ESE 318-02, Spring 2016 Homework Set #9 Due Tuesday, Mar. 29 1. Zill 9.3.7. 2. Zill 9.3.12. 3. Zill 9.4.17. (There is a typo in the answer in the back of the book.) 4. Zill 9.4.30. 5. Zill 9.4.33. 6. Zill 9.4.50. 7. Zill 9.4.52. Note: at t = 0, all values (x, y, w) are numbers, not variables. 8. Consider the function u(x), where x = x(t). Find 22dtud at the point where dudxx=x t0( )= 5,d2udx2x=x t0( )= 2,dxdtt=t0= −2 andd2xdt2t=t0= 4. 9. Zill 9.5.2 and 9.5.3. 10. Zill 9.5.14. 11. Zill 9.5.16. 12. Zill 9.5.17. 13. Zill 9.5.24. 14. Zill 9.5.30. 15. Zill 9.5.31. Hint: Find the tangent by knowing its slope is y’ = dy/dx. So u is a unit vector with that direction in the x-y plane. 16. Zill 9.5.33. 1. Zill 9.3.7. ( )( )( )( ) ( )( ) ( )22224124120021022241422222tttkjitkjktkjvavatttkjktkjvavaktratkjtrvkttjitrNT+=+=+×+=×=+=+⋅+=⋅==ʹʹ=+=ʹ=++=ESE 318-02, Spring 2016 2. Zill 9.3.12. ( )( )( )( )( )( ) ( )( )( ) ( )( ) ( )[ ]( ) ( )( )( ) ( ) ( )( ) ( )( ) ( )( )( )( )( )2/322/12222232232222222222/3232/522/123322222222222/122121222221111111211011120111111211211211211111111lntan−−−−−−−−−−−−−+=++=×=+=⎥⎥⎦⎤⎢⎢⎣⎡+++−=+−+−++=×+−=−−+=+−+−⋅+=⋅=−+−+=+−+++−=ʹʹ=+=++=++=+++=ʹ=++=tttvavaktktttttttttttkjiavttttttttttvavajttitjtttitttratttvtjitjttittrvjttitrNT 3. Zill 9.4.17. (There is a typo in the answer in the back of the book.) z = 4x1/23y2+1( )−1∂z∂x= 2 x−1/23y2+1( )−1=2x 3y2+1( )∂z∂y= −4x1/23y2+1( )−26y( )=−24y x3y2+1( )2 4. Zill 9.4.30. ( )zxyxzxxyzwxzxywxzyxzyyxzyxzzxyxwxzxyw==∂∂=∂∂+=+=+=∂∂=lnln1lnlnlnESE 318-02, Spring 2016 5. Zill 9.4.33. ( )( )( ) ( )( ) ( )( )( ) ( )( ) ( )( ) ( )0222222222222ln22222222222222222222222222222222222222222222=+−++−−=∂∂+∂∂+−=+−+=∂∂+=∂∂+−−=+−+=∂∂+=∂∂+=yxyxyxyxyzxzyxyxyxyyyxyzyxyyzyxyxyxxxyxxzyxxxzyxz 6. Zill 9.4.50. ( ) ( )334255334245tan5sec5355tan5sec5543uvuttvvuedtdvvzdtduuzdtdzttdtdvedtduuvuvzvvuuztt−+−−=∂∂+∂∂==−=−=∂∂−=∂∂−− 7. Zill 9.4.52. Note: at t = 0, all values (x, y, w) are numbers, not variables. ( )( )( ) ( ) ( )2020202022812404385540312853124eeeedtdwyxtxetyedtdyywdtdxxwdtdwtytxewxyxyxy−=+−=+−===⇒=+⎥⎦⎤⎢⎣⎡+−=∂∂+∂∂=+=+==ESE 318-02, Spring 2016 8. Consider the function u(x), where x = x(t). Find 22dtud