FALL 2021 MATH 147 Lecture Notes
1 Multivariable Functions
1.1 12.1 Introduction to Multivariable Functions
For each of the followings find the range and domain of the functions 1. f (x, y) = x2+ y2
2. g(x, y, z) = x2+y12+z2
3. h(x1, x2, ..., xn) = x1+ x2+ x3+ ... + xn 4. t(x, y) = √ 1
1−x2−y2
1.2 Graphing Functions of two variables
To best understand a function of two variables z = f (x, y) we start by understanding its level curves, that is f (x, y) = c for different values of c.The level curves do not give a complete picture of the graph of f (x, y).
for example
1. The level curves of f (x, y) = x2+y2are circles of increasing radii (Desmos) but the graph is a paraboloid
2. The level curves of f (x, y) = p
x2+ y2 are also circles of increasing radii but the graph is a cone
Note that the difference between them can be seen by setting x = c and in particular x = 0.
Example:
Graph the function z = x2+y1 2
1.3 Level Surfaces
It is very difficult to produce a meaningful graph of a function of three variables. A function f one variable is a curve drawn in 2 dimensions; a function of two variables is a surface drawn in 3 dimensions; a function of three variables is a hypersurface drawn in 4 dimensions
level surfaces: Given w = f (x; y; z), the level surface at w = c is the surface in space formed by all points (x; y; z) where f (x; y; z) = c.
Example: The level surfaces of w = x2+ y2+ z2 are ?
Homework 1a: Problems indicated by section number refer to our main text Apex Calculus II by G.
Hartman. Problems assigned from Marsden and Weinstein will be preceded by MW. For problems assigned from either textbook, turn in the boldface numbers only. Problems assigned from the Openstax calculus 3 textbook will be preceded by OS.
1. Experiment with graphing functions of two variables with the online graphing function Desmos. Turn in a couple of pictures of interesting surfaces you created
2. Section 12.1: 11,13,15,20,21,23,14,17,22
Before moving to limits let’s looked at pictures of graphs the six basic quadric surfaces: elliptic paraboloid, hyperbolic paraboloid, ellipsoid, double cone, hyperboloid of one sheet, and hyperboloid of two sheets,(See here).
Definition 1 The distance between (a, b) and (c, d) in R2 is d =p
(a − c)2+ (b − d)2.
We use the notation for the distance as |(a, b)−(c, d)|. Similarly the distance from (x1, x2, ..., xn) to (y1, y2, ..., yn) in Rn is
d =p
(x1− x2)2+ (x2− y2)2+ ... + (xn− yn)2.
1.4 Limits and Continuity of Multivariable Functions
Before calculating some examples, note that the following rules for taking limits hold, and that these are the same rules that apply when taking limits of function of one variable.
Now is time to look at an example.
Example 1 Evaluate the limit
lim
(x,y)→(1,2)
3x2y + 4xy x + y Example 2 For the function
f (x) =
(x, if x 6= 1 2, if x = 1 evaluate lim(x,y)→(1,2)f (x).(x2+ 3xy).
The above examples makes limits in two variables looks easy and nice, well this is quite far away from the truth. To that end, consider the example below
Example 3 Evaluate the limit
lim
(x,y)→(0,0)
xy2 x2+ y2.
Now we will look at an example where the limit was shown not to exist by approaching the target in question along different paths.
Example 4 Evaluate the limit by first taking the limit along the lines y = 0 and y = x lim
(x,y)→(0,0)
y2 x2+ y2. What is your conclusion?
Note that having a uniform limit as (x, y) approaches (0, 0) along every straight line does not imply existence of limit.
Example 5 Evaluate the limit
lim
(x,y)→(0,0)
4xy2 x2+ 3y4 along y = mx. What is your conclusion?
Example 6 Evaluate the limit
lim
(x,y)→(0,0)
x2− xy
√x −√ y Next we very limits using the , δ definition.
