"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]
Thursday, May 29, 2008
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]
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