Discovering Prince of Calculand: Solutions

Episode 6: Solutions

2006-05-15T19:30:00.000-05:00

a) dT/dh = k(T-5050)
Separate the variables.
dT(1/(T-5050)) = k dh
Antidifferentiate both sides of the equation.
ln(T-5050) = kh +C
T -5050 = e^(kh+C)
T = e^(kh+C) + 5050

b) Using the general solution, and the points (0,5000) and (150,3610.5405), solve for the exact equation.
T = e^(kh+C) + 5050
First use the point (0,5000) to solve for one of the variables.
5000 = (e^(k(0)))(e^(C)) +5050
-50 = e^(C)
T = -50e^(kh) + 5050
Now using the other point (150,3610.5405), solve for the last unknown.
3610.5405 = -50(k(150)) +5050
-1439.4595 = -50e^(150k)
Solve for k.
-1439.4595/-50 = e^(150k)
28.7892 = e^(150k)
ln(28.7892) = 150k
k = (1/150)ln(28.7892)
k = 0.0224
T = -50e^(0.0224h) +5050

c) Using the fact that the element can only be used at 4444 Kelvin, you have to insert that into the equation and solve for the height, h.
T = -50e^(0.0224h) +5050
4444 = -50e^(0.0224h) +5050
-606 = -50e^(0.0224h)
-606/-50 = e^(0.0224h)
12.12 = e^(0.0224h)
ln(12.12) = 0.0224h
h = (1/0.0224)ln(12.12)
h = 111.3775 cm
The height that the element has to be held at in order to reforge the sword is 111.3775 cm.

Solutions to Episode Five

2006-04-01T13:27:00.000-06:00

a) f(x)= 4x-x^2 and g(x)=5-2x
in this case 4x-x^2 lagrer than or equal to 5-2x. The graph interest where 4x-x^2=5-2x
solving this equation,
4x-x^2=5-2x
x^2-6x+5=0
(x-1)(x-5)=0
x=1 or x=5
we substitute in either formula and find that graphs intersect at (1,3) and (5,-5)
then,
Area(x) = S (from 1 to 5) ((4x-x^2) - (5-2x)) dx

=(3x^2- (1/3 )x^3- 5x) (from 1 to 5)

=75-125/3-25)-(3-1/3-5)

=32/3

b)1/14S (from 6 to 14) -(80-10 cos (pai t/12) dt

=-1/8(697.2957795)

=-87 F

c) -(80 -10 cos (pai t/12) >/ -78

80 - 10 cos (pai t/12) /<>

-5.230>/t>/-18.769

between this time it will have the storm

The Archrival

2006-03-03T06:00:00.000-06:00

Problem Translation:

A rectangle is inscribed in a circle with a radius of 4.

What is the major equation required for this problem?
Find the critical values of x
What is the largest area the rectangle can have and what are the dimensions?

ANSWERS:

The major equation required for this problem is

The area of the rectangle A = length * width

Since a circle is not a function, we use a semi-circle so that our calculations will be satisfied and more valid.

The equation of this circle is x2 + y2 = 4
So, the coordinates of the rectangle with respect to x are:

Length = 2x width = √ 16 – x2 formula – A(x) = x √ 16 – x2

Thus, the domain for this function would be [ 0 , 4 ]

Now to find the critical values of x

A’(x) = 2 √ 16 – x2 + .5(16 – x2)- .5 (-2x)(2x)

= A’(x) = (2 √ 16 – x2) – 2x2
√ 16 – x2

The function is not defined at x = 4

(2 √ 16 – x2) – 2x2 = 0
√ 16 – x2
Then,

2 ( 16 – x2 ) – 2x2 = 0

32 = 4x2

X = + or – 2 √ 2

So the length of the rectangle is

2x = 2 ( 2 √ 2 ) = 4 √ 2

The width of the rectangle is

Y = √ 16 – x2

Y = √ 16 – ( 2 √ 2 )2

Y = 2 √ 2

Therefore,

A rectangle with a length of 4 √ 2 and width of 2 √ 2 with an area of 16 units can be inscribed in a circle with a radius of 4

Solution to Episode Two: The Trip Across Styx

2005-12-21T01:15:00.000-06:00

A)What is the function that represents the velocity of the boat?