Example 7 Verify using the , δ definition
•
lim
(x,y)→(a,b)x = a
•
lim
(x,y)→(0,0)
x4 x2+ y2 = 0 Finally we give the definition of continuity.
And also the following properties
Homework 1b: Section 12.2: 10,12,15,17,21,11,14,18,20.
Turn in the solutions of the following problems:
• For
lim
(x,y)→(0,0)
x3y x6+ 2y2,
show that the limit along any line through the origin exists and equals 0, but if we take the limit along the curve y = x3, the limit is not zero. What conclusion can you draw from this?
• Determine if the limit exists, and if so what is its value:
lim
(x,y)→(0,0)
x5+ y4− 3x4y + 2x2+ 2y2 x2+ y2
• Determine if the function
f (x, y) =
(x3+x2+xy2+y2
x2+y2 (x, y) 6= (0, 0)
2 (x, y) = (0, 0)
is continuous at (0, 0)
• At what values of (x, y) is the function f(x, y) = cos (xx22+1−y2) continuous ?
• Optional Bonus Problem. 3 points. Use the items below to show that for f(x, y) = xyxx22−y+y22, lim(x,y)→(0,0) exists, and then find its value
– Prove that if a, b are real numbers, then 2|ab| ≤ a2+ b2
– Use the just proven fact above to show that if 0 < |(x, y)| < δ, then |f (x, y)| < δ22. Homework 1a and 1b are due on Monday August 30th in class.
1.5 Partial Derivatives:
Geometrically we can think of ∂f∂y(a, b) as the slope of the line passing through the point (a, b, f (a, b)) tangent to the curve z = f (a, y) lying on the graph of z = f (x, y)
Example 8 Using the limit definition to calculate the followings and then verify your results by direct calculation
1. fx(1, 1) where f (x, y) = 2x + 3y 2. gx(2, −1) where g(x, y) = x3y + 3xy2
Example 9 Find all the first order partial derivatives of h(x, y) = 6x2y3ex2+2y2+ 5 cos (xy)
Before we move on to talk about higher partial derivatives, we will describe a line in R3. If P = (x0, y0, z0) ∈ R3 is a point and ~D = (d1, d2, d3) is a direction vector, then the line through P in the di- rection of ~D is given parametrically by
L(t) = P + t. ~D = (x0+ td1, y0+ td2, z0+ td3).
We will use this to describe a tangent line of the graph z = f (x, y) in the directions given by the x and y axes.
We can extend the definition of partial derivatives to n− dimensions. To that end Definition 2 The partial derivative of f (x1, ..., xn) with respect to xi at (a1, ..., an) is
∂f
∂xi(a1, a2, ..., an) = lim
h→0
f (a1, ..., ai−1, ai+ h, ai+1, ..., an) − f (a1, a2, ..., an) h
if the limit exists.
Example 10 Find all the partial derivatives of the function h(x, y, z, u, w) = x3y2z5+3exu2w2+cos (u2+ w2+ x2) Just like in single variable, all the rules of differentiation’s also applies to the partial derivatives except chain rule which will be treated separately.
Next we consider second order partial derivatives, that are defined as follows:
For most functions fxy= fyx. We will study the theorem that guarantees such equality.
Example 11 Find all the second order partial derivatives of f (x, y) = x3y2+ 7xy5+ exy
1.6 Tangent Lines and Tangent Planes
We begin by recalling that to find a parametric equation of a line in R3 passing through a point P in the direction of the vector ~D is L(t) = P +t. ~D. Also recall that a plane in R3is determined by a point P = (a, b, c) and a normal vector ~n = (n1, n2, n3) and the equation of the plane is given by
~
n. ~P X = n1(x − a) + n2(y − b) + n3(z − c) = 0.
We know that ∂f∂x(a, b) gives the slope of the line tangent in the x direction to the graph of z = f (x, y) at (a, b, f (a, b)). So in the plane y = b, the equation of the tangent line is z = ∂f∂x(a, b)(x − a) + f (a, b). If we replace x by t and taking into account that the tangent line is in R3 we obtained the equation
Lx(t) = (t, b,∂f
∂x(a, b)(t − a) + f (a, b)) = (a, b, f (a, b)) + (t − a)(1, 0,∂f
∂x(a, b)).
