#$&*
course MTH 271
oct 20 2011 11:14 pm
clock t
4.2
8.4
12.6
16.8
21
25.2
depth
80.2
72.9
67.5
63.3
59.3
55
y=a*t^2+bt+c
So we take three points and plug them in to get our system of equations: (4.2,80.2),(8.4,72.9),(12.6,67.5)
Giving us:
[1]: 80.2=a(4.2^2)+4.2b+c=17.64a+4.2b+c
[2]: 72.9=70.56a+8.4b+c
[3]: 67.5=158.76a+12.6b+c
Now we solve for b by first subtracting the first equation from the second
[4]: -7.3=(70.56-17.64)a+(8.4-4.2)b=52.92a+4.2b
Now we subtract the first from the third:
[5]: -12.7=(158.76-17.64)a+(12.6-4.2)b=141.12a+8.4b
And the then multiply the fourth equation by 141.12 and the fifth by 52.92 giving:
141.12(-7.3)=-1030.176=52.92*141.12a+141.12*4.2b = 7468.07a+592.704b
and
52.92-(-12.7)=-672.084=7468.07a+52.92*8.4b=7468.07a+444.53b
Now we subtract these equations giving:
-358.092=148.174b
b=-2.417
Now we plug this value for b into the equation [5]:
-12.7=141.12a+8.4(-2.417)
-12.7=141.12a-20.303
7.603=141.12a
a=0.0539
next we plug these two values into equation [3]:
67.5=158.76(.0539)+12.6(-2.417)+c
67.5=8.557-30.454+c
67.5=-21.8968+c
c=89.3968
So we have an equation:
y=.0539t^2-2.417t+89.397
Now we can plug in a point to test the equation: (8.4,72.9)
72.9=.0539(8.4^2)-2.417(8.4)+89.397
72.9=.0539(70.56)-20.303+89.397
72.9=3.8032+69.094
72.9=72.9
and now we test another point to be sure: (16.8,63.3)
63.3 ?=? .0539(282.24)-2.417(16.8)+89.397
63.3 ?=? 15.2127-40.6056+89.397
63.3 ?=? 64 The answer is close so maybe the problem is just that I picked three points at the beginning so I will go back and try again using the same first two equations but for the third use the point (25.2,55) instead:
[1]: 80.2=a(4.2^2)+4.2b+c=17.64a+4.2b+c
[2]: 72.9=70.56a+8.4b+c
[3]: 55=635.04a+25.2b+c
Now subtract [2] from [3] to get a new equation [4]:
55-72.9=(635.04-70.56)a+16.8b
-17.9=564.48a+16.8b
Next we subtract [1] from [2] to get a new equation [5]:
72.9-80.2=-7.3=(70.56-17.64)a+4.2b=52.92a+4.2b
And now we can solve for a by multiplying equation [5] by 4 and subtracting the two:
-29.2-(-17.9)=(211.68-564.48)a
11.3=-352.8a
a=-.032
Next we plug this a into equation [4]:
-17.9=564.48(-.032)+16.8b
-17.9=-18.06336+16.8b
0.16336=16.8b
b=0.00972
and finally we plug these values into [2] to get c:
72.9=70.56(-.032)+8.4(.00972)+c
72.9=-2.2579+ 0.0817+c
72.9=-2.1762+c
c=75.0762
So we have a new equation for the model:
y=-.032t^2+.00972t+75.0762
Now we must test this equation using one of the original points:
(8.4,72.9)
72.9=-.032(70.56)+0.00972(8.4)+75.0762
72.9=-2.2579+0.08165+75.0762
72.9=72.8183+.08165
72.9=72.89 CHECK
And now lets test a point that was not used for an equation: (21,59.3)
59.3 ?=? -.032(441)+.00972(21)+75.0762
59.3 ?=? -14.112+0.20412+75.0762
59.3 ?=? -14.112+75.2803
59.3 ?=? 61.16832
CLOSE
Now for the excercizes, we first plug in 46 seconds:
y=-.032(2116)+.00972(46)+75.0762
y=-67.712+0.44712+75.0762
y=-67.712+75.52332
y=7.81132 cm when t=46 seconds
And the second question about this model asks what at what time the depth will be 14cm
14=-.032t^2+.00972t+75.0762
0=-.032t^3+.00972t+61.0762 (next step is to use the quadratic formula)
x=(-.00972{+/-}sqrt(0.0000945-4(-.032)(75.0762)))/(-0.064)
x=(-.00972{+/-}sqrt(.0000945+9.6097536))/-.064
x=(-.00972{+/-}sqrt(9.6098481))/-.064
x=(-.00972{+/-}3.099975)/-.064
x=3.0902/-.064 or x=-3.10962/-.064
x= -43.536 or x= 43.84
since the negative root is irrelevant when dealing with time, obviously the depth will be 14cm at time=43.8 seconds. This model seems to fit the data pretty well but its suspicious that over 2.2 seconds there will be 6.2cm of drainage
- - - - -
- - - - -
In order to calculate a model I must select three points; the first the last and one in between are what I decided: (0,1), (50,2.76776),(100,3.5)
Now for the equations:
[1]: 1=0a+0b+c
[2]: 2.76776=2500a+50b+c
[3]: 3.5=10000a+100b+c
Since we know from the first equation where the a and b are canceled out by the 0, that c, which seems like the y-intercept, is 1.
c=1
Therefore we can double equation [2] and subtract it from equation [3] to calculate a.
3.5-5.53552=(10000-5000)a-1
-1.03552=5000a
a=-0.000207104
Now we plug c and a into [2] to calculate b.
