Monday, May 15, 2006

Episode 6: Solutions

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.

Saturday, April 01, 2006

Solutions to Episode Five

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

Friday, March 03, 2006

The Archrival

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

Wednesday, December 21, 2005

Solution to Episode Two: The Trip Across Styx

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

Tuesday, November 22, 2005

Solution to Episode 1: THe Cerebus

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).