Linear Algebra - Least Squares At time t = 0 an object is dropped from a height
ID: 3143176 • Letter: L
Question
Linear Algebra - Least Squares
At time t = 0 an object is dropped from a height of 1 meter above a fluid. A recording device registers the height of the object above the surface of the fluid at 1/2 second intervals, with a negative value indicating that the object is below the surface of the fluid. The following table of data is the result. (a) Determine the least squares quadratic polynomial. (b) Estimate the depth at t = 5 and t = 6 seconds. (c) Estimate the time the object breaks through the surface of the fluid the second time.Explanation / Answer
Using Matlab
x = [1 0 0
1 0.5 0.25
1 1 1
1 1.5 2.25
1 2 4
1 2.5 6.25
1 3 9
1 3.5 12.25
1 4 16
1 4.5 20.25
];
>> x
x =
1.0000 0 0
1.0000 0.5000 0.2500
1.0000 1.0000 1.0000
1.0000 1.5000 2.2500
1.0000 2.0000 4.0000
1.0000 2.5000 6.2500
1.0000 3.0000 9.0000
1.0000 3.5000 12.2500
1.0000 4.0000 16.0000
1.0000 4.5000 20.2500
>> y = [1
0.88
0.54
0.07
-0.42
-0.8
-0.99
-0.94
-0.65
-0.21
];
>> y
y =
1.0000
0.8800
0.5400
0.0700
-0.4200
-0.8000
-0.9900
-0.9400
-0.6500
-0.2100
>> b= inv(x'*x)*(x'*y)
b =
1.3378
-1.2974
0.2006
hence Depth^ = 1.337818182 -1.297393939*t +0.200606061 *t^2
a) Depth^ = 1.337818182 -1.297393939*t +0.200606061 *t^2
b) t = 5
Depth^ = 1.337818182 -1.297393939*5 +0.200606061 *25
=-0.133999988
t = 6
Depth^ = 1.337818182 -1.297393939*6+0.200606061 *36
=0.775272744
c) solving y = 0
we get t = 1.28745 and t = 5.17992
since we are asked when y = 0 second time it is
5.17992 sec
Please don't forget to rate positively if you found this response helpful. Feel free to comment on the answer
if some part is not clear or you would like to be elaborated upon. Thanks and have a good day!
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.