Now if the line passes through the point (a, b, f (a, b)) when t = 0, gives Lx(t) = (a, b, f (a, b)) + t(1, 0,∂f
∂x(a, b)).
Example 12 Calculate the tangent line Lx(t) for f (x, y) = x3y2+ 2xy + 7 at the point (−1, 2, f (−1, 2)).
Similarly, the tangent line in the y direction to the graph of z = f (x, y) at (a, b, f (a, b)) is given by Ly(t) = (a, b, f (a, b)) + t(0, 1,∂f
∂y(a, b)).
Example 13 Find Ly(t) for the example above.
The general analysis above yields two tangent vectors to the graph of z = f (x, y) at (a, b, f (a, b)), namely, (1, 0,∂f∂x(a, b)) and (0, 1,∂f∂y(a, b)). Since cross product of these vectors gives a normal vector to the tangent plane, we calculated
(1, 0,∂f
∂x(a, b)) × (0, 1,∂f
∂y(a, b)) =
i j k
1 0 ∂f∂x(a, b) 0 1 ∂f∂y(a, b)
= −∂f
∂x(a, b)~i −∂f
∂y(a, b)~j + ~k.
Thus an equation for the plane tangent to the graph of z = f (x, y) at (a, b, f (a, b)) is given by
−∂f
∂x(a, b)(x − a) −∂f
∂y(a, b)(y − b) + 1.(z − f (a, b)) = 0 Rewriting this gives
z = ∂f
∂x(a, b)(x − a) +∂f
∂y(a, b)(y − b) + f (a, b).
Example 14 Find the tangent plane for the above example.
1.7 Directional derivatives and Gradient
Definition 3 The directional derivative of the function f (x, y), at the point (a, b) in the direction of the unit vector ~u = u1~i + u2~j is
D~uf (a, b) = lim
h→0
f (a + hu1+ u2h) − f (a, b)
h .
Remark 1 1. D~uf (a, b) is the rate of change of f (x, y) at the point (a, b) along the line (a, b) + t.~u, assuming the limit exists.
2. Taking unit vector is very important so that the quantity calculated only depends upon the function f and the direction of the direction vector, and not the magnitude of the direction vector.
Example 15 Using the above definition to calculate D~uf (1, 2) for f (x, y) = x2y in the direction of ~u =
√2
2~i + √22~j
Example 16 Show that D~uf (x, y) = (2xy~i + x2~j).~u = (∂f∂x~i +∂f∂y~j).~u From the above example we can generalize the directional derivatives as
D~uf (a,b)= (∂f
∂x(a, b)~i +∂f
∂y(a, b)~j).~u This lead to the definition of gradient ∇f of a scalar vector f.
Definition 4 For a scalar function f the gradient function is defined as:
• For f(x, y), ∇ = ∂f∂x~i + ∂f∂y~j
• For f(x, y, z), ∇ = ∂f∂x~i + ∂f∂y~j + ∂f∂z~k
• For f(x1, x2, ..., xn), ∇ = ∂x∂f
1e~1+ ∂x∂f
2e~2+ ... +∂x∂f
ne~n, where the vector ~ei is the vector in Rn all of whose coordinates are zero, except the ith coordinate, which is 1.
Remark 2 The ∇ is a differential operator that turns scalar value functions into a vector valued function function through differentiation process. As such, one can expect ∇ to have similar properties that hold upon differentiation. Indeed, the following properties hold:
1. ∇(f + g) = ∇f + ∇g.
2. ∇(cf ) = c∇f, for the constant c.
3. ∇(f g) = g∇f + f ∇g.
4. For h(t), ∇(h(f )) = h0(f )∇f.
Example 17 Calculate the directional derivative of f = x2y3+ 4xz2+ 17, at (1, −1, 1) in the direction of the vector 2~i + 3~j + 4~k.
