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- Thread starter jk8985
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- Jan 29, 2012

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We can think of the two surfaces as "equi-potential" surfaces of [tex]f(x,y,z)= z- x^2- y^2[/tex] and [tex]g(x,y,z)= x+ y+ z- e^{xyz}[/tex]. Their gradients, [tex]\nabla f= -2x\vec{i}- 2y\vec{j}+ \vec{k}[/tex] and [tex]\nabla g= (1- yze^{xyz})\vec{i}+ (1- xze^{xyz})\vec{j}+ (1- xye^{xyz})\vec{k}[/tex] are perpendicular to the surfaces and perpendicular to the tangent planes at each point.

In particular, at (0, 0, 1), [tex]\nabla f(0, 0, 1)= \vec{k}][/tex] so the tangent plane is [tex]0(x- 0)+ 0(y- 0)+ 1(z- 1)= 0[/tex] or z= 1. And [tex]\nabla g(0, 0, 1)= \vec{i}+ \vec{j}+ \vec{k}[/tex] so the tangent plane is [tex]1(x- 0)+ 1(y- 0)+ 1(z- 1)= x+ y+ z- 1= 0[/tex] or [tex]x+ y+ z= 1[/tex]. The angle between those two planes is the same as the ange between the two perendicular vectors and you can use [tex]\vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta)[/tex] to find that angle.

The length of [tex]\vec{k}[/tex] is 1, the length of [tex]\vec{i}+ \vec{j}+ \vec{k}[/tex] is [tex]\sqrt{3}[/tex], and their dot product is 1. So [tex]1(\sqrt{3})cos(\theta)= 1[/tex].

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This is exactly what I ended up getting, with theta equalling ~1 radian. How would I go about finding the tangent line? I would always think it's a plane, so I didn't know how to approach it.We can think of the two surfaces as "equi-potential" surfaces of [tex]f(x,y,z)= z- x^2- y^2[/tex] and [tex]g(x,y,z)= x+ y+ z- e^{xyz}[/tex]. Their gradients, [tex]\nabla f= -2x\vec{i}- 2y\vec{j}+ \vec{k}[/tex] and [tex]\nabla g= (1- yze^{xyz})\vec{i}+ (1- xze^{xyz})\vec{j}+ (1- xye^{xyz})\vec{k}[/tex] are perpendicular to the surfaces and perpendicular to the tangent planes at each point.

In particular, at (0, 0, 1), [tex]\nabla f(0, 0, 1)= \vec{k}][/tex] so the tangent plane is [tex]0(x- 0)+ 0(y- 0)+ 1(z- 1)= 0[/tex] or z= 1. And [tex]\nabla g(0, 0, 1)= \vec{i}+ \vec{j}+ \vec{k}[/tex] so the tangent plane is [tex]1(x- 0)+ 1(y- 0)+ 1(z- 1)= x+ y+ z- 1= 0[/tex] or [tex]x+ y+ z= 1[/tex]. The angle between those two planes is the same as the ange between the two perendicular vectors and you can use [tex]\vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta)[/tex] to find that angle.

The length of [tex]\vec{k}[/tex] is 1, the length of [tex]\vec{i}+ \vec{j}+ \vec{k}[/tex] is [tex]\sqrt{3}[/tex], and their dot product is 1. So [tex]1(\sqrt{3})cos(\theta)= 1[/tex].

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- Feb 7, 2012

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The curve is the intersection of the two graphs, and the tangent line to the curve (at the point (0,0,1)) will be the intersection of the tangent planes to the graphs at that point. HallsofIvy has already found the equations of the two planes, so all you need to do now is to find the equation of their intersection.How would I go about finding the tangent line? I would always think it's a plane, so I didn't know how to approach it.

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How would I make that at the point (0,0,1) [i think it was that point]The curve is the intersection of the two graphs, and the tangent line to the curve (at the point (0,0,1)) will be the intersection of the tangent planes to the graphs at that point. HallsofIvy has already found the equations of the two planes, so all you need to do now is to find the equation of their intersection.

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My apologies for the bad grammar, by the way, I'm on my phone.How would I make that at the point (0,0,1) [i think it was that point]

- Jan 29, 2012

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What tangent line are you talking about? The tangent line to the curve of intersection? "I would always think it's a plane" confuses me.This is exactly what I ended up getting, with theta equalling ~1 radian. How would I go about finding the tangent line? I would always think it's a plane, so I didn't know how to approach it.

The tangent planes of the two surfaces, at (0, 0, 1), are, as I said before, z= 1 and x+ y+ z= 1. Since z= 1, x+ y+ z= x+ y+ 1= 1 so x+ y= 0 and y= -x. Taking x as parameter, we can write x= t, y= -t, z= 1 as parametric equations for the tangent line to the curve of intersection at (0, 0, 1).