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CALCULUS II MATH 120 WINTER 2019 LECTURE NOTES

M. P. COHEN

CARLETON COLLEGE

Remark 0.1. These lecture notes sprung forth not only from the author’s imagination but also from acombination of Rogawski’s Calculus Early Transcendentals, Briggs’ and Cochran’s Calculus, and Stew-art’s Calculus Early Transcendentals, and many of the examples included were taken from these sources.

1 Inverse Trigonometric Functions

Definition 1.1. For x satisfying −1 ≤ x ≤ 1, we define the arcsine and arccosine functions as follows.

y = arcsinx if and only if x = sin y and −π2 ≤ y ≤π2 .

y = arccosx if and only if x = cos y and 0 ≤ y ≤ π.

For x satisfying x ≤ −1 or x ≥ 1, we define arcsecant as follows.

y = arcsec x if and only if x = sec y and 0 ≤ y ≤ π.

Lastly, for any real number x, we define arctangent as follows.

y = arctanx if and only if x = tan y and −π2 < y < π2 .

Remark 1.2 (How To Think About Inverse Trigonometric Functions). • Think of the input x asa given ratio of side lengths of some triangle (opposite/hypotenuse for sine; opposite/adjacentfor tangent; etc.). Verify that the given input x makes sense when interpreted as this ratio (i.e.it must satisfy −1 ≤ x ≤ 1 for sine and cosine, etc.).

• Think of the output y as the corresponding angle of the triangle whose side length ratio is x.Although there are infinitely many such y, we always choose the one which lies in the rangespecified in the definition.

Remark 1.3 (Notation For Inverse Trig Functions). I always write arcsinx instead of sin−1 x, andsimilarly for the other inverse trig functions. This is to avoid confusion: the notation sin−1 x calls tomind the function 1

sin x , or cscx, which is not the same function.

Example 1.4. Make sense of the following, if they make sense.

(1) arctan√

3

(2) arccos(−√22 )

(3) arcsin 4

(4) cos(arccos 0.7)

(5) arccos(cos 9π4 )

(6) arccos(cos 7π4 )

1

2 M. P. COHEN CARLETON COLLEGE

(7) sec(arcsin 12 )

Fact 1.5 (Right Inverse Property But Not Left). We always have sin(arcsinx) = x, but not necessarilyarcsin(sinx) = x. Similarly for the other inverse trig functions.

Example 1.6. Let f(x) = arcsinx. Find f ′(x).

Solution. We use a trick which is common for computing the derivatives of inverse functions. By theprevious fact, we have

sin(arcsinx) = x

Take the derivative of both sides above, and use the chain rule on the left side.

cos(arcsinx) · f ′(x) = 1

Solving for f ′, we get

f ′(x) = sec(arcsinx).

The above is the correct derivative, but we can write it in terms which are easier to understand. Ifθ = arcsinx, then imagine θ as the base angle of a right triangle with opposite/hypotenuse ratio x = x

1 .Label the opposite leg x and the hypotenuse 1. Then by the Pythagorean theorem, the base leg haslength

√1− x2. Then we finish the problem by finding sec θ, to obtain

f ′(x) =1√

1− x2.

�

Fact 1.7 (Inverse Trigonometric Derivatives). Because they come up sometimes.

• d

dxarcsinx =

1√1− x2

• d

dxarccosx = − 1√

1− x2

• d

dxarctanx =

1

1 + x2

• d

dxarcsec x =

1

|x|√x2 − 1

Example 1.8 (Challenge Problem). Verify the derivative formulas above for arctangent and arcsecant.

2 Limits of Indeterminate Form and L’Hospital’s Rule

Definition 2.1 (Informal Definition). We say that the limit limx→a

f(x) is of indeterminate form if the

initial attempt to evaluate f(a) yields an undefined expression of the type 00 or ∞∞ .

Example 2.2 (Examples of Limits of Indeterminate Form). What are the following limits?

(1) limx→3

x2 − 4x+ 3

x2 + x− 12

(2) limx→π/2−

tanx

secx

CALCULUS II MATH 120 WINTER 2019 LECTURE NOTES 3

(3) limx→∞

3x7 − 22x5 + 19x

8x7 − 100x6 + 22

(4) limx→0

sinx

x

(5) limx→∞

lnx

x− 1

(6) limx→1

lnx

x− 1

Remark 2.3 (Limits Can Be Hard To Find). In Calculus I the student has probably seen severaltricks for computing limits of indeterminate form (like factoring and dividing polynomials; pulling outsinx/x terms; etc.), but of course these tricks only work on special examples. There is no absolutegeneral method for computing limits of indeterminate form, and many limits turn out to be difficult orimpossible to identify exactly.

This difficulty accounts for much of the beauty and value inherent in the careful study of limits inthe first place, which are the foundation of modern calculus and analysis.

Example 2.4 (TTYNP). Give examples of limits of indeterminate form 00 . Give (1) one where the

limit is 0; (2) one where the limit is ∞; and (3) one where the limit is 7. Can you find (4) one wherethe limit doesn’t exist?

Theorem 2.5 (L’Hospital’s Rule). Suppose f and g are functions differentiable at x = a, and limx→a

f(x)

g(x)is of indeterminate form 0

0 or ∞∞ , i.e. either

limx→a

f(x) = limx→a

g(x) = 0,

or else

limx→a

f(x) = ±∞ and limx→a

g(x) = ±∞.

Then

limx→a

f(x)

g(x)= limx→a

f ′(x)

g′(x)

provided the limit on the right-hand side exists (or is equal to ±∞).

Example 2.6. Use L’Hospital’s rule to find the limits in the examples above.

Example 2.7. Evaluate the following limits:

(1) limx→1

x3 + x2 − 2x

x− 1

(2) limx→0

√9 + 3x− 3

x

(3) limx→2

x3 − 3x2 + 4

x4 − 4x3 + 7x2 − 12x+ 12

Example 2.8 (L’Hospital’s Rule is Inapplicable). Evaluate limx→1

x2 + cosπx

x2 − 4x+ 2.

Example 2.9. Evaluate limx→2

4− x2

sinπx.

Example 2.10. Evaluate limx→0

ex − x− 1

cosx− 1.


Example 2.11 (Indeterminate Form ∞−∞). Evaluate limx→0

(1

sinx− 1

x

).

Example 2.12 (Indeterminate Form ∞ · 0). Evaluate limx→∞

x2 sin

(1

4x2

).

Example 2.13 (Indeterminate Form 00). Evaluate limx→0+

xx.

Example 2.14 (Indeterminate Form 1∞). Evaluate limx→0

(1 + 4x)12x .

Definition 2.15. Let f and g be two functions which satisfy limx→∞

f(x) = limx→∞

g(x) =∞. We say that

f(x) grows faster than g(x) if

limx→∞

f(x)

g(x)=∞, or equivalently, lim

x→∞

g(x)

f(x)= 0.

We denote this relationship by writing g(x) << f(x).

Example 2.16. Which function grows faster, f(x) = x2 or g(x) = x lnx?

Example 2.17 (TTYNP). Rank the following functions in order of growth rate:

(1)√x;

(2) lnx;

(3) (lnx)3;

(4) ex; and

(5) 1.0001x.

3 Review: Substitution Rule

Theorem 3.1 (Substitution Rule). If f and g are functions, u is differentiable, and F is any antideriv-ative of f , then ∫

f(u(x))u′(x)dx = F (u(x)) + C.

Remark 3.2 (Procedure for the substitution rule). (1) Given an indefinite integral with a compos-ite function f(u(x)) appearing in the integrand, identify the “inner function” u(x) whose de-rivative u′(x) also appears as a factor in the integrand. (i.e. Look for something of the formf(u(x))u′(x).)

(2) Compute du = g′(x)dx.

(3) Evaluate the integral with respect to u.

(4) Un-substitute u = u(x) to finish the problem.

Example 3.3.∫

2(2x+ 1)3dx.

Example 3.4.∫

cos3 x sinxdx.

Example 3.5.∫

x√x+1

dx.

Theorem 3.6 (Substitution Rule for Definite Integrals (with Change of Parameters)). If f and u arefunctions, u is differentiable, and F is any antiderivative of f , then∫ b

af(u(x))u′(x)dx =

∫ u(b)u(a)

f(u)du.


Remark 3.7. In the above theorem we are using a common but perhaps slightly confusing “abuse ofnotation”: we are using u to denote both a function, and a dummy variable inside the integral on theright-hand side of the equality. In practice, this usually doesn’t cause confusion, because we typicallywrite u(x) in terms of its explicit formula in x, and u occupies solely its latter role as a dummy variablewhen we work through problems.

Example 3.8. Compute∫ 2

0dx

(x+3)3 .


−1x2dx

(x3+2)3 .

Example 3.10. Compute∫ π

2

0sin4 x cosxdx.

