A Discussion in Intermediate Calculus
AVOID TO USE
SPHERICAL COORDINATE SYSTEM IN PARAMETRIC REPRESENTATION FOR DOUBE
INTEGRALS
Thomas Nguyen
Abstract:
One common technique that is used to evaluate integrals is changing coordinate systems. As a principle, students are familiar with using Polar coordinate for double integrals (two-dimensional) and Cylindrical/Spherical coordinate for triple integrals (three-dimensional). However, James Stewart, in his Calculus book (Early Transcendentals Multivariable 5e), applied several times the spherical coordinate system to demonstrate the use the parametric representation for calculation of double integrals. One particular demonstration is example 5 in pages 1117-1118. This is not an appropriate way to demonstrate the use of parametric representation. It might cause confusing for students not only by inappropriate coordinate system choice but also by difficulty of interpretation of the transformation: dA --> dФ.dθ (*). A suggested solution will be proposed.
Problem:
Find the flux of vector field F(x,y,z) = z.i + y.j + x.k across the unit sphere:
x2 +
y2 + z2 =1.
Stewart�s
solution:
Using the parametric representation
r(Ф,θ) = sinФ cosθ i + sinФ sinθ j + cosФ k
0 ≤ Ф ≤ л 0 ≤ θ ≤
2л
We
have
And
Therefore
and by formula 9, the flux
is
(*)
Suggested solution:
Of course the best solution for this problem is using The Divergence Theorem
However, we want to use parametric representation to solve it. Therefore I am going to solve this problem by using Polar coordinate system (2-D). I prefer to rewrite equation 8 from page 1116 (James Stewart�s Calculus 5e) as the following:
Apply this formula for our problem: 
and F = <z,y,x>
By symmetry, we can compute the flux across semi sphere first (we will double answer later), therefore from:
We have
Using Polar parametric representation: 
--->
where 0 ≤ r ≤ 1 and 0 ≤ θ ≤
2л
Rewrite
Let u = 1-
r2
Hence total flux
crosses entire of the sphere is:
Conclusion:
Although the
problem asks for the flux across a space surface S (3-D) and vector field F is
given with three components x, y, and z; however, because the problem�s domain
is transferred from S to a plane surface D; furthermore, because of constrain
x2+y2+z2=1, we can consider this problem as a
problem with two variables x and y. Therefore, it makes sense for students to
use Polar in double integral and to avoid a confusing setting: dA = dФ.dθ (*) of using sphere parametric
representation. We can use Polar instead of Sphere for example 10 in page 1104,
example 2 in page 1112, and example 5 in page 1117 with
the same technique that has shown above.
Second discussion
A DISCUSSION
IN POLAR COORDINATE SYSTEM
Thomas Nguyen
Abstract:
Isaac Newton is a great
scientist. He had invented many useful things for life. Polar coordinate system
is one of his inventions (written 1671). Polar coordinate system is very helpful
in many cases. It is used in many fields, including mathematics, physics,
engineering, navigation, and robotics. However, I am still wondering why when he
invented this coordinate system, he set 
not
(zero or positive
numbers).
I prefer the following
setting:
where:
r ≥ 0
and
There are two main reasons for
my claim:
-
When we convert
from Cartesian to Polar, we do not need to worry about choosing r to be positive or
negative.
-
We do not have to
worry about whether we were missing solutions or not.
Let consider one example
(Example 3 in page 671, Early Transcendentals Multivariable Calculus 5e of James
Stewart)
Problem:
Represent the point with
Cartesian coordinate system (1, -1) in Polar coordinate
system.
Stewart�s solution:
If we choose r to be positive,
the Equation 2
give:
Since the point lies in the
fourth quadrant, we can choose: 
Thus one possible answer
is 
and
another is 
Discussion:
First, as you can see
the author first has to make the statement: �If we choose r to be positive,
...�; Students can wonder such as �Why we choose r to be positive?� or �What if r to be chosen
negative?�
In negative case of r,
there are two more basic possible solutions, which are:
and 
These two additional solutions
came up because of the rule ( i.e. r can be
negative or positive). It seems like we put ourself into troubles (ex:
forgetting solutions sometimes or some students might be lost some points if in
their solutions, they declair these two new solutions!). Why we don�t set r to be positive because angle
θ can cover all positions in the plane.
