Calculus II

Vladimir V. Kisil

Chapter 1
General Information

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Course Outline

Chapter 8
Infinite Series

8.5   A brief review of series

We refer to the chapter Infinite Series of the course Calculus I for the review of the following topics.

1. Sequences of numbers

2. Convergent and Divergent Series

3. Positive Term Series

4. Ratio and Root Test

5. Alternating Series and Absolute Convergence

8.6  Power Series

It is well known that polynomials are simplest functions, particularly it is easy to differentiate and integrate polynomials. It is desirable to use them for investigation of other functions. Infinite series reviewed in the previous sections are very important because they allow to represent functions by means of power series, which are similar to polynomials in many respects. An example of such representations is harmonic series

∞ ∑ n=0  rn =  1 1−r .

Definition 1 Let x be a variable. A power series in x is a series of the form

∞ ∑ n=0  bn xn = b0+b1x+b2x2+…+bnxn+…,

where each b_k is real number.

A power series turns to be infinite (constant term) series if we will substitute a constant c instead of the variable x. Such series could converge or diverge. All power series converge for x=0. The convergence of power series described by the following theorem.

Theorem 2

1. If a power series ∑b_nxⁿ converges for a nonzero number c, then it is absolutely convergent whenever | x | < | c |.

2. If a power series ∑b_n xⁿ diverges for a nonzero number d, then it diverges whenever | x | > | d |.

PROOF. The proof follows from the Basic Comparison Test of the power series for | x | and convergent geometric series with r=| [ x/c] |. ^[¯]

From this theorem we could conclude that

Theorem 3 If ∑b_n xⁿ is a power series, then exactly one of the following true:

1. The series converges only if x=0.

2. The series is absolutely convergent for every x.

3. There is a number r such that the series is absolutely convergent if x is in open interval (−r,r) and divergent if x < −r or x > r.

The number r from the above theorem is called radius of convergence. The totality of numbers for which a power series converges is called its interval of convergence. The interval of convergence may be any of the following four types: [−r,r], [−r,r), (−r,r], (−r,r).

There is a more general type of power series

Definition 4 Let b be a real number and x is a variable. A power series in x−d is a series of the form

∞ ∑ n=0  bn (x− d)n = b0 + b1 (x−d) +b2 (x−d)2 + …+ bn (x−d)n +…,

where each b_n is a real number.

This power series is obtained from the series in Definition 9.6.1 by replacement of x by x−d. We could obtain a description of convergence of this series by replacement of x by x−d in Theorem 9.6.3.

The following exercises should be solved in the following way:

1. Determine the radius r of convergence, usually using Ratio test or Root Test .

2. If the radius r is finite and nonzero determine if the series is convergent at points x=−r, x=r. Note that the series could be alternating at one of them and apply Alternating Test .

Exercise 5 Find the interval of convergence of the power series:

∑  1 n2+4 xn; ∑  1 ln(n+1) xn; ∑  10n+1 32n xn; ∑  (3n)! (2n)! xn; ∑  10n n! xn; ∑  1 2n+1 (x+3)n; ∑  n 32n−1 (x−1)2n; ∑  1 √ 3n+4 (3x+4)n;

8.7  Power Series Representations of Functions

Example 1 Find function represented by ∑(−1)^k x^k.

The following theorem shows that integration and differentiations could be done with power series as easy as with polynomials:

Theorem 2 Suppose that a power series ∑b_n xⁿ has a radius of convergence r > 0, and let f be defined by

f(x) = ∞ ∑ n=0  bn xn = b0+b1x+b2x2+…+bn xn+…

for every x ∈ (−r,r). Then for −r < x < r

f′(x) = b1+b2x+b3x2+…+nbn xn−1+… (8.1) = ∞ ∑ n=1  n bn xn−1; ⌠ ⌡ x 0  f(t) dt = b0 x +b1  x2 2 +b2  x3 3 +…+bn  xn+1 n+1 +… (8.2) = ∞ ∑ n=0   bn n+1 xn+1.

Example 3 Find power representation for

1. [ 1/((1+x)²)].

2. ln(1+x) and calculate ln(1.1) to five decimal places.

3. arctanx.

Theorem 4 If x is any real number,

ex=1+  x 1 +  x2 2! +  x3 3! +… = ∞ ∑ n=0   xn n! .

PROOF. The proof follows from observation that the power series f(x)=∑ [(xⁿ)/n!] satisfies to the equation f′(x)=f(x) and the only solution to this equation with initial condition f(0)=1 is f(x)=e^x. ^[¯]

Corollary 5

e = 1+  1 1! +  1 2! +  1 3! +….

Example 6 Find a power series representation for sinhx, xe^−2x.

Exercise 7 Find a power series representation for f(x), f′(x), ∫₀^x f(t) dt.

f(x)=  1 1+5x ;        f(x)=  1 3−2x .

Exercise 8 Find a power series representation and specify the radius of convergence for:

x 1−x4 ;         x2−3 x−2 .

Exercise 9 Find a power series representation for

f(x)=x2 e(x2);        f(x)=x4 arctan(x4).

8.8  Maclaurin and Taylor Series

Theorem 1 If a function f has a power series representation

f(x)= ∞ ∑ k=0  bn xn

with radius of convergence r > 0, then f^(k)(0) exists for every positive integer k and

f(x) = f(0)+  f′(0) 1! x+  f′′(0) 2! x2+ …+  f(n)(0) n! xn + … = ∞ ∑ n=0   f(n)(0) n! xn

Theorem 2 If a function f has a power series representation

f(x)= ∞ ∑ k=0  bn (x−d)n

with radius of convergence r > 0, then f^(k)(d) exists for every positive integer k and

f(x) = f(d)+  f′(d) 1! (x−d)+  f′′(d) 2! (x−d)2+ …+  f(n)(d) n! (x−d)n + … = ∞ ∑ n=0   f(n)(d) n! (x−d)n

Exercise 3 Find Maclaurin series for:

f(x)=sin2x;        f(x) =  1 1−2x .

Remark 4 It is easy to see that linear approximation formula is just the Taylor polynomial P_n(x) for n=1.

