Lecture 11 Appendix B: Some sample problems from Boas Here are some solutions to the sample problems assigned for Chapter 4. 4.1: 3 Solution: Consider a function of 3 (independent) variables,, ln z u v w u v w. Using the chain rule we find the partial derivatives are given by 4.1: 4 z dz u v w 1 u u, u d u v w u u v w u v w z dz u v w 1 v v, v d u v w v u v w u v w z dz u v w 1 w w. w d u v w w u v w u v w Solution: Now consider a problem in -D where we want the second derivatives at points where the first derivatives vanish. First consider the vanishing first derivatives to find w x y x y xy 3 3, 6 w w x y 3x y, 3y x, w w 3 7 7 x y 4 8 4 3 0 y x x x x x 1 0 8 x x x 7 3 0 or 3 treating as real. Physics 7 Lecture 11 Appendix B 1 Autumn 008
Thus the first derivatives both vanish at the points, 0,0, 3, 3 x0 y0. (Note that the x y symmetry of the initial expression guarantees the symmetry for these zeros.) Now we can evaluate the second derivatives at these points, w 0 w x 0,0 6 x, x w 4 x 3, 3 w 0 w y 0,0 6 y. y w 4 y 3, 3 4.1: 9 Solution: If we have z x y, x rcos and y rsin, then to obtain the desired partial derivative we need to express z in terms of x and. Then we find x sin r, y x x tan z x, x x tan cos cos z y x x x 1 tan x. 4.1: 15 Solution: This is similar to the previous exercise but now treating r and as independent. We have in this case Physics 7 Lecture 11 Appendix B Autumn 008
x r cos, y r sin z r, r cos sin r 1 sin z r sin cos r sin xy. r 4.: 5 Solution: We want to perform a two-variable expansion of 1 xy about the origin. Proceeding straightforwardly, and using the fact that only the combination xy appears, we find the familiar expansion of a square root 1 1 1 3 1 3 35 1 4 1 z 1 z z z z 4! 8 3! 16 4! 1 1 1 3 5 4 1 xy 1 xy xy xy xy 8 16 18 4.: 8 Solution: Now we try using our knowledge of complex variables to simplify finding a -variable expansion. We start with e xiy n0 x iy n! n. Now take the real and imaginary parts to find Physics 7 Lecture 11 Appendix B 3 Autumn 008
xiy x Re e e cos y n0 n0 Re n! 3 x y x 3xy 1 x, 6 xiy x Im e e sin y Im 3 y 3x y y xy 6 x iy x iy n! n n We could also use polar notation, z i re, to obtain e xiy n0 n re n! n0 n x sin sin sin cos, n0 in, n x r cos n r e cos y 1 r cos cos sin n! r sin n e y r r n! i.e., the same result. 4.4: 4 Solution: We want to use our understanding of expansions to approximately solve the thin lens formula when we slightly change one of the parameters. The focal length of the given lens is fixed by 1 1 1 1 1. f i o 1 18 Now consider a slightly different situation with o = 17.5. To first order in the change we have (note that the focal length does not change) Physics 7 Lecture 11 Appendix B 4 Autumn 008
0.5 18 1 o 1 i 1 1,, o o i i o i i 1 4 io 0.5 0.5 0. o 18 9 9 i 1 i1.. 4.4: 6 Solution: This exercise is similar to the last one where we use expansions to find first-order relative uncertainties. We have 4 l 4 l 1l l 4 l g 1 l l T T T T 1T T T g g l l T T 5% % 9%. The fact that we simply added the uncertainties (as if they had the same sign) is why this is a worse-case estimate. If the uncertainties in the length and the tension are independent, a more sensible estimate would be to add them in quadrature, g g l l T T 6.4%. 4.5: 3 Solution: We want to use the chain rule to find dr r dp r dq dp dq p q e p q ds p ds q ds ds ds s s e e s s p q s s p q e pe qe e p q r p q e e e. Physics 7 Lecture 11 Appendix B 5 Autumn 008