When we solve the
equation: 
we just have to
look at where the point located in Cartesian as James Stewart did and we have to
write only one general solution:
In this case, we will not to
worry about all solutions with negative r.
Life would be easier then for students.
I have tried all times
to apply this setting into other applications of Polar coordinate system (and
Cylindrical coordinate system-3D) such as solving
double, triple integrals without any problems. It is very convinient to use this
setting to determine is in which direction (θ) and how far (r) to move for a
robot. It makes sense in avitation, too; for example, an aircraft traveling 5
nautical miles due East will be travelling 5 units at heading 90. However, I am
worrying that perhaps in some cases we need the setting r to be negative to cover all possible
options in some special problems. That�s why I really appreciate it if someone
can explain why
Conclusion:
I am always looking for
simpler/better methods/techniques to solve problems. This is one of my own
experiences that have helped me went through my studying.
A Discussion in Geometry
DISCUSS ON OPTIMIZATION PROBLEMS IN GEOMETRY
Thomas Nguyen
Abstract:
Calculus is a very powerful tool to solve optimization problems in Geometry. Quite commonly, students have been learned and paid respect to Calculus much more than Algebra. They usually think Intermediate Algebra or College Algebra is �elementary school� stuff.
Using Algebra to solve optimization problems in Geometry has not been seen much in college. In this article, I am going to demonstrate the beauty of some of Algebra�s tools (Cauchy�s Formula, Ploting technique, etc). I think after reading this article, students will have a new look on Algebra.
Example 1: (Example 6, page 958, Early Transcendental Multivariable Calculus 5e., James Stewart)
This example demonstrates how to use partial derivatives to solve optimization problems.
A rectangular box without a lid is to be made from 12 m2 of cardboard. Find the maximum volume of such a box.
Book�s Solution: Let the length, width, and height of the box (in meters) be x, y, and z, as shown in Figure 10. Then the volume of the box is V = xyz
Figure 10
We can express V as a function of just two variables x and y by using the fact that the area of the four sides and the bottom of the box is
2xz + 2yz + xy = 12 (m2)
Solving this equation for z, we get z = (12 � xy)/[2(x+y)], so the expression for V becomes
We compute the partial derivatives:
If
V is maximum, then 
but x = 0 or y = 0 gives V = 0, so we must solve the
equations
There imply that x2 = y2 and so x = y. (Note that x and y must both be positive in this problem.) If we put x = y in either equation we get
12 � 3x2 = 0 , which gives x = 2, y = 2, and z = (12 � 2.2)/[2(2+2)] = 1.
We could use the Second Derivatives Test to show that this gives a local maximum of V, or we could simply argue from the physical nature of this problem that these must be an absolute maximum volume, which has to occur at a critical point of V, so it must occur when x = 2, y = 2, and z = 1. Then V = 2.2.1 = 4 (m3), so the maximum volume of the box is 4m3.
Discussion:
As you can see in this solution, the author only shows details in finding critical points, how to prove these critical values are absolute maximum values is just a suggestion. Although partial derivatives can solve this problem, their solution is �pretty� tough, isn�t it?
Before showing Algebra�s solution, I�d like to introduce one of Algebra�s tools that I am going to use it to solve Example 1.
Cauchy�s Formula: The geometric mean is smaller than the arithmetic mean. (That�s it. Very neat!)
We can express it for three variables as follows:
For any a > 0 , b > 0, and c >
0, we have
with
equality in the case a
= b = c.
Solving Example by using Cauchy�s Formula:
Call x, y, and z are three sides of the box as
shown in Figure 10. Because they are 3 sides of the box, they must be positive
(i.e x > 0, y > 0, and
z > 0 ---> Let
a = (2xz) > 0, b =
(2yz) > 0, and c = (xy) > 0
----> a, b, and c satisfies Cauchy�s condition, therefore, we
have:
The right side is always smaller or equal to the left side. The equality
case happens only when a = b = c.
