Which sketch is relevant?

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Today An optimizatoion example Residuals, SSR, Least squares Reminders: No class on Monday (Thanksgiving) OSH 4 due Wednesday Regular PLQs. Wednesday - bring laptop/tablet for spreadsheet practice

A boat leaves a dock at 2:00 P.M. and travels due south at a speed of 5 km/h. Another boat has been heading due east at 5 km/h and reaches the same dock at 3:00 P.M. How many minutes past 2:00 P.M. were the boats closest together? (A) (B) Which sketch is relevant? (C) (D)

A boat leaves a dock at 2:00 P.M. and travels due south at a speed of 5 km/h. Another boat has been heading due east at 5 km/h and reaches the same dock at 3:00 P.M. How many minutes past 2:00 P.M. were the boats closest together? (A) (B) Which sketch is relevant? (C) (D)

A boat leaves a dock at 2:00 P.M. and travels due south at a speed of 5 km/h. Another boat has been heading due east at 5 km/h and reaches the same dock at 3:00 P.M. How many minutes past 2:00 P.M. were the boats closest together? Two quantities relevant to solving this problem are: (A) x = 5/60 t, y = 5/60 (60-t). (B) x = 5(t-2), y=5(3-t). (C) x = 5-2, y=5+3. (D) x = 5t-2, y=5t-3.

A boat leaves a dock at 2:00 P.M. and travels due south at a speed of 5 km/h. Another boat has been heading due east at 5 km/h and reaches the same dock at 3:00 P.M. How many minutes past 2:00 P.M. were the boats closest together? Two quantities relevant to solving this problem are: (A) x = 5/60 t, y = 5/60 (60-t). (B) x = 5(t-2), y=5(3-t). (C) x = 5-2, y=5+3. (D) x = 5t-2, y=5t-3.

A boat leaves a dock at 2:00 P.M. and travels due south at a speed of 5 km/h. Another boat has been heading due east at 5 km/h and reaches the same dock at 3:00 P.M. How many minutes past 2:00 P.M. were the boats closest together? Objective function to be minimized: (A) f(t) = 25 t + 25 60-t (B) f(t) = 5/60 sqrt( 2t 2 ) (C) f(t) = t 2 + (60-t) 2 (D) f(t) = sqrt( 25(t-2) 2 + 25(3-t) 2 )

A boat leaves a dock at 2:00 P.M. and travels due south at a speed of 5 km/h. Another boat has been heading due east at 5 km/h and reaches the same dock at 3:00 P.M. How many minutes past 2:00 P.M. were the boats closest together? Objective function to be minimized: (A) f(t) = 25 t + 25 60-t (B) f(t) = 5/60 sqrt( 2t 2 ) (C) f(t) = t 2 + (60-t) 2 Figure it out - this is a homework problem, after all. (D) f(t) = sqrt( 25(t-2) 2 + 25(3-t) 2 )

A boat leaves a dock at 2:00 P.M. and travels due south at a speed of 5 km/h. Another boat has been heading due east at 5 km/h and reaches the same dock at 3:00 P.M. How many minutes past 2:00 P.M. were the boats closest together? When minimizing the function f(t), if the derivatives are easier to calculate, we can minimize the function instead. (A) g(t) = f(t) 2 (B) h(t) = 1/f(t) (C) k(t) = f(t) 3 (D) You have to minimize f(t).

A boat leaves a dock at 2:00 P.M. and travels due south at a speed of 5 km/h. Another boat has been heading due east at 5 km/h and reaches the same dock at 3:00 P.M. How many minutes past 2:00 P.M. were the boats closest together? When minimizing the function f(t), if the derivatives are easier to calculate, we can minimize the function instead. (A) g(t) = f(t) 2 <--- if f(t) 0. (B) h(t) = 1/f(t) <--- no, maximize (if f(t) 0). (C) k(t) = f(t) 3 <--- yes (D) You have to minimize f(t).

A boat leaves a dock at 2:00 P.M. and travels due south at a speed of 5 km/h. Another boat has been heading due east at 5 km/h and reaches the same dock at 3:00 P.M. How many minutes past 2:00 P.M. were the boats closest together? Expectation: The boats will be closest together... (A) at 2 pm. (B) at 3 pm. (C) sometime between 2 pm and 3 pm. (D) before 2 pm. (E) after 2 pm.

