It is only two weeks into the term that, in a calculus class, a student raises his hand and asks: "Will we ever need this stuff in real life?"
The professor gently smiles at him and says: "Of course not - if your real life will consist of flipping hamburgers at MacDonald's!"
...
Source:
[2]
Saturday, September 13, 2008
A drink too many
A cone-shaped paper drinking cup is to be made to hold 27 cu cm of water. Find the height and radius of the cup that will use the smallest amount of water. [4]
This is one of the problems which, by taking the usual road, can make the solution more complicated than the problem.
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Source:
[4]
Problem 36, page 338.
This is one of the problems which, by taking the usual road, can make the solution more complicated than the problem.
...
Source:
[4]
Problem 36, page 338.
Friday, September 12, 2008
Horseplay {*_*}
A mathematician and a stock broker go to the races to bet on horses. The broker suggests a bet of $10,000. That's too much for the mathematician's taste: First, he wants to understand the rules, have a look at the horses, etc.
"Don't worry", the broker says. "I know an empirical algorithm that allows me to find the number of the winning horse with absolute certainty."
This does not convince the mathematician.
"You are too theoretical!" the broker exclaims and puts his $10,000 on a horse.
The horse comes in first - making the broker even richer than he already is. The mathematician is baffled.
"What is your algorithm?" he wants to know.
"It's rather easy. I have two children, three and five years old. I add up their ages and bet on that number."
"But three plus five is eight - and that horse had number nine!"
"I told you that you're too theoretical! Didn't I just experimentally prove that my calculation is correct?!"
Source:
[2]
"Don't worry", the broker says. "I know an empirical algorithm that allows me to find the number of the winning horse with absolute certainty."
This does not convince the mathematician.
"You are too theoretical!" the broker exclaims and puts his $10,000 on a horse.
The horse comes in first - making the broker even richer than he already is. The mathematician is baffled.
"What is your algorithm?" he wants to know.
"It's rather easy. I have two children, three and five years old. I add up their ages and bet on that number."
"But three plus five is eight - and that horse had number nine!"
"I told you that you're too theoretical! Didn't I just experimentally prove that my calculation is correct?!"
Source:
[2]
Losing weight
Most stocks in the US and in the Philippines are taking a beating recently. The biggest drops (so far) were early this March. The past few days have mellowed but still there are some lingering erratic movements.
There are 4 possible outcomes and so, each outcome has a probability of 1 of 4 or 25%. After day 2, note that there are two $975 coming from outcome 2 and 4. Hence, the probability of getting $975 after day 2 is the sum of the probability of outcome 2 and 3. This is 0.25 + 0.25 = 0.5 or 50%.
To wise investors, there's always a good opportunity even when the stock drops. For one, the stocks become cheap during this slide and so buying is better.
While at this, I thought of posting this problem 4.65 from page 305 of Introduction to the Practice of Statistics, 5th edition, by David S. Moore and George P. McCabe. This is a simple problem on stocks using basic statistical probability. We have our final exam these days so this was also part of my review for Statistics.
4.65/305). You buy a hot stock for $1,000. The stock either gains 30% or loses 25% each day, each with probability 0.5. Its returns on consecutive days are independent of each other. You plan to sell the stock after two days.
a) What are the possible values of the stock after two days, and what is the probability for each value? What is the probability that the stock is worth more after two days than the $1,000 you paid for it?
b) What is the mean value of the stock after two days?
The problem can be solved starting with a tabular presentation below:
For a), the possible outcomes are:
There are 4 possible outcomes and so, each outcome has a probability of 1 of 4 or 25%. After day 2, note that there are two $975 coming from outcome 2 and 4. Hence, the probability of getting $975 after day 2 is the sum of the probability of outcome 2 and 3. This is 0.25 + 0.25 = 0.5 or 50%.
There is only one outcome of getting more than $1000 after two days. This is 1690 in outcome 1 which has probability 25%.
The mean value of the stock, say ยตX, is the summation of the all individual outcome with its specific probability. This is the same as the weighted value with probability serving as the weights.
There is a small gain on the average which is possible if one has several investments of this kind. However, if there is only one account, the law of large number is not at work. Given only one stock with an equal chance of losing 25% or gaining 30%, the chance of losing is actually greater. As shown in the table, there are three possible losing outcomes (out of four) that would shrink the stock below $1000.
In other words, this stock has a losing chance of 75%. Not a good one. In the stock market, there are several companies with better lose-gain attribute.
Historical daily performance can be checked from the company profile. This simple statistical analysis is only one of the tools that can be used as guide in investing. In particular, this gives an insight on what may happen should one redeem or sell the stock.