That means when 2xz = 2yz =
xy -----> x = y =
2z
(1)
In this case, the right side will become 
(2)
From (2), we can say that the volume V is maximum, when condition (1) is
satisfied, but: a + b + c =
2xz + 2yz + xy = 12 m2 (3)
Therefore, by using (1) and (3) we can easily
have: x = y
= 2
and z =1 ------> Vmax= xyz =
2.2.1 = 4 m3.
Note: Now you have seen the beauty of Cauchy�s Formula (one of Algebra�s tool). In this solution, we know very clear that x = y = 2 and z = 1 are absolute critical values, and V = 4 m3 is the absolute maximum value that it can have (Cauchy�s equality satisfied). The solution is short, compact, and clear by comparison to Calculus�s partial derivatives tool.
Example 2: (Example 1, page 332, Early Transcendentals Single Variable Calculus 5e, James Stewart)
A farmer has 2400 ft fencing and wants to fence off a rectangular field that borders a straight river. He needs no fence along the river. What are the dimensions of the field that has the largest area?
river
Book�s Solution: We wish to maximize the area A of the rectangle. Let a and b be the depth and width of the rectangle (in feet). Then we express A in term of x and y:
A = xy
We want to express A as a function of just one variable, so we eliminate y by expressing it in terms of x. To do this we use the given information that the total length of the fencing is 2400 ft. Thus
2x + y = 2400
From this equation we have y = 2400 � 2x, which gives
A = x(2400 � 2x) = 2400x � 2x2
Note that x ≥ 0 and x ≤ 1200 (otherwise A < 0). So the function that we wish to maximize is
A(x) = 2400x � 2x2 0 ≤ x ≤ 1200
The derivative is A�(x) = 2400 � 4x, so to find the critical numbers we solve the equation 2400 � 4x = 0
Which give x = 600. The maximum value of A must occur either at this critical number or at an endpoint of the interval. Since A(0) = 0, A(600) = 720,000 and A(1200) = 0, the Closed Interval Method gives the maximum value as A(600) = 720,000.
[Alternately, we could have observed that A�(x) = - 4 < 0 for all x, so A is always concave downward and the local maximum at x = 600 must be an absolute maximum.]
Algebraic Solution: Let x = 2a and y = b (x > 0, y > 0
satisfied Cauchy�s condition). Apply Cauchy�s Formula for two
variables:
Hence the area of the field A will be maximized
only when equality�s hold, i.e. x = y or 2a = b
(1)
We have given information: 2a + b =
2400
(2)
From (1) and (2), we have:
a = 600 and b = 1200 ------> Amax = 720,000
ft2.
*******************
Example 3: (Example 5, page 336, Early Transcendentals Single Variable Calculus 5e, James Stewart)
Find the area of the largest rectangle that can be inscribed in a semicircle of radius r.
Book�s Solution:
Let�s take the semicircle to be the upper half of the circle: x2 + y2 = r2 with center the origin. Then the word inscribed means that the rectangle has two vertices on the semicircle and two vertices on the x-axis as shown in Figure 9.
Let (x,y) be the vertex that lies in the first quadrant. Then the rectangle has sides or lenghs 2x and y, so its area is
A = 2xy
To eliminatey we use the fact that (x,y) lies on the
circle, so
. Thus
The domain of this function is 0 ≤ x ≤r. Its derivative is
Which is 0
when 2x2 =
r2, that is, 
(since x ≥ 0). This value of x gives a maximum
value of A since A(0) = 0 and A(r) = 0. Therefore, the
area of the largest inscribed rectangle is
Figure 10.
Trigonometric Solution(special): A simpler solution is possible if we think of using an angle as a variable. Let θ be the angle shown in Figure 10. Then the area of the rectangle is
Wi know that sin(2θ) has a maximum value of 1 and it occurs when 2θ = л/2.
So A(θ) has a maximum value of r2 and it occurs when θ = л/4
Algebraic Solution: We have x > 0, y
> 0 ---> x2 > 0 and y2 > 0 satified Cauchy�s
condition. Apply Cauchy�s Formula for two variables
= half of
rectangle area.
If this area is maximized, then entire of
rectangle area will be maximized. This happens when equality�s hold: x2 = y2
or x =
y
(1)
Furthermore, we have x2 + y2 =
r2
(Pythagorean Theorem) (2).