4 Integration by Parts

Remark 4.1 (Reversing the Product Rule). At this point in our development of the calculus, thestudent should recognize that our differentiation techniques (chain rule, product rule, quotient rule,etc.) give us a tremendous amount of power to compute derivatives, but that our integration techniquesare comparatively lacking. So far the only major integration tool we have developed is the substitutionrule, which one should think of as a “reverse chain rule.” We now wish to expand the number ofintegration tools at our disposal by introducing a “reverse product rule.”

So consider a pair of differentiable functions u(x) and v(x). The product rule tells us that

ddxu(x)v(x) = u′(x)v(x) + u(x)v′(x)

and hence

u(x)v′(x) = ddxu(x)v(x)− v(x)u′(x)

Now we antidifferentiate both sides (and invoke the Fundamental Theorem of Calculus) to obtain anintegration rule, which we call integration by parts:

∫u(x)v′(x)dx = u(x)v(x) =

∫v(x)u′(x)dx

Note that in the above expression we need not include any +C term, since there is an indefinite inte-gral on both sides of the equation. We may also write the rule for integration by parts in the followingmemorable compact fashion, by using just a slight abuse of notation:

∫udv = uv −

∫vdu

Example 4.2. Compute the following antiderivatives using integration by parts.

(1)∫xexdx

(2)∫x sinxdx

Solution to part (a). We see that the integrand xex is a product of two functions (x and ex). We needto choose an appropriate u and dv to apply integration by parts. There is no hard and fast rule for howto choose this, but a general guideline is that we should choose u to be whichever function “becomessimpler” when we derive it. Since x derives to 1 (a “simpler function”), we choose

u = x and dv = exdx.

Deriving u and integrating dvdx , we get

du = 1 · dx and v = ex.


Plugging in u and v into the integration by parts formula, we get

∫xexdx = xex −

∫exdx

= xex − ex.

�

Example 4.3. Compute the following antiderivatives.

(1)∫x2exdx

(2)∫

arcsinxdx

Example 4.4. Compute∫ex cosxdx.

Solution. First the student should notice why the problem is tougher than usual. Normally when weapply integration by parts, we want to choose u to be whichever function “becomes simpler” when wederive it. But in this case the derivative of ex is ex, which is not any simpler, and the derivative of cosxis − sinx, which is also not any simpler! So the usual strategy for integration by parts actually will notbe enough in and of itself. To compute this integral, we use integration by parts twice, and then a clevertrick at the end to finish the problem.

To start, we choose u = ex and dv = cosxdx. This gives du = exdx and v = sinxdx. So applyingintegration by parts, we get

∫ex cosxdx = ex sinx−

∫ex sinxdx.

Now we apply integration by parts to the integral on the right-hand side above. Taking u = ex anddv = sinx, we get du = exdx and v = − cosx, so

∫ex cosxdx = ex sinx−

∫ex sinxdx

= ex sinx− [−ex cosx+

∫ex cosxdx]

= ex sinx+ ex cosx−∫ex cosxdx.

Now we have applied integration by parts twice, and at first glance it seems we have not made anyprogress: we started out wanting to compute

∫ex cosxdx, and on the right-hand side above we have

gotten an expression involving∫ex cosxdx, the very thing we are wanting to compute!

While this initially seems frustrating, in fact it is good news that we have∫ex cosxdx appearing on

both sides of the equation. Now we can be clever and just use algebra to finish the problem. Take theequality above, and add

∫ex cosxdx to both sides to get:

∫ex cosxdx+

∫ex cosxdx = ex sinx+ ex cosx+ C.

Collecting like terms, we get

2∫ex cosxdx = ex sinx+ ex cosx+ C.


Now we simply divide by 2 to find the value of∫ex cosxdx.

∫ex cosxdx = 1

2 (ex sinx+ ex cosx) + C.

�

Remark 4.5 (General Strategy for Integrals of the Form∫eax sin(bx)dx or

∫eax cos(bx)dx). (1) First

use integration by parts twice in a row. This should give you an expression which has the integralyou started with appearing on both the left- and the right-hand sides of the equality. Then

(2) use algebra to solve for the integral.

Theorem 4.6 (Integration by Parts for Definite Integrals). .∫ bau(x)v′(x)dx = [u(x)v(x)]ba −

∫ bav(x)u′(x)dx

Example 4.7. Compute the following definite integrals.

(1)∫ 2

1lnxdx

(2)∫ π/20

x cos 2xdx

5 Integration of Rational Functions by the Method of Partial Fractions

Example 5.1. Evaluate

∫3x

x2 + 2x− 8dx.

Solution. The key to evaluating this integral is knowing that every rational function may be rewrittenas a partial fraction expansion. Note that the denominator x2 +2x−8 in the integrand above factorsinto (x+ 4)(x− 2). Then it is possible to find two real numbers A and B for which

3x

x2 + 2x− 8=

3x

(x+ 4)(x− 2)=

A

x+ 4+

B

x− 2.

The only trick is to find just what the values of A and B should be. To find the values of A and B,multiply both sides of the equation above by (x + 4)(x − 2) to clear all denominators. Then combinelike terms:

3x = A(x− 2) +B(x+ 4)

= Ax− 2A+Bx+ 4B

= (A+B)x+ (−2A+ 4B)

Now we have obtained the equality 3x = (A + B)x + (−2A + 4B); it follows that we must have3 = A + B and 0 = −2A + 4B. So we have a system of two equations in two variables; now we mayresort to our algebra techniques. If we carefully solve the given system, we should get A = 2 and B = 1.Thus we may convert our original integral into the following, much easier problem:

∫3x

x2 + 2x− 8dx =

∫2

x+ 4dx+

∫1

x− 2dx

= 2 ln |x+ 4|+ ln |x− 2|+ C.

�


∫3x2 + 7x− 2

x3 − x2 − 2xdx.


Example 5.3 (Denominator Has Linear Factor With Degree > 1). Evaluate

∫5x2 − 3x+ 2

x3 − 2x2dx.

Solution. The same techniques we have used above will work for the last example above, if we keep trackof one important detail: Notice that the denominator x3−2x2 factors into x2(x−2). Since x appears asa factor with degree 2; we must represent x twice in the partial fraction expansion: once as an x term,and once as an x2 term. In other words, we will be able to find a partial fraction expansion of the form

5x2 − 3x+ 2

x2(x− 2)=A

x+B

x2+

C

x− 2,

for some real numbers A, B, and C. Otherwise we may proceed as usual. We will check the remainingdetails in class. �

Example 5.4 (Denominator Has Irreducible Quadratic as a Factor). Evaluate

∫7x2 − 13x+ 13

(x− 2)(x2 − 2x+ 3)dx.

Solution. To solve the above, we must again keep track of an important detail. Notice that the poly-nomial x2 − 2x + 3 which appears in the denominator is an irreducible quadratic, i.e. it cannotbe factored into binomials with real coefficients. We may still find a partial fraction expansion for theintegrand, but it will take the form

7x2 − 13x+ 13

(x− 2)(x2 − 2x+ 3)=

A

x− 2+

Bx+ C

x2 − 2x+ 3,

where A, B, and C are some real numbers. A similar principle holds whenever an irreducible quadraticappears in the denominator of a rational expression. These integrals can in general be challenging tocompute, and may involve both the method of completing the square and the use of the arctangentfunction. The following fact will be helpful to remember:

Fact 5.5.

∫1

u2 + a2du =

1

aarctan

u

a+ C

Again, we work through the details in class. �


∫z + 1

z(z2 + 4)dz.

Example 5.7 (Denominator Has Irreducible Quadratic Factor With Degree > 1). Evaluate∫x4 − x3 + 6x2 − 2x+ 4

x(x2 + 2)2dx.

Example 5.8 (Use Long Division First). Evaluate

∫x4 − 5x3 + 9x2 − 3x+ 5

x2 − 5x+ 6dx.

Fact 5.9 (Fundamental Theorem of Algebra). Every polynomial (with real coefficients) can be factoreduniquely into a product consisting of: (1) a constant; (2) linear factors of the form (x − a); and (3)irreducible quadratics of the form (x2 + bx+ c) where b2 − 4c < 0.

Corollary 5.10. Every rational function (quotient of polynomials) admits a unique partial fractionexpansion, i.e. every rational function can be written as the sum of (1) a polynomial; (2) rationalfunctions of the form A

(x−a)n ; and (3) rational functions of the form Bx+C(x2+bc+c) where the denominator is

an irreducible quadratic.


6 Improper Integrals

Example 6.1. Let b be any real number greater than 1. Compute the following integrals.

(1)∫ b1

1dx

(2)∫ b1

1x2 dx

Solution. We have

∫ b

1

1dx = [x]b1

= b− 1

and

∫ b

1

1

x2dx =

[− 1

x

]b1

= 1− 1

b.

