Math 326A Exercise 4. Due Wednesday, October 24, 2012

Math 326A Exercise 4. Due Wednesday, October 24, 2012.

Implicit Function Theorem: Suppose that F (x, y, z) has continuous partial derivatives, and that the gradient ∇P F is nonzero at a point P . Consider the level surface of F through P . There is a cylinder, centered at P and normal to the tangent plane of the surface at P , such that the part of the level surface inside the cylinder can be mapped from the base of the cylinder.

Special version of the implicit function theorem:

∂F If ∂z (P ) = 0, then there is a cylinder, centered at P and parallel to the z-axis, such that the part of the level surface inside the cylinder can be graphed from a disk in the x-y plane.

∂F If ∂y (P ) = 0, then the part of the surface near P can be graphed from the x-z plane, and if ∂F ∂x (P ) = 0, then the part of the surface near P can be graphed from a disk in the y-z plane.

1. Consider the continuously diﬀerentiable function F (x, y, z)= z3 + z2 +(x2 + y2)z.

(a). Compute the gradient of F (x, y, z).

Solution: ∇F = (2xz)i + (2yz)j + (3z2 +2z + x2 + y2)k.

(b). Does the implicit function theorem tell us that the equation of the level surface can be solved for z as a function of x and y near P = (3, 0, −1); for y as a function of x and z; or for x as a function of y and z?

Solution: ∇F |(3,0,−1) = −6i +0j +10k. Since the j-component of the gradient equals 0, the implicit function theorem does not tell us that we can solve the level surface equation for y as a function of x and z. Since the x-component of the gradient is nonzero, the implicit function theorem tells us that we can solve for x as a function of y and z.

Solution: By part (a), we can solve for x as a function x(y, z) of y and z near P0. We have the formulas: ∂F ∂F 2 2 2 ∂x ∂y 2yz y ∂x ∂z 3z +2z + x + y = − ∂F = − = − and = − ∂F = − . ∂y ∂x 2xz x ∂z ∂x 2xz A linear perspective of the implicit function theorem. For a ﬁxed point (x0,y0, z0) and a variable point (x, y, z), the vector from (x0,y0, z0) to (x, y, z) is

(x − x0)i +(y − y0)j +(z − z0)k.

Let F (x, y, z) be a continuously diﬀerentiable function. Suppose that P = (x0,y0, z0) is a point on the level surface F (x, y, z)= c. ∂F ∂F ∂F If the gradient vector ∇P F = ∂x |P i + ∂y |P j + ∂z |P k at P is not the zero vector, then ∂F ∂F ∂F ∇ F ((x − x )i +(y − y )j +(z − z )k)= | (x − x )+ | (y − y )+ | (z − z )=0 P 0 0 0 ∂x p 0 ∂y P 0 ∂z P 0 is the linear equation of the tangent plane to the level surface at P . The implicit function theorem says that if the linear equation ∂F ∂F ∂F | (x − x )+ | (y − y )+ | (z − z )=0 ∂x P 0 ∂y P 0 ∂z P 0 for the tangent plane can be solved for z − z0 as a function of x − x0 and y − y0, then the nonlinear equation F (x, y, z)= c can be solved for z as a function of x and y near (x0,y0, z0), i.e., the part of the surface near P can be displayed as a graph of a function z = f(x, y) deﬁned on a disk in the x-y plane centered at (x0,y0).

2. Consider the function F (x, y, z)= x2 + y2 + z2 + cos z. (a). Write down the linear equation for the tangent plane to the level surface of F through a point (x0,y0, z0).

Solution: The equation is 2x0(x − x0)+2y0(y − y0)+(2z − sin(z))(z − z0) = 0.

π π (b). Let P = (1, 1, 2 ). Find out if the linear equation can be solved for z − 2 as a function of x − 1 and y − 1. Can the nonlinear equation F (x, y, z)= c be solved for z as a function π of x and y near (1, 1, 2 ), according to the implicit function theorem?

Solution:

π π ∇F |(1,1, 2 ) =2xi +2yj + (2z − sin z)k|(1,1, 2 ) =2i +2j +(π − 1)k. Hence, the equation of the tangent plane is π 2(x − 1)+2(y − 1)+(π − 1)(z − )=0. 2 π Since the coeﬃcient of z − 2 is nonzero, we can solve the linear equation for the tangent π plane for z − 2 as a function of x − 1 ane y − 1. Hence, by the implicit function theorem, the π nonlinear equation F (x, y, z)= c can be solved for z as a function of x and y near (1, 1, 2 ). Two equations: Suppose that F (x, y, z) and G(x, y, z) are functions with continuous ﬁrst partial derivatives. The gradient vector ∇P F is normal to the level surface of F (x, y, z) through P , and the gradient vector ∇P G are normal to the level surface of G(x, y, z) through P . The crossproduct ∇P F and ∇pG is a vector tangent to the curve of intersection of the level surfaces at P .