To formally derived the above formula for directional derivative, first we will assume the simplest version of the chain rule
df (x(t), y(t), z(t))
dt =∂f
∂x.dx dt +∂f
∂y.dy dt +∂f
∂z.dz dt.
Now if we write ~u = u1~i + u2~j + u3~k, and set g(t) := f (a + tu1, b + tu2, c + tu3), then g0(0) = lim
h→0
g(0 + h) − g(0)
h = D~uf (a, b, c) But then note that by chain rule we also have
g0(0) = fx(a, b, c).u1+ fy(a, b, c).u2+ fz(a, b, c).u3= ∇f (a, b, c).~u and thus
D~uf (a, b, c) = ∇f (a, b, c).~u
Some geometrical properties of Directional derivatives: Note that since ∇f (a, b, c).~u = k∇f (a, b, c)k.k~uk cos θ = k∇f (a, b, c)k cos θ where θ is the angle between ∇f (a, b, c) and ~u, then D~uf (a, b, c) achieves its highest value
when ~u points in the same direction as ∇f (a, b, c) and more over the rate of change in that direction is k∇f (a, b, c)k. Similarly, −∇f (a, b, c) points in the direction in which the rate of change at (a, b, c) is the least, and that rate of change is −k∇f (a, b, c)k.
Homework 2
Read section 15.1 and 15.2 in Marsden and Weinstein and section 12.6 in Hartman and (i) MW section 15.1 : 9 − 41, odd;8,7,15,33,41,71
(ii) MW section 15.2:5,7,9,26,27.
(iii) Section 12.6: 13,15,17,19,21,23,27
(iv) Turn in solutions to the following problems:
(a) For f (x, y) = x2y + xy2, use the limit definitions to find fx(2, 3) and fy(2, 3).
(b) For f (x, y) = 3x2+ 2xy + 4, use the limit definition to find the directional derivative of f (x, y) at (1, 1). in the direction of the vector a~i + b~j.
(c) For
f (x, y) = ( xy
x2+y2 (x, y) 6= (0, 0) 2 (x, y) = (0, 0) determine if fx(x, y) is continuous at (0, 0).
This assignment is due in class on Tuesday, September 7.
1.7.1 Equality of Mixed partials
Theorem 1 Given f (x, y) and a point (a, b) ∈ R2, suppose that there is an open disk D about (a, b) such that fxy, fyx exist and are continuous at every point in D. Then
fxy(x0, y0) = fyx(x0, y0) for all (x0, y0) ∈ D.
Note that the continuity assumptions in the theorem above cannot be relaxed . To see that consider the following example
Example 18 Suppose
f (x, y) =
(xyxx22−y+y22 (x, y) 6= (0, 0) 0 (x, y) = (0, 0) We can show
(i) f (x, y) is continuous at (0, 0), even at (x, y) 6= (0, 0). Recall the bonus problem in the first homework.
(ii) fx(x, y) and fy(x, y) exist and are continuous everywhere, including at (0, 0).
(iii) fxy(x, y) and fyx(x, y) exist everywhere, and that fxy(0, 0) = −1, while fyx(0, 0) = 1.
(iv) fxy(x, y) is not continuous at (0, 0) since lim(x,y)→(0,0)fxy does not exist.
And this in light of the above theorem, (iv) explains why for this example fxy(0, 0) 6= fyx(0, 0).
1.7.2 Linear approximations
We will discuss what should constitute a good linear approximation to f (x, y) at (a, b). Geometrically, this should mean that there is well defined tangent plane to the graph of z = f (x, y) at (a, b, f (a, b)). That is does the proposed tangent plane z = fx(a, b)(x − a) + fy(a, b)(y − b) + f (a, b) do a good job pf approximating f (x, y) in the vicinity of (a, b)?