Let us consider the geometric meanings of the expressions we have computed above. If we sketch the

picture, the first integral∫ b1

1dx may be interpreted as the area of the box with the x-axis for a base,the line x = 1 for a left side, the line y = 1 for a top side, and the line x = b for a right side. If weallow b to grow large, i.e. we allow the right side of the box to slide further right, the integral confirmsour intuition that the area of the box becomes larger and larger. It grows unboundedly, i.e. we have

limb→∞

∫ b

1

1dx = limb→∞

(b− 1) =∞.

On the other hand, the second integral∫ b1

1x2 dx = 1− 1

b behaves remarkably differently. If we compute

the area bounded above by the graph of 1x2 between, say, 1 and 2, we get 1− 1

2 = 12 . If we compute from

1 to 3 we get 23 ; if we compute from 1 to 4 we get 3

4 , and so on. In general the area is increasing, whichconfirms our intuition. But what happens as b becomes very large? Notice that no matter how large bis, the area 1− 1

b will never exceed 1, i.e. the area does not grow unboundedly as in the previous example.

Thus the region bounded by the curve 1x2 , the x-axis, and the line x = 1 is infinitely long but has

finite area. Since limb→∞

∫ b

1

1

x2dx = lim

b→∞1− 1

b= 1, we say that the region has area 1. �

Definition 6.2. Let f be a continuous function. We define improper integrals with infinite boundsof integration as below, provided the given limits exist:∫ ∞

a

f(x)dx = limb→∞

∫ b

a

f(x)dx for any real number a;

∫ b

−∞f(x)dx = lim

a→−∞

∫ b

a

f(x)dx for any real number b;

∫ ∞−∞

f(x)dx = lima→−∞

∫ c

a

f(x)dx+ limb→∞

∫ b

c

f(x)dx for any real number c.

If the limit exists and is finite, then we say the integral converges. If the limit does not exist, or isequal to ∞ or −∞, then we say the integral diverges.

Example 6.3. Evaluate each integral.


(1)∫∞0e−3xdx

(2)∫∞0

11+x2 dx

(3)∫∞1x−1dx

Example 6.4. Compute∫∞0xe−xdx.

We may also define improper integrals for functions with vertical asymptotes, as we see in the nextexample.

Example 6.5. What should the value of∫ 1

01√xdx be?

Definition 6.6. Let f be a function continuous at x 6= p and with a vertical asymptote at x = p. Definethe improper integrals (with unbounded integrand) as follows:∫ p

a

f(x)dx = limc→p−

∫ c

a

f(x)dx for a < p;

∫ b

p

f(x)dx = limd→p+

∫ b

d

f(x)dx for p < b;

∫ b

a

f(x)dx = limc→p−

∫ c

a

f(x)dx+ limd→p+

∫ a

d

f(x)dx for a < p < b.

Again, we say the integral converges if the limit exists and is finite; otherwise we say the integraldiverges.


−31√

9−x2dx.

Example 6.8. State whether the following integrals converge or diverge. If they converge, give thevalue of the integral. (Be careful!)

(1)∫ 1

−11

x1/3 dx

(2)∫ 1

−11x3 dx

Example 6.9 (TTYNP). (1) Determine for what values of p the integral∫∞1

1xp dx converges or

diverges.

(2) Determine for what values of p the integral∫ 1

01xp dx converges or diverges.

7 Differential Equations

Definition 7.1. A differential equation is an equation that contains an unknown function, and oneor more of its derivatives. The order of the equation is the order of the highest derivative that occursin the equation.

Example 7.2 (Modeling a Spring). We consider the motion of an object with mass m at the end ofa vertical spring. Hooke’s law states that if the spring is stretched (or compressed) y units from itsequilibrium position, then it exerts a force F (y) that is proportional to y, i.e.

F (y) = −ky for some constant k.

Now imagine the variable y as being a function of time t representing the position of the object (rel-ative to the spring’s equilibrium state). Since force equals mass times acceleration, if we ignore externalforces (like friction, air resistance) we get

my′′ = −ky


because y′′ represents the acceleration of the object. Thus we arrive at the following second-order dif-ferential equation:

y′′ = − kmy

To solve the differential equation would be to find a function x = x(t) which satisfies the conditionabove. Such a function x would model the movement of the object at the end of the spring.

Example 7.3 (TTYNP). Find a function y = y(t) which satisfies the differential equation y′′ = −5y.

Example 7.4. Consider the first-order differential equation y′ = 12 (y2 − 1). Show that every member

of the family of functions y = 1+cet

1−cet is a solution.

Example 7.5 (Slope Fields). Try to sketch a solution to the first-order differential equation y′ = x2

y2 .

Definition 7.6. A first-order differential equation is called separable if it can be written in the follow-ing form:

y′ = g(x)f(y)

where g(x) is a function in only the variable x, and f(y) is a function in only the variable y.

Example 7.7. Find all solutions to the separable differential equation y′ = x2

y2 .

Solution. Note that the equation is separable. The trick to solving separable equations is to put all thex terms on one side of the equation, and all the y terms on the other:

y2y′ = x2.

Now integrate both sides. Recall that y is a function of x, so we can integrate y2y′ by reversing thechain rule, to get the following.

13y

3 = 13x

3 + C.

Solving for y, we get

y = 3√x3 + C.

Verify that each y as above is a solution to the given differential equation. �

Example 7.8 (Solutions Are Defined Only Implicitly). Solve dydx = 6x2

2y+cos y .

Example 7.9. Solve y′ = x2y.

Example 7.10 (The Law of Natural Growth). The law of natural growth is the assumption thatthe growth rate of a given population is always proportional to its size. In other words, if P = P (t) isthe population at time t, then

P ′ = kP .

Solve this differential equation.

Fact 7.11 (Solutions to the Law of Natural Growth). The functions P = P (t) which satisfy P ′ = kPare given by

P (t) = P0ekt for constant P0.


Example 7.12 (The Logistic Model). In practice, populations often increase exponentially when theyare small, but then plateau as they become large enough to create a scarcity of resources in theirenvironment. In addition, populations can decrease if they are so large that they exceed the carryingcapacity M of the environment.

A common way to model these ideas is to modify the law of natural growth (P ′ = kP ) by replacingthe constant k with a term k(1− P

M ), which is close to k for small populations but becomes near 0 whenP is near M . This model for population growth is called the logistic differential equation:

P ′ = kP (1− PM )

Consider slope fields given by the logistic equation. What would solutions to the logistic equationlook like?

Example 7.13. Solve the logistic equation.

Fact 7.14 (Solutions to the Logistic Model). The functions P = P (t) which satisfy P ′ = kP (1 − PM )

are given by

P (t) =M

1 +Ae−ktwhere A =

M − P0

P0for a constant P0.

8 Vectors in the Plane

Remark 8.1. In the following, we will try to develop some intuition for what vectors are supposed torepresent, and how they are meant to be manipulated geometrically. For this reason, initially we willonly work with informal definitions, and save the rigorous definitions for a bit later on.

Definition 8.2 (Naive Definition of a Vector). A vector is an object which consists of both a (strictlypositive) length (or magnitude) and a direction. We typically denote vectors as lower-case letters

with an arrow hat, e.g. ~u, ~v, ~w, ~a, ~b, etc. We denote the length of a vector ~v by ‖~v‖. We consider twovectors ~u and ~v to be equal if they have both the same length and the same direction, and we write~u = ~v. (For equality, note that we do not require ~u and ~v to have the same location!) We also allow the

existence of a unique zero vector, denoted ~0, which we consider to have 0 length and no direction.We may represent a vector ~v pictorially by drawing an arrow. We refer to the pointy part at the end

of the arrow as the terminal point, and the base of the arrow as the initial point. If P and Q are two

points in (two-dimensional or three-dimensional) space, then we denote by−−→PQ the vector which has its

initial point at P and its terminal point at Q. (Note that unless P = Q, we always have−−→PQ 6= ~QP .)

A scalar is just a magnitude, with no direction. In other words, a scalar is just a real number (amember of the set R).

Example 8.3. Using the intuitive definition above, classify the following physical quantities as either avector or a scalar.

(a) A wind blowing southwest at 22 miles per hour.(b) The mass of an apple.(c) The force exerted by gravity on the apple.(d) The temperature in Fargo, ND at 2pm today.(e) The velocity of a parked car at rest.

Definition 8.4 (Naive Definition of Scalar Multiplication). Let ~v be a vector and let c be a scalar. Thescalar product of c and ~v, denoted c~v, is defined as follows:

(1) if c > 0 then c~v is the vector which points in the same direction as ~v, and whose length is c timesthe length of ~v.

(2) if c < 0 then c~v is the vector which points in the opposite direction from ~v, and whose length is|c| times the length of ~v.

(3) if c = 0 then c~v = ~0.

We call the vector c~v a scalar multiple of ~v.


