Recall from Section 2.6: A curve in R2 can be defined by f(x,y) = b (which is a level curve of the function z = f(x,y)). If c(t) = (x(t) , y(t)) describes a path on the curve f(x,y) = b, then f(x(t) , y(t)) = f(c(t)) =   d — f(x(t) , y(t)) = dt b f(c(t)) • c  (t) = 0 x Consider the surface in R3 defined by f(x,y,z) = b (which is a level surface of the function w = f(x,y,z)). If c(t) = (x(t) , y(t) , z(t)) describes a path on the surface f(x,y,z) = b, then f(x(t) , y(t) , z(t)) = f(c(t)) =   d — f(x(t) , y(t) , z(t)) = dt b Under what conditions will it be possible to describe a curve/surface in terms of one variable as a function of the other variables? f(c(t)) • c  (t) = 0

Suppose F(x,y) = 0 describes a relation, and we are interested in a part of the graph where y = f(x). Then, we must have F(x, f(x)) = 0  F(c(x)) = 0 where c(x) = (x, f(x))  dy — dx d —F(x, f(x)) = F(c(x)) • c  (x) = dx F F — , — x y • 1 , = dy Solving for — , we find dx F F dy — + — — = 0. x y dx dy F / x f  (x) = — = – ——— . dx F / y We have y = f(x) provided that F / dy = Fy  0. Suppose F(x,y,z) = 0 describes a relation, and we are interested in a part of the graph where z = f(x,y). Then, similar to the previous derivation, we find z F / x z F / y fx(x,y) = — = – ——— and fy(x,y) = — = – ——— . x F / z y F / z We have z = f(x,y) provided that F / z = Fz  0.

The Implicit Function Theorem for defining one variable as a function of n other variables (Theorem 11 on page 247) can be stated as follows: Suppose F(x1 , x2 , …, xn , z) = F(x , z) : Rn+1  R has continuous partial derivatives. For any point (x0 , z0) where F(x0 , z0) = 0 and Fz(x0 , z0)  0, we can find a “ball” around (x0 , z0) in Rn where F(x0 , z0) = 0 is defined by a function z = f(x), that is, F(x , f(x)) = 0; furthermore, z F / xi f (x) = — = – ——— for each i = 1, 2, …, n . xi F / z xi Consider the relation x2 + y2 = 1 in the xy plane. (a) (b) (c) Find all points on the graph where can we say y = f(x) in some surrounding neighborhood, and find a formula for dy/dx = f  (x). Find formula for y = f(x) in a neighborhood surrounding (4/5 , –3/5). Find the equation of the line tangent to y = f(x) (i.e., tangent to x2 + y2 = 1) at (4/5 , –3/5).

Consider the relation x2 + y2 = 1 in the xy plane. (b) (c) Find all points on the graph where can we say y = f(x) in some surrounding neighborhood, and find a formula for dy/dx = f  (x). Find formula for y = f(x) in a neighborhood surrounding (4/5 , –3/5). Find the equation of the line tangent to y = f(x) (i.e., tangent to x2 + y2 = 1) at (4/5 , –3/5). (0,1) By inspecting the graph, we see that the points where we can say y = f(x) in some surrounding neighborhood are (–1,0) (1,0) all points except where x = –1 or x = 1. To apply the Implicit Function Theorem, we write the equation of the graph as x2 + y2 – 1 = 0 to see that F(x,y) = x2 + y2 – 1. The points where we can say y = f(x) in some surrounding neighborhood are those where (0,–1) Fy(x,y)  0  2y  0  x  –1 and x  1.

Consider the relation x2 + y2 = 1 in the xy plane. (b) (c) Find all points on the graph where can we say y = f(x) in some surrounding neighborhood, and find a formula for dy/dx = f  (x). Find formula for y = f(x) in a neighborhood surrounding (4/5 , –3/5). Find the equation of the line tangent to y = f(x) (i.e., tangent to x2 + y2 = 1) at (4/5 , –3/5). (0,1) Applying the Implicit Function Theorem with F(x,y) = x2 + y2 – 1, we have dy F / x — = – ———— = dx F / y (–1,0) (1,0) 2x – — = 2y x – — . y Observe that we may find a formula for dy/dx from x2 + y2 = 1 by using implicit differentiation with respect to x as follows: (0,–1) dy 2x + 2y — = 0  dx dy x — = – — . dx y

Consider the relation x2 + y2 = 1 in the xy plane. (b) (c) Find all points on the graph where can we say y = f(x) in some surrounding neighborhood, and find a formula for dy/dx = f  (x). Find formula for y = f(x) in a neighborhood surrounding (4/5 , –3/5). Find the equation of the line tangent to y = f(x) (i.e., tangent to x2 + y2 = 1) at (4/5 , –3/5). (0,1) The formula for y = f(x) in a neighborhood surrounding (4/5 , –3/5) is y = – 1 – x2 . (–1,0) (1,0) One way to find the slope of the line tangent to y = – 1 – x2 at (4/5 , –3/5) is use the derivative directly: (4/5 , –3/5) (0,–1) dy x 4/5 — = ———— , and the slope is —————— = dx  1 – x2  1 – (4/5)2 4 — . 3

Consider the relation x2 + y2 = 1 in the xy plane. (b) (c) Find all points on the graph where can we say y = f(x) in some surrounding neighborhood, and find a formula for dy/dx = f  (x). Find formula for y = f(x) in a neighborhood surrounding (4/5 , –3/5). Find the equation of the line tangent to y = f(x) (i.e., tangent to x2 + y2 = 1) at (4/5 , –3/5). (0,1) The formula for y = f(x) in a neighborhood surrounding (4/5 , –3/5) is y = – 1 – x2 . (–1,0) (1,0) From the Implicit Function Theorem, we found that dy x — = – — , dx y (4/5 , –3/5) from which we find that the slope of the line tangent to y = f(x) at (4/5 , –3/5) is (0,–1) 4/5 4 – —— = — . – 3/5 3

Consider the relation x2 + y2 = 1 in the xy plane. (b) (c) Find all points on the graph where can we say y = f(x) in some surrounding neighborhood, and find a formula for dy/dx = f  (x). Find formula for y = f(x) in a neighborhood surrounding (4/5 , –3/5). Find the equation of the line tangent to y = f(x) (i.e., tangent to x2 + y2 = 1) at (4/5 , –3/5). (0,1) The equation of the line tangent to y = f(x) (i.e., tangent to x2 + y2 = 1) at (4/5 , –3/5) is given by (–1,0) (1,0) F(4/5 , –3/5) • (x – 4/5 , y + 3/5) = 0 with F(x,y) = x2 + y2 – 1, F(x,y) = (2x , 2y) (4/5 , –3/5) F(4/5 , –3/5) = (8/5 , –6/5) (0,–1) The equation of the line tangent is (Note that the equation of the line could be obtained by using the point-slope method.) (8/5 , –6/5) • (x – 4/5 , y + 3/5) = 0  4x – 3y = 5

Consider the relation x3 + xy2 + 8x sin y = 0 in the xy plane. (b) Use the Implicit Function Theorem to find all points on the graph where can we say y = f(x) in some surrounding neighborhood, and find a formula for dy/dx = f  (x), Using the Implicit Function Theorem with F(x,y) = x3 + xy2 + 8x sin y , we have dy F / x — = – ———— = dx F / y 3x2 + y2 + 8 sin y – ———————— . 2xy + 8x cos y A differentiable function y = f(x) can be defined whenever 2xy + 8x cos y  0 , that is whenever x  0 and y + 4 cos y  0. Use implicit differentiation to find a formula for dy/dx = f  (x), dy dy 3x2 + (y2 + 2xy — ) + (8 sin y + 8x cos y — ) = 0 , dx dx from which we obtain dy 3x2 + y2 + 8 sin y — = – ———————— . dx 2xy + 8x cos y

Consider the relation x3 + 3y2 + 8xz2 – 3z3y = 1 in R3 Consider the relation x3 + 3y2 + 8xz2 – 3z3y = 1 in R3. Use the Implicit function Theorem to (a) (b) find points near which the graph can be represented as a differentiable function z = f(x,y), To apply the Implicit Function Theorem, we write the equation of the graph as x3 + 3y2 + 8xz2 – 3z3y – 1 = 0 in R3. The points near which the graph can be represented as a differentiable function z = f(x,y), are all points where F(x,y,z) Fz(x,y,z)  0  16xz – 9z2y  0  z(16x – 9zy)  0  z  0 and 16x – 9zy  0 find a formula for each of fx and fy when they exist. z F / x — = – ——— = x F / z z F / y — = – ——— = y F / z 3x2 + 8z2 – ————— , 16xz – 9z2y 6y – 3z3 – ————— . 16xz – 9z2y