Implicit Function Theorem: If the crossproduct ∇P F and ∇P G is a nonzero vector, then there is a cylinder, centered at P and parallel to the line tangent to the curve of intersection of the level surfaces at P , such that the part of the curve of intersection inside the cylinder may be mapped from the line tangent to the curve at P .

Special version of the implicit function theorem: If the i-component of the crossproduct of the gradient vectors at P is nonzero, then there is a cylinder, centered at P and parallel to i, such that the part of the curve of intersection in the cylinder can be graphed from the x-axis, i.e., the curve can be expressed in parametric form as (x, f(x),g(x)) for some diﬀerentiable functions f(x) and g(x).

Similarly, if the y-component of the crossproduct is nonzero, then the part of the curve of intersection near P can be graphed from the y-axis, and if the z-component of the crossproduct is nonzero, then the part of the curve of intersection near P can be graphed from the z-axis.

3. Consider F (x, y, z)= z3 + z2 +(x2 + y2)z and G(x, y, z)= x − y3 − z3.

(a). Compute the cross product of the gradients of F and G.

Solution: ∇F ×∇G = ((2xz)i + (2yz)j + (3z2 +2z + x2 + y2)k) × (i − 3y2j − 3z2k).

(b). Find out whether or not the curve of intersection of the level surfaces near (1, −1, −1) can be graphed from each of the coordinate axes.

Solution: ∇F ×∇G|(1,−1,−1) =(−2i +2j +3k) × (i − 3j − 3k)=3i − 3j +4k. Hence, the intersection of the level surfaces near (1, −1, −1) can be graphed from each coordinate axis, since each coordinate of the crossproduct is nonzero.

(c). Answer the question in problem 2(b) for the same functions near the point (3, 0, −1).

Solution: ∇F |((3,0,−1) ×∇G|(3,0,−1) =

2 2 2 2 2 (2xzi +2yzj + (3z +2z + x + y )k)|(3,0,−1) × (i − 3y j − 3z k)|(3,0,,−1) = (−6i +0j + 10k) × (i +0j − 3k)=0i − 8j +0k. The implicit function theorem tells us that we can graph the curve of intersection near (3, 0, −1) from the y-axis. The theorem does not tell us if we can or cannot graph the curve near (3, 0, −1) from the x-axis or from the z-axis. A linear perspective on solving two equations. Let F (x, y, z) and G(x, y, z) be two continuously diﬀerentiable functions. Suppose that P = (x0,y0, z0) is a point on the level ∂F surfaces F (x, y, z) = c and G(x, y, z) = d. Suppose that ∂z (P ) = 0. Then we can express the level surface F (x, y, z) = c near P as the graph of a function z = f(x, y). The expression G(x, y, f(x, y)) then gives the value of G on the surface F (x, y, z) = c. The points (x, y, f(x, y)) where G(x, y, f(x, y)) = c are the points that are on both level surfaces. We will show in class that if the Jacobian matrix of the y and z partials of F and G is nonzero at P , then we can solve G(x, y, f(x, y)) = c for y as a function of x, i.e., we can express the curve G(x, y, f(x, y)) = c on the level surface F (x, y, z) = c as a graph y = g(x). Altogether, the curve of intersection of the two level surfaces is described by (x, g(x), f(x, g(x))=(x, g(x), h(x)), where h(x)= f(x, g(x)). The implicit function theorem says that if the Jacobian matrix of the y and z partials of F and G has nonzero determinant at P , then we can express the intersection of the level surfaces F (x, y, z)= c and G(x, y, z)= d as a graph (x, y(x), z(x)) near P .

Consider the linear equations ∂F ∂F ∂F | (x − x )+ | (y − y )+ | (z − z )=0, ∂x P 0 ∂y P 0 ∂z P 0

and ∂G ∂G ∂G | (x − x )+ | (y − y )+ | (z − z )=0 ∂x P 0 ∂y P 0 ∂z P 0 for the tangent planes to the level surfaces F (x, y, z) = c and G(x, y, z) = d. If those two linear equations are independent in the sense of linear algebra, then the common solutions to those equations (the intersection of the tangent planes) form the tangent line to the intersection of the level surfaces, at the point P . Write those linear equations in matrix form as ∂F ∂F ∂F |P |P y − y | ∂y ∂z 0 = − ∂x P (x − x ). ∂G ∂G z z ∂G 0 ∂y |P ∂z |P ! − 0 ∂x |P

We can solve the matrix equation for z − z0 and y − y0 as functions of x − x0 if and only if

the Jacobian matrix ∂F ∂F ∂y ∂z ∂G ∂G ∂y ∂z ! of the y and z partials of F and G has nonzero determinant at P . Hence, by the implicit function theorem, if we can solve the linear system of equations for z − z0 and y − y0 as functions of x − x0, then we can solve the nonlinear system of equations F (x, y, z)= c and G(x, y, z)= d for z and y as functions of x near P . 4. Consider the functions F (x, y, z)= x2 + y2 + z2 + cos zand G(x, y, z)= xyz + sin z. (a). Write down the linear equations for the tangent planes to the level surfaces of F and G through a point (x0,y0, z0).