Example 8.5. Draw any non-zero vector ~v. Then draw the vectors 3~v, 12~v, −~v = (−1)~v, − 5

2~v, 0~v, and−π~v.

Remark 8.6. Observe that another way of writing part (b) in the definition above is that for any vector~v and any scalar ~c,

‖c~v‖ = |c| · ‖~v‖.

This is a helpful identity we will use many times.

Definition 8.7 (Naive Definition of Vector Addition and Subtraction). Let ~u and ~v be vectors. If theinitial point of ~v is placed at the terminal point of ~u, then the vector sum of ~u and ~v, denoted ~u+ ~v,is the vector whose initial point coincides with the initial point of ~u and whose terminal point coincideswith the terminal point of ~v. (Picture helps here.)

The vector difference of ~u and ~v, denoted ~u− ~v, is defined to be the vector sum ~u+ (−1)~v.

Example 8.8. (a) In general is ‖~u+ ~v‖ = ‖~u‖+ ‖~v‖?(b) Is the inequality ‖~u+ ~v‖ > ‖~u‖+ ‖~v‖ possible?

Fact 8.9 (Triangle Inequality). Let ~u and ~v be any vectors. Then

‖~u+ ~v‖ ≤ ‖~u‖+ ‖~v‖.

Next we will give a more rigorous definition of vectors and vector operations by coordinatizing themin R2.

Definition 8.10. A vector in the plane is an ordered pair ~v = (v1, v2) in R2. (We think of theinitial point of ~v as being the origin (0, 0) and the terminal point as the point (v1, v2).) Two vectors~u = (u1, u2) and ~v = (v1, v2) are considered equal if both u1 = v1 and u2 = v2, in which case we write~v = ~u. The number v1 is called the x-component of ~v and v2 is called the y-component of ~v.

The magnitude ‖~v‖ of a vector ~v = (v1, v2) is the number

‖~v‖ =√v21 + v22 .

Of course by the Pythagorean theorem, the magnitude of ~v is its genuine geometric length.If c is a scalar, we define the scalar product of c and ~v to be the vector c~v = (cv1, cv2). We define

the vector sum of ~u and ~v to be the vector ~u + ~v = (u1 + v1, u2 + v2), and the vector difference tobe the vector ~u− ~v = (u1 − v1, u2 − v2).

Remark 8.11. Note that according to our definition, the collection R2 may be imagined informally asboth the “set of all points in the plane” as well as the “set of all vectors in the plane.” This dualityis intentional. Given a pair (x, y) in R2, its geometric “role” as either a point or a vector must bedistinguished from the context in which its given.

Definition 8.12. Let P = (x1, y1) and Q = (x2, y2) be two points in R2. By the notation−−→PQ, we mean

the vector−−→PQ = (x2 − x1, y2 − y1).

Example 8.13. Let P = (x1, y1) and Q = (x2, y2) be two points in R2.

(a) Sketch a picture of−−→PQ.

(b) Compute the magnitude ‖−−→PQ‖.

Definition 8.14. Let ~u = (u1, u2) and ~v = (v1, v2) be vectors in R2 and let c be a scalar. Define thevector sum ~u+ ~v, the vector difference ~u− ~v, and the scalar product c~v as follows:

~u+ ~v = (u1 + v1, u2 + v2);~u− ~v = (u1 − v1, u2 − v2);

c~v = (cv1, cv2).

Definition 8.15. A unit vector is a vector of length 1. In particular we single out the coordinateunit vectors, which we permanently denote by ~i = (1, 0) and ~j = (0, 1).

Note that if ~v = (v1, v2) is any vector, then we may write


~v = (v1, v2) = (v1, 0) + (0, v2) = v1(1, 0) + v2(0, 1) = v1~i+ v2~j.

So in general ~v = (v1, v2) = v1~i+ v2~j; this gives us another notation for writing vectors.

Fact 8.16. If ~v is a non-zero vector, then the vector ~v‖~v‖ = 1

‖~v‖~v has the same direction as ~v, and

magnitude 1. In other words ~v‖~v‖ is the unique unit vector which points in the same direction as ~v.

9 Vectors in Three Dimensions

Definition 9.1. The set of all ordered triples (x, y, z) where x, y, and z are all real numbers is calledxyz-space or three-dimensional space, and denoted permanently by R3.

Remark 9.2 (How To Draw R3). We always draw three coordinate axes according to the followingconvention: The z-axis runs vertically (up and down). The y-axis runs horizontally (left and right). Thex-axis is drawn at an angle (down-left to up-right) in such a way that suggests the positive x-axis isshooting out of the page toward the reader, or shooting out of the blackboard toward the student.

Sometimes this traditional arrangement of the axes is described using the right-hand rule: if youstand at the origin (0, 0, 0) and stick your right hand out in the direction of the positive x-axis, thencurl your fingers toward the positive y-axis, then stick your thumb up, then your thumb should point inthe direction of the positive z-axis.

Example 9.3. (1) Plot the points P = (3, 4, 5) and Q = (−2,−3, 4) in xyz-space.(2) Compute the distance between P and Q.

Definition 9.4. A vector in three dimensions is an ordered triple in R3. The magnitude of avector ~v = (v1, v2, v3) in three dimensions is the quantity

‖~v‖ =√v21 + v22 + v23 .

The real numbers v1, v2, and v3 are called the x-component, y-component, and z-component of~v respectively. The notions of vector addition, vector subtraction, and scalar multiplication aredefined analogously to the two-dimensional case.

Definition 9.5. When working in R3, the unit coordinate vectors are the following three distin-guished vectors:

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

Note that for any vector ~v = (v1, v2, v3) we have ~v = v1~i+ v2~j + v3~k.

Example 9.6. Let ~u = (2,−4, 1) and ~v = (3, 0,−1).

(a) Find −4~u+ 2~v.(b) Find ‖~u− ~v‖.(c) Find the unique unit vector with the same direction as ~u− ~v.

(d) Write the unit vector from part (c) as a sum of scalar multiples of ~i, ~j, and ~k.

10 Dot Products

Definition 10.1. Given two vectors ~u = (u1, u2, u3) and ~v = (v1, v2, v3), we define the dot product,denoted ~u · ~v, to be the scalar given by

~u · ~v = u1v1 + u2v2 + u3v3.

Example 10.2. (a) Let ~u = (√

3, 1, 0) and ~v = (1,√

3, 0). Compute ~u · ~v.

(b) Let ~u = (2,−1,−2√

3) and ~v = (2,−2,√

3). Compute ~u · ~v.

Fact 10.3 (Properties of the Dot Product). For any vectors ~u, ~v, and ~w, and any scalar c,

(i) ~0 · ~v = ~v ·~0 = 0;


(ii) ~u · ~v = ~v · ~u (commutative property);(iii) (c~u) · ~v = ~u · (c~v) = c(~u · ~v);(iv) ~u · (~v + ~w) = ~u · ~v + ~u · ~w (distributive property of the dot product over vector addition); and(v) ~v · ~v = ‖~v‖2.

Fact 10.4 (Law of Cosines). If a triangle has angle measures A, B, and C and corresponding oppositeside lengths a, b, and c, then the following equalities hold:

c2 = a2 + b2 − 2ab cosC; b2 = a2 + c2 − 2ac cosB; a2 = b2 + c2 − 2bc cosA.

Theorem 10.5 (Dot Products Compute Angles). Let ~u = (u1, u2, u3) and ~v = (v1, v2, v3) be non-zerovectors, and let θ be the angle between ~u and ~v with 0 ≤ θ ≤ π. Then

~u · ~v = ‖~u‖ · ‖~v‖ cos θ.

Proof. Consider the triangle which has ~u and ~v for two of its sides. The third side is equal as a vectorto ~u− ~v, and hence the Law of Cosines (previous fact) implies that

‖~u− ~v‖2 = ‖~u‖2 + ‖~v‖2 − 2‖~u‖||~v‖ cos θ.

Now solving for ‖~u‖ · ‖~v‖ cos θ in the above, we get:

‖~u‖ · ‖~v‖ cos θ = 12 (‖~u− ~v‖2 − ‖~u‖2 − ‖~v‖2).

Now by the definition of magnitude in R3, we have ‖~u‖2 = (√u21 + u22 + u23)2 = u21+u22+u23. Similarly

we have ‖~v‖2 = v21 + v22 + v23 and

‖~u− ~v‖2 = (u1 − v1)2 + (u2 − v2)2 + (u3 − v3)2.

In that case, expanding out our equality from earlier and simplifying, we get:

‖~u‖||~v‖ cos θ =1

2(|~u− ~v|2 − ‖~u‖2 − ‖~v‖2)

=1

2((u1 − v1)2 + (u2 − v2)2 + (u3 − v2)2 − u21 − u22 − u23 − v21 − v22 − v23)

=1

2(2u1v1 + 2u2v2 + 2u3v3)

= u1v1 + u2v2 + u3v3

= ~u · ~v.

