Hurkyl

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What is it that you did?Is this all I need to do?

- #3

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Equalised the equations, with z = 0.

- #4

Hurkyl

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(You can't set two equations equal... but I think I know what you meant)Equaled the equations, with z = 0.

(And you were looking for a geometric object, not an equation... but I assume you meant to say the object described by that equation)

Does that process have anything to do with finding the intersection of the two surfaces? Or with projecting that intersection to the x-y plane? If so, then you're done. If not, then you're not.

- #5

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How steep is the path C (the intersection between the curves) at the point S over the (1, 1/4)?

My plan is to find the directional derivative in the direction of the tangent of the projection of C in the xy-plane. What do I do next then?

- #6

HallsofIvy

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Since 1= 4x^2y, x^2= 1/(4y). Replacing x^2 by that in the first equation, z= 4- x^2- y^2= 4- 1/(4y)+ y^2. Now project by letting z= 0: y^2- 1/(4y)+ 4= 0.

However, there really is no reason to project to the xy-plane. What you want is the derivative of z= 4- x^2- y^2

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