From (1) and (2), we have 
and Amax
= 2xmaxymax = r2.
***************
Example 4: (Example 2, page 333, Early Transcendentals Single Variable Calculus 5e, James Stewart)
A cylindrical can is to be made to hold 1 L of oil. Find the dimensions that will minimize the cost of the metal to manufacture the can.
Book�s Solution: Diagram as in Figure 3, where r is the radius and h the height (both in centimeters). In order to minimize the cost of the metal, we minimize the total surface area of the cylinder (top, bottom, and sides). We see that the sides are made from a rectangular sheet with dimension 2л.r and h. So the surface area is
A = 2л.r2 + 2л.r.h
To eliminate h we use the fact that the volume is given as 1 L, where we take to be 1000 cm3. Thus: л.r2.h = 1000
Which gives: h = 1000/(л.r2). Substitution of this into the expression for A gives
Therefore, the function that we want to minimize is
To find the critical numbers, we differentiate
Then A�(r) = 0
when лr2 = 500, so the only critical number is 
Since the domain of A is (0,∞), we can�t use the
argument of Example 1 concerning endpoints. But we can observe that A�(r) < 0
for 
and A�(r)
> 0 for 
, so A is
decreasing for all r to the left of the critical number and increasing for all r
to the right. Thus
must give rise to an absolute minimum.
[Alternately, we could argue that A(r)--> ∞ as r
--> 0+ and A(r)
--> ∞ as r --> ∞, so there
must be a minimum value of A(r), which must occur at the critical number]. The
value of h corresponding to is
Thus, to minimize the cost of the can, the radius should be cm and the
height should be equal to twice the radius, namely, the diameter.
The total surface will be
Discussion: This solution is also
pretty tough for most of students at this level, especially the argument at the
end of the solution (argue that the value is the
minimum value). Let�s see how Algebra deals with this problem
Algebraic solution : Let R
be the radius, h be the height, S be the total surface, V be the volue of the
cylinder can. We have
S = 2* лR2 + 2лRh (1) and V = лR2h =
1000 cm3
(2)
Solve h from (2) and substitude h into (1),
which gives
(3)
Plot (3) in Cartesian coordinate system, we
have
Note:
R = (1.7* pi) is equivalent to 
. This
is the critical value for minimum value of total surface
area.
We can have
and S ≈ 554
m2 the same as
Calculus solution (we don�t need derivatives).
***************
From now on, I
just show problems and Algebraic solutions only.
**************
Example 5: If the length of the diagonal of a rectangular box must be L, what if the largest possible volume?
Algebraic Solution: Let x, y, and z be the width, length, and height of the rectangular box with diagonal�s length L. Set a = x2, b = y2, and c = z2. Then a > 0, b > 0, and c > 0 satisfied Cauchy�s condition. Apply Cauchy�s Formula for 3 variables:
The right side is 
Similarly, Volume V is
maximized only when equality�s hold. That means:
a = b = c -----------> x2 = y2 = z2 (1)
Furthermore, we have given information: x2 + y2 + z2 = L2 (2)
From (1) , (2) and condition: x > 0, y > 0, z > 0, we
have:
*****************
Example 6: Find the dimensions of a rectangular box of maximum volume such that the sum of the lengths of its 12 edges is a constant c.
Algebraic Solution: Setting 3 dimensions as Example 4. Then the sum of all edges is: 4(x + y + z) = c (1)
Similar argument in Example 4, the volume of the rectangular box is maximized only when: x = y = z (2)
Solve (1) & (2), we have x = y = z = c/12 hence Vmax = (c/12)3
******************
Example 7: Find the dimensions of the rectangle of largest area that has its base on the x-axis and its other two vertices above the x-axis and lying on the parabola: y = 8 � x2.
Algebraic Solution: Let x be the width and y be the length of a half of inscribed rectangle (they are also two coordinates of point P).
