Math 263 Assignment #3 Solutions. 1. A function z = f(x, y) is called harmonic if it satisfies Laplace s equation:

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1 Math 263 Assignment #3 Soltions 1. A fnction z f(x, ) is called harmonic if it satisfies Laplace s eqation: z 2 0 Determine whether or not the following are harmonic. (a) z x We se the one-variable chain rle to compte the partials of z with respect x and. x and x x We onl needed to compte one of these partials and then se the smmetr of z x to compte the other. Appling the one-variable qotient rle we see that ( ) 2 x x B smmetr, we see that x x x x x (x2 + 2 ) x 2 (x ) 3/2 2 x 2 (x ) 3/2. 2 (x ) 3/2. Therefore / 2 + / 2 1/ x , so z x is not harmonic. (b) z e x sin. These partials are easier to compte than in (a): / e x sin, / 2 e x sin, / e x cos, and / 2 e x sin. Therefore the fnction is harmonic. (c) z 3x 2 3. Again we compte the partials: / 6x, / 2 6, / 3x 2 3 2, and / 2 6. Since 6 6 0, this fnction is also harmonic. 2. Give an example of a fnction with the indicated properties or show that no sch fnction exists. (a) A fnction f(x, ) with continos first and second order partial derivatives in the x-plane and which satisfies f/ 6x 2 and f/ 8x 2. The simplest wa to solve this problem is with Theorem 1 from section 12.4, becase then the mixed second-order partial derivatives mst be eqal, i.e., we shold have 2 f/ 2 f/. For the given f, we have that so there can be no sch fnction f. 2 f 16x 12x 2 f, (b) A fnction g(x, ) satisfing the eqations g/ g/ 2x. For part (b), we cold make the same argment as in (a), as long as we verif that g has continos first and second-order partials everwhere. Since 2x 2, we cold conclde that

2 no g exists. I don t feel like verifing the hpotheses of Theorem 1 of section Let s tr a different method. Note that g g(x, ) dx 2xdx x 2 + h(). Now taking the partial of g(x, ) with respect to gives that g/ x 2 + h (). Bt b or hpothesis, we have that g/ 2x. However, x 2 + h () cannot eqal 2x, since h is a fnction of onl. Ths there are no fnctions g with the indicated properties. 3. Use the appropriate version of the chain rle to compte the following. (a) dw/dt at t 3, where w ln(x z 2 ), x cos t, sin t, and z 4 t. Therefore dw dt 16 t3 49. dw dt w dx dt + w d dt + w dz dt 2x sin t x z cos t x z cos t sin t + 2 sin t cos t + 16 cos 2 t + sin 2 t + 16t t 4zt 1/2 x z 2 (b) / and / v, where z x, x cos v, and sin v. and + cos v + x sin v 2 sin v cos v v v + v sin v + x cos v 2 (cos 2 v sin 2 v) 4. Sppose that a dck is swimming in the circle x cos t, sin t and that the water temperatre is given b the formla T x 2 e x 3. Find the rate of change in temperatre the dck might feel. We want to compte dt/dt. This is jst a simple application of the chain-rle. dt dt T dx dt + T d dt (2xe 3 )( sin t) + (x 2 e 3x 2 )(cos t) 2 cos t sin te sin t + sin 4 t + cos 3 te sin t 3 cos 2 t sin 2 t 5. A boat is sailing northeast at 20 km/h. Assming that the temperatre drops at a rate of 0.2 C/km in the northerl direction and 0.3 C/km in the easterl direction, what is the rate of change of temperatre with respect to time as observed on the boat? First we need to parametrize the motion of the boat. Since the boat is sailing northeast, we know that it is travelling along the line x. Hence we have the parameterization

3 r(t) αt, αt, where α is a constant. Since v(t) 20 2α 2, we find that α 10 2 and r(t) x(t), (t) 10 2 t, t. Now we appl the chain-rle. dt dt T dx dt + T d dt ( 0.3) ( 0.2)10 2 ( 5 2) C/h 6. Compte the following sing implicit differentiation. (a) / if e z x 2 z ln π. Method 1. Appl the operator / to both sides of the above eqation, considering (x, z) and x fixed. ( e z x 2 z ln ) (π) e z ( + z ) x2 (z ln ) z ln (x2 ) 0 e z ( + z ) x2 ( z + ln ) 0 (ze z x2 z ) x2 ln e z x2 ln e z ze z x2 z Method 2. Let F (x,, z) e z x 2 z ln π. We consider (x, z) and x fixed and implicitl differentiate F (x,, z) 0 with respect to z. F,, 0 2xz ln, ze z x2 z, ez x 2 ln 0,, 1 0 (b) d/dx if F (x,, x 2 2 ) 0. (ze z x2 z ) + ez x 2 ln 0 x2 ln e z ze z x2 z Since we are looking for d/dx, we are assming that (x). Hence z x 2 ((x)) 2. As in method 2 above, we have,, + 2x + ( F dx dx, d dx, dz dx 0 d d 1,, 2x 2 dx dx 0 2 ) d dx 0 d dx ( ( + 2x ) 2 )