The last formula could be split to two parts: the nth-degree Taylor polynomial P_n(x) of f at d:

Pn(x) = f(d)+  f′(d) 1! (x−d)+  f′′(d) 2! (x−d)2+ …+  f(n)(d) n! (x−d)n

and the Taylor remainder

Rn(x) =  f(n+1)(z) (n+1)! (x−d)n+1,

where z ∈ (d,x). Then we could formulate a sufficient condition for the existence of power series representation of f.

Theorem 5 Let f have derivatives of all orders throughout an interval containing d, and let R_n(x) be the Taylor remainder of f at d. If

lim n→ ∞  Rn(x) = 0

for every x in the interval, then f(x) is represented by the Taylor series for f(x) at d.

Example 6 Let f be the function defined by

f(x)=    e−1/x2 if x ≠ 0; 0 if x = 0,

then f cannot be represented by a Maclaurin series.

Exercise 7 Show that for function f(x)=e^−x

lim n→ ∞  Rn (x)=0

and find the Maclaurin series.

The important Maclaurin series are:

Function	Maclaurin series	Convergence
ex	∑n=0∞ [(xn)/n!]	(−∞,∞)
ln(1+x)	∑n=0∞ [((−1)n xn+1)/(n+1)]	(−1,1]
sinx	∑n=0∞ [((−1)n x2n+1)/((2n+1)!)]	(−∞,∞)
cosx	∑n=0∞ [((−1)n x2n)/(2n)!]	(−∞,∞)
sinhx	∑n=0∞ [(x2n+1)/((2n+1)!)]	(−∞,∞)
coshx	∑n=0∞ [(x2n)/(2n)!]	(−∞,∞)
arctanx	∑n=0∞ [((−1)n x2n+1)/(2n+1)]	[−1,1]

Exercise 8 Find Maclaurin series for sin² x.

Exercise 9 Find a series representation of lnx in powers of x−1.

Exercise 10 Find first three terms of the Taylor series for f at d:

f(x)=arctanx,     d=1;        f(x)=cscx,    d=π/3.

8.9  Applications of Taylor Polynomials

Exercise 1 Find the Maclaurin polynomials P₁(x), P₂(x), P₃(x) for f(x), sketch their graphs. Approximate f(a) to four decimal places by means of P₃(x) and estimate R₃(x) to estimate the error.

f(x)=ln(x+1)       a=0.9.

Exercise 2 Find the Taylor formula with remainder for the given f(x), d and n.

f(x)=e−1; d=1,     n=3. f(x) = 3 √   x   ; d=−8,     n=3.

Chapter 10
Vectors and Surfaces

10.2  Vectors in Three Dimensions

Similarly to two dimensional case we have the following formulas

Theorem 1

1. The distance between P₁(x₁,y₁,z₁) and P₂(x₂,y₂,z₂) is
   

   d(P1,P2) = √   (x1−x2)2+(y1−y2)2+(z1−z2)2   .

2. The midpoint of the line segment P₁(x₁,y₁,z₁) to P₂(x₂,y₂,z₂) is
   

      x1+x2 2 ,  y1+y2 2 ,  z1+z2 2  

3. An equation of a sphere of radius r and center P₀(x₀,y₀,z₀) is
   

   (x−x0)2+(y−y0)2+(z−z0)2=r2.

We define vector a=(a₁, a₂, a₃) in the three dimensional case as a transformation which maps point (x,y,z) to (x+a₁,y+a₂, z+z₃). Vectors could be added and multiplied by a scalar according to the rules:

a+b = (a1+b1, a2+b2, a3+b3); ca = (ca1, ca2, ca3);

There is a special null vector 0=(0, 0, 0) and inverse vector −a=(−a₁, −a₂, −a₃) for any vector a.

We have the following properties:

1. a+b=b+a.

2. a + (b + c) = (a + b) + c.

3. a +0 = a.

4. a + −a = 0.

5. c(a + b) = c a + a b.

6. (c+d)a = c a + d a.

7. (cd)a = c(da) = d(ca).

8. 1a = a.

9. 0a = 0 = c0.

We define subtraction of vectors (or difference of vectors ) by the rule:

a−b=a+(−b).

Definition 2 Nonzero vectors a and b have

1. the same direction if b=ca for some scalar c > 0.

2. the opposite direction if b=ca for some scalar c < 0.

Definition 3 We define vectors:

i=(1, 0, 0),        j=(0, 1, 0),        k=(0, 0, 1).

It is obvious that

a=(a1, a2, a3)=a1i + a2j + a3k.

The magnitude of vector is defined to be

||a||= √   a12+a22+a32   .

10.3  Dot Product

Besides addition of vectors and multiplication by the scalar there two different operation which allows to multiply vectors.

Definition 1 The dot product (or scalar product, or inner product) a·b is

a·b=a1b1+a2b2+a3b3.

Theorem 2 Properties of the dot product are:

1. a·a=||a||².

2. a·b = b·a.

3. a ·(b+c)=a·b+a·c.

4. (ma)·b=ma·b = a·(mb).

5. 0·a=0.

Definition 3 Let a and b be nonzero vectors.

1. If b ≠ ca then angle θ between a and b is the angle of triangle defined by them.

2. If b=ca then θ = 0 if c > 0 and θ = π if c < 0.

Vectors are orthogonal or perpendicular if θ = π/2. By a convention 0 is orthogonal and parallel to any vector.

Theorem 4 For nonzero a and b:

a·b=||a||||b||cosθ.

Corollary 5 For nonzero a and b:

cosθ =  a·b ||a||||b|| .

Corollary 6 Two vectors a and b are orthogonal if and only if ab=0.

Corollary 7 [Cauchy-Schwartz-Bunyakovskii Inequality]

| a·b | ≤ ||a||||b||

Theorem 8 [Triangle Inequality]

||a+b|| ≤ ||a||+||b||.

We define component of a along b

compb a=a·  1 ||b|| b

Definition 9 The work done by a constant force a as its point of application moves along the vector b is a·b.

10.4  Vector Product

Definition 1 A determinant of order 2 is defined by

   a1 a2 b1 b2    = a1 b2 −a2b1.

A determinant of order 3 is defined by

     c1 c2 c3 a1 a2 a3 b1 b2 b3      =    a2 a3 b2 b3    c1 −    a1 a3 b1 b3    c2 +    a1 a2 b1 b2    c1.

Definition 2 The vector product (or cross product) a×b is

a×b =      i j k a1 a2 a3 b1 b2 b3      =    a2 a3 b2 b3    i −    a1 a3 b1 b3    j +    a1 a2 b1 b2    k.