4.6: Solution: Let s practice with implicit differentiation. We simply take derivatives and solve for the desired form. We find xy d xy dy xy dy xy xy dy cos x y e ye sin x e x ye y e cos x xy dx dx dx dx e 1 xy d xy d dy xy xy ye sin x e 1 xy cos x y e dx dx dx d y xy dy xy dy xy 1 e xy xe xy ye xy dx dx dx dy xy 3 xy sin x y xy e y e dx dy xy dy xy 3 xy xe xy ye xy sin x y e d y dx dx xy dx e 1 xy So evaluating these expressions at the origin we find xy dy cos x y e 10 1, dx e xy xy 1 1 10 0,0 0,0 dy xy dy xy 3 xy xe xy ye xy sin x y e d y dx dx xy dx e 0,0 1 xy. 11 0 101 0 101 0 0 0 0. 0,0 Physics 7 Lecture 11 Appendix B 6 Autumn 008
4.6: 5 Solution: Again we can use implicit differentiation to find d 3 3 3 dy 3 dy xy yx 6 y 3xy x 3yx 0 dx dx dx 3 dy 3yx y 3 dx 3xy x dy yx y dx 3xy x 1 1 11 3 3 6 8 3. 1, 1, Thus the tangent line passes through the point (1,) and is parallel to the line 11y +x = 0 and thus is given by 11(y ) +(x 1) = 0 (or 11y + x = 4). Using ContourPlot in Mathematica we find the indicated plot ContourPlot[{x y^3 - y x^3 6,11 y+ x4},{x,-10,10},{y,-10,10}, Axes->True, FrameFalse, AxesLabel{x,y}, LabelStyleDirective[Large]] y 10 5 10 5 5 10 x 5 10 4.7: 4 Solution: Now we use the chain rule with two independent variables. We have (just following our noses) r s r uv w r, s e s u v w w r w s w r v s wrv s u r u s u w w r w s w r u s wru s. v r v s v Physics 7 Lecture 11 Appendix B 7 Autumn 008
4.9: 4 Solution: Now some practice with Lagrange multipliers. This exercise is a specific version of Example 3, with a =, b = 3 and c = 5 describing the ellipsoid within which we want to inscribe a box. With the coordinates x, y, z given the corner of the box in the first octant we want to maximize the volume V = 8xyz subject to the ellipsoid constraint. In Langrage notation we maximize the function x y z F 8xyz 4 9 5 F x 0 8yz 0 x F y 0 8xz 0 y 9 F z 0 8xy 0. z 5 Solving these 3 simultaneous equations as in the example (multiply the first by x, the second by y and the third by z, add, and use the equation for the ellipsoid) we find 1xyz and thus (from the x equation) 8yz 6x yz 0 x y 0, z 0. 3 Correspondingly y 3 3 3 and z 5 3. Hence the desired volume is given by V 5 80 8xyz 8 3. 3 3 3 4.9: 6 Solution: This is a similar problem to maximize the volume in a box but the x,y,z vertex is in the plane x +3y + 4z = 6 and the opposite 3 faces are in the coordinate planes (x =0, y = 0, z =0). So now we want to maximize (using Lagrange) Physics 7 Lecture 11 Appendix B 8 Autumn 008
F xyz x 3y 4z F 0 yz 0 x F 0 xz 3 0 y F 0 xy 4 0. z As above we multiply the first equation by x, the second by y and the third by z, add, and use the equation for the (flat) plane to obtain Thus the volume is 3xyz x 3y 4z 3xyz 6 0 1 xyz 1 x : yz xyz 0 x 1 1 y : xz 3 xyz 0 y 3 1 1 z : xy 4 xyz 0 z. V xyz 1. 3 4.10: 10 Solution: Here we want to practice finding extrema in situations where the boundaries matter. Consider a unit sphere (ball) in 3-D, x y z 1, whose interior temperature is given by the distribution T y xz. We want to find the maximum and minimum temperatures in the following regions. Physics 7 Lecture 11 Appendix B 9 Autumn 008
a) On the circle y 0, x z 1 T xz x 1 x 1 x 1, a 1-D problem. The temperature clearly vanishes at the points where the axes intersect the circle, T 0 x 1, z 0; x 0, z 1. In between the extremal values occur where the derivative vanishes we have dt x 1 1 x 0 x z dx 1 x 1 1 1 1 TMAX x z, TMIN x z. The former corresponds to the maximum on the circle, while the latter is the minimum. b) On the surface of the sphere, x y z 1, we have a -D problem that we can write as T x z xz x z x z derivatives we have 1 1 1, 1 1, 1 T x z 0 z x x T T 0 x z 0 T x z z x 0 x z z T T T T, x z x z T x 0, z 0, y 1 1. MAX 0,0 0,0. In terms of Thus we conclude that the temperature is a maximum (with value 1) at the antipodal points y 1. We also know from part a) that on the corresponding equator (y = 0), which is also part of the surface of the sphere, there are points, x z 1, where the temperature takes on its minimum value T 1, but where these individual -D derivatives do not vanish. Physics 7 Lecture 11 Appendix B 10 Autumn 008
