Math Quiz 2 - Tuesday, October 4 Your name here:
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1 Math Quiz 2 - Tuesday, October 4 Your name here: 1. Let f (x, y) =sin x sin y. (a) Find r f (x, y). (1 point) r f (x, y) =hcos x sin y, sin x cos yi (b) Find all critical points of f and use the second derivative test to identify their types. (5 points) The system cos x sin y = 0 and sin x cos y = 0 is solved if either cos x = cos y = 0 or sin y = sin x = 0. Notice that we cannot have cos x = 0 and sin x = 0 since cos 2 x + sin 2 y = 1. We have: or x = p cos x = cos y = 0 ) 2 + np n is any integer y = p 2 + mp m is any integer (1) x = np sin y = sin x = 0 ) y = mp n is any integer m is any integer (2) To use the second derivative test, we compute f xx = sin x sin y, f yy = sin x sin y, and f xy = cos x cos y. Call the Hessian D. We then have D = f xx f yy ( f xy ) 2 = (sin x sin y) 2 (cos x cos y) 2. In case (2) above D = (cos x cos y) 2 = 1 < 0 (remember we have cos 2 x + sin 2 x = 1 so if sin x = 0 then cos 2 x = 1). So all the points in (2) are saddle points. In case (1) above D =(sin x sin y) 2 = 1 > 0, so we then analyze f xx = sin x sin y for x = p 2 + np and y = p 2 + mp. Notice that sin(p/2 + np) is 1 if n is even and 1 if n is odd. This can be stated compactly by writing sin(p/2 + np) =( 1) n. We have f xx (p/2 + np, p/2 + mp) = sin(p/2 + np) cos(p/2 + mp) = ( 1) n ( 1) m =( 1) m+n+1 Hence f xx > 0 when m + n is odd and f xx < 0 when m + n is even. The table below summarizes these findings:
2 Point (p/2 + np, p/2 + mp) with m + n odd (p/2 + np, p/2 + mp) with m + n even (np, mp) min / max / saddle min max saddle For instance, if we restrict to the box p/2 apple x apple p/2 and p/2 apple y apple p/2 then we have 5 critical points: ( p/2, p/2) and (p/2, p/2) are both maximums, ( p/2, p/2) and (p/2, p/2) are both minimums, and (0, 0) is a saddle point. (OVER)
3 2. Use the method of Lagrange multipliers to find the closest point to P =( 3, 1) on the line 4x 3y = 5. (4 points) Minimize D =(x + 3) 2 +(y 1) 2, the function measuring distance from P to (x, y), subject to the restraint g = 0 where g = 4x 3y 5. We have rd = h2(x + 3), 2(y 2)i and rg = h4, 3i. Using the method of Lagrange multipliers we get the system of equations: (x + 3) =2l 2(y 2) = 3l Solving for x and y in terms of l we get x = 2l 3 and y = 3/2l + 2. Plugging these values for x and y into 4x 3y 5 = 0 we obtain that l = 8/5 and so x = 1/5 and y = 7/5.
4 Math Quiz 2 - Tuesday, October 4 Your name here: 1. Let f (x, y) =cos x cos y. (a) Find r f (x, y). (1 point) r f (x, y) =h sin x cos y, cos x sin yi (b) Find all critical points of f and use the second derivative test to identify their types. (5 points) The system sin x cos y = 0 and cos x sin y = 0 is solved if either cos x = cos y = 0 or sin y = sin x = 0. Notice that we cannot have cos x = 0 and sin x = 0 since cos 2 x + sin 2 y = 1. We have: or x = p cos x = cos y = 0 ) 2 + np n is any integer y = p 2 + mp m is any integer (1) x = np sin y = sin x = 0 ) y = mp n is any integer m is any integer (2) To use the second derivative test, we compute f xx = cos x cos y, f yy = cos x cos y, and f xy = sin x sin y. Call the Hessian D. We then have D = f xx f yy ( f xy ) 2 = (cos x cos y) 2 (sin x sin y) 2. In case (1) above D = (sin x sin y) 2 = 1 < 0 (remember we have cos 2 x + sin 2 x = 1 so if cos x = 0 then sin 2 x = 1). So all the points in (1) are saddle points. In case (2) above D =(cos x cos y) 2 = 1 > 0, so we then analyze f xx = cos x cos y for x = np and y = mp. Notice that cos(np) is 1 if n is even and 1 if n is odd. This can be stated compactly by writing cos(np) =( 1) n. We have f xx (np, mp) = cos(np) cos(mp) = ( 1) n ( 1) m =( 1) m+n+1 Hence f xx > 0 when m + n is odd and f xx < 0 when m + n is even. The table below summarizes these findings:
5 Point (np, mp) with m + n odd (np, mp) with m + n even (p/2 + np, p/2 + mp) min / max / saddle min max saddle For instance, if we restrict to the box 0 apple xp and 0 apple y apple p then we have 5 critical points: (0, 0) and (p, p) are both maximums, (0, p) and (p,0) are both minimums, and (p/2, p/2) is a saddle point. (OVER)
6 2. Use the method of Lagrange multipliers to find the closest point to P =( 3, 1) on the line 4x 3y = 5. (4 points) Minimize D =(x + 3) 2 +(y 1) 2, the function measuring distance from P to (x, y), subject to the restraint g = 0 where g = 4x 3y + 5. We have rd = h2(x + 3),2(y 2)i and rg = h4, 3i. Using the method of Lagrange multipliers we get the system of equations: (x + 3) =2l 2(y 2) = 3l Solving for x and y in terms of l we get x = 2l 3 and y = 3/2l + 2. Plugging these values for x and y into 4x 3y + 5 = 0 we obtain that l = 4/5 and so x = 7/5 and y = 1/5.
7 Math Quiz 2 - Tuesday, October 4 Your name here: 1. Let f (x, y) =cos x sin y. (a) Find r f (x, y). (1 point) r f (x, y) =h sin x sin y, cos x cos yi (b) Find all critical points of f and use the second derivative test to identify their types. (5 points) The system sin x sin y = 0 and cos x cos y = 0 is solved if either cos x = sin y = 0 or sin x = cos y = 0. Notice that we cannot have cos x = 0 and sin x = 0 since cos 2 x + sin 2 y = 1. We have: x = p cos x = sin y = 0 ) 2 + np y = mp n is any integer m is any integer (1) or x = np sin x = cos y = 0 ) y = p/2 + mp n is any integer m is any integer (2) To use the second derivative test, we compute f xx = cos x sin y, f yy = cos x sin y, and f xy = sin x cos y. Call the Hessian D. We then have D = f xx f yy ( f xy ) 2 = (cos x sin y) 2 (sin x cos y) 2. In case (1) above D = (sin x cos y) 2 = 1 < 0 (remember we have cos 2 x + sin 2 x = 1 so if cos x = 0 then sin 2 x = 1). So all the points in (1) are saddle points. In case (2) above D =(cos x sin y) 2 = 1 > 0, so we then analyze f xx = cos x sin y for x = np and y = p/2 + mp. Notice that cos(np) is 1 if n is even and 1 if n is odd and sin(p/2 + mp) is 1 if m is even and 1 if m is odd. This can be stated compactly by writing cos(np) =( 1) n and sin(p/2 + mp) =( 1) m. We have f xx (np, p/2 + mp) = cos(np) sin(p/2 + mp) = ( 1) n ( 1) m =( 1) m+n+1 Hence f xx > 0 when m + n is odd and f xx < 0 when m + n is even. The table below summarizes these findings:
8 Point (np, p/2 + mp) with m + n odd (np, p/2 + mp) with m + n even (p/2 + np, mp) min / max / saddle min max saddle For instance, if we restrict to the box 0 apple xp and p/2 apple y apple p/2 then we have 5 critical points: (0, p/2) and (p, p/2) are both maximums, (0, p/2) and (p, p/2) are both minimums, and (p/2, 0) is a saddle point. (OVER)
9 2. Use the method of Lagrange multipliers to find the closest point to P =( 3, 1) on the line 4x 3y = 10. (4 points) Minimize D =(x + 3) 2 +(y 1) 2, the function measuring distance from P to (x, y), subject to the restraint g = 0 where g = 4x 3y We have rd = h2(x + 3),2(y 2)i and rg = h4, 3i. Using the method of Lagrange multipliers we get the system of equations: (x + 3) =2l 2(y 2) = 3l Solving for x and y in terms of l we get x = 2l 3 and y = 3/2l + 2. Plugging these values for x and y into 4x 3y + 10 = 0 we obtain that l = 4/5 and so x = 11/5 and y = 7/5.
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