1. 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

2. 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.

3. 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).

4. 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.

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)
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

6. 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 – x  1 – (4/5)2 4
— . 3

7. 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/
– —— = — .
– 3/

8. 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

9. 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 x2 + y2 + 8 sin y
— = – ———————— .
dx xy + 8x cos y

10. 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

11. 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

12. (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

13. 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

14. 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

15. 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) — + 2 — = 0
y y y
u v
(3) — (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

16. 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.

17. 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