Statistics is a good exercise to keep our stocks and other investments in good shape.
...
[written 03.14.2007]
Wednesday, August 13, 2008
Anti-terrorism {*_*}
In a speech to a gathering of mathematics professors from throughout the United States, George W. Bush warned the academics not to misuse their position to force their often extremist political views on young Americans. "It is my understanding", the president said, "that you are frequently teaching algebra classes in which your students learn how to solve equations with the help of radicals. I can't say that I approve of that..."
...
Source:
[3]
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Source:
[3]
Root of the problem: Negative infinity
I came across this interesting problem which is quite simple in appearance yet, at the same time, it possesses some qualities of a quiz bee material.
The solution starts with removing the radical by multiplying the expression with its conjugate. From here, there are algebraic routes to the same destination of simplifying the expression which has become a fraction. The denominator now contains the other half of the conjugate.
Here, the gerater trick lies on the concept of absolute value.
The x-squared inside the radical sign can be extracted as x. If x goes to negative infinity, x must be negative. Therefore the root of this term must be negative.
Here is another interesting part. If the true value of x is negative becasue x approaches negative infinity, it is very tempting to over-emphasize the absolute value concept by changing the sign of all x including all the other terms with no root extraction required. Again, one may argue, if x goes to negative infinity as indicated in the limit, then each x in the whole expression is supposed to be covered by that condition. It means x is less than zero or negative. Hence, it is so tempting to rewrite the expression as:
Note that, when simplifying towards the end, x is factored out from the denominator and it cancels with the x in the numerator. As this happens, the rewriting of x outside the radical sign to reflect its true sign becomes somewhat moot.
On the other hand, all x outside the radical sign before extraction, in this case, the terms -2x in the numerator and the first term x in the denominator, are not normally rewritten in texts that I've seen at least, to reflect the true location of x in the number line. There must be some good reason. Before going there, consider:
In this second case, when one changes the sign of all x terms by invoking the blanket effect of absolute value as x goes to negative infinity, the limit is positive 3. On the contrary, when absolute value is invoked only to the roots of extracted term, the limit is negative 3.
It is wise therefore to confine the absolute value effect on root extraction. The other terms outside the radical sign may appear the way they do because they may have acquired their sign from previous mathematical operations. This shouldn't discount that they are essentially negative from the start when the condition says the limit goes to negative infinity. Likewise, when the expression is yet to be simplified, changing the sign of all the other unextracted x can lead the expression to wayward paths during succeeding mathematical operations.
The limit going to positive infinity makes similar problems very easy. On the other hand, when it goes to negative infinity, the concept of absolute value plays a very important role. If not taken into consideration very well, the answer usually becomes either zero or infinity which is wrong.
In the first case, the answer is negative 1.
Thursday, May 29, 2008
Gen. Calculus { *_* }
"Students nowadays are so clueless", the math professor complains to a colleague. "Yesterday, a student came to my office hours and wanted to know if General Calculus was a Roman war hero..."
...
Reference
[3]
...
Reference
[3]
Shadow of doubt
A street light is mounted at the top of a 15-ft tall pole. A man 6 ft tall walks away from the pole with a speed of 5 ft/s along a straight path. How fast is the tip of his shadow moving when he is 40 ft from the pole?
The figure below can represent the problem:
[ a ]
To solve this, the variables must be reflected into the diagram.
[ b ]
[ c ]
Figure 1. Illustration of the problem.
The movement of the shadow at any given time is represented by dy/dt. The problem is, which diagram is correct? Are these the same?
From Figure 1b, the basic equation becomes:
The derivative is:
On the other hand, from Figure 1c, the equation becomes:
By differentiation,
Now, it is obvious that the two figures are different because the ensuing equations are not the same.
Figure 1b is the change of the shadow from the point 40 ft from the post. On the other hand, Figure 1c is the movement of the shadow from the post.
Which one is correct?
Taking another look, Figure 1b presents the rate of change of the shadow initially from 40 ft from the post but this reference point apparently moves as the man walks. It is not a fixed point. On the other hand, the post as a common fixed reference point both for the man and his shadow is presented in Figure 1c. However, it is interesting to note also that there is no shadow behind the man or at the length x. Is it right to use Figure 1b where the shadow is confined in front of the man?
This could have been easier to resolve if the question asks: How fast is the tip of his shadow moving from the post when he is 40 ft from it? The original sentence in the problem essentially casts a shadow of doubt.
Only the final answer was given in the book which implies Figure 1c must be correct.