We have A = 2* x.y (1) Furthermore, because P belongs to the parabola, so we have y = 8 � x2. (2)
Substitute (2) into (1), which gives A = 16x � 2x3. (3)
Note: Since the rectangle
must be inscribed
---> Conditions
for x and y:
Plot (3) in the Cartesian coordinate system with interval for x is 
, we
got the following graph:
We can see that xmax = sqrt(8/3) ; therefore y = 8 � x2 = 8 � (8/3) = 16/3 and Amax =2xmaxy ≈ 17.4 (we don�t need derivatives)
*******************
Example 8: A rectangular building is being designed to minimize heat loss. The east and west walls lose heat at a rate of 10 units/m2 per day, the north and south walls at a rate of 8 units/m2 per day, the floor at a rate of 1 unit/m2 per day, and the roof at a rate of 5 units/m2 per day. Each wall must be at least 30 m long, the height must be at least 4 m, and the volume must be exactly 4000 m3. Find the dimensions that minimize heat loss?
Algebraic Solution:
Let L, W, and H be the length, the width, and the height of the building, then boudary conditions are: W ≥ 30 ; From V = LWH = 4000 m3 (1)
---> H = 4000/(WL) ≥ 4 so 30 ≤ L ≤ 1000/W (*) We can plot these conditions as follows:
100/3 B1 W
Total heat loss is: T = (2WH)*10 + (2LH)*8 + (WL)*1 + (WL)*5 = 20WH + 16LH + 6WL (2)
Let a = 20WH; b = 16LH; c = 6WL; a, b, c > 0 satisfied Cauchy�s Formula, hence
(3)
The left side (total heat loss T) will be minimized only when equality�s hold; i.e. only when a = b = c or 20WH = 16LH = 6WL (4)
By solving (1) & (4), we got a critical point:
Heat loss in this case is about 9,396 units (this is the minimum heat loss that we might have)
(No derivatives needed!)
This critical point is the only one we can have, but it�s not satisfied condition (*); However, this is a bounday value problem (an absolute max/min problem), therefore we continue checking the boundary to find the minimum heat loss which satisfied condition (*). Solve the boundary value problem, we have:
On B1 Tmin(W,L) = Tmin(30, 30) = 10,200 units (minimum heat loss)
On B2 Tmin(W,L) = Tmin(100/3, 30) ≈ 10,533 units
On B3
Tmin(W,L) = Tmin(30, 30) =
10,200 units (minimum heat
loss)
=====> Dimensions of the building that minimized heat loss (10,200 units) are: L = W = 30 m and H = 40/9 ≈ 4.44 m
Note: Because of boundary conditions,
dimensions (W, L, H) = (30, 30, 4.44) become the chosen dimensions with the
minimum heat loss of 10,200 units. If there were no boundary condition, then the
chosen dimensions would be (W, L, H) = ( 20.43, 25.54,
7.66) with the minimum heat loss of 9, 396 units.
***************
A Discussion in Algebra
DISCUSS ON MASTER PRODUCT AND SUM
Thomas Nguyen
(This article was certified by
Register of Copyright,
Abstract:
In Algebra, when dealing with trinomials in general form ax2 + bx + c there is a very efficient factoring method called �Master Product and Sum�. The main focus of this method is trying to split the middle term (bx) into two parts ( bx = n1x + n2x ), then using grouping method to factor again.
In general, there are 4 steps as the following:
- Step 1: Form the product (ac)
- Step 2: Find a pair of numbers whose product is (ac) and whose sum is (b)
- Step 3: Rewrite the trinomial to be factored so that the middle term (bx) is written as the sum of two terms whose coefficients are the two numbers found in step 2
- Step 4: Continue to factor by grouping method
Example: Factor 3x2 � 10x � 8
Solution: The above trinomial has the form ax2 + bx + c, where a = 3, b = -10, and c = - 8
- Step 1: The product (ac) = 3. (-8) = - 24
- Step 2: We need to find two numbers whose product is � 24 and whose sum is � 10, Let�s list all the pairs of number whose product is � 24 to find the pair whose sum is � 10
|
Product = -24 |
Sum of two numbers = -10 |
|
1(-24) = -24 -1(24) = -24 2(-12) = -24 -2(12) = -24 3(-8) = -24 -3(8) = -24 4(-6) = -24 -4(6) = -24 |
1+(-24) = -23 -1+(24) = 23 2+(-12) = -10 <----- here -2+(12) = 10 3+(-8) = -5 -3+(8) = 5 4+(-6) = -2 -4+(6) = 2 |
As you can see, of all the pairs of numbers whose product is � 24, only 2 and � 12 have a sum of � 10
- Step 3: We now rewrite our original trinomial so the middle term -10x is written as the sum of � 12x and 2x (splitting step)
3x2 � 10x � 8 = 3x2 � 12x + 2x � 8
- Step 4: Factoring by grouping method, we have
3x2 � 12x + 2x � 8 = (3x2 � 12x) + (2x � 8)
= 3x(x � 4) + 2(x � 4)
= (x � 4).(3x + 2)
You can check this answer by multiplying (x � 4) and (3x + 2) to get (3x2 � 10x � 8)
Discussion:
I can have a similar method to factor trinomial, but after step 1 and 2, students can write out the answer in general form. Then they usually need one more step to rewrite the answer in the simplest form. My method is actually not save much space in the solution space; however, it helps students to avoid the splitting step and the grouping step which are pretty �tough� for most of them at this level. Because they have a tough time to deal with rewrite two splitting terms and decide which term will go with what in order to do grouping.