Theorem 3 The vector a×b is orthogonal to both a and b.

Theorem 4 If θ is the angle between nonzero vectors a and b, then

||a×b||=||a||||b||sinθ.

Corollary 5 Two vectors a and b are parallel if and only if a×b=→0.

Exercise 6 Compile the multiplication table for vectors i, j, k.

Be careful, because:

i×j ≠ j×i (i×j) ×j ≠ i×(j×j).

Theorem 7 Properties of the vector product are

1. a×b=−(b×a).

2. (ma)×b=m(a×b) = a×(mb).

3. a×(b+c) = a×b + a×c.

4. (a+b)×c = a×c + b×c.

5. (a×b)·c = a·(b×c).

6. a×(b×c) = (a·c)b − (a·b)c.

Dot and vector products related to geometric properties.

Exercise 8 Prove that the distance from a point R to a line l is given by

d= || → PQ   × → PR   || || → PQ   || .

Exercise 9 Prove that the volume of the oblique box spanned by three vectors a, b, c is | (a×b)· c |.

10.5  Lines and Planes

Theorem 1 Parametric equation for the line through P₁(x₁,y₁,z₁) parallel to a=(a₁, a₂, a₃) are

x=x1+a1t,    y=y1+a2t,    z=z1+a3t;       t ∈ R.

Note that we obtain the same line if we use any vector b=ca, c ≠ 0.

Corollary 2 Parametric equation for the line through P₁(x₁,y₁,z₁) and P₂(x₂,y₂,z₂) are

x=x1+(x2−x1)t,    y=y1+(y2−y1)t,    z=z1+(z2−z1)t;        t ∈ R.

Exercise 3 Find equations of the lines:

1. P(1,2,3); a=i+2j+3k.

2. P₁(2,−2,4), P₂(2,−2,−3).

Exercise 4 Determine whether the lines intersect: x=2−5t, y=6+2t, z=−3−2t; x=4−3v, y=7+5v, z=1+4v.

Definition 5 Let θ be the angle between nonzero vectors a and b and let l₁ and l₂ be lines that are parallel to the position vectors of a and b.

1. The angles between lines l₁ and l₂ are θ and π−theta.

2. The lines l₁ and l₂ are parallel iff b=ca for c ∈ R.

3. The lines l₁ and l₂ are orthogonal iff a·b=0 for c ∈ R.

The plane through P₁ with normal vector →P₁P₂ is the set of all points P such that →P₁P is orthogonal to →P₁P₂.

Theorem 6 An equation of the plane through P₁(x₁,y₁,z₁) with normal vector a=(a₁, a₂, a₃) is

a1(x−x1)+a2(y−y1)+a3(z−z1)=0.

Theorem 7 The graph of every linear equation ax+by+cz+d=0 is a plane with normal vector (a, b, c).

Exercise 8 Find an equation of the plane through P(4,2,−6) and normal vector →OP.

Exercise 9 Sketch the graph of the equation

1. y=−2;

2. 3x−2z−24=0;

Definition 10 Two planes with normal vectors a and b are

1. parallel if a and b are parallel;

2. orthogonal if a and b are orthogonal;

Exercise 11 Find an equation of the plane through P(3,−2,4) parallel to −2x+3y−z+5=0.

Theorem 12 [Symmetric Form for a Line]

x−x1 a1 =  y−y1 a2 =  z−z1 a3 .

Exercise 13 Show that distance from a point P₀(x₀,y₀,z₀) to the plane ax+by+cz+d=0 is

h=   compn → P0P1     ,

where n=(a, b, c) and P₁-any point on the plane.

Exercise 14 Show that planes 3x+12y−6z=−2 and 5x+20y−10z=7 are parallel and find distance between them.

Exercise 15 Find an equation of the plane that contains the point P(4,−3,0) and line x=t+5, y=2t−1, z=−t+7.

Exercise 16 Show that distance between two lines defined by points P₁, Q₁ and P₂, Q₂ is given by the formula

d=   compn → P1P2     ,        n= → P1Q1   × → P2Q2   || → P1Q1   × → P2Q2   || .

Exercise 17 Find the distance between point P(3,1,−1) and line x=1+4t, y=3−t, z=3t.

10.6  Surfaces

There are several classic important types of surfaces. To follows given examples you need to remember equations of conics in Cartesian coordinates.

Example 1 z=x²+y² define circular paraboloid or paraboloid of revolution.

Definition 2 Let C be a curve in a plane, and let l be a line that is not in a parallel plane. The set of points on all lines that are parallel to l and intersect C is a cylinder. The curve C called is called directrix of the cylinder .

Example 3 The right circular cylinder is given by the equation x²+y²=r².

Similarly to quadratic equations equations defining conics the equation

Ax2+By2+cz2+Dxy+Exz+Fyz+Gx+Hy+Iz+J=0

defines quadric surface. We consider simplest cases with D=E=F=G=H=I=0.

Definition 4 Ellipsoid :

x2 a2 +  y2 b2 +  z2 c2 = 1.

Definition 5 The hyperboloid of one sheet:

x2 a2 +  y2 b2 −  z2 c2 = 1.

Definition 6 The hyperboloid of two sheets:

−  x2 a2 −  y2 b2 +  z2 c2 = 1.

Definition 7 The cone:

x2 a2 +  y2 b2 −  z2 c2 = 0.

Definition 8 The paraboloid:

x2 a2 +  y2 b2 = cz .

Definition 9 The hyperbolic paraboloid:

y2 b2 −  x2 a2 = cz.

Chapter 11
Vector-Valued Functions

Definition 10 Let D be a set of real numbers. A vector-valued function r with domain D is a correspondence that assigns to each number t in D exactly one vector r(t) in R³.

Theorem 11 If D is a set of real numbers, then r is a vector-valued function with domain D if and only if there are scalar function f, g, and h such that

r(t)=f(t) i+g(t) j+h(t) k.

Exercise 12 Sketch the two vectors

r(t)=t i+3sint j + 3cost k,     r(0), r(π/2).

Set of endpoints of all vectors →OP=r(t) define a space curve C. A parameter equation of the curve C is

x=f(t),     y=g(t),     z=z(t).

The orientation of C is the direction determined by increasing values of t.

