Math  /  Algebra

QuestionThe Data: Here are three times and the measured height of the plane at those times. \begin{tabular}{|c|c|} \hline \begin{tabular}{c} Time (t)(t) in \\ seconds \end{tabular} & \begin{tabular}{c} Height (h)(h) in \\ feet \end{tabular} \\ \hline 2 & 24205 \\ \hline 20 & 32305 \\ \hline 40 & 33705 \\ \hline \end{tabular}
To find the coefficients for your model, plug the data into the equation: h=at2+bt+ch=a t^{2}+b t+c
The data points are just like xx and yy values, where the xx value is the time tt in seconds and the yy value is the altitude hh in feet. Plug these into the model to get three linear equations with variables a,ba, b, and cc. \begin{tabular}{|c|c|c|} \hline Data Values & Enter the resulting equation & \\ \hline(2,24205)(2,24205) & \\ \hline(20,32305)(20,32305) & \\ \hline(40,33705)(40,33705) & \\ \hline \end{tabular}

Studdy Solution

STEP 1

1. We are given three data points that represent time t t and height h h .
2. We need to find the coefficients a a , b b , and c c of a quadratic equation h=at2+bt+c h = at^2 + bt + c .
3. Each data point will be used to create a system of linear equations.

STEP 2

1. Substitute each data point into the quadratic equation to form three equations.
2. Solve the system of linear equations to find the coefficients a a , b b , and c c .

STEP 3

Substitute the first data point (2,24205)(2, 24205) into the quadratic equation h=at2+bt+c h = at^2 + bt + c :
24205=a(2)2+b(2)+c 24205 = a(2)^2 + b(2) + c 24205=4a+2b+c 24205 = 4a + 2b + c

STEP 4

Substitute the second data point (20,32305)(20, 32305) into the quadratic equation:
32305=a(20)2+b(20)+c 32305 = a(20)^2 + b(20) + c 32305=400a+20b+c 32305 = 400a + 20b + c

STEP 5

Substitute the third data point (40,33705)(40, 33705) into the quadratic equation:
33705=a(40)2+b(40)+c 33705 = a(40)^2 + b(40) + c 33705=1600a+40b+c 33705 = 1600a + 40b + c

STEP 6

Now we have a system of three linear equations:
1. 4a+2b+c=24205 4a + 2b + c = 24205
2. 400a+20b+c=32305 400a + 20b + c = 32305
3. 1600a+40b+c=33705 1600a + 40b + c = 33705

Solve this system using methods such as substitution, elimination, or matrix operations (e.g., Gaussian elimination).

STEP 7

To eliminate c c , subtract the first equation from the second:
(400a+20b+c)(4a+2b+c)=3230524205 (400a + 20b + c) - (4a + 2b + c) = 32305 - 24205 396a+18b=8100 396a + 18b = 8100

STEP 8

Subtract the second equation from the third:
(1600a+40b+c)(400a+20b+c)=3370532305 (1600a + 40b + c) - (400a + 20b + c) = 33705 - 32305 1200a+20b=1400 1200a + 20b = 1400

STEP 9

Now solve the two equations:
1. 396a+18b=8100 396a + 18b = 8100
2. 1200a+20b=1400 1200a + 20b = 1400

Use elimination or substitution to find a a and b b .

STEP 10

Divide the second equation by 20:
60a+b=70 60a + b = 70
Now solve the system:
1. 396a+18b=8100 396a + 18b = 8100
2. 60a+b=70 60a + b = 70

STEP 11

Substitute b=7060a b = 70 - 60a into the first equation:
396a+18(7060a)=8100 396a + 18(70 - 60a) = 8100 396a+12601080a=8100 396a + 1260 - 1080a = 8100 684a=6840 -684a = 6840 a=10 a = -10

STEP 12

Substitute a=10 a = -10 back into b=7060a b = 70 - 60a :
b=7060(10) b = 70 - 60(-10) b=70+600 b = 70 + 600 b=670 b = 670

STEP 13

Substitute a=10 a = -10 and b=670 b = 670 into the first equation to find c c :
4(10)+2(670)+c=24205 4(-10) + 2(670) + c = 24205 40+1340+c=24205 -40 + 1340 + c = 24205 c=242051300 c = 24205 - 1300 c=22905 c = 22905
The coefficients for the model are a=10 a = -10 , b=670 b = 670 , and c=22905 c = 22905 .

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