The discussion revolves around finding the tangent line at the point (0,0,1) for the intersection of two surfaces defined by the equations z=x^2+y^2+1 and x+y+z=e^(xyz). Participants explore the tangent planes at this point and the angle between them, as well as the method to derive the tangent line of the curve formed by the intersection of these surfaces.
Participants generally agree on the method to find the tangent planes and the angle between them. However, there is some confusion and disagreement regarding the concept of the tangent line, with differing interpretations of how to approach it.
Some participants express uncertainty about the definitions and concepts involved, particularly regarding the distinction between tangent lines and tangent planes. There are also unresolved questions about the specific steps needed to derive the tangent line from the tangent planes.
This discussion may be useful for students or individuals interested in multivariable calculus, particularly those exploring concepts related to tangent lines, tangent planes, and surface intersections.
We can think of the two surfaces as "equi-potential" surfaces of f(x,y,z)= z- x^2- y^2 and g(x,y,z)= x+ y+ z- e^{xyz}. Their gradients, \nabla f= -2x\vec{i}- 2y\vec{j}+ \vec{k} and \nabla g= (1- yze^{xyz})\vec{i}+ (1- xze^{xyz})\vec{j}+ (1- xye^{xyz})\vec{k} are perpendicular to the surfaces and perpendicular to the tangent planes at each point.jk8985 said:The graph of z=x^2+y^2+1 and the graph of x+y+z=e^(xyz) have a common point (0,0,1). Find the angle between the two corresponding tangent planes respect to these two graphs at point (0,0,1). In addition, find the tangent line of the curve intersected by these two graphs at point (0,0,1).
HallsofIvy said:We can think of the two surfaces as "equi-potential" surfaces of f(x,y,z)= z- x^2- y^2 and g(x,y,z)= x+ y+ z- e^{xyz}. Their gradients, \nabla f= -2x\vec{i}- 2y\vec{j}+ \vec{k} and \nabla g= (1- yze^{xyz})\vec{i}+ (1- xze^{xyz})\vec{j}+ (1- xye^{xyz})\vec{k} are perpendicular to the surfaces and perpendicular to the tangent planes at each point.
In particular, at (0, 0, 1), \nabla f(0, 0, 1)= \vec{k}] so the tangent plane is 0(x- 0)+ 0(y- 0)+ 1(z- 1)= 0 or z= 1. And \nabla g(0, 0, 1)= \vec{i}+ \vec{j}+ \vec{k} so the tangent plane is 1(x- 0)+ 1(y- 0)+ 1(z- 1)= x+ y+ z- 1= 0 or x+ y+ z= 1. The angle between those two planes is the same as the ange between the two perendicular vectors and you can use \vec{u}\cdot\vec{v}= |\vec{u}||\vec{v}|cos(\theta) to find that angle.
The length of \vec{k} is 1, the length of \vec{i}+ \vec{j}+ \vec{k} is \sqrt{3}, and their dot product is 1. So 1(\sqrt{3})cos(\theta)= 1.
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.jk8985 said: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.
Opalg said: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.
jk8985 said:How would I make that at the point (0,0,1) [i think it was that point]
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. What do you always think is a plane"?jk8985 said: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.