2.76776=-0.51776+50b+1
2.76776=0.48224+50b
2.28552=50b
b=0.0457104
So we have an equation:
y=-0.000207104x^2+0.0457104x+1
Time to test it by plugging in a number, lets try 100:
y=-2.07104+4.57104+1
y=2.5+1
y=3.5 check
Now lets plug in another point to check(50,2.767767)
2.767767 ?=? -0.000207104(2500)+0.0457104(50)+1
2.767767 ?=? -0.51776+2.28552+1
2.767767 ?=? 2.76776 CLOSE
Now the assignment says to check the percent which must be reviewed to achieve grades 3.0 and 4.0:
3.0=-0.000207104x^2+0.0457104x+1
0=-0.000207104x^2+0.0457104x-2
x=(-.0457104{+/-}sqrt(0.00208944066816-4(-0.000207104)(1)))/(2(-0.000207104))
x=(-.0457104{+/-}sqrt(0.00208944066816+0.000828416))/-0.000414208
x=(-.0457104{+/-}sqrt(0.00291785666816))/-0.000414208
x=(-.0457104{+/-}0.0540171886)/-0.000414208
x=0.0540171886/-0.000414208 or x=-0.0997275886/-0.000414208
x= 60.142 or x=160.57
since obviously the second root is too high, we can reasonably assume this model shows 60.142% of assignments must be reviewed to get a 3.0, which varies slightly from what you would expect from plugging in 60, which would be the 2.9364 on the table.
Now to solve for when y=4.0:
4=-0.000207104x^2+0.0457104x+1 (now we get 0 on one side to use the quadratic equation
0=-0.000207104x^2+0.0457104x-3
x=(-.0457104{+/-}sqrt(0.00208944066816-4(-0.000207104)(-3)))/(2(-0.000207104))
x=(-.0457104{+/-}sqrt(-0.00039580733184))/-0.000414208
Now we can see there is no real solution because the square root of a negative number will result in a number containing an imaginary number which will not be applicable to this problem as it will not be possible to study an imaginary percentage of the assignments. This makes sense because 100% reviewed only yields a 3.5 grade average, meaning its impossible to make a 4.0
Now we must determine the projected grade for someone who reviews 80% of the notes
y=-0.000207104(6400)+0.0457104(80)+1
y=-1.3254656+3.656832+1
y=-1.3254656+4.656832
y=3.3313664
so if a student reviews 80% of the assignments we project from this model a grade average of 3.33
The model seems to fit fairly well but there are a few decimals off, as could be expected since no mathematical model can actually calculate the grade a student will receive, it can only predict a close guess.
The curve I obtained would be similar to what a real group of students would invoke; a grade average which increases as the percentage of assignments is reviewed.
- - - - - -
- - - - - -
To obtain a model of the distance from the sun versus the illumination of the comet, I must select three points; I will use (1,935.1395), (5,43.06238), and (10,9.484465)
So we have these equations:
[1]: 935.1395=a+b+c
[2]: 43.06238=25a+5b+c
[3]: 9.484465=100a+10b+c
Now we remove the c by subtracting equation [2] from equation [1]:
935.1395-43.06238=-24a-4b
[4]: 892.07712=-24a-4b
And we do the same to get another equation by subtracting equation [2] from [3]:
9.484465-43.06238=75a+5b
[5]: -33.577915=75a+5b
Now we can eliminate b by multiplying equation [4] by 5 and adding it to equation [5] after multiplying it by 4 to solve for a
-134.31166+4460.3856=-120a+300a
4326.07394=180a
a=24.03374411
Now we can plug this a value into equation [5] to get a value for b:
-33.577915=75(24.03374411)+5b
-33.577915=1802.53080833+5b
-1836.1087233=5b
b=-367.22174467
And finally to get our equation we plug both of these into one of the original equations then solve for c. lets use [1]
935.1395=13.5189810625-367.22174467+c
935.1395=-353.7027636075+c
c=1288.8422636075
So we get our equation:
y=24.03374411x^2-367.22174467x+1288.8422636075
Now for the exercises, first we plug in 1.6 for x to get the illumination of the comet at 1.6 earth distances
y=24.03374411(2.56)-367.22174467(1.6)+1288.8422636075
y=61.5263849216-587.554791472+1288.8422636075
762.8138570571W/m^2 is the illumination when x=1.6AU
Lastly, we must calculate the range of illumination which would be comfortable for reading by calculating the Distance from the Sun when Y=25 and Y=100. Begin by plugging in y=25:
25=24.03374411x^2-367.22174467x+1288.8422636075
0=24.03374411x^2-367.22174467x+1263.8422636075
x=10.044 or x=5.2357 Since we know that the illumination is only9.484465 at 10 AU we can assume that the 5.2357AU begins the range of illumination that makes it comfortable for reading. Now we plug in 100 to get the second value:
100=24.03374411x^2-367.22174467x+1288.8422636075
0=24.03374411x^2-367.22174467x+1188.8422636075
X=10.623 or 4.6565 Now we can assume the lower number is correct again using the same logic as before and we get that the range of distance the comet is from the earth before its illumination makes reading comfortable is from 4.6565 to 5.2357 AU
The model seems to fit the data fairly well but from the points we are given we would infer that 25W/m^2 would be reached just barely over 6AU and 100 would be just under 3 but the model does not exactly give those numbers. This means the model does not exactly fit the data."
@& Very good.
However that's an awful lot of typing for you.
The Query is designed to reduce the typing load by asking selected questions. You don't need to submit the entire problem set, though you should work it out before doing the Query.
No need to submit this part of the assignment using the Query, and I don't mind this format at all. So you can do as you wish. But you will probably prefer the Query for future assignments.*@