It is also interesting to note that the problem can be solved even without using the point 40 ft from the post. Therefore, the movement of the shadow is constant or the same wherever the man is.
The point 40 ft can be useful if the solution took the longer route of using the quotient rule of differentiation from the basic equation:
The derivative becomes:
where y should be solved first from similar triangles.
In all cases,
The answer is
.
I checked the book again. Calculus: Early Transcendentals Volume 1 by Stewart.
His first name?
James.
Ok, I believe him.
...
Reference
[4]
The figure below can represent the problem:
[ a ]
To solve this, the variables must be reflected into the diagram.
[ b ]
[ c ]
Figure 1. Illustration of the problem.
The movement of the shadow at any given time is represented by dy/dt. The problem is, which diagram is correct? Are these the same?
From Figure 1b, the basic equation becomes:
The derivative is:
On the other hand, from Figure 1c, the equation becomes:
By differentiation,
Now, it is obvious that the two figures are different because the ensuing equations are not the same.
Figure 1b is the change of the shadow from the point 40 ft from the post. On the other hand, Figure 1c is the movement of the shadow from the post.
Which one is correct?
Taking another look, Figure 1b presents the rate of change of the shadow initially from 40 ft from the post but this reference point apparently moves as the man walks. It is not a fixed point. On the other hand, the post as a common fixed reference point both for the man and his shadow is presented in Figure 1c. However, it is interesting to note also that there is no shadow behind the man or at the length x. Is it right to use Figure 1b where the shadow is confined in front of the man?
This could have been easier to resolve if the question asks: How fast is the tip of his shadow moving from the post when he is 40 ft from it? The original sentence in the problem essentially casts a shadow of doubt.
Only the final answer was given in the book which implies Figure 1c must be correct.
It is also interesting to note that the problem can be solved even without using the point 40 ft from the post. Therefore, the movement of the shadow is constant or the same wherever the man is.
The point 40 ft can be useful if the solution took the longer route of using the quotient rule of differentiation from the basic equation:
The derivative becomes:
where y should be solved first from similar triangles.
In all cases,
The answer is
.I checked the book again. Calculus: Early Transcendentals Volume 1 by Stewart.
His first name?
James.
Ok, I believe him.
...
Reference
[4]
Monday, May 19, 2008
Math inverse { *_* }
Teacher: "Who can tell me what 7 times 6 is?"
Student: "It's 42!"
Teacher: "Very good! - And who can tell me what 6 times 7 is?"
Same student: "It's 24!"
...
Source
[2]
Student: "It's 42!"
Teacher: "Very good! - And who can tell me what 6 times 7 is?"
Same student: "It's 24!"
...
Source
[2]
Crossing an intersection
2.2.8.85] Show that the curve
intersects the plane
at a right angle.
My intuitive solution is to use the tangent to the curve and set it perpendicular to the given plane, as given in the problem. As such, there are two perpendicular vectors to the plane: the normal of the plane and the tangent of the curve.
The normal of the plane is immediately obvious from the equation of the plane.
The tangent vector is actually more complicated in form because of the scalar denominator representing the length or magnitude of the tangent vector. It will be seen later that this scalar part of the tangent looks complex because of the presence of variable t. Good thing, this scalar is obviously non-zero, regardless of the value of t.
These two vectors, being both perpendicular to the same plane, must be parallel to each other. And so, the cross product of these two vectors must be zero.
Getting the cross product doesn't give us a clear answer immediately because of the presence of the variable t. If the cross product is zero, the scalar part must be zero or the resulting vector is in itself zero. Or both. In this case, the scalar is non-zero which leads us to set up the individual vector components (of the resulting cross product) to be zero. This way, we can solve for t which gives us zero vector components.
At this value of t, we can get the point x,y,z at which the curve intersects the plane perpendicularly. This turns out to be (2,8,8).
The other solution to this is to use the velocity vector of the curve, instead of the tangent vector. They are very similar but the velocity vector is simpler in form. I did in fact use the velocity vector the first time I solved this problem a long time ago. When I came back to this problem this morning, I spontaneously used the tangent vector.
What is the difference between the velocity and tangent approaches? Not much, actually. In my view however, the velocity vector is simpler but the tangent vector approach is more intuitive. Tangent is a concept that has a long history of supporting drills from lower levels of mathematics so I suppose most students will be more comfortable with the tangent concept. Velocity is generally regarded as a physics concept rather than a math concept, unless otherwise a given math problem has expressly used the term velocity. In this particular case, velocity isn't mentioned at all.
...
Reference
[1]
intersects the plane
at a right angle.