This proves the equality in the theorem. �

Corollary 10.6. Let ~u and ~v be non-zero vectors, and let θ be the angle between ~u and ~v with 0 ≤ θ ≤ π.Then

cos θ =~u · ~v‖~u‖ · ‖~v‖

.

Example 10.7. (a) Let ~u = (√

3, 1, 0) and ~v = (1,√

3, 0). Compute the angle θ between ~u and ~v.

(b) Let ~u = (2,−1,−2√

3) and ~v = (2,−2,√

3). Compute the angle θ between ~u and ~v.

Corollary 10.8. If θ is the angle between two vectors ~u and ~v, then θ = π2 (90 degrees) if and only if

~u · ~v = 0.

Definition 10.9. Two non-zero vectors ~u and ~v are called orthogonal if ~u · ~v = 0. (Orthogonal andperpendicular are synonyms in two dimensions.)


Definition 10.10. Given two vectors ~u and ~v, define the othogonal projection of ~u onto ~v, denotedproj~v ~u, to be the vector component of ~u which lies in the direction of ~v.

We have already observed that ~v‖~v‖ is the unit vector which points in the same direction as ~v. Ele-

mentary geometric considerations show that the magnitude of proj~v ~u is ‖~u‖ cos θ, where θ is the angleformed by ~u and ~v. So we immediately have

proj~v ~u = ‖~u‖ cos θ(

~v‖~v‖

).

Theorem 10.11. For any two vectors ~u and ~v,

proj~v ~u =

(~u · ~v~v · ~v

)~v.

Example 10.12. Find the projection of ~u = (5, 1,−3) onto ~v = (4, 4, 2).

11 Cross Products

Definition 11.1. Let ~u and ~v be vectors in R3. Define the cross product ~u× ~v to be the vector withmagnitude given by

‖~u× ~v‖ = ‖~u‖ · ‖~v‖ sin θ,

where θ is the angle between ~u and ~v (0 ≤ θ ≤ π), and with direction given by the right-hand rule:when you position the vectors so that they have the same initial point, point your right hand toward ~uand curl your fingers toward ~v, the direction of ~u×~v is the direction of your thumb, orthogonal to both~u and ~v. (Note that the right-hand rule only makes sense if ~u and ~v are not parallel; if they are parallel

and θ = 0 or θ = π, check that the given magnitude is 0 and hence ~u× ~v = ~0.)

Example 11.2 (TTYNP). Find the direction and magnitude of ~u × ~v, where ~u = (1, 1, 0) and ~v =

(1, 1,√

2).

Example 11.3. (a) In general is ~u · ~v = ~v · ~u?(b) In general is ~u× ~v = ~v × ~u?

Fact 11.4 (Properties of the Cross Product). Let ~u, ~v, and ~w be any vectors in R3 and let c be anyscalar. The following properties all hold.

(a) ~u× ~v = −(~v × ~u) (anticommutative property);(b) (c~u)× ~v = ~u× (c~v) = c(~u× ~v);(c) ~u× (~v+ ~w) = (~u×~v) + (~u× ~w) (left distributive property of cross product over vector addition); and(d) (~u+ ~v)× ~w = (~u× ~w) + (~v × ~w) (right distributive property of cross product over vector addition).

Example 11.5 (TTYNP). Evaluate all possible cross products of the coordinate vectors ~i, ~j, and ~k.

Example 11.6. Let ~u = (u1, u2, u3) and ~v = (v1, v2, v3) be any two vectors in R3. Find a closed formulafor the coordinates of ~u× ~v.

Theorem 11.7 (The Cross Product as a Determinant). Let ~u = (u1, u2, u3) and ~v = (v1, v2, v3). Then

~u× ~v = det

~i ~j ~ku1 u2 u3v1 v2 v3

= det

[u2 u3v2 v3

]~i− det

[u1 u3v1 v3

]~j + det

[u1 u2v1 v2

]~k.

Remark 11.8. Note that the “matrix” written in the formula above is not really a matrix in the sensethat students will have seen previously, as some its entries are vectors rather than real numbers. Hencethe formula above should be read as a formal determinant; that is, equality holds if one ignoresthe unusual context and just performs the usual determinant computation (using the Laplace cofactor

expansion), pretending that ~i, ~j, and ~k are real numbers, and treating each scalar multiplication asthough it were just multiplication of real numbers.

Example 11.9. Find a unit vector orthogonal to ~u = −~i+ 6~k and ~v = 2~i− 5~j − 3~k.


12 Equations of Planes in R3

Definition 12.1. Given a fixed point P0 = (x0, y0, z0) in R3 and a non-zero vector ~n = (a, b, c) in R2,the set of all points P = (x, y, z) in R3 for which P − P0 is orthogonal to ~n is called a plane. (Thevector ~n is called the normal vector to the plane.)

Example 12.2. Given a point P0 and a normal vector ~n as in the above definition, find an equation inthree variables whose graph is the plane determined by P0 and ~n.

Fact 12.3. The plane passing through P0 = (x0, y0, z0) with normal vector ~n = (a, b, c) is the set of allpoints (x, y, z) which satisfy the equation

a(x− x0) + b(y − y0) + c(z − z0) = 0.

Example 12.4. Find an equation of the plane that passes through the points (2,−1, 3), (1, 4, 0), and(0,−1, 5).

Solution. Let P , Q, R denote the points listed above, respectively. Compute the vectors−−→PQ =

(−1, 5,−3) and−→PR = (−2, 0, 2), which are parallel to the given plane.

All we need to find for the equation of a plane is a normal vector ~n; it suffices to find a vector ~n which

is orthogonal to−−→PQ and

−→PR. For this, compute the cross product:

~n =−−→PQ×

−→PR = (10, 8, 10).

Thus one possible equation for the plane is given by 10(x−2)+8(y+1)+10(z−3) = 0. This simplifiesto z = −x− 4

5y + 215 ; the student can verify that all three points indeed satisfy this equation. �

Definition 12.5. Two distinct planes are called parallel if their respective normal vectors are parallel.Two planes are orthogonal if their respective normal vectors are orthogonal.

Example 12.6 (TTYNP). Which of the following distinct planes are parallel and which are orthogonal?

Q: 2x− 3y + 6z = 12R: −x+ 3

2y − 3z = 14S: 6x+ 8y + 2z = 1

T : −9x− 12y − 3z = 7

13 Functions of Several Variables

Definition 13.1 (Set-Builder Notation). At this point we wish to familiarize the student with set-builder notation which is ubiquitous in mathematics. For convenience we will describe informallyusing examples, rather than give a formal definition. For an example, suppose we wish to formallydescribe the set D of all ordered pairs (x, y) for which x is twice y. Then we may write

A = {(x, y) : x = 2y}.

The above notation should be read as The set of all (x, y) in R2 such that x = 2y, which clearly andprecisely defines our set A. For another example, we could write

Y = {x : 5 ≤ x < 10},

which reads The set of all x in R such that 5 is less than or equal to x and x is strictly less than 10.Thus we are talking about the set Y = [5, 10) (written in interval notation). In general, given a set Aand a precise mathematical sentence P (x) about a variable x, the set-builder notation should be readas follows.

{ x : P (x)}“The set of all elements x such that sentence P (x) is true for the element x.


Note that for our present purposes we implicitly assume that variables x range over real numbers inR, pairs (x, y) range over R2, triples (x, y, z) range over R3, etc., depending on context.

Example 13.2 (TTYNP). What do the following sets look like? Sketch a picture or describe in words.

(a) {(x, y, z) ∈ R3 : 0 ≤ x ≤ 1, y ≥ 0}

(b) {(x, y, z) ∈ R3 : x2 + y1 = 1, z = 3}

(c) {(x, y, z) ∈ R3 : x2 + y2 = 1}

(d) {(x, y, z) ∈ R3 : (x− 2)2 + (y − 2)2 + (x+ 5)2 = 1}

Definition 13.3. A function in two variables f(x, y) is a function which assigns a unique (real)output to each input pair (x, y) from a particular set D in R2. The set D is called the domain of fand is often implicitly (rather than explicitly) described. The range of f is the set of all real numbersz = (x, y) which are assumed as (x, y) ranges over the domain D.

Example 13.4. Let g(x, y) =√

4− x2 − y2. Find the domain and range of g.

Definition 13.5. Let f(x, y) be a function in two variables. Define the graph of f to be the followingset:

{(x, y, z) : f(x, y) = z}.

Example 13.6. Sketch the graph of g(x, y) =√

4− x2 − y2.

Example 13.7. Sketch the graph of f(x, y) = 2x2 + 5y2.

Definition 13.8. If f(x, y) is a function in two variables, then a trace of the graph of f is the intersectionof the graph with any plane parallel to a coordinate plane. The traces which lie in the coordinate planesare called the xy-trace, the xz-trace, and the yz-trace respectively.