Because Charon keeps his boat at a constant speed of 6 m/s, but the boat causes it to speed up by another 4 m/s every other second, the equation is v(t)=4t+6

B)Given the time interval of the trip, how far will the boat travel before letting the prince off?

It takes the ferryman 29 seconds to cross the river, so the time interval of the entire trip would be from t=0 to t=29. Using the fundamental theorem, we find that the distance travelled is 1856 meters.

C)What will his final fare at the end of the trip be?

Because Charon charges 1 gold coins for every 4 meters, by taking the total distance and dividing it by 4, it give the total cost of the trip.
1856/4=464 gold coins

Solution to Episode 1: THe Cerebus

2005-11-22T18:28:00.000-06:00

1) When is the heads of the dog not moving?

It is where the velocity function is equal to 0.
0.5x^3 - 3x^2 + 0.25x + 5 = 0
It is 0 where x = 1.5577 and 5.5906.

2) At what time does our young man have to be most careful? (*hint* Steep tangent slopes on parent funtion.)

In other words, when is the heads of the dog moving the fastest? At a very high tangent slope on the parent function. Steep tangent slopes on parent function is where the parent function have inflection points. To find inflection points of a parent function you look at its second derivative, in this case its acceleration function. Find zeroes of the acceleration function and create a line analysis to show whether it is a inflection point or not. (Changes sign at the root).
Our young man have to be most careful when x=3.9579.

3) What interval(s) is the dog's heads slowing down?
The heads are slowing down when a(x) and v(x) are opposite signs from each other. Find the zeroes of each function and create a line analysis shadowing the other. Find where a(x) and v(x) have opposite signs, this is where the dog's heads are slowing down.
This occurs when (0.0421, 1.5577) u (3.9579, 5.5906).

The Grading Rubric

2005-11-21T15:58:00.000-06:00

These are the criteria by which your work will be assessed. Please share your thoughts regarding if and how we should change this rubric in the comments below this post. This instrument can be made better only if you participate in its development -- it's your mark on the line; it's up to you ...

Creativity (25%)
4
The problem contains many creative details and/or descriptions that contribute to the reader's enjoyment. The student has really used his/her imagination.

3
The problem contains a few creative details and/or descriptions that contribute to the reader's enjoyment. The student has used his/her imagination.

2
The problem contains a few creative details and/or descriptions, but they distract from the problem(s). The student has tried to use his/her imagination.

1
There is little evidence of creativity in the problem. The student does not seem to have used much imagination.

Mathematical Concepts (30%)
4
Problem requires understanding three or more mathematical concepts or representations to solve the problem(s).

3
Problem requires understanding two mathematical concepts or representations to solve the problem(s).

2
Problem requires understanding one mathematical concept or representation to solve the problem(s).

1
Problem requires understanding mathematical concepts or representations unrelated to this course to solve the problem(s).

Mathematical Terminology and Notation (20%)
4
Correct terminology and notation are always used, making it easy to understand what is required.

3
Correct terminology and notation are usually used, making it fairly easy to understand what is required.

2
Correct terminology and notation are used, but it is not easy to understand what is required.

1
There is little use, or a lot of inappropriate use, of terminology and notation.

Mathematical Challenge (25%)
4
The problem(s) presented can reasonally be solved by a knowledgable student in 10 to 15 minutes.

3
The problem(s) presented can reasonally be solved by a knowledgable student in 5 to 10 minutes.

2
The problem(s) presented can reasonally be solved by a knowledgable student in 5 minutes or less.

1
The problem(s) presented does not offer a reasonable challenge to the average student in this course.

Solution (20%)
4
All calculations are shown, brief explanations given where appropriate, and the results are correct and labeled appropriately.

3
All calculations are shown and the results are correct and may or may not be labeled appropriately.

2
Some calculations are shown and the results labeled appropriately.

1
No calculations are shown OR results are inaccurate or mislabeled.

Bonus Marks! - Humour or Drama
+0%
No humour or drama used in the problem presentation.

+2%
The scenario made me smile.

+4%
This work elicited a big grin/a sense of suspence.

+6%
I chuckled/was very moved when I read this.

+8%
I was laughing/crying out loud!