To answer this question we first explore the one variable case. Suppose f0(a) exists and we set L(x) = f0(a)(x − a) + f (a), The tangent line to the graph of y = f (x) at x = a. Since f0(a) = limx→af (x)−f (a)
x−a we
can show
x→alim
f (x) − L(x) x − a = 0.
Since the term x−a1 tends to infinity as x tends to a, this suggests that the difference f (x) − L(x) tends to zero so quickly that it overcomes the tendency of x−a1 to tend to infinity. Therefore the closer x is to a, the better L(x) approximates f (x).
This idea can be extended to f (x, y) at (a, b). Assume fx(a, b) and fy(a, b) exist, we set L(x, y) :=
fx(a, b)(x − a) + fy(a, b)(y − b) + f (a, b), the linear function we expect to be a good approximation of f (x, y) for (x, y) near (a, b). we then said that f (x, y) is differentiable at (a, b) if
lim
(x,y)→(0,0)
f (x, y) − L(x, y)
|(x, y) − (a, b)| = 0,
where |(x, y) − (a, b)| =p(x − a)2+ (y − b)2. Similar to the single variable, since the term |(x,y)−(a,b)|1
tends to infinity as (x, y) tends to (a, b), this suggests that the difference f (x, y) − L(x, y) tends to zero so quickly that it overcomes the tendency of |(x,y)−(a,b)|1 to tend to infinity. Therefore the closer (x, y) is to (a, b) the better L(x, y) approximates f (x, y). Thus
Definition 5 if f (x, y) is differentiable at (a, b) and ∆x and ∆y are small, then f (a + ∆x, b + ∆y) ∼= L(a + ∆x, b + ∆y)
= fx(a, b)(a + ∆x − a) + fy(a, b)(b + ∆y − b) + f (a, b)
=fx(a, b)∆x + fy(a, b)∆y + f (a, b)
Example 19 Use the linear approximation to estimate f (1.01, 2.02), for f (x, y) = x2+ 2xy Assuming f is differentiable at (1, 2).
Example 20 Show that the followings are differentiable at the given point.
(i) f (x, y) = x, g(x, y) = y at any (a, b)
(ii) If f (x, y), g(x, y) are differentiable at (a, b) the so are f + g, f g,fg, g 6= 0.
(iii) f (x, y) = xy at (a, b).
Example 21 Consider the function
f (x, y) =
(√2xy
x2+y2 (x, y) 6= (0, 0) 0 (x, y) = (0, 0) Show that
(i) f (x, y) is continuous at (0, 0).
(ii) fx(0, 0) = 0 = fy(0, 0) (iii) L(x, y) = 0.
(iv) f (x, y) is not differentiable at (0, 0).
Theorem 2 (Differentiability Criterion) Given f (x, y) and (a, b) in the domain of f (x, y), then f (x, y) is differentiable at (a, b) if the partial derivatives fx(x, y), fy(x, y) exist and are continuous in an open disk containing (a, b).
Example 22 Using the above theorem verify that the function given in example 21 is not differentiable at (0, 0).
Remark 3 If f (x, y) is differentiable at (a, b) then f is continuous at (a, b).
1.8 Chain Rule
Example 23 Find all the first order partial derivative for the following (i) f (x, y) = x2y + exy, with x = sin t, y = cos t
(ii) f (x, y) = x2y, with x = uv and y = u2+ v2
Homewor 3Read section 15.3 and 15.4 in Marsden and Weinstein and section 12.5 of HArtman. And work the following
(i) MW, 15.1,71,72
(ii) MW, 15.2, 17-27,17,22,27 (iii) MW, 15.3 9-21, odd, 9,11,22 (iv) MW, 15.4, 25,26,29
Also, turn in the following problems:
(a) Use the limit definition to prove that f (x, y) = x2y + xy2 is differentiable at (1, 1).
(b) Consider the function
f (x, y) = ( x2y
x2+y2 (x, y) 6= (0, 0) 0 (x, y) = (0, 0) (i) Determine if f (x, y) is continuous at (0, 0).