Example 13.9. Use traces to sketch the graphs of the following functions.

(a) f(x, y) = x− y2(b) f(x, y) = x sin y

Definition 13.10. Let f(x, y) be a function. Given any fixed real number z0, the level curve of f atz0 is the set of all (x, y) in R2 for which f(x, y) = z0.

Example 13.11. Sketch some of the level curves of the following functions.

(a) f(x, y) = y − x2 − 1

(b) f(x, y) = e−x2−y2

14 Limits and Continuity for Functions of Several Variables

Definition 14.1. A disk in R2 is the set

{(x, y) : ‖(x, y)− (a, b)‖ < r}

for some fixed point (a, b) and some fixed radius r.A punctured disk in R2 is a disk with its center point removed, i.e. a set of the form

{(x, y) : 0 < ‖(x, y)− (a, b)‖ < r}.

Definition 14.2. Let f(x, y) be a function in two variables. The function f has the limit L as (x, y)approaches (a, b), denoted

lim(x,y)→(a,b)

f(x, y) = L,


if for every ε > 0, there exists a punctured disk P centered at (a, b) such that

|f(x, y)− L| < ε whenever (x, y) is in P .

A function f(x, y) is continuous at a point (a, b) provided f(a, b) exists, lim(x,y)→(a,b)

f(x, y) exists,

and lim(x,y)→(a,b)

f(x, y) = f(a, b). We say f is continuous if f is continuous at every point in its domain.

Fact 14.3 (Familiar Functions Are Continuous). • Every polynomial in two variables (i.e. sumof functions of the form cxmyn) is defined and continuous at every point (a, b) in R2.

• Sums, differences, products, and quotients of continuous functions are continuous (provided thedenominator is nonzero.

• Compositions of continuous functions are continuous. Note that since we are dealing with mul-tivariable functions now, there are more kinds of function compositions available—for examplethe function sin(x2 + y) can be viewed as the composition f(g(x, y)) where f(x) = sinx andg(x, y) = x2 + y; or it can be viewed as the composition h(k(x), y) where h(x, y) = sin(x + y)and k(x) = x2; etc.

Remark 14.4 (Plug-n-Chug to Evaluate Limits). The previous fact can be taken to mean: functionswhich are built up in familiar ways from familiar functions we have encountered before are generallycontinuous, at any point where they are defined. And similarly to Calc 1, we can evaluate limits ofcontinous functions simply by plugging in the ordered pair at which we are evaluating the limit. (Seenext example.)

Example 14.5. Evaluate lim(x,y)→(2,8)

(3x2y +√xy), if it exists.

Remark 14.6 (Limits Along Paths). On the other hand, limits of multivariable functions can fail toexist. One way to see this is to consider a given limit at (a, b) by approaching (a, b) along different paths.

To understand this better, recall from Calc 1 a limit of the form limx→a

f(x) exists if and only if both

the left- and right-hand limits limx→a−

f(x) and limx→a+

f(x) both exist, and take the same value. So for

example, you can show that limx→01x does not exist because its left-hand limit is −∞ and its right-hand

limit is ∞.A related principle holds for multivariable limits: if the limit of f(x, y) at (a, b) takes different values

when computed along different paths toward (a, b), then lim(x,y)→(a,b)

f(x, y) does not exist.

Example 14.7. Show that lim(x,y)→(0,0)

x2 + 2y2

x2 + y2does not exist.

Solution. First check the limit at (a, b) by approaching along the x-axis. This corresponds to settingy = 0 and taking the limit as x→ 0. We get

limx→0

x2 + 0

x2 + 0= 1.

On the other hand let’s look at the limit approaching along the y-axis. We set x = 0 and let y → 0,to get

limy→0

0 + 2y2

0 + y2= 2.

Since the limits along two different paths are not equal, we get that lim(x,y)→(0,0)

x2 + 2y2

x2 + y2does not

exist. �

Example 14.8 (TTYNP). Show that lim(x,y)→(0,0)

xy

x2 + y2does not exist.


Example 14.9 (Seeing 0/0 as an Indeterminate Form). Consider the function f(x, y) = xy with domain

D = {(x, y) : x 6= 0}. Does lim(x,y)→(0,0)

x

yexist, and if so what is its value?

Solution Strategy. Consider approaching (0, 0) along lines of the form y = mx. Do you get the samelimit regardless of the choice of m? �

Example 14.10 (How Should We Define 00?). Consider the function f(x, y) = xy with domain D ={(x, y) : x > 0}. Does lim

(x,y)→(0,0)xy exist, and if so what is its value?

Solution Strategy. Consider approaching (0, 0) along the graph of y = logxA for different positive con-stants A. �

15 Partial Derivatives

Definition 15.1. Let f(x, y) be a function in two variables. The partial derivative of f with respectto x is

fx(x, y) = ∂∂xf(x, y) = lim

h→0

f(x+ h, y)− f(x, y)

h,

provided the limit exists. The partial derivative of f with respect to y is

fy(x, y) = ∂∂yf(x, y) = lim

h→0

f(x, y + h)− f(x, y)

h,

provided the limit exists.

Example 15.2. Let f(x, y) = x2 − y2 + 4.

(a) Compute ∂f∂x and ∂f

∂y .

(b) Evaluate each derivative at (2,−4).

Solution to (a). We can compute the partial derivative ∂f∂x by just holding y constant and deriving on

the x variable. So

fx(x, y) = 2x.

Similarly, fy(x, y) = −2y. �

Example 15.3 (TTYNP). Compute the partial derivatives of the following functions.

(a) f(x, y) = sinxy(b) g(x, y) = x2exy

Definition 15.4. Let f(x, y) be a function in two variables. The second-order partial derivativesof f are the following four functions (written with both available notations):

∂∂x

(∂f∂x

)= ∂2f

∂x2 or (fx)x = fxx;

∂∂y

(∂f∂y

)= ∂2f

∂y2 or (fy)y = fyy;

∂∂x

(∂f∂y

)= ∂2f

∂x∂y or (fy)x = fyx;

∂∂x

(∂f∂y

)= ∂2f

∂y∂x or (fx)y = fxy.

The latter two derivatives above are called mixed partial derivatives.

Example 15.5. Compute the second-order partial derivatives of f(x, y) = 3x4y − 2xy + 5xy3.

Theorem 15.6 (Clairaut’s Theorem). Assume that f is defined on a disk D of R2, and fxy and fyxare continuous at every point of D. Then fxy = fyx at every point of D.


16 Differentiability and Tangent Planes

Definition 16.1. A multivariable function f(x, y) is differentiable at a point (a, b) only if there existsa tangent plane L to f at (a, b). The tangent plane at (a, b) should give a good approximation tothe secant lines connecting (a, b, f(a, b)) and (a+ h1, b+ h2, f(a+ h1, b+ h2)) for all choices of (small)~h = (h1, h2).

More precisely, we say f(x, y) is differentiable at (a, b) if the following conditions hold:

• there is a function L(x, y) = m1(x− a) +m2(x− b) + z0 whose graph is a plane containing thepoint (a, b, f(a, b)); and

• lim(h1,h2)→0

f(a+ h1, b+ h2)− f(a, b)− (L(a+ h1, b+ h2)− L(a, b))

‖(h1, h2)‖= 0.

Example 16.2. Suppose f and L are as above. What must the values of z0, m1, and m2 be?

Fact 16.3 (Equation of the Tangent Plane). If f(x, y) is differentiable at (a, b) and L(x, y) is its tangentplane at (a, b), then

L(x, y) = fx(a, b)(x− a) + fy(a, b)(y − b) + f(a, b).

Fact 16.4. Let f(x, y) be a function in two variables. If fx and fy exist and are continuous at everypoint of some open set D, then f(x, y) is differentiable at every point of D.

Example 16.5. Show that f(x, y) = 5x + 4y2 is differentiable, and find the equation of the tangentplane at (2, 1).

Example 16.6. Find the equation of the tangent plane to f(x, y) = 1 + x ln(xy − 5) at (2, 3).

17 Multivariable Chain Rules

Theorem 17.1 (Chain Rule (Case 1)). Let f be a differentiable function in two variables; and let x(t)and y(t) be differentiable functions of a single variable. Then the composition f(x(t), y(t)) is differen-tiable, and has derivative

ddtf(x(t), y(t)) = (fx(x(t), y(t)), fy(x(t), y(t))) · (x′(t), y′(t)) = fx(x(t), y(t))x′(t) + fy(x(t), y(t))y′(t).

Written in shorthand with the Leibniz notation, we have

df

dt=∂f

∂x

dx

dt+∂f

∂y

dy

dt=

(∂f

∂x,∂f

∂y

)·(dx

dt,dy

dt

).