Exercise 13 Sketch the curve and indicate orientation:

r(t)=t3 i+t2 j+3 k;    0 ≤ t ≤ 4.

The following theorem is completely analogous to arc length of a plane curve :

Theorem 14 If a curve C has a smooth parameterization

x=f(t),     y=g(t),     z=z(t),     a ≤ t ≤ b

and if C does not intersect itself, except possibly for t=a and t=b, then the length L of C is

L= ⌠ ⌡ b a  √   [f′(t)]2+[g′(t)]2+[h′(t)]2    dt.

Exercise 15 Find the arc length:

x=et cost , y = et, z=etsint;     0 ≤ t ≤ 2π; x=2t, y = 4 sin3t , z = 4cos3t;     0 ≤ t ≤ 2π;

11.1  Limits, Derivatives and Integrals of Vector-valued Functions

All definitions and results in this section are in close relation with the theory of scalar-valued function Calculus I. We advise to refresh Chapters on Limits and Derivative from Calculus I course.

Definition 1 Let r(t)=f(t)i+g(t)j+h(t)k. The limit r(t) as t approaches a is

lim t→ a  r(t) =   lim t→ a  f(t)   i +   lim t→ a  g(t)   j +   lim t→ a  h(t)   k.

provides f, g, and h have limits as t approaches a.

The next definition coincides with definition of continuity for scalar-valued function :

Definition 2 A vector valued function r is continuous at a if

lim t→ a  r(t) = r(a).

Particularly r(t) is continuous iff f(t), g(t), and h(t) are continuous. Similarly we define derivative

Definition 3 Let r be a vector-valued function. The derivative is the vector-valued function r′ defined by

r′(t) = lim ∆t → 0   1 ∆t [r(t + ∆t) −r(t)]

for every t such that the limit exists.

Exercise 4 Find the domain, first and second derivatives of the functions:

r(t) = 3 √   t    i +  1 t  j + e−t k; r(t) = ln(1−t) i + sint j + t2 k.

Theorem 5 Let r(t)=f(t)i+g(t)j+h(t)k and f, g, and h are differentiable, then

r′(t)=f′(t) i+g′(t) j+h′(t) k.

The geometric meaning is as expected-this is tangent vector to the curve defined by r.

Exercise 6 Find parameter equation for the tangent line to C at P:

x=et,     y=tet,     z=t2+4;       P(1,0,4).

The properties of the derivative are as follows:

Theorem 7 If u and v are differentiable vector-valued functions and c is a scalar, then

1. [u(t) + v(t)]′ = u′(t) + v′(t);

2. [cu(t)]′ = cu′(t);

3. [u(t) ·v(t)]′ = u′(t) · v(t) + u(t) ·v′(t);

4. [u(t) ×v(t)]′ = u′(t) × v(t) + u(t) ×v′(t);

As a consequence of these properties we could easily prove the following

Theorem 8 If r is differentiable and ||r|| is constant, then r′ is orthogonal to r′(t) for every t in the domain of r′.

Finally we define integrals of vector-valued functions using integrals of scalar-valued functions:

Definition 9 Let r(t)=f(t)i+g(t)j+h(t)k and f, g, and h are integrable, then

⌠ ⌡ b a  r(t)  dt =   ⌠ ⌡ b a  f(t) dt   i +   ⌠ ⌡ b a  g(t) dt   j +   ⌠ ⌡ b a  h(t) dt   k.

If R′(t)=r(t), then R(t) is an antiderivative of r(t).

Theorem 10 If R(t) is an antiderivative of r(t) on [a,b], then

⌠ ⌡ b a  r(t) dt = R(t)]ab = R(b)−R(a).

Exercise 11 Find r(t) subject to the given conditions:

r′(t) = 2i−4t3 j+6√tk,     r(0)=i+5j+3k.

Chapter 12
Partial Differentiation

12.1  Functions of Several Variables

Definition 1 Let D be a set of ordered pairs of real numbers. A function of two variables f is a correspondence that assigns to each pair (x,y) in D exactly one real number, denoted by f(x,y). The set D is the domain of f. The range of f consists of all real numbers f(x,y), where (x,y) ∈ D.

Exercise 2 Describe domain of f and find its values:

f(r,s) = √   1−r   −er/s;     f(1,1), f(0,4), f(−3,3) f(x,y,z) = 2 +tanx + ytanz;     f(π/4,4,π/6), f(0,0,0).

Exercise 3 Sketch graph of f:

f(x,y)= √   2−2x−x2−y2   ,        f(x,y)=3−x−3y.

Exercise 4 Sketch the level curves for f:

f(x,y)=xy,        k=−4, 1, 4.

Exercise 5

1. Find the equation of level surface of f that contains the point P.
   

   f(x,y,z) = z2y+x;     P(1,4,−2).

2. Describe the level surface of f for given k:
   

   f(x,y,z) = z+ x2+4y2,        k=−6,6,12.

12.2  Limits and Continuity

Definition 1 Let a function f of two variables be defined throughout the interior of a circle with center (a,b), except possibly at (a,b) itself. The statement

lim (x,y) → (a,b)  f(x,y) = L     or     f(x,y)→ L as (x,y)→ (a,b)

means that for every ε > 0 there is a δ > 0 such that if

0 < √   (x−a)2+(y−b)2   < δ, then | f(x,y)−L | < ε.

Exercise 2 Find limits

lim (x,y) → (2,1)   4+x 2−y ,        lim (x,y) → (−1,3)   y2+x (x−1)(y+2) .

Theorem 3 [Two-Path Rule] If two different paths to a point P(a,b) produce two different limiting values for f, then lim_{(x,y)→(a,b)} f(x,y) does not exist.

Exercise 4 Show that the limit does not exist

lim (x,y) → (0,0)   x2−2xy+5y2 3x2+4y2 ,        lim (x,y) → (0,0)   3xy 5x4+2y4 .

Definition 5 A function f of two variables is continuous at an interior point (a,b) of its domain if

lim (x,y)→(a,b)  f(x,y) = f(a,b).

Exercise 6 Describe the set of all points at which f is continuous

f(x,y)=  xy x2−y2 ,        f(x,y)= √   xy   tanz.