My Method:
- Step 1: Same as Step 1 above
- Step 2: Same as Step 2 above
- Step 3: After Step 2, we found two numbers n1 and n2 that satisfied the product and sum condition. Now, we just write out the answer in the following form
(1)
Redo above example: After Step 2 (with table), we have found two numbers 2 and -12 that satisfied the product and sum condition. Now let�s try to write out the answer:
Of course, you will ask me �Can you prove your formula?�
Proof of formula (1):
As we know, two solutions of the general trinomial are:
therefore
(2)
and
(3)
There is a theorem that states (2) & (3) of
two solutions of a quadratic equation, but it does not have any relations to
n1 and n2
The next simple trick is
very important in order to find the formula (1).
Multiply both sides of (2) by a a(x1 + x2) = - b or (- ax1 ) + (- ax2 ) = b (4)
Multiply both sides of (3) by a2 a2(x1*x2) = a2 (c/a) or (- ax1)*(- ax2) = ac (5)
From (4) and (5), we can see that (- ax1) and (- ax2) are the two numbers that satisfied the product and sum condition (product is ac and sum is b).
Hence
(6)
We all know that we can write the trinomial in the form:
(7)
Solve x1 and x2 from (6), and plug them into (7):
That is the formula (1).
Conclusion: Trying to find simpler methods/techniques to solve problems is one of my hobbies. I think this is a better way to do �Master Product and Sum� and it should be a benefit to students. If anybody has seen the formula (1) before (i.e. someone else did create this formula before me), please let me know.
AN EXAMPLE NEEDED TO BE MODIFIED
IN JAMES STEWART'S CALCULUS BOOK
Thomas Nguyen
Example 7 (page 812, Early Transcendental Multivariable Calculus 5e - James Stewart - ISBN: 0-534-41778-7)
A crate is hauled 8 (m) up a ramp under a constant force of 200 (N) applied at an angle of 25o to the ramp. Find the work done.
James Stewart's solution:
If F and D are the force and displacement vectors, as pictured in Figure 7, then the work done is
W = F. D = |F|.|D|. cos(α) = (200)(8)cos(25o) ≈ 1450 Nm = 1450 J
Figure 7
Suggested Modification:
There are two ways to modify this problem:
- First, easiest way, to keep James' solution, we should change the question: "A crate is hauled 8 m up a ramp under a constant force of 200 N applied at an angle of 25o to the ramp. Find the work done by this force (F force)"
However, in Physics, this is not a "normal" way to have a question like this. We have to change problem a little bit: no more up ramp, instead of that, the crate should be moved on horizontal platform as the following figure:
- Second, with the formal up ramp form (as in Figure 7), the problem needs two more given parameters: the mass of the crate (m) and the deep of the ramp (angle "θ"). Then, in this case, the problem should be modified as the following: "A crate is hauled 8 m up a ramp under a constant force of 200 N applied at an angle of 25o to the ramp. Find the work done on this crate."
Solution, in this case, is modified, too. Work will be done by two force: F force and gravitational force (g force).
W = F. D - [m.g.sin(θ)]. D where "θ" is the deep of the ramp.
Last but not least, frictionless should be stated.