Example 17.2. Let f(x, y) = x2 − 3y2 + 20 and let x(t) = 2 cos t, y(t) = 2 sin t. Find ddtf(x(t), y(t))

and evaluate it at t = π4 .

Theorem 17.3 (Chain Rule (Case 2)). Let z be a differentiable function of x and y, where x and y aredifferentiable functions of s and t. Then

∂z

∂s=∂z

∂x

∂x

∂s+∂z

∂y

∂y

∂sand

∂z

∂t=∂z

∂x

∂x

∂t+∂z

∂y

∂y

∂t.

Written with dot products, we have:

∂z

∂s=

(∂z

∂x,∂z

∂y

)·(∂x

∂s,∂y

∂s

)and

∂z

∂t=

(∂z

∂x,∂z

∂y

)·(∂x

∂t,∂y

∂t

).

Example 17.4. Let z = sin 2x cos 3y, where x = s+ t and y = s− t. Evaluate ∂z∂s and ∂z

∂t .


Example 17.5 (Special Exercise - Implicit Differentiation). Use a function in two variables and the

chain rule to find dydx , where sin(xy) + πy2 = x.

Theorem 17.6. Let F be a differentiable function in two variables, and suppose a relationship betweeny and x is defined implicitly by the rule F (x, y) = 0. If Fy 6= 0, then

dydx = −FxFy .

Proof. Derive both sides of F (x, y) = 0 using the multivariable chain rule to get

(Fx, Fy) · (1, y′) = 0.

Now solve for y′. �

Example 17.7. Find dydx , where x4 + 3x2y2 − y = 10.

18 Directional Derivatives and the Gradient

Definition 18.1. Let f be differentiable at (a, b) and let ~u = (cos θ, sin θ) be a unit vector in R2. Thedirectional derivative of f at (a, b) in the direction of ~u is

D~uf(a, b) = limh→0

f(a+ h cos θ, b+ h sin θ)− f(a, b)

h,

provided the limit exists.

Theorem 18.2. Let f be differentiable at (a, b) and let ~u = (u1, u2) be a unit vector in R2. Then

D~uf(a, b) = (fx(a, b), fy(a, b)) · (u1, u2).

Proof. Define a new function g(s) (real inputs and real outputs) by the rule

g(s) = f(a+ su1, b+ su2).

Geometrically, the graph of g is a “cross-section” of the graph of f passing through the point (a, b)parallel to vector ~u. It is clear from the definition of the directional derivative that

D~uf(a, b) = g′(0).

Now setting x(s) = a+ su1 and y(s) = b+ su2 (so g(s) = (x(s), y(s)) and applying the chain rule, weget

D~uf(a, b) = g′(0)

= fx(x(0), y(0))x′(0) + fy(x(0), y(0))y′(0)

= fx(a, b)u1 + fy(a, b)u2)

= (fx(a, b), fy(a, b)) · (u1, u2).

�

Example 18.3. Consider the paraboloid z = f(x, y) = 14 (x2+2y2)+2 and the unit vectors ~u = ( 1√

2, 1√

2)

and ~v = ( 12 ,−

√32 ).

(a) Find the directional derivative of f at (3, 2) in the directions of ~u and ~v.(b) Graph the surface and interpret the directional derivatives.

Definition 18.4. Let f(x, y) be differentiable. The gradient of f at (x, y) is the function

∇f(x, y) = (fx(x, y), fy(x, y)).

Remark 18.5. The symbol ∇ is spoken “nabla,” “atled,” or “del.”


Example 18.6. Find ∇f and ∇f(3, 2) for f(x, y) = x2 + 2xy − y3.

Example 18.7. Let f(x, y) = 3− x2

10 + xy2

10 .

(a) Compute ∇f(3,−1).(b) Compute D~uf(3,−1) where ~u = ( 1√

2, 1√

2).

(c) Compute the directional derivative of f at (3,−1) in the direction of the vector (3, 4).

Theorem 18.8. Let f be differentiable at (a, b).

(1) f has its maximum rate of increase at (a, b) in the direction of the gradient ∇f(a, b). The rateof increase in this direction is ‖∇f(a, b)‖.

(2) f has its maximum rate of decrease at (a, b) in the direction of −∇f(a, b). The rate of decreasein this direction is −‖∇f(a, b)‖.

(3) If ~u is orthogonal to ∇f(a, b), then D~uf(a, b) = 0.

Proof. For any unit vector ~u, we have

D~u = ∇f(a, b) · ~u= ‖∇f(a, b)‖‖~u‖ cos θ

= ‖∇f(a, b)‖ cos θ,

where θ is the angle between ∇f(a, b) and ~u. Then cos θ is maximized when θ = 0 and minimized whenθ = π, which proves statements (1) and (2) above. If ∇f(a, b) and ~u are orthogonal then θ = π

2 andhence cos θ = 0; this shows statement (3). �

Example 18.9. Consider the bowl-shaped paraboloid z = f(x, y) = 4 + x2 + 3y2.

(a) If you are located at the point (2,− 12 ,

354 ) on the paraboloid, in which direction should you move in

order to ascend the surface at the maximum rate? How quickly will you ascend?(b) If you are at the point (3, 1, 16), in which directions may you walk in order to neither gain nor lose

height?

19 Optimization in Several Variables

Definition 19.1. Let f be a function in two variables. We say f has a local maximum at (a, b) isthere is some disk D containing (a, b) such that f(a, b) ≥ f(x, y) for all (x, y) in D. We say that f hasa local minimum at (a, b) if there is some disk D containing (a, b) such that f(a, b) ≤ f(x, y) for all(x, y) in D. In either case, we say that f has a local extremum at (a, b).

A point (a, b) is a critical point of f if either

(1) fx(a, b) = fy(a, b) = 0 or(2) one (or both) of fx and fy does not exist at (a, b).

Fact 19.2. If f has a local maximum or minimum at (a, b), then (a, b) is a critical point of f .

Example 19.3. Find the critical points of f(x, y) = xy(x− 2)(y + 3).

Example 19.4 (TTYNP). Find all local maxima and/or minima of the following functions.

(a) f(x, y) = 4− x2 − y2(b) f(x, y) = 4− x2 + y2

Definition 19.5. Let f be a function in two variables. We say that f has a saddle point at (a, b)if (a, b) is a critical point, but for every disk D containing (a, b) there are points (x, y) in D for whichf(x, y) > f(a, b) and points (x, y) in D for which f(x, y) < f(a, b) (in other words f has neither a minnor a max at (a, b)).


Theorem 19.6 (Second Derivative Test). Suppose that the second partial derivatives of f(x, y) arecontinuous in a disk containing (a, b), where fx(a, b) = fy(a, b) = 0. Set

D(x, y) = [fxxfyy − f2xy](x, y).

(1) If D(a, b) > 0 and fxx(a, b) < 0, then f has a local maximum value at (a, b).(2) If D(a, b) > 0 and fxx(a, b) > 0, then f has a local minimum value at (a, b).(3) If D(a, b) < 0, then f has a saddle point at (a, b).(4) If D(a, b) = 0, the test is inconclusive.

Definition 19.7. The quantity D(x, y) in the above Theorem 19.6 is called the discriminant of f .

D(x, y) is the determinant of the Hessian matrix

[fxx fxyfyx fyy

].

Example 19.8. Classify all the critical points of f(x, y) = x2 + 2y2 − 4x+ 4y + 6.

Example 19.9. Classify all the critical points of f(x, y) = xy(x− 2)(y + 3).

Example 19.10. A shipping company handles rectangular boxes provided the sum of the length, width,and height of the box does not exceed 96 in. Find the dimensions of the box that meets the conditionat has the largest volume.

Definition 19.11. If f(x, y) ≤ f(a, b) for all (x, y) in the domain of f , then f has an absolutemaximum at (a, b). If f(x, y) ≥ f(a, b) for all (x, y) in the domain of f , then f has an absoluteminimum at (a, b).

Example 19.12. Find the absolute maximum and minimum values, if they exist, of f(x, y) = 4−x2−y2on the open disk R = {(x, y) : x2 + y2 < 1}.

Definition 19.13. Let A be a subset of R2. A point (a, b) is called a boundary point of A if everydisk centered at (a, b) contains both a point in A and a point outside of A.

A is called a closed set if A contains all of its boundary points.

Fact 19.14 (Extreme Value Theorem). If f is continuous on a closed bounded domain D in R2, thenf obtains an absolute maximum and absolute minimum value on D.

Example 19.15. Find the absolute maximum and minimum values of f(x, y) = x2 + y2 − 2x+ 2y + 5on the set D = {(x, y) : x2 + y2 ≤ 4}.

Solution. First find the critical points by solving the system of equations:

0 = fx(x, y) = 2x− 2

0 = fy(x, y) = 2y + 2

We get only one solution: the sole critical point (1,−1). (We make a mental note that (1,−1) reallydoes lie within our domain D, so we do not discard it.) If f obtains its maximum or minimum withinthe interior of D, then it must occur at this point (1,−1).