Definition 7 Let a function f of two variables be defined throughout the interior of a circle with center (a,b,c), except possibly at (a,b,c) itself. The statement

lim (x,y,z) → (a,b,c)  f(x,y,z) = L     or     f(x,y,z)→ L as (x,y,z)→ (a,b,c)

means that for every ε > 0 there is a δ > 0 such that if

0 < √   (x−a)2+(y−b)2 + (z−c)2   < δ, then | f(x,y,z)−L | < ε.

Theorem 8 [Composition of Continuous Functions] If a function f of two variables is continuous at (a,b) and a function g of one variables is continuous at f(a,b), then the function h(x,y)=g(f(x,y)) is continuous at (a,b).

Exercise 9 Use Theorem on Composition of Continuous Functions to determine where h is continuous.

f(x,y)=3x+2y−4,        g(t)=ln(t+5).

12.3  Partial Derivatives

Definition 1 Let f be a function of two variables. The first partial derivatives of f with respect to x and y are functions f′_x and f′_y such that

∂ ∂x f (x,y) = f′x(x,y) = lim h→ 0   f(x+h,y)−f(x,y) h ,  ∂ ∂x f (x,y) = f′x(x,y) = lim h→ 0   f(x,y+h)−f(x,y) h .

Exercise 2 Find first partial derivatives of f

f(x,y)=(x3−y2)5; f(x,y) = exlnxy; f(r,s,v,p)=r3tans + √se(v2)−vcos2p; f(x,y,z)=xyz exyz.

This notion has a geometrical meaning which is very close to geometrical meaning of usual derivative derivative .

Theorem 3 Let S be the graph of z=f(x,y), and let P(a,b,f(a,b)) be a point on S at which f′_x and f′_y exists. Let C₁ and C₂ be the traces of S on the planes x=a and y=b, respectively, and let l₁ and l₂ be the tangent lines to C₁ and C₂ at P.

1. The slope of l₁ in the plane x=a is f′_y(a,b).

2. The slope of l₁ in the plane y=b is f′_x(a,b).

We could define second partial derivatives by repetition. There are four of them:

f′′xx =  ∂2 f ∂x2 =  ∂ ∂x    ∂f ∂x   ; f′′yy =  ∂2 f ∂y2 =  ∂ ∂y    ∂f ∂y   ; f′′xy =  ∂2 f ∂y ∂x =  ∂ ∂y    ∂f ∂x   ; f′′yx =  ∂2 f ∂x ∂y =  ∂ ∂x    ∂f ∂y   .

Exercise 4 If v=yln(x²+z²), find v′′′_zzy.

Theorem 5 Let f be a function of two variables x and y. If f, f′_x, f′_y, f′′_xy, and f′′_yx are continuous on an open region R, then f′′_xy=f′′_yx through R.

Exercise 6 Verify that f′′_xy=f′′_yx.

f(x,y)=  x2 x+y ;        f(x,y)= √   x2+y2+z2   .

Review

Exercise 7 Find the interval of convergence of the power series:

∑ (−1)n  3n n! (x−4)n;        ∑ (−1)n  en+1 nn (x−1)n.

Exercise 8 Obtain a power series representation for the function

f(x) = x2 ln(1+x2);        f(x) = arctan√x.

Exercise 9 Find all values of c such that a and b are orthogonal a = 4 i + 2 j + ck, and b = i + 22 j − 3ck.

Exercise 10 Find the volume of the box having adjacent sides AB, AC, AD: A(2,1,−1), B(3,0,2), C(4,−2,1), D(5,−3,0).

Exercise 11 Find an equation of the plane through P(−4,1,6) and having the same trace in xz-plane as the plane x+4y−5z=8.

Exercise 12 Find arc length of the curve: x=2t, y=4sin3t, z=4cos3t; 0 ≤ t ≤ 2π.

Exercise 13 Find a parametric Al equation of the tangent line to curve x=tsin t, y=tcost, z=t; at P(π/2,0,π/2).

Exercise 14 Show that limit does not exist.

lim (x,y,z)→ (2,0,0)   (x−2)yz2 (x−2)4+y4 .

12.4  Increments and Differentials

Definition 1 Let w=f(x,y), and let ∆x and ∆ be increments of x and y, respectively. The increment of function w is

∆w = f(x+∆x, y + ∆y) − f(x,y).

Theorem 2 Let w=f(x,y), where the function f is defined on a rectangular region R={(x,y): a < x < b,  c < y < d}. Suppose f′_x and f′_y exist throughout R and are continuous at (x₀,y₀). Then

∆w = f′x (x0,y0) ∆x + f′y (x0,y0) ∆y + ε1 ∆x + ε2 ∆y.

A function w is differentiable if its increment could be represented as above.

Definition 3 The differential of function w is

d w = f′x (x0,y0) ∆x + f′y (x0,y0) ∆x.

12.5  Chain Rules

Theorem 1 [Chain rules] If w=f(u,v), with u=g(x,y), v=h(x,y), and if f, g, and h are differentiable, then

∂w ∂x =   ∂w ∂ u   ∂u ∂x +   ∂w ∂ v   ∂v ∂x ;   ∂w ∂y =   ∂w ∂ u   ∂u ∂y +   ∂w ∂ v   ∂v ∂y .

PROOF. It follows from the Theorem on Increment . ^[¯]

This formulas could be better understood and remembered if we will draw a tree representing dependence of variables.

Exercise 2 Find ∂w/∂x, ∂w partial y if w=uv+v², u=xsiny, v=ysinx.

Similar formulas are true for different number of variables

Exercise 3 Find ∂z/∂x, ∂z/∂y if z=pq + qw, p=2x−y, q=x−2y, w=−2x+2y.

Chain rules could be used to derive already known formulas in a new way.

Exercise 4 Derive formula (uv)′=u′v+uv′ using chain rules.

Exercise 5 Derive from chain rules the following formula for implicit derivatives of y defined by F(x,y)=0:

y′=−  F′x(x,y) F′y(x,y) .

12.6  Directional Derivatives

Definition 1 Let w=f(x,y) and u=u₁i+u₂j be a unit vector. The directional derivative of f at P(x,y) in the direction u, denoted D_u f(x,y), is

Du = lim s→ 0   f(x+su1, y+su2)−f(x,y) s .

Partial derivatives are particular cases of directional derivatives: ∂/∂x = D_i and ∂/∂y = D_j. It is interesting that we could calculate any directional derivative if we know only partial ones.

Theorem 2 If f is a differentiable function of two variables, then

Du f(x,y) = f′x(x,y) u1 + f′y(x,y)u2.

PROOF. It is follows from the Chain Rules . ^[¯]

Exercise 3 Find directional derivative

f(x,y)=x3 −3x2y−y3,     P(1,−2),     u =  1 2 (−i+ √3j).

Definition 4 Let f be a function of two variables. The gradient of f is the vector valued function

∇f(x,y) = f′x(x,y) i + f′y(x,y) j.

Directional derivative in gradient form is

Du f(x,y) = ∇f(x,y) ·u.

Exercise 5 Find gradient

f(x,y)=e3x tany,     P(0,π/4).

From gradient form of directional derivative easily follows the following theorem:

Theorem 6 Let f be a function of two variables that is differentiable at the point P(x,y).

1. The maximum value of D_u is ||∇f(x,y)||.

2. The maximum rate of increase of f(x,y) occurs in direction of ∇f(x,y).

3. The minimum value of D_u is −||∇f(x,y)||.

4. The minimum rate of increase of f(x,y) occurs in direction of −∇f(x,y).

Similarly directional derivatives and gradients could be defined for functions of three variables.

Exercise 7 Find directional derivative at P in the direction to Q. Find directions of maximal and minimal increase of f.

f(x,y,z)=  x y −  y z ,     P(0,−1,2),     Q(3,1,−4).

12.7  Tangent Planes and Normal Lines

Theorem 1 Suppose that F(x,y,z) has continuous first partial derivatives and that S is the graph of F(x,y,z)=0. If P₀ is a point on S and if F′_x, F′_y, F′_z are not all 0 at P₀, then the vector ∇F ]_P0 is normal to the tangent plane to S at P₀. And equation of the tangent plane is

F′x(x0,y0,z0) (x−x0) + F′y(x0,y0,z0) (y−y0) + F′z(x0,y0,z0) (z−z0) = 0.

Theorem 2 An equation for the tangent plane to the graph of z=f(x,y) at the point (x₀,y₀,z₀) is

z−z0=f′x(x0,y0) (x−x0) + f′y(x0,y0) (y−y0)

Exercise 3 Find equation for the tangent plane and normal line to the graph.

9x2 −4y2 − 25 z2=40;     P(4,1,−2).

12.8  Extrema of Functions of Several Variables

Definition 1 Let f be a function of two variables. A pair (a,b) is a critical point of f if either

1. f′_x(a,b)=0 and f′_y(a,b)=0, or

2. f′_x(a,b) or f′_y(a,b) does not exist.

Definition 2 Let f be a function of two variables that has continuous second partial derivatives. The discriminant D of f is given by

D(x,y)=f′′xx f′′yy− [f′′xy]2 =    f′′xx f′′xy f′′yx f′′yy    .

The following result is similar to Second Derivative Test .

Test 3 [Test for Local Extrema] Let f be a function of two variables that has continuous second partial derivatives throughout an open disk R containing a critical point (a,b). If D(a,b) > 0, then f(a,b) is

1. a local maximum of f if f′′_xx(a,b) < 0.

2. a local minimum of f if f′′_xx(a,b) > 0.

If a critical point with existent partial derivatives is not a local extrema then it is called saddle point. We could determine them by determinant:

Theorem 4 Let f have continuous second partial derivatives throughout an open disk R containing an critical point (a,b) with existent derivatives. If D(a,b) is negative, then (a,b) is a saddle point.

Exercise 5 Find extrema and saddle points.

f(x,y) = x2−2x+y2−6y+12 f(x,y) = −2x2−2xy−  3 2 y2−14x−5y f(x,y) = −  1 3 x3 +xy+  1 2 y2−12y.

Exercise 6 Find the max and min of f in R.

f(x,y)=x2−3xy−y2+2y−6x;     R={(x,y) | | x | ≤ 3, | y | ≤ 2 }.

Exercise 7 Find three positive real numbers whose sum is 1000 and whose product is a maximum.

12.9  Lagrange Multipliers

Theorem 1 Suppose that f and g are functions of two variables having continuous first partial derivatives and that ∇g ≠ 0 throughout a region. If f has an extremum f(x₀,y₀) subject to the constraint g(x,y)=0, then there is a real number λ such that

∇f(x0,y0) = λ∇g (x0,y0).

By other words they are among solution of the system

     f′x (x,y) = λg′x (x,y) f′y (x,y) = λg′y (x,y) g(x,y) = 0 .

Exercise 2 Find the extrema of f subject to the stated constrains

f(x,y)=2x2+xy−y2+y;     2x+3y=1.

Chapter 13
Multiply Integrals

13.1  Double Integrals

Definition 1 Let f be a function of two variables that is defined on a region R. The double integral of f over R, is

⌠ ⌡ ⌠ ⌡ R  f(x,y) dA= lim ||P||→ 0  ∑ k  f(xk,yk) ∆A,

provided the limit exists for the norm of the partition tensing to 0.

The following is similar to geometrical meaning of definite integral

Definition 2 [Geometrical Meaning of Double Integral] Let f be a continuous function of two variables such that f(x,y) is nonnegative for every (x,y) in a region R. The volume V of the solid that lies under the graph of z=f(x,y) and over R is

V = ⌠ ⌡ ⌠ ⌡ R  f(x,y) dA.

Double integral has the following properties (see one variable case ).

Theorem 3

1. 
   

   ⌠ ⌡ ⌠ ⌡ R  cf(x,y) dA = c ⌠ ⌡ ⌠ ⌡ R  f(x,y) dA.

2. 
   

   ⌠ ⌡ ⌠ ⌡ R  [f(x,y) + g(x,y)] dA = ⌠ ⌡ ⌠ ⌡ R  f(x,y) dA + ⌠ ⌡ ⌠ ⌡ R  g(x,y) dA

3. If R=R₁ ∪R₂ and R₁ ∩R₂ = ∅
   

   ⌠ ⌡ ⌠ ⌡ R  f(x,y) dA = ⌠ ⌡ ⌠ ⌡ R1  f(x,y) dA + ⌠ ⌡ ⌠ ⌡ R2  f(x,y) dA

4. If (x,y) ≥ 0 throughout R, then ∫∫_R f(x,y) dA ≥ 0.

Practically double integrals evaluated by means of iterated integrals as follows:

Theorem 4 Let R be a region of R_x type. If f is continuous on R, then

⌠ ⌡ ⌠ ⌡ R  f(x,y) dA = ⌠ ⌡ b a  ⌠ ⌡ g2(x) g1(x)  f(x,y) dy dx.

Exercise 5 Evaluate

⌠ ⌡ 3 0  ⌠ ⌡ −1 −2  (4xy3+y) dx dy        ⌠ ⌡ 1 −1  ⌠ ⌡ x+1 x3  (3x+2y) dy dx.

Exercise 6 Evaluate ∫∫_R e^x/y dA if R bounded by y=2x, y=−x, y=4.

Exercise 7 Sketch the region x=2√y, √3x=√y, y=2x+5 and express the double integral as iterated one.

Exercise 8 Sketch the region of integration for the iterated integral

⌠ ⌡ 2 −1  ⌠ ⌡ x−2 x2−4  f(x,y) dy dx.

Exercise 9 Reverse the order of integration and evaluate

⌠ ⌡ e 1  ⌠ ⌡ lnx 0  y  dy dx.

13.2  Area and Volume

Exercise 1 Describe surface and region related to

⌠ ⌡ 1 0  ⌠ ⌡ 1−x2 3−x  (x2+y2) dy dx.

Exercise 2 Find volume under the graph z=x²+4y² over triangle with vertices (0,0), (1,0), (1,2).

Exercise 3 Sketch the solid in the first octant and find its volume z=y³, y=x³, x=0, z=0, y=1.

13.3  Polar Coordinates, Double Integrals in Polar Coordinates

1. circle (O,R): r=R.

2. circle (a,a): r=2asinθ.

3. cardioid: r=a(1+cosθ).

4. limacons: r=a+bcosθ.

5. n-leafed rose: r=a sinnθ.

6. spiral of Archimedes: r=aθ.

Exercise* 1 Find equation of a straight line in polar coordinates.

Connection between the Cartesian coordinates and polar coordinates is as follows:

Theorem 2 The rectangular coordinates (x,y) and polar coordinates (r,θ) of a point P are related as follows:

1. x=rcosθ, y=rsinθ;

2. r²=x²+y², tanθ = y/x if x ≠ 0.

Theorem 3 [Test for Symmetry]

1. The graph of r=f(θ) is symmetric with respect to the polar axis if f(−θ)=f(θ).

2. The graph of r=f(θ) is symmetric with respect to the vertical line if f(π−θ)=f(θ) or f(−θ)=−f(θ).

3. The graph of r=f(θ) is symmetric with respect to the pole if f(π+θ)=f(θ).

Theorem 4 The slope m of the tangent line to the graph of r=f(θ) at the point P(r,θ) is

m=  dr dθ sinθ+ rcosθ  dr dθ cosθ− rsinθ

The element of area in polar coordinates equal to ∆A = [ 1/2] (r₂²−r²₁)∆θ = r∆r∆θ, where r=[ 1/2] (r₂−r₁). Thus double integral in polar coordinates could be presented by iterated integral as follows:

⌠ ⌡ ⌠ ⌡ R  f(r,θ) dA = ⌠ ⌡ β α  ⌠ ⌡ g2(θ) g1(θ)  f(r,θ)r dr dθ. = ⌠ ⌡ β α  ⌠ ⌡ h2(r) h1(r)  f(r,θ)r dθ dr.

Exercise 5 Use double integral to find the area inside r=2−2cosθ and outside r=3.

Exercise 6 Use polar coordinates to evaluate the integral

⌠ ⌡ ⌠ ⌡ R  x2(x2+y2)3 dA

R is bounded by semicircle y=√{1−x²} and the x-axis.

Exercise 7 Evaluate

⌠ ⌡ a 0  ⌠ ⌡ √{a2−x2} 0  (x2+y2)3/2 dy dx.

Exercise 8 Find volume bounded by paraboloid z=4x²+4y², the cylinder x²+y²=3y, and plane z=0.

13.4  Surface Area

Theorem 1 The surface area of the graph z=f(x,y) over the region R is given by

A= ⌠ ⌡ ⌠ ⌡ R  √   [f′x(x,y)]2+[f′y(x,y)]2+1    dA.

Exercise 2 Setup a double integral for the surface area of the graph x²−y²+z² = 1 over the square with vertices (0,1), (1,0), (−1,0), (0,−1).

Exercise 3 Find the area of the surface z = y² over the triangle with vertices (0,0), (0,2), (2,2).

Exercise 4 Find the area of the first-octant part of hyperbolic paraboloid z = x²−y² that is inside the cylinder x²+y²=1.

13.5  Triple Integrals

There is no any principal differences to introduce triple integral, it could be done using ideas on definite integrals and double integrals .

Definition 1 Triple integral of f over 3d-region Q is defined by Riemann sums :

⌠ ⌡ ⌠ ⌡ ⌠ ⌡ Q  f(x,y,z) dV = lim ||P||→ 0  ∑ k  f(xk,yk,zk) ∆Vk.

To evaluate triple integrals we reduce them by iteration to double integrals:

Theorem 2

⌠ ⌡ ⌠ ⌡ ⌠ ⌡ Q  f(x,y,z) dV = ⌠ ⌡ ⌠ ⌡ R    ⌠ ⌡ k2(x,y) k1(x,y)  f(x,y,z) dz   dA = ⌠ ⌡ b a  ⌠ ⌡ h2(x) h1(x)  ⌠ ⌡ k2(x,y) k1(x,y)  f(x,y,z) dz  dy dz.

Exercise 3 Evaluate the iterated integral

⌠ ⌡ 1 0  ⌠ ⌡ 2 −1  ⌠ ⌡ 3 1  (6x2z+5xy2) dz dx dy;        ⌠ ⌡ 2 −1  ⌠ ⌡ z2 1  ⌠ ⌡ x−z x+z  z dy dx dz.

Exercise 4 Describe region represented by integrals

⌠ ⌡ 1 0  ⌠ ⌡ √z z3  ⌠ ⌡ 4−x 0   dy dx dz,        ⌠ ⌡ 1 0  ⌠ ⌡ 3x x  ⌠ ⌡ xy 0   dz dy dx.