Solution. The equation for the tangent plane to the level surface of F at (x0,y0, z0) is

2x0(x − x0)+2y0(y − y0)+(2z0 − sin(z0))(z − z0)=0,

and the equation for the tangent plane to the level surface of G at (x0,y0, z)) is

(y0z0)(x − x0)+(x0z0)(y − y0)+(x0y0 + cos(z0))(z − z0)=0.

π π (b). Let P = (1, 1, 2 ). Decide if those linear equations can be solved for z − 2 and y − 1 as functions of x − 1. Can the nonlinear equations F (x, y, z)= c and G(x, y, z)= d be solved π for z and y as a function of x near (1, 1, 2 ), according to the implicit function theorem?

π Solution. ∇F = (2x, 2y, 2z − sin z) and ∇G = (yz,xz,xy + cos z). Hence, ∇F |1,1, 2 = , , π G π π , π , (2 2 − 1) and ∇ |1,1, 2 =( 2 2 1). Hence, the equations of the tangent planes are: π 2(x − 1)+2(y − 1)+(π − 1)(z − )=0, 2

and π π π (x − 1)+ (y − 1)+(z − )=0. 2 2 2 When we write those linear equations as a single system and separate the variable x−1 from π the variables y − 1 and z − 2 , we get the matrix equation 2 (π − 1) y − 1 −2 = (x − 1). π z − π − π 2 1 2 2 π That equation can be solved for y − 1 and z − 2 in terms of x − 1 because the coeﬃcient matrix 2 (π − 1) π 2 1 is invertible. By the implicit function theorem, since the linear equations for the tangent spaces can be π solved for y − 1 and z − 2 in terms of x − 1, the nonlinear equations F (x, y, z) = c and π G(x, y, z)= d can be solved for z and y as a function of x near (1, 1, 2 ). 5. Suppose that we have a system of two equations in 4 variables, F (x,y,u,v) = c and G(x,y,u,v) = d, and a point P = (x0,y0,u0, v0) satisfying both equations. Again, we can write down the system of linear equations

∇F |P ((x − x0)i +(y − y0)j +(u − u0)k +(v − v0)l)=0, and ∇G|P (x − x0)i +(y − y0)j +(u − u0)k +(v − v0)l)=0.

The implicit function theorem tells us that if we can solve those linear equations for x−x0 and y−y0 in terms of u−u0 and v−v0, then we can solve the nonlinear equations F (x,y,u,v)= c and G(x,y,u,v)= d for x and y near (x0,y0,u0, v0). To see if you can solve the linear equations, write the system in matrix form as

∂F ∂F ∂F ∂F ∂x |P ∂y |P x − x0 ∂u |P ∂v |P u − u0 ∂G ∂G = − ∂G ∂G . y − y |P v − v ∂x |P ∂y |P ! 0 ∂u|P ∂v ! 0

That matrix equation can be solved for x − x0 and y − y0 in terms of u − u0 and v − v0 if and ∂F ∂F ∂x ∂y only if the Jacobian matrix ∂G ∂G has nonzero determinant at P , i.e., if the matrix is ∂x ∂y ! invertible at P .

Consider the functions F (x,y,u,v)= xv + yu and G(x,y,u,v)= xu + yv. (a). Write down the system of linear equations in matrix form corresponding to the level surface equations of F and G at a general point (x0,y0,u0, v0).

Solution: The linear equations are

v0(x − x0)+ u0(y − y0)+ y0(u − u0)+ x0(v − v0)=0,

and u0(x − x0)+ v0(y − y0)+ x0(u − u0)+ y0(v − v0)=0.

(b). Consider the point (x0,y0,u0, v0) = (1, −1, 2, 1). Decide if the matrix equation can be solved for x − 1 and y + 1 as functions of u − 2 and v − 1. Can the nonlinear equations F (x,y,u,v)= −1 and G(x,y,u,v) = 1 be solved for x and y as a function of u and v near (1, −1, 2, 1)?

Solution: ∇F = (v,u,y,x) and ∇G = (u,v,x,y). Hence, ∇F |1,−1,2,1 = (1, 2, −1, 1) and ∇G|1,−1,2,1 = (2, 1, 1, −1). The equations for the tangent planes are

(x − 1)+2(y + 1) − (u − 2)+(v − 1)=0, 2(x − 1)+(y +1)+(u − 2) − (v − 1)=0. That system of two linear equations can be written as a single matrix equation

1 2 x − 1 −1 1 u − 2 = − . y v 1 1 +1 1 −1 − 1 Because the matrix 1 2 1 1 is invertible, we can solve that matrix equation for x − 1 and y +1 in terms of u − 2 and v − 1. But because the matrix −1 1 1 −1 is noninvertible, we cannot solve for u − 2 and v − 1 in terms of x − 1 and y + 1. By the implicit function theorem, we can solve the nonlinear equations F (x,y,u,v) = −1 and G(x,y,u,v)=1 for x and y as a function of u and v near (1, −1, 2, 1).