On the other hand, it is possible that f obtains its max/min along the boundary of the disk D. Sowe search for critical values along the boundary as well. All points on the boundary of D have the form:

(x, y) = (2 cos θ, 2 sin θ) for 0 ≤ θ ≤ 2π.

Since we would like to find the max/min of f(x, y) along the boundary, we consider the function:


g(θ) = f(2 cos θ, 2 sin θ)

= 4 cos2 θ + 4 sin2 θ − 4 cos θ + 4 sin θ + 5

= −4 cos θ + 4 sin θ + 9.

Setting 0 = g′(θ) = 4 sin θ + 4 cos θ, we get tan θ = −1, so θ = 3π4 and θ = 7π

4 are the two critical

points of g. These correspond to two points in R2, namely:

(2 cos 3π4 , 2 sin 3π

4 ) = (−√

2,√

2) and (2 cos 7π4 , 2 sin 7π

4 ) = (√

2,−√

2).

We finish by evaluating f at each of our three test points:

f(1,−1) = 3,

f(−√

2,√

2) = 9 + 4√

2,

f(√

2,−√

2) = 9− 4√

2.

Comparing the size of the three outputs above, we see that f obtains its absolute maximum at(−√

2,√

2), and its absolute minimum at (1,−1). �

Example 19.16. Find the absolute maximum and minimum values of f(x, y) = 6−x2− 4y2 on the setR = {(x, y) : −2 ≤ x ≤ 2,−1 ≤ y ≤ 1}.

20 Optimizing with a Constraint: The Method of Lagrange Multipliers

Fact 20.1 (Method of Lagrange Multipliers). Assume that f(x, y) and g(x, y) are differentiable func-tions. If f(x, y) has a local minimum or a local maximum on the constraint curve g(x, y) = 0 at the

point (a, b), and if ∇g(a, b) 6= ~0, then there is a scalar λ (called a LaGrange multiplier) such that

∇f(a, b) = λ∇g(a, b).

Definition 20.2. A point (a, b) satisfying the equation ∇f(a, b) = λ∇g(a, b) for some scalar λ is saidto be a critical point for optimizing f(x, y) with respect to the constraint g(x, y) = 0.

Example 20.3. Find the extreme values of f(x, y) = 2x+ 5y on the ellipse (x4 )2 + (y3 )2 = 1.

Solution. Set g(x, y) = (x4 )2 + (y3 )2 = 1, so the constraint in the problem is given by g(x, y) = 0. Toapply the method of Lagrange multipliers, we want to consider the equation ∇f(a, b) = λg(a, b) whereλ is a scalar. Computing it out, we get:

(2, 5) = λ(a8 ,2b9 ).

Taken together with the condition g(a, b) = 0, we have really obtained a system of three equations inthree variables a, b, λ:


2 =λa

8

5 =2λb

9a2

16+b2

9= 1

The easiest way to solve this system is to write a = 16λ , b = 45

2λ and plug a, b into the third equation.This yields

2894 = λ2 or λ = ±

√2892 .

So our two critical points for this optimization problem are

(a, b) = ( 32√289

, 45√289

) and (a, b) = (− 32√289

,− 45√289

).

Plugging these critical points into the function f(x, y), we can see that the former gives a maximum

of√

289, while the latter gives a minimum of −√

289. �

Remark 20.4. For optimization problems involving functions of two variables (f(x, y) subject to theconstraint g(x, y) = 0) it is debatable whether the method of Lagrange multipliers is more or lesswork than the optimization methods we used in the previous section (See Example 19.15 for example).However, a big advantage of Lagrange multipliers is that the method easily extends to optimizationproblems involving 3 or more variables– see example below.

Example 20.5. Find the point on the plane x2 + y

4 + z4 = 1 which is closest to the origin in R3.

21 Double Integrals over Rectangular Regions

Definition 21.1. A region R in R2 is called a rectangle or rectangular region if R is of the form

R = {(x, y) : a ≤ x ≤ b, c ≤ y ≤ d}

for some real numbers a ≤ b, c ≤ d. We also denote R by

R = [a, b]× [c, d].

Definition 21.2. Let f be a function in two variables, and let R = [a, b]× [c, d] be a rectangular regionin R2. Given any positive integer n, set ∆x = b−a

n and ∆y = d−cn . For each integer k with 0 ≤ k ≤ n,

set

xk = a+ k∆y and yk = c+ k∆y.

For each pair of integers j, k with 1 ≤ j ≤ n and 1 ≤ k ≤ n, let (xj , yk) be a point chosen arbitrarilyfrom the rectangular region [xj−1, xj ]× [yk−1, yk].

We define the double integral, or double definite integral, of f over R to be∫ ∫Rf(x, y)d(x, y) = lim

n→∞

n∑j=1

n∑k=1

f(xj , yk)∆x∆y,


provided the limit exists and is independent of the choices of (xk, yk). (Note: Our notation differsslightly from Stewart here.) If the limit exists we say that f is integrable over R. If f is non-negativeon R, then the double definite integral corresponds to the volume of the solid bounded by the graph off over R.

Theorem 21.3 (Fubini’s Theorem). Let f be continuous on the rectangular region R = [a, b] × [c, d].Then f is integrable, and∫ ∫

Rf(x, y)d(x, y) =

∫ dc

∫ baf(x, y)dxdy =

∫ ba

∫ dcf(x, y)dydx.

Example 21.4. Find the volume of the solid bounded by the surface z = 4 + 9x2y2 over the regionR = [−1, 1]× [0, 2]. Use both possible orders of integration.

Example 21.5. Evaluate∫ ∫

Rxexyd(x, y), where R = [0, 1]× [0, ln 2].


R1

(x+y)2 d(x, y), where R = [1, 2]× [0, 1].

Example 21.7. Calculate∫ 2

0

∫ π/20

ex cos ydydx.

22 Double Integrals over General Regions

Fact 22.1. Let g and h be continuous functions in one variable. Suppose R is a region in R2 boundedbelow and above by the graphs of y = g(x) and y = h(x) respectively, and the lines x = a and x = b. Iff is continuous on R, then ∫ ∫

Rf(x, y)d(x, y) =

∫ ba

∫ h(x)g(x)

f(x, y)dydx.

Alternatively, if R is bounded on the left and right by the graphs of x = g(y) and x = h(y) respectively,and the lines y = c and y = d, and f is continuous on R, then

∫ ∫Rf(x, y)d(x, y) =

∫ dc

∫ h(y)g(y)

f(x, y)dxdy.

Example 22.2. Compute the integral∫ ∫

R2x2yd(x, y), where R is the region bounded by the parabolas

y = 3x2 and y = 16− x2.

Example 22.3. Compute the volume of the solid below the surface f(x, y) = 2+ 1y and above the region

R in the xy-plane bounded by the lines y = x, y = 8− x, and y = 1.


Rey

2

d(x, y), where R is the triangle bounded by the lines x = 0, y = 12x,

and y = 2.

Example 22.5. Evaluate∫√π0

∫√πy

sinx2dxdy.

23 Integration in Polar Coordinates

Definition 23.1. To each pair (r, θ) of real numbers with r ≥ 0, we (implicitly) associate the pair (x, y)in R2 where x = r cos θ and y = r sin θ. We refer to (r, θ) as the polar coordinates of (x, y). Noticethat any pair (x, y) has infinitely many polar coordinatizations, as (r, θ + 2πn) gives the same (x, y) forany choice of integer n.

A polar rectangle is a set of the form {(r, θ) : 0 ≤ a ≤ r ≤ b, α ≤ θ ≤ β}, where a, b, α, β are realnumbers with β − α ≤ 2π.

Theorem 23.2. Let f be continuous on the region R = {(r, θ) : 0 ≤ a ≤ r ≤ b, α ≤ θ} (where (r, θ)corresponds to a point (x, y) = (r cos θ, r sin θ) as in the above definition). Then

∫ ∫Rf(r, θ)d(x, y) =

∫ βα

∫ baf(r, θ)rdrdθ.

Example 23.3. Find the volume of the solid bounded by the paraboloid z = 9 − x2 − y2 and thexy-plane.


Example 23.4. Find the volume of the region bounded beneath the surface z = xy+ 10 and above theannular region R = {(r, θ) : 2 ≤ r ≤ 4, 0 ≤ θ ≤ 2π}.

Fact 23.5. Let f be continuous on the region R = {(r, θ) : 0 ≤ g(θ) ≤ r ≤ h(θ), α ≤ θ ≤ β}, where gand h are continuous functions of θ. Then

∫ ∫Df(r, θ)d(x, y) =

∫ βα

∫ h(θ)g(θ)

f(r, θ)rdrdθ.


0

∫√9−x2

0

√x2 + y2dydx.

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