#$&* course Mth 277 11/29 2 11.5
.............................................
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): I think f_yy is 2 not 5. Also its hard for me to see that f_xy is -5. do you factor out the constant of -5 and fill in the other values x and y with 0 ? Also I dont understand why you did that last equation. ------------------------------------------------ Self-critique rating:
.............................................
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): ------------------------------------------------ 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: x values are ( 4, 3, 0, -1, -2, -3 ) y values are ( -2, -1, 0, 3, 1, 2 ) get to equations 4a + b + 2 3a + b + 1 b -a + b -3 -2a + b - 1 -3a + b -2 square them all and sum together to get 39a^2 + 2ab + 44a + 6b^2 -6b + 24 take both partial derivatives and get a = 0.6 ; b = -.4 so y = 0.6x - 0.4 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): for b i'm getting -0.4 and a = 0.6 solve for a with f_b. a = 3-6b substitute into other equation to get 234 - 470b = -44 b = -0.4 fill back into small a equation a = 0.6 would not have been able to do this without solution ------------------------------------------------ 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: f(x,y) True; has CP at point (0,0) has saddle p because fxx*fyy-fxy^2 is negative. which is also at (0,0) g(x,y) TRUE has CP (0,0) and fxx*fyy-fxy^2 is pos. with all positive numbers so relative min. h(x,y) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: `q005. We can use the exponential and logarithm to help us to linearize data that does not tend to change linearly. The following problems will demonstrate this. Suppose we have y = kx^m. Show that by taking the natural logarithm of this equation we obtain a linear relationship Y = K + mX. Explain these new variables and the constant K. The data that follows relates the periods of revolution, t (in days), of the six inner planets and to their semimajor axes a (in 10^6 km). Kepler conjectured that the relationship is t = ka^m for some k and m. Transform the data as in the first part and find k and m. t-data : (87.97, 224.7, 365.26, 686.98, 4332.59, 10759.2) a-data : (58, 108, 149, 228, 778, 1426). log(y) = log( k x^m) = log(k) + m log(x), which is of the form Y = K + m X for Y = log(y), K = log(k) and X = log(x). It follows that the graph of Y vs. X is a straight line. Logs of the a data are 1.76342799356294 2.03342375548695 2.17318626841227 2.35793484700045 2.89097959698969 3.15411952551585 Logs of the b data are 1.94433459197078 2.35160307241913 2.56260211477846 2.83694409365915 3.63674759298951 4.0317799805774 Both are given above to way too many significant figures. We can find the best-fit line for these data. The slope of the best-fit line will be our power m. Since K = log(k), k = 10^K. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since a slope is a change in Y over a change in X, we can use change in Y= log(y) which = log(k) + m log(x) / (change in X which = log(x) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating:" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!