My intuitive solution is to use the tangent to the curve and set it perpendicular to the given plane, as given in the problem. As such, there are two perpendicular vectors to the plane: the normal of the plane and the tangent of the curve.
The normal of the plane is immediately obvious from the equation of the plane.
The tangent vector is actually more complicated in form because of the scalar denominator representing the length or magnitude of the tangent vector. It will be seen later that this scalar part of the tangent looks complex because of the presence of variable t. Good thing, this scalar is obviously non-zero, regardless of the value of t.
These two vectors, being both perpendicular to the same plane, must be parallel to each other. And so, the cross product of these two vectors must be zero.
Getting the cross product doesn't give us a clear answer immediately because of the presence of the variable t. If the cross product is zero, the scalar part must be zero or the resulting vector is in itself zero. Or both. In this case, the scalar is non-zero which leads us to set up the individual vector components (of the resulting cross product) to be zero. This way, we can solve for t which gives us zero vector components.
At this value of t, we can get the point x,y,z at which the curve intersects the plane perpendicularly. This turns out to be (2,8,8).
The other solution to this is to use the velocity vector of the curve, instead of the tangent vector. They are very similar but the velocity vector is simpler in form. I did in fact use the velocity vector the first time I solved this problem a long time ago. When I came back to this problem this morning, I spontaneously used the tangent vector.
What is the difference between the velocity and tangent approaches? Not much, actually. In my view however, the velocity vector is simpler but the tangent vector approach is more intuitive. Tangent is a concept that has a long history of supporting drills from lower levels of mathematics so I suppose most students will be more comfortable with the tangent concept. Velocity is generally regarded as a physics concept rather than a math concept, unless otherwise a given math problem has expressly used the term velocity. In this particular case, velocity isn't mentioned at all.
...
Reference
[1]
Friday, May 2, 2008
Getting a Tan { *_* }
Q: Why do you rarely find mathematicians spending time at the beach?
A: Because they have sine and cosine to get a tan and don't need the sun!
...
Reference(s):
[2]
A: Because they have sine and cosine to get a tan and don't need the sun!
...
Reference(s):
[2]
Distance education
Here are two problems about distance from a point and a line in space.
1.12.13.52] Find the distance from the point (5,7,14) to the line passing through (2,3,8) and (3,6,12).
1.12.14.52] Determine the shortest distance from the point (3,4,5) to the line through the origin parallel to the vector
Both of these problems can be solved in at least two ways. One is by using pythagorean theorem and the other is by using vector properties, particularly parallel-perpendicular decomposition.
The answers are:
...
For solution, email me at agriengineering at yahoo.com
Reference(s):
[1]
1.12.13.52] Find the distance from the point (5,7,14) to the line passing through (2,3,8) and (3,6,12).
1.12.14.52] Determine the shortest distance from the point (3,4,5) to the line through the origin parallel to the vector
Both of these problems can be solved in at least two ways. One is by using pythagorean theorem and the other is by using vector properties, particularly parallel-perpendicular decomposition.
The answers are:
...
For solution, email me at agriengineering at yahoo.com
Reference(s):
[1]
Tuesday, April 8, 2008
The lake
Let me share something about a lake problem which has showed up in the Math Webwork 11, problem 12.
A man with a boat is located at point P on the shore of a circular lake of radius 6 miles. He wants to reach the point Q on the shore diametrically opposed to P as quickly as possible. He plans to paddle his boat at an angle t to PQ to some point R on the shore, then walk along the shore to his Q. If he paddle 3.1 miles per hour and walk 4 miles per hour, what is the shortest time it will take him to reach Q?
To solve this problem, we need to minimize the following function of the angle t:
f(t) = ____________________
A stationary point for f(t) is
t = __________.
(Write DNE if there is none.)
We conclude that the minimal possible time for the trip is = _________.
The maximal possible time for the trip is = __________.
The basic equation for the travel time is:
The crux of the problem lies on the expression of travel time as a function of angle t. It should also be noted that the rate of moving by boat is not the same as the speed of walking. If these rates are the same, the problem becomes very easy.
Give the restrictions, the easiest solution for me is the use of the right angle conjecture when a triangle is inscribed in a semicircle, there is always a right angle anywhere along the perimeter when the two other points of the triangle are diametrically opposed to each other.
This is shown in the figure below. Angle PRQ must be 90 degrees.
From here, length PR can be expressed solely in terms of t.