(ii) Determine if fx(0, 0), fy(0, 0) exist, and if so, what are their values.
(iii) Show that f (x, y) is not differentiable at (0, 0).
(iv) Explain, with justification, why f (x, y) fails to be differentiable at (0, 0).
This assignment is due in class on Monday, September 13.
1.9 Extreme Values
In single variable calculus we know that a maximum or minimum values of f (x, y) should occur when the tangent plane to the graph of z = f (x, y) is parallel to the xy−plane. Since such plane is of the form z = c, with c a constant, this implies that fx(a, b) = 0 = fy(a, b). We then define the critical point of f (x, y).
Example 24 Verify that (0, 0) is the critical point of f (x, y) = x2+ y2 both graphically and algebraically.
Example 25 Find the CP of the following (i) f (x, y) = x2+ y2− xy − x − 2.
(ii) f (x, y) = −p
x2+ y2+ 2 (iii) f (x, y) = x3− 3x − y2+ 4y.
We then defined, what it means for a point to be a relative maximum or relative minimum of f (x, y), or a saddle point on the graph of f (x, y)
This enabled us to then state the Second Derivative Test for critical points of f (x, y) :
Example 26 Using the second derivative test, classify the critical points we found in the examples above:
Example 27 Using the second derivative test, classify the critical points of f (x, y) = x2y + y2+ xy.
Example 28 Among all rectangular boxes with volume V , find the dimensions of the box with least surface area.
Remark 4 we noted that one requires that the first and second order partial derivatives of f (x, y) must be continuous in order for the test to be applicable.
Next We begin a discussion concerning absolute maxima and absolute minima for a function of two variables. We noted that given a subset X ⊂ R2 f (x, y) has an absolute maximum (respectively, absolute minimum) at (a, b) if f (x, y) ≤ f (a, b) (respectively, f (a, b) ≤ f (x, y)), for all (x, y) ∈ X. In this case, f (a, b) is the absolute maximum (respectively, absolute minimum) value of f (x, y) on X. We recalled, that for a function f (x) of one variable, in order to guarantee that f (x) has absolute extreme values, one must assume that f (x) is continuous on a closed interval [c, d] . To find these values, one must find critical points on the interior of the interval [c, d] and evaluate f (x) at each of these points, and then one must calculate f (c) and f (d). The largest of these values is the absolute maximum of f (x) on [c, d] and smallest is the absolute minimum of f (x) on [c, d]. We explained that one needs similar condition for functions of two variables. This lead to the following definition.
Example 29 Find the absolute maximum and absolute minimum values for f (x, y) = x2− y2+ 5 on the closed triangle X in R2 with endpoints (0, 1), (−1, −2), (2, −10).
1.9.1 Why do the Second derivatives works
We then began a discussion of why the second derivative test works. Recalling Taylor’s Theorem from Calculus I, note that, given f (x) and x = a, under suitable differentiability conditions, a good quadratic approximation of f (x) near x = a is given by
f (x) ≈ f (a) + f0(a)(x − a) +1
2f00(a)(x − a)2.
Note that if f0(a) = 0, i.e., f (x) has a critical point at x = a, then the parabola y = f (a) +12f00(a)(x − a)2 approximates f (x) near x = 0. This shows that if f00(a) > 0, then the graph of f (x) at x = a is concave up, while the graph is concave down if f00(a) < 0.
Now we state a multivariable version of Taylor’s Theorem guarantees that, under suitable differentiability conditions, a good quadratic approximation of f (x, y) for (x, y) sufficiently close to (a, b) is given by (∗) f (x, y) ≈ f (a, b)+fx(a, b)(x−a)+fy(a, b)(y−b)+1
2{fxx(a, b)(x−a)2+2fxy(a, b)(x−a)(y−b)+fyy(a, b)(y−b)2}.
What about Second derivative test for function of three variables? First not that P = (a, b, c) is a critical point if f (x, y, z) if it is either a solution to the system of equations
fx(x, y, z) = 0 fy(x, y, z) = 0 fz(x, y, z) = 0.