A boat leaves a dock at 2:00 P.M. and travels due south at a speed of 5 km/h. Another boat has been heading due east at 5 km/h and reaches the same dock at 3:00 P.M. How many minutes past 2:00 P.M. were the boats closest together? Constraint: (A) The minimum distance must occur between 2 pm and 3 pm. (B) x(t) 2 +y(t) 2 = t 2 /6. (C) x(t) = 60-y(t). (D) There isn t really a constraint for this problem.

Least squares model fitting How do we find the best line to fit through the data?

Least squares model fitting y=ax (x n,y n ) (x i,y i ) i=1,2,3,...,n (x 1,y 1 ) Each red bar is called a residual. We want all the residuals to be as small as possible.

The residuals are... (A) r i = y i 2 + x i 2 (B) r i = a 2 (y i 2 + x i2 ) y=ax (C) r i = y i - ax i (x i,y i ) (D) r i = y i - x i (E) r i = x i - y i

The residuals are... (A) r i = y i 2 + x i 2 (B) r i = a 2 (y i 2 + x i2 ) y=ax (C) r i = y i - ax i (x i,y i ) (D) r i = y i - x i (E) r i = x i - y i

To minimize the residuals, we define the objective function... (A) f(a) = y 1 -ax 1 + y 2 -ax 2 +...+ y n -ax n (B) f(a) = (y 1 -ax 1 ) 2 + (y 2 -ax 2 ) 2 +...+ (y n -ax n ) 2 (C) f(a) = (y 1 -ax 1 )(y 2 -ax 2 )...(y n -ax n ) (D) f(a) = (ay 1 -x 1 ) 2 + (ay 2 -x 2 ) 2 +... + (ay n -x n ) 2

To minimize the residuals, we define the objective function... (A) f(a) = y 1 -ax 1 + y 2 -ax 2 +...+ y n -ax n (B) f(a) = (y 1 -ax 1 ) 2 + (y 2 -ax 2 ) 2 +...+ (y n -ax n ) 2 (C) f(a) = (y 1 -ax 1 )(y 2 -ax 2 )...(y n -ax n ) (D) f(a) = (ay 1 -x 1 ) 2 + (ay 2 -x 2 ) 2 +... + (ay n -x n ) 2 (B) is called the sum of squared residuals. (A) is also reasonable but not as good (take a stats class to find out more).

Find a so that y=ax fits (4,5), (6,7) in the least squares sense. Define f(a): (A) SSR(a) = 5-4a + 7-6a (B) SSR(a) = (4-5a) 2 + (6-7a) 2 (C) SSR(a) = (5-4a) 2 + (7-6a) 2 (D) SSR(a) = (5-4-a) 2 + (7-6-a) 2

Find a so that y=ax fits (4,5), (6,7) in the least squares sense. Define f(a): (A) SSR(a) = 5-4a + 7-6a (B) SSR(a) = (4-5a) 2 + (6-7a) 2 (C) SSR(a) = (5-4a) 2 + (7-6a) 2 (D) SSR(a) = (5-4-a) 2 + (7-6-a) 2 Recall: f(a) = (y 1 -ax 1 ) 2 + (y 2 -ax 2 ) 2

Find a so that y=ax fits (4,5), (6,7) in the least squares sense. Find the a that minimizes SSR(a): (A)a = 7/6 (B) a = 5/4 (C) a = (7/6 + 5/4) / 2 (D) a = 31/26

Find a so that y=ax fits (4,5), (6,7) in the least squares sense. Find the a that minimizes SSR(a): SSR(a) = (5-4a) 2 + (7-6a) 2 (A)a = 7/6 (B) a = 5/4 (C) a = (7/6 + 5/4) / 2 (D) a = 31/26 = 5 2-2 4 5a + 4 2 a 2 + 7 2-2 6 7a + 6 2 a 2 SSR (a) = -2 4 5 + 2 4 2 a -2 6 7+2 6 2 a = 0 a = ( 2 4 5 + 2 6 7) / (2 4 2 +2 6 2 ) = ( 4 5 + 6 7 ) / (4 2 +6 2 ) = 62/52 = ( x 1 y 1 + x 2 y 2 ) / ( x 1 2 + x 2 2 )

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