The other problem is to get the relationship between angle a and t. Another figure clarifies this. When angle a becomes 90 degrees, t is pushed upward by the moving leg of angle a. The line connecting the vertex of t and the endpoint of that leg at the perimeter of the circle becomes shorter. This is illustrated below:
This essentially simplifies the equation into:
So, the answers are:
>stationary point t = 0.8867 rad
>shortest travel time is at f(t) = f(0) = 3.87 hrs
>longest travel time is at the critical point t which makes f(t) = 5.11 hrs.
This problem actually took me quite a while to solve. My lack of familiarity with the right angle conjecture led me to 7 equations that only complicated the problem.
In another piece (Winter wonders), I wrote about my adventure at Mirror Lake. This time, solving this problem gives me another conquest of the lake.
A man with a boat is located at point P on the shore of a circular lake of radius 6 miles. He wants to reach the point Q on the shore diametrically opposed to P as quickly as possible. He plans to paddle his boat at an angle t to PQ to some point R on the shore, then walk along the shore to his Q. If he paddle 3.1 miles per hour and walk 4 miles per hour, what is the shortest time it will take him to reach Q?
To solve this problem, we need to minimize the following function of the angle t:
f(t) = ____________________
A stationary point for f(t) is
t = __________.
(Write DNE if there is none.)
We conclude that the minimal possible time for the trip is = _________.
The maximal possible time for the trip is = __________.
The basic equation for the travel time is:
The crux of the problem lies on the expression of travel time as a function of angle t. It should also be noted that the rate of moving by boat is not the same as the speed of walking. If these rates are the same, the problem becomes very easy.
Give the restrictions, the easiest solution for me is the use of the right angle conjecture when a triangle is inscribed in a semicircle, there is always a right angle anywhere along the perimeter when the two other points of the triangle are diametrically opposed to each other.
This is shown in the figure below. Angle PRQ must be 90 degrees.
From here, length PR can be expressed solely in terms of t.
The other problem is to get the relationship between angle a and t. Another figure clarifies this. When angle a becomes 90 degrees, t is pushed upward by the moving leg of angle a. The line connecting the vertex of t and the endpoint of that leg at the perimeter of the circle becomes shorter. This is illustrated below:
This essentially simplifies the equation into:
So, the answers are:
>stationary point t = 0.8867 rad
>shortest travel time is at f(t) = f(0) = 3.87 hrs
>longest travel time is at the critical point t which makes f(t) = 5.11 hrs.
This problem actually took me quite a while to solve. My lack of familiarity with the right angle conjecture led me to 7 equations that only complicated the problem.
In another piece (Winter wonders), I wrote about my adventure at Mirror Lake. This time, solving this problem gives me another conquest of the lake.
Saturday, April 5, 2008
Zero is nothing { *_* }
Family members came down from Fairbanks, Alaska, to visit us in Anchorage just as the thermometer dropped to zero. I was freezing, but not the. “We're used to cold weather,” my brother-in-law said.
“Sure,” I replied. “To you folks, zero is nothing.”
...
Source:
[3]
“Sure,” I replied. “To you folks, zero is nothing.”
...
Source:
[3]
Monday, March 31, 2008
Center of attention
[1.7 | 11 | 24] A treasure map has n villages marked on it, and it contains the following instructions: Start at village A, go 1/2 of the way to B, 1/3 of the way to village C, 1/4 of the way to village D, and so forth. The treasure is buried at the last stop. Problem: You lose the instructions, and don't know in what order to select the villages. Show that it doesn't matter! You can still find the treasure.
The solution is in the title.
...
Reference:
[1]
The solution is in the title.
...
Reference:
[1]
Saturday, March 15, 2008
What's the point?
If the vector
represents the segment AB, and the midpoint of AB is (2,1), find A and B.
This is one of the problems which is easier solved by utilizing the properties of vectors.
The other solution is the conventional distance formula to solve for the coordinates of A and B but this actually takes longer because there are four unknowns in this case. At least two simultaneous equations may be required.
On the other hand, the vector method hinges on the vector property that its components represent the difference between the coordinates at endpoints. With this, the coordinates of A and B can be solved using a simple linear equation, one unknown at a time.
...
...
Reference:
[1]
represents the segment AB, and the midpoint of AB is (2,1), find A and B.This is one of the problems which is easier solved by utilizing the properties of vectors.
The other solution is the conventional distance formula to solve for the coordinates of A and B but this actually takes longer because there are four unknowns in this case. At least two simultaneous equations may be required.
On the other hand, the vector method hinges on the vector property that its components represent the difference between the coordinates at endpoints. With this, the coordinates of A and B can be solved using a simple linear equation, one unknown at a time.
...
...
Reference:
[1]
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