at the point where dudxx=x t0( )= 5,d2udx2x=x t0( )= 2,dxdtt=t0= −2 andd2xdt2t=t0= 4. dudt=dudxdxdtd2udt2=ddtdudxdxdt⎛⎝⎜⎞⎠⎟=dudxddtdxdt⎛⎝⎜⎞⎠⎟+ddtdudx⎛⎝⎜⎞⎠⎟dxdt⎛⎝⎜⎞⎠⎟=dudxd2xdt2⎛⎝⎜⎞⎠⎟+ddxdudx⎛⎝⎜⎞⎠⎟dxdt⎛⎝⎜⎞⎠⎟dxdt⎛⎝⎜⎞⎠⎟=dudxd2xdt2⎛⎝⎜⎞⎠⎟+d2udx2dxdt⎛⎝⎜⎞⎠⎟2= 5 4( )+ 2 4( )= 28 9. Zill 9.5.2 and 9.5.3. f = y − e−2 x2y∇f =∂f∂x,∂f∂y∇f = 4xye−2 x2y,1+ 2x2e−2 x2y f =xy2z3= xy2z−3∇f =∂f∂x,∂f∂y,∂f∂z=y2z3,2xyz3, −3xy2z4 10. Zill 9.5.14. f =xyx + y; 2, −1( ); 6i + 8 j∇f =x + y( )y − xyx + y( )2,x + y( )x − xyx + y( )2=y2x + y( )2,x2x + y( )2∇f 2, −1( )= 1, 4u =6, 836 + 64=610,810=35,45Du= ∇f ⋅ u = 1, 4 ⋅35,45=3+165=195ESE 318-02, Spring 2016 11. Zill 9.5.16. We can simplify by noting that all we need is the i term of the gradient, or the negative of the partial wrt x. f x, y( )= x2tan y;12,π3( ); − x axis∇f =∂f∂x= 2 x tan y ⇒ −∇f12,π3( )= −tanπ3= − 3 12. Zill 9.5.17. ( ) ( )( ) ( ) ( )( ) ( ) ( ) ( )2322626212218,,012,18,18,,0233,3,0993,3,012,18,181218181241221221,1,11241221223,3,01,1,1122121212122222222222−=−=−=+−=⋅−=⋅∇===+=−=+−=+++−+=−∇+++++=∇−+=ufDukjikjifkzyxjzyxizxyfzyxFu 13. Zill 9.5.24. Direction of maximum increase is the gradient. The corresponding rate is its magnitude. f x, y( )= xyex−y5, 5( )∇f = xyex−y+ yex−y, −xyex−y+ xex−y= ex−yy x +1( ), −x y −1( )∇f 5, 5( )= 5 6( ), −5 4( )= 30,−20rate = ∇f 5, 5( )= 900 + 400 = 1300 = 10 13direction = 30,−20 or 3, −2 14. Zill 9.5.30. Direction of maximum decrease is the negative gradient. The corresponding rate is the negative of its magnitude. F = lnxyz∂F∂x=1xyzyz⎛⎝⎜⎞⎠⎟=1x∂F∂y=zxyxz⎛⎝⎜⎞⎠⎟=1y∂F∂z=zxy−xyz2⎛⎝⎜⎞⎠⎟= −1z∇F =1x,1y, −1z− ∇F12,16,13( )= − 2, 6,−3 = −2, −6, 3rate = − ∇F12,16,13( )= − 22+ 62+ 32= − 4 + 36 + 9 = − 49 = −7ESE 318-02, Spring 2016 15. Zill 9.5.31. Find the tangent by knowing its slope is y’ = dy/dx. So u is a unit vector with that direction in the x-y plane. slope =dydx2x2+ y2= 9 4x + 2ydydx= 0 ⇒dydx=−2xy→−2 2( )1= −4∴u = ±1, −41+16= ±1, −417∇f = 1, 2y ∇f 3, 4( )= 1,8∇f ⋅ u = ± 1,8 ⋅ 1, −4 / 17 = ±31 / 17 16. Zill 9.5.33. ∇f = 4i + 3 j u = xi + yj x2+ y2=1a( )∇f ⋅ u = 4x + 3y = 0 ⇒ y = −43x ⇒ u =1,−431+169=3 1,−435=35,−45also−35,45b( )max :4, 316 + 9=45,35c( )min :