c) Finally consider the full surface and interior of the sphere. Now we have a 3-D problem with 3 derivatives. We have T T T z 0, y 0, x 0 x y z T T T 0,. x z y From this we conclude that there are no true extrema inside the sphere and that the extrema must occur on the surface, i.e., we are back to the results of part b), T 1, T 1. MAX MIN 4.13: Solution: Here we are back to finding the distance from a point, r 0 xˆ yˆ, to a line, 3x + y = 4, in -D. In part a) we use the previous geometry method. With the line parallel to the vector v xˆ 3yˆ and going through the point r 1 0xˆy ˆ (by inspection), we have r 1r 0v 1 1 4 d xˆ yˆ xˆ 3yˆ 6. v 13 13 13 Now in b) we use the methods of Lagrange. We combine the distance from the point and the line constraint to find the minimum of Satisfying the equation for the line means 1 3 F x y x y F 3 x 3 0 x x F y1 0 y 1. y Physics 7 Lecture 11 Appendix B 11 Autumn 008
3 3 1 8 13 4 1 14 x 8 13 13. 13 8 5 y 1 13 13 Thus the distance from the point, as in a), is d 1 8 144 64 08 4 13 4. 13 13 13 13 13 13 c) Finally let us consider the general case of the point, x0 y 0 and the line ax by c. Then the vector parallel to the line is v bxˆ ayˆand a point on the line can be chosen as 0,cb. Thus, using vector methods the distance can be written as d x0xˆ c b y0 yˆ bxˆ ayˆ a b ax0 c by0 ax0 by0 c, a b a b as desired. Using the same general expressions in the Lagrange multiplier case we obtain the same result, Physics 7 Lecture 11 Appendix B 1 Autumn 008
F x x y y ax by 0 0 F a x x0 a 0 x x0 x F b y y0 b 0 y y0 y a b ax0 by0 c a x0 b y0 c a b a b x x0 ax 0 by0 c y y0 ax 0 by0 c a b a b, a b d x x y y ax by c a b 0 0 0 0 0 ax by a 0 b c. 4.13: 9 Solution: Here we again practice using the chain rule and implicit differentiation. Our starting point yields 3 3 dx dy x y 3t 6x 6y 6t dt dt z xy. dx dy 3x 3y 6t 6x 6y 6 dt dt Solving the simultaneous equations for the derivatives of x and y we find dx dx t y dt dt x x y x xy t y dy dy t x dt dt y y x y xy t x Thus the desired derivative is given by (assuming z 0 ). Physics 7 Lecture 11 Appendix B 13 Autumn 008
dz z dx z dy t y t x y x dt x dt y dt x x y y y x 1 t x y x xy y xy x y z z y t y x t x 1 1 1 x x t y y t 1 t t x y z z t 1 x y. z 4.13: 18 Solution: Now we want a minimum distance from the origin to the line defined by the intersection of two planes. a) Using vector methods we find the direction of the line of intersection by taking the cross product of the normal to the two planes. We find N xˆ 3 yˆ zˆ, N 3xˆ yˆ zˆ 1 xˆ yˆ zˆ v N N 3 1 xˆ 6 1 yˆ 3 4 zˆ 9 1 3 1 7 xˆ yˆ zˆ. By inspection the point (0,-3,-4) is on both planes and thus on the line of intersection. Thus the distance from the origin is given by d xˆ yˆ zˆ 3yˆ 4zˆ xˆ yˆ zˆ 1 0 3 4 xˆ yˆz ˆ 3 1 1 1 1 116 9 6 xˆ 3 4 yˆ 4 zˆ 3. 3 3 3 Physics 7 Lecture 11 Appendix B 14 Autumn 008
b) Now consider solving this exercise using Lagrange multipliers, where here we need to ensure having a point on both planes. We want to find an extreme value of 3 3 F x y z 1 x y z x y z F 3 x : x 1 3 0 x 1 x F 3 1 y : y 31 0 y 1 y F 1 z : z 1 0 z 1. z The two equations for the planes tell us that 3 3 1 1 7 1 3 1 1 71 5 3 3 1 1 7 3 1 1 1 1 7 11 34 1,. 1 1 So finally we have (the same result) 51 7 3 17 1 34 5 x, y, z 1 1 3 1 1 3 1 1 3 1 1 6 d 49 4 5 78. 3 3 3 Which approach is easier and quicker? In fact Mathematica is very handy here. See the notebook. Physics 7 Lecture 11 Appendix B 15 Autumn 008