The tangent line at the intersection of the surfaces defined by the equations z=x^2+y^2+1 and x+y+z=e^(xyz) at the point (0,0,1) can be derived from the tangent planes at that point. The gradients of the surfaces, \nabla f and \nabla g, are calculated to find the tangent planes, which are z=1 and x+y+z=1, respectively. The angle between these planes is approximately 1 radian, determined using the dot product of their normal vectors. The parametric equations for the tangent line at the intersection are x=t, y=-t, z=1.
PREREQUISITESStudents and professionals in mathematics, particularly those focused on calculus and geometry, as well as engineers and physicists working with three-dimensional surfaces and their 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.