Or one of the first order partials is undefined at P . We then defined D(P ) = fxx(P )fyy(P ) − fxy(P )2and
H(P ) =
fxx(P ) fxy(P ) fxz(P ) fyx(P ) fyy(P ) fyz(P ) fzx(P ) fzy(P ) fzz(P ) H(P ) is called the Hessian of f (x, y, z) at P.
Theorem 3 (Second Derivative Test) Suppose P = (a, b, c) critical point that is a solution to the system of equations above and all second order partial derivatives of f (x, y, z) are continuous in a disk about P. Then:
(i) If fxx(P ), D(P ), H(P ) are all greater than zero, f (x, y, z) has a relative minimum at P.
(ii) If fxx(P ) < 0, D(P ) > 0, H(P ) < 0, then f (x, y, z) has a relative maximum at P.
(iii) If H(P ) 6= 0, and neither (i) nor (ii) holds, then f (x, y, z) has a saddlle point at P.
(iv) If H(P ) = 0, the test is inconclusive.
Example 30 Find and classify all the critical point of f (x, y, z) = x3+ xy + x2+ y2+ 3z2.
We can generalized the second derivative test to a function of n variables as follows. The critical points are found in a similar fashion, e.g., by solving the system of equations
fx1 = 0 fx2 = 0 ... fxn= 0.
For a critical point P = (a1, a2, . . . , an), one forms the matrix
H(P ) =
fx1x1(P ) fx1x2(P ) . . . fx1xn(P ) fx2x1(P ) fx2x2(P ) . . . fx2xn(P )
... ... ...
fxnx1(P ) fxnx2(P ) . . . fxnxn(P )
We then set d1 to be the upper left 1 × 1 determinant of H(P ), d2 to be the upper left 2 × 2 determinant of H(P ), . . . , dn to be the full n × n determinant of H(P ). The Second Derivative tests then states, that if the second order partials of f (x1, x2, . . . , xn) are continuous in an open ball around P, then:
(i) f (P ) is a local minimum if d1> 0, d2> 0, . . . , dn > 0.
(ii) f (P ) is a local maximum if d1< 0, d2> 0, d3< 0, d4> 0, . . .
(iii) f has a saddle point at P if dn6= 0 and neither case (i) nor case (ii) holds.
(iv) The test is inconclusive if dn6= 0.
Homework 4
Read Section 12.8 in Hartman and Section 16.3 in Marsden and Weinstein and work the following problems:
(i) Section 12.8, 1-17, odd, and 9, 12, 13, 15, 17 (ii) MW, Section 16.3,24, 28, 32, 35.
Optional Bonus Problem. Read the prelude to problems 43-48 in MW Section 16.3. Then use the techniques of this homework section to find the equation of the line of best fit passing through data points (x1, y1), . . . , (xn, yn).
This assignment is due in class on Monday, September 20.
1.9.2 Lagrange multipliers
We begin our discussion of constrained maxima and minima problems via the technique of Lagrange multi- pliers. We start by stating the theorem which says the following:
Theorem 4 Given f (x, y) (or f (x, y, z)) subject to the constraint g(x, y) = c( or g(x, y, z) = c), if : (i) P ∈ R2(or P ∈ R3) is in the domain of f and g(P ) = c
(ii) f and g have continuous partials at P (iii) ∇g(P ) 6= 0
(iv) f (P ) is a maximum value or minimum value of f subject to g = c.
Then there exists λ ∈ R such that ∇f (P ) = λ∇g(P ).
The theorem tells us that if the maximum or minimum value we seek exists, then it lies among the points P satisfying ∇f (P ) = λ∇g(P ), for some λ ∈ R. Now we work the following examples
Example 31 Find the minimal surface area among rectangular boxes with volume 27
Remark 5 Note that the next example shows how Lagrange multipliers can be used when one cannot use the constraint equation to solve for one of the variables in terms of the others.
