#$&* course Mth 277 December 2 around 12pm. 11.7*********************************************
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Given Solution: f_x = 6 x - 5 y and f_y = 5 x + 2 y. Our critical point therefore occurs when 6x - 5y = 0 and 5x + 2y = 0. Solving simultaneously we get x = 0, y = 0 so our critical point is (0, 0). The second derivatives are f_xx = 6, f_yy = 5 and f_xy = -5. f_xx and f_yy are both positive, indicating that if the critical point is not a saddle point, it is a minimum f_xx * f_yy - f_xy^2 = 6 * 5 - (-5)^2 = 5, which is > 0 and indicates that the critical point is not a saddle point. The one critical point for this function therefore corresponds to a relative minimum, which occurs at the point (0, 0, f(0, 0)) = (0, 0, 3). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK! Got everything about right except for finding a point corresponding to this relative min. Also, our f_yy’s are different?!
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Given Solution: f_x = 2 x + y / 16 and f_y = -2 y + x / 16. Setting the two equal to zero and solving we find that (0, 0) is our only critical point. f_xx = 2 and f_yy = -2. This indicates that the y = 0 'slice' of the graph has a minimum while the x = 0 'slice' has a maximum. The critical point therefore yields a saddle point. We could evaluate f_xy (which is 1/16) and test for a saddle point, but we've already seen that we do have a saddle point. So we normally wouldn't bother. However, just to illustrate that the test for a saddle point works in this case, let's do the test. f_xx * f_yy - f_xy ^ 2 = 2 * (-2) - (1/16)^2 = -4 - 1/16. This is clearly < 0, indicating what we already knew, that the point (0, 0) is a saddle point. Note how f_xx * f_yy came out negative, since the signs of the two second derivatives differed. As soon as this happened the left-hand side was doomed to be negative, since the only other term - f_xy ^ 2 (being the negative of a square) cannot be positive. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `q003. Find the least squares regression line for the set of points {(4,-2), (3,-1), (0,0), (-1,3), (-2,1), (-3,2)}. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: **** Plug in “x” values in y=mx+b: (4m+b), (3m+b), (b), (-m+b), (-2m+b), (-3m+b) Plug in “y” values in mx+b-y=0: (4m+b+2), (3m+b+1), (b), (-m+b-3), (-2m+b-1), (-3m+b-2) Square each term and find the sum: (39m^2)+(2mb)+(44m)+(6b^2)-(6b)+19 Find partial derivatives (m,b): fm=(78m+2b+44) fb=(12b+2m-6) Solving simultaneously to put values back into equation: m=-0.6 b=0.6 y=(-0.6)m+0.6 #$&* confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: If y = a x + b, then the points on the line corresponding to x = 4, 3, 0, -1, -2, -3 have respective y coordinates 4 a + b, 3 a + b, b, -a + b, -2 a + b and -3 a + b. These differ from the given y coordinates by respective amounts 4 a + b - (-2), 3 a + b - (-1), b, -a + b - 3, -2 a + b - 1 and -3a + b - 2. The sum of the squares of these differences is thus (4 a + b - (-2))^2 + (3 a + b - (-1))^2 + b^2 + (-a + b - 3)^2 + ( -2 a + b - 1 )^2 + (-3a + b - 2))^2 = 39 a^2 + 2 a b + 44 a + 6 b^2 - 6 b + 19 In order to find the regression line we will find the values of a and b that minimize this sum. For the sake of convenient notation let f(a, b) = 39 a^2 + 2 a b + 44 a + 6 b^2 - 6 b + 19. f_a = 78 a + 2 b + 44 f_b = 12 b + 2 a - 6 Solving simultaneously we obtain approximate values a = -.58 and b = .60. We conclude that the best-fit equation, with a and b accurate to 2 significant figures, is y = -.58 x + .60. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): PERFECT! ------------------------------------------------ self-critique rating #$&*: ********************************************* Question: `q004. Consider these following functions, at each of which D = 0 at a critical point. Show whether each of the following is true or false: • f(x,y) = x^4 - y^4 has a saddle point at (0,0). • g(x,y) = x^2*y^2 has a relative minimum at (0,0). • h(x,y) = x^3 + y^3 has a relative maximum at (0,0). f_xx is positive and f_yy is negative. The intersection of the graph of f(x, y) with the x-z plane is a curve with a relative min at (0, 0); the intersection with the x-y plane has a relative max at the same point. Consider the line parameterized by x = t cos(theta), y = t sin(theta). Along this line we have z = t^4 cos^2(theta) sin^2(theta). For any value of theta except a multiple of 2 pi the function z vs. t has a relative minimum at t = 0. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: **** 1st equation: f(x,y)=x^4-y^4 fx=4x^3, fy=-4y^3, fxx=12x^2, fyy=-12y^2 fxy=0 fyx=0 D=fxx*fyy-fxy^2=(-144x^2y^2) 2nd equation: g(x,y)=x^2y^2 gx=2xy^2, gy=2yx^2, gxx=2y^2, gyy=2x^2 gxy=4xy gyx=4xy D=gxx*gyy-gxy^2=(-12x^2y^2) 3rd equation: h(x,y)=x^3+y^3 hx=3x^2, hy=3y^2, hxx=6x, hyy=6y hxy=0 hyx=0 D=hxx*hyy-hxy^2=(36xy) Where do I go from here, since D=0 at a critical point? The text says if D(x0,y0)=0 then the test in INCONCLUSIVE. A saddle point occurs at P0 if D(x0,y0) <0. A relative minimum occurs at P0 if D(x0,y0) >0 and fxx(x0,y0)>0. A relative maximum occurs at P0 if D(x0,y0)>0 and fxx(x0,y0)<0. What do I do with the calculations I have? #$&*
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ self-critique rating #$&*: