MATH 209 Lab Solutions

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1 MATH 9 Lab Solutions Richard M. Slevinsky 1 November 1, 13 1 Contact: rms8@ualberta.ca

2 Contents 1 Multivariable Functions and Limits Partial Derivatives 6 3 Directional Derivatives and Gradients 15 4 Maximum and Minimum Problems 5 Lagrange Multipliers 7 6 Double Integrals 36 7 Polar Coordinates, Centre of Mass, Surface Area 4 8 Triple Integrals 46 9 Line Integrals 5 1 Green s Theorem Surface Integrals 57 1 Stokes Theorem and the Divergence Theorem 6 1

3 Lab 1 Multivariable Functions and Limits Problems 1. Find the domain and range of the following functions and sketch their graphs: (a) z f(x, y) x + y ; The domain is {(x, y) : x R, y R}. The range is {z : z R, z }. (b) z f(x, y) 3 x ; The domain is {(x, y) : x R, y R}. The range is {z : z R, z }. (c) x f(y, z) 4 y z. The domain is {(y, z) : y R, z R, y + z 4}. The range is {x : x R, x }.. Find each of the following limits, or show that the limit does not exist: (a) lim (x,y) (,) x y x y ;

4 Let y m x. Then we have: lim (x,y) (,) x y x y lim x m x 4 x m x 4, m lim x + 3 m m + 3 m. But this is different for every value m we choose, so the limit DNE. (b) lim (x,y) (,) x y 3 x + 3 y ; Let y m x. Then we have: lim (x,y) (,) x y 3 x + 3 y lim x m 3 x 5 x + 3 m x, m 3 lim x + 3 m x3. Clearly, it works for any line y m x. However, how can we be sure? Switch to polar coordinates: lim (x,y) (,) x y 3 x + 3 y lim r 3 e y (c) lim (x,y) (e,1) 1 ln x. Sometimes it is as easy as plugging in the values! r cos θr 3 sin 3 θ r cos θ + 3 r sin θ, cos θ sin 3 θ lim r cos θ + 3 sin θ r3, cos θ sin 3 θ lim r + sin θ r3. Exercises lim (x,y) (e,1) 3 e y 1 ln x 3 e1 1 ln e.3e1. 1. Find the domain and range and sketch the graph for each of the following functions: (a) z f(x, y) y; The domain is {(x, y) : x