Class notes from 8/29
Try same ideas with
3x + 7 y = 9
81 x 35 y = 17
and with
2 x + 32 y = 7
3 x 5 y = 8.
Solutions
(291/106, 5/106)
(1.55, .62)
(2.745,.047)
3 * [ 2x + 32 y = 7 ]
-2 * [ 3 x 5 y = 8 ]
6 x + 96 y = 21
-6 x + 10 y = - 16
Sum of equations is
0 x + 106 y = 5 so y = 5 / 106
Plugging y into 2x + 32 y =7 we get
2x + 32 ( 5/106) = 7
Multiplying both sides by 106 we get
106 ( 2 x ) + 32 ( 5 / 106) * 106 = 106 * 7
212 x + 160 = 742
212 x = 742 170 = 582
x = 582 / 212 = 291 / 106
Should also verify in both equations to be sure.
Solution is (291 / 106, 5 / 106) or approx. (2.75, .047).
Next: get model by simul eq, compare, etc..
The form of a quadratic function is
If we use the data points (10,80), (46, 50) and (99,20) we can get three simultaneous linear equations.
(10, 80) means y = 80 when t = 10. In this case our equation will be
The point (46, 50) gives us the equation
The point (99, 20) gives the equation
Squaring and rearranging we see that we end up with the equations
100 a + 10 b + c = 80
2116 a + 46 b + c = 50
9801 a + 99 b + c = 20.
How we gonna solve this system?
Eliminate one of the variables to get 2 equations in 2 unknowns. Pick on c:
1st eqn + (2d) eqn gives:
100 a + 10 b + c = 80
-2116 a 46 b c = -50
sum is
-2016 a 36 b = 30
Do the same with the first and third equations:
100 a + 10 b + c = 80
-9801 a - 99 b - c = -20.
Sum is 9701 a 89 b = 60.
Now we have the system
-2016 a 36 b = 30
9701 a 89 b = 60.
To solve eliminate b. Multiply 1st by 89 and 2d by 36 to get
89 [ -2016 a 36 b = 30 ]
-36 [ 9701 a 89 b = 60 ]
so
-179,424 a 3204 b = 2670
349,232 a + 3204 b = -2160.
Sum is
169,812 a = 510 so
a = 510 / 169,812 = .003003 (approx.).
Now plug a into -2016 a 36 b = 30 to get
-2016 (.003003) 36 b = 30
-6.054 - 36 b = 30
-36 b = 36.054
b = -1.0015.
Now plug a and b into the first equation:
100 a + 10 b + c = 80 so
100 ( .003003) + 10 (-1.0015) + c = 80
.3003 10.015 + c = 80
-9.712 + c = 80
c = 89.71.
So our y = a t^2 + b t + c model is now
The Excel solution, which used ALL the data points and not just the three we chose, was
so we're pretty close.
According to the rounded-off model y = .003 t^2 t + 90:
What should be the depth at t = 60?
When should the depth be 25?
25 = .003 t^2 t + 90.
Solve this for t. This is a quadratic equation in t. Rearrange to get
.003 t^2 t + 65 = 0 and use the quadratic formula.
When should the depth reach 0?
What should be the minimum depth?