Consider the relation x2z + xy2z = 5 in R3 Consider the relation x2z + xy2z = 5 in R3. Observe that it is easy to write z = f(x,y), but it does not appear as easy to write x = f(y,z). (a) (b) Specify all points where we can write x = f(y,z), and confirm that (1,2,1) is one of these points. F(x,y,z) = x2z + xy2z – 5 Fx(x,y,z)  0  2xz + y2z  0  z  0 and y2  –2x; we see that (1,2,1) is one of these points. Find the equation of the plane tangent to x = f(y,z) at (1,2,1). F(1,2,1) • (x – 1 , y – 2 , z – 1) = 0 F(x,y,z) = ( 2xz + y2z , , ) 2xyz x2 + xy2 (6,4,5) • (x – 1, y – 2, z – 1) = 0 6x + 4y + 5z = 19 That is, the differentiable function x = f(y,z) near the point (1,2,1) is approximated by 19 – 4y – 5z x = ————— 6

(c) Find a formula for each of fy and fz when they exist. x F / y — = – ——— = y F / x x F / z — = – ——— = z F / x 2xy – ————— , 2x + y2 x2 + xy2 – ————— . 2xz + y2z

The General Implicit Function Theorem for defining each of m variables z1 , z2 , … , zm as a function of n other variables x1 , x2 , … , xn (Theorem 12 on page 251) can be stated as follows: Suppose each of F1 , F2 , … , Fm is a function from Rn+m  R with continuous partial derivatives. Then the equations F1(x1 , x2 , …, xn , z1 , z2 , … , zm) = F(x , z) = 0 F2(x1 , x2 , …, xn , z1 , z2 , … , zm) = F(x , z) = 0 . Fm(x1 , x2 , …, xn , z1 , z2 , … , zm) = F(x , z) = 0 will have a unique solution z1 = f1(x1 , x2 , …, xn) , z2 = f2(x1 , x2 , …, xn) , … , zm = fm(x1 , x2 , …, xn) in a “ball” around any point (x0 , z0) at which F1 / z1 … F1 / zm Fm / z1 … Fm / zm . . det  0

Consider the relation defined by the equations xu + yvu2 = 2 xu3 + y2v4 = 2 (a) (b) Specify all points near which we can write u = f1(x,y) and v = f2(x,y), and confirm that (x,y,u,v) = (1,1,1,1) is one of these points. F1(x,y,u,v) = F2(x,y,u,v) = xu + yvu2 – 2 xu3 + y2v4 – 2 F1 / u F1 / v F2 / u F2 / v x + 2yvu yu2 3xu2 4y2v3 det = det = (x + 2yvu)(4y2v3) – (3xu2)(yu2) = y(4xyv3 + 8y2uv4 – 3xu4) We can write u = f1(x,y) and v = f2(x,y) near points where y(4xyv3 + 8y2uv4 – 3xu4)  0, and (1,1,1,1) is one of these points. Compute u/y at the point (x,y) = (1,1). To compute u/y at (x,y) = (1,1), we use implicit differentiation: u v u x — + vu2 + yu2 — + 2yvu — = 0 y y y u v 3xu2 — + 2yv4 + 4y2v3 — = 0 y y

Consider the relation defined by the equations xu + yvu2 = 2 xu3 + y2v4 = 2 (b) Compute u/y at the point (x,y) = (1,1). To compute u/y at (x,y) = (1,1), we use implicit differentiation: u v u x — + vu2 + yu2 — + 2yvu — = 0 y y y u v 3xu2 — + 2yv4 + 4y2v3 — = 0 y y Setting (x,y,u,v) = (1,1,1,1), we have u v u (1) — + 1 + (1) — + 2 — = 0 y y y u v (3) — + 2 + (4) — = 0 y y u v (3) — + — = –1 y y u v (3) — + (4) — = –2 y y u — = and y v — = . y 2 – — 9 1 – — 3 Solving this system of equations, we have

Consider the special situation where y1 = f1(x1 , x2 , …, xn) f1(x1 , x2 , …, xn) – y1 = 0 y2 = f2(x1 , x2 , …, xn) f2(x1 , x2 , …, xn) – y2 = 0 .  . yn = fn(x1 , x2 , …, xn) fn(x1 , x2 , …, xn) – yn = 0 We can apply the General Implicit Function Theorem in order to identify all points near which we can have x1 = g1(y1 , y2 , …, yn) , x2 = g2(y1 , y2 , …, yn) , … , xn = gn(y1 , y2 , …, yn); these will be all points where f1 / x1 … f1 / xn fn / x1 … fn / xn . . det  0 This is called the Jacobian determinant.

Consider the equations Near what points can we have x = g1(u,v) and y = g2(u,v)? y3 u = ——— x6 + y4 v = ey – ln(x2 + y2 + 1) – 6x5y3 ———— (x6 + y4)2 3x6y2 – y6 ———— (x6 + y4)2 These will be all points where det  0 – 2x ———— x2 + y2 + 1 2y ey – ———— x2 + y2 + 1