Physical meaning of triple integrals is given by

Theorem 5 Mass of a solid with a mass density δ(x,y,z) is given by

m= ⌠ ⌡ ⌠ ⌡ ⌠ ⌡ Q  δ(x,y,z) dV

Theorem 6 Mass of a lamina with an area mass density δ(x,y) is given by

m= ⌠ ⌡ ⌠ ⌡ ⌠ ⌡ R  δ(x,y) dA

Exercise 7 Using triple integrals find volume bounded by

1. x²+z²=4, y²+z²=4.

2. z=x²+y², y+z=2.

13.7  Cylindrical Coordinates

Theorem 1 The rectangular coordinates (x,y,z) and the cylindrical coordinates (r,θ,z) of a point are related as follows:

x = rcosθ,    y=rsinθ,     z=z, r2 = x2+y2,     tanθ =  x y .

Exercise 2 Describe the graph in cylindrical ccordinates:

1. r=−3secθ.

2. z=2r.

Exercise 3 Change the equation to cylindrical coordinates:

1. x²+y²=4z.

2. x²+z²=9.

Theorem 4 Evaluation of triple integral in cylindrical coordinates:

⌠ ⌡ ⌠ ⌡ ⌠ ⌡ Q  f(r,θ,z) dV= ⌠ ⌡ β α  ⌠ ⌡ g2(θ) g1(θ)  ⌠ ⌡ k2(r,θ) k1(r,θ)  f(r,θ,z) dz dr dθ.

Exercise 5 A solid is bounded by the cone z=√{x²+y²}, the cylinder x²+y²=4, and the xy-plane. Find its volume.

13.8  Spherical Coordinates

Theorem 1 The rectangular coordinates (x,y,z) and the spherical coordinates (ρ, φ, θ) of a point related as follows:

x = ρsinφcosθ, x=ρsinφsinθ,     z=ρcosθ ρ2 = x2+y2+z2.

Exercise 2 Change coordinates

1. spherical (1, 3π/4,2π/3) to rectangular and cylindrical.

2. rectangular (1, √3,0) to spherical and cylindrical.

Exercise 3 Describe graphs

1. ρ = 5.

2. φ = 2π/3.

3. θ = π/4.

Exercise 4 Change the equation to spherical coordinates.

x2+y2=4z;        x2+(y−2)2=4;        x2+z2=9.

Theorem 5 [Evaluation theorem]

⌠ ⌡ ⌠ ⌡ ⌠ ⌡ Q  f(ρ,φ,θ) dV= ⌠ ⌡ n m  ⌠ ⌡ d c  ⌠ ⌡ b a  f(ρ,φ,θ) ρ2sinφ dρ dφ dθ.

Exercise 6 Find volume of the solid that lies outside the cone z²=x²+y² and inside the sphere x²+y²+z²=1.

Exercise 7 Evaluate integral in spherical coordinates:

⌠ ⌡ √2 0  ⌠ ⌡ √{4−y2} y  ⌠ ⌡ √{4−x2−y2} 0  √   x2+y2+z2    dz dx dy.

Chapter 14
Vector Calculus

14.1  Vector Fields

Definition 1 A vector field in three dimensions is a function F whose domain D is a subset of R³ and whose range is is a subset of V³. If (x,y,z) is in D, then

F(x,y,z) = M(x,y,z)i + N(x,y,z)j + P(x,y,z)k.

where M, N, and P are scalar functions.

Exercise 2 Plot the vector field F(x,y)=−yi+xj.

Example of vector field is as follows:

Definition 3 Let r=xi+yj+zk. A vector field F is an inverse square field if

F(x,y,z) =  c ||r||3 r.

Examples of inverse square field are given by Newton's law of gravitation and Coulom's law of charge interaction.

Definition 4 A vector filed F is conservative if

F(x,y,z)=∇f(x,y,z)

for some scalar function f. Then f is potential function and its value f(x,y,z) is potential in (x,y,z).

Exercise 5 Find a vector field with potential f(x,y,z)=sin(x²+y²+z²).

Theorem 6 Every inverse square vector filed is conservative.

PROOF. The potential is given by f(r)=[ c/r]. ^[¯]

Definition 7 Let F(x,y,z) = M(x,y,z)i + N(x,y,z)j + P(x,y,z)k. The curl of F is given by

curl F = ∇×F =        i j k  ∂ ∂x  ∂ ∂y  ∂ ∂z M N P        =    ∂P ∂y −  ∂N ∂z   i +    ∂M ∂z −  ∂P ∂x   j    ∂N ∂x −  ∂M ∂y   k

Definition 8 Let F(x,y,z) = M(x,y,z)i + N(x,y,z)j + P(x,y,z)k. The divergence of F is given by

div F=∇·F =  ∂M ∂x +  ∂N ∂y +  ∂P ∂z .

Exercise 9 Find curl F and div F for

F(x,y,z)=(3x+y)i + xy2zj + xz2k.

Exercise 10 Prove that for a constant vector a

1. curl (a×r) = 2a;

2. ÷(a×r) = 0.

Exercise 11 Verify the identities:

curl (F+G) = curl F+ curl G; div (F+G) = div F+ div G; curl (fF) = f(curl F)+ (∇f)×F;

14.2  Line Integral

Definition 1 The line integrals along a curve C with respect to s, x, y, respectively are

⌠ ⌡ C  f(x,y) ds = lim ||P||→ 0  ∑ k  f(uk,uk) ∆sk ⌠ ⌡ C  f(x,y) dx = lim ||P||→ 0  ∑ k  f(uk,uk) ∆xk ⌠ ⌡ C  f(x,y) dy = lim ||P||→ 0  ∑ k  f(uk,uk) ∆yk

Let a curve C be given parametrically by x=g(t) and y=h(t). Because

dx=g′(t) dt, dy=h′(t) dt, ds= √   (dx)2+(dy)2   = √   (g′(t))2+(h′(t))2    dt.

we obtain

Theorem 2 [Evaluation formula for line integrals] If a smooth curve C is given byx=g(t) and y=h(t); a ≤ t ≤ b and f(x,y) is continuous in a region containing C, then

⌠ ⌡ C  f(x,y) ds = ⌠ ⌡ C  f(g(t),h(t)) √   (g′(t))2+(h′(t))2    dt ⌠ ⌡ C  f(x,y) dx = ⌠ ⌡ C  f(g(t),h(t))(g′(t) dt ⌠ ⌡ C  f(x,y) dy = ⌠ ⌡ C  f(g(t),h(t))h′(t)) dt