Example 32 Find the extreme values of f (x, y) = x42 + y2 subject to the constraint g(x, y) = x2+ y2= 1.
Example 33 Find the extreme values of f (x, y) = 2x + y, subject to the constraint g(x, y) =√ x +√
y = 3.
Example 34 Find the extreme values of f (x, y, z) = x − y + z subject to g(x, y, z) = x2+ y2+ z2= R2. Remark 6 Why does the LM works?
Example 35 Find the absolute maximum and absolute minimum values of f (x, y) = xy over the region D : 0 ≤ x2+ y2≤ 1.
If one wants to find the extreme values of f subject to constraints, g1= c1, . . . , gn = cn, then one has to consider the system of equations arising from the vector equation ∇f = λ1∇g1+ . . . + λn∇gn, together with the constraint equations. Consider following example
Example 36 Find the extreme values of f (x, y, z) = x + y + z subject to g1(x, y, z) = x2+ y2 = 2 and g2(x, y, z) = x + z = 1.
Homework 5
Read section 16.4 in Marsden and Weinstein and wrk the following problems.:
(i) MW, Section 16.4, 2-19, odd, and7, 9, 11, 19, 20
(a) Find and classify the critical points of f (x, y, z) = x3+ xz2− 3x2+ y2+ 2z2 using the second derivative test for functions of three variables.
(b) Use Lagrange multipliers to find and classify the extreme values f (x, y, z) = x2+ y2+ z2, subject to x + y − z = 1.
(c) Use Lagrange multipliers to find and classify the extreme values of f (x, y, z) = x6+ y6+ z6, subject to x2+ y2+ z2= 6.
This assignment is due in class on Monday, September 27.
2 Integrals
Definition 6 The average value of f (x, y) over the region bounded R as
AV = 1
area(R) Z Z
R
f (x, y)dA
Example 37 Calculate the average value of f (x, y) = x2+ 2xy over the rectangle R = [−1, 1] × [−2, 2]
Example 38 Using double integrals calculate the volume of a sphere of radius R.
Example 39 CalculateRR
Rx + ydA where R is given below.
Example 40 Calculate the volume of inverted cone z = h −p
x2+ y2, with z ≥ 0.
Remark 7 Why is dA = rdrdθ?
Example 41 CalculateRR
DydA where D is the set of points lying above the x-axis and inside the cardioid r = 1 + cos θ :
Example 42 CalculateRR
D(x2+ y2)−2dA where D is the shaded region
2.1 Volumes
Example 43 Find the volume bounded by the paraboloids z = 8 − x2− y2 and z = x2+ y2, over the region D : [−1, 1] × [−1, 1].
Example 44 Find the volume of the region in R3enclosed by the graphs of z = 8 − x2− y2and z = x2+ y2.
2.1.1 Convergence of double integrals
Example 45 Discuss the convergence of the following integrals 1. R √1
xdx 2. R∞
0 e−2xdx 3. RR
D
√ 1
1−x2−y2dA where D : 0 ≤ x2+ y2≤ 1 Example 46 RR
Dxye−x2−y2dA where D is the first quadrant of R2
Example 47 RR
D ey
ydA where D is:
Example 48 ConsiderRR
D x2+1
√3
y−1 where D is the rectangle 0 ≤ x ≤ 1 and 0 ≤ y ≤ 2.
Example 49 This example shows how things can go wrong in Fubini’s Theorem if f (x, y) is not non-negative on the domain of integration. Consider: RR f (x, y)dA, for D : 0 ≤ x ≤ 2, 0 ≤ y ≤ 1. and
f (x, y) =
(xy(x2−y2)
(x2+y2)3 (x, y) 6= (0, 0) 0 (x, y) = (0, 0)
Example 50 CalculateR∞ 0 e−x2dx
2.1.2 Application of Double Integrals
Example 51 Find the Surface area of a sphere centered at the origin with radius R.
Homework 6a
Homework 6b
both due on after the fall break Wednesday 10/13