4 (c) ( /) if xv 1 and x v 0. [Hint: Yo shold consider (x, ) and v v(x, ).] Let F (x,,, v) xv 1 and G(x,,, v) x+++v. Then we implicitl differentiate F (x,,, v) 0 and G(x,,, v) 0 with respect to x considering fixed, (x, ) and v v(x, ). and ( F, ( v, xv, xv, x 1, v + xv ( ), ( ) v, 0 ) ( ) v, 0, 0 ) ( ) v + x 0 (1) ( ) G,, ( ) v, 0 ( ) ( ) v 1, 1, 1, 1 1,, 0, 0 ( ) ( ) v Now we have two eqations and two nknowns. If we mltipl the bottom row of (2) b x and add it to the bottom row of (1). Then we have ( ) (v x) + (xv x) ( ) 0 x v xv x 7. Given a srface F (x,, z) 0 that defines x f(, z), g(x, z), and z h(x, ). Use implicit differentiation to verif that ( ) ( ) ( ) 1. (i) We consider g(x, z) and z fixed and implicitl differentiate F (x,, z) 0 with respect to x. + 0 (ii) We consider z h(x, ) and x fixed and implicitl differentiate F (x,, z) 0 with respect to. (2) (3) + 0 (4)

5 (iii) We consider x f(, z) and fixed and implicitl differentiate F (x,, z) 0 with respect to z. Combining (3), (4), and (5), we see. + 0 (5) ( ) ( ) ( ) 1 8. Find second-degree (Talor) polnomial approximations to the given fnctions at the points provided. (a) f(x, ) tan 1 (x + x) at the point ( 1, 0). First we need to compte all the first and second order partials. Recall that d dx (tan 1 x) 1, with this in mind the first partials are straightforward. 1 + x2 f x (x, ) (x + x) 2 and f (x, ) x 1 + (x + x) 2 Once we have the first partials we jst appl the qotient rle to compte the second partials. f xx (x, ) 2(1 + )2 (x + x) (1 + (x + x) 2 ) 2 ; f x (x, ) 1 (x + x)2 (1 + (x + x) 2 ) 2 ; f (x, ) 2x2 (x + x) (1 + (x + x) 2 ) 2. Or qadratic approximation to f(x, ) near the point ( 1, 0) is given b, (recall that tan 1 ( 1) π/4) P 2 (x, ) f( 1, 0) + f x ( 1, 0)(x + 1) + f ( 1, 0) f xx( 1, 0)(x + 1) 2 + f x ( 1, 0)(x + 1) f ( 1, 0) 2 π (x + 1) (x + 1) (b) g(x, ) x 3 + 2x + x 2 at the point (1, 1). The process is the same as in (a) bt the comptation is simpler. We have g/ 3x , g/ 2x + 2x, 2 g/ 2 6x, 2 g/ 2 + 2, and finall 2 g/ 2 2x. So we have a qadratic approximation of g(x, ) near (1, 1) P 2 (x, ) g(1, 1) + g x (1, 1)(x 1) + g (1, 1)( 1) g xx(1, 1)(x 1) 2 + g x (1, 1)(x 1)( 1) g (1, 1)( 1) (x 1) + 4( 1) + 3(x 1) 2 + 4(x 1)( 1) + ( 1) 2 (c) Use part (b) to estimate (1.1) 3 + 2(1.1)(.9) + (1.1)(.9) 2. [Note that the actal vale is ] Finall an eas qestion! All we have to do is compte P 2 (1.1, 0.9) from part (b). (1.1) 3 + 2(1.1)(.9) + (1.1)(.9) (0.1) + 4( 0.1) + 3(0.1) 2 + 4(0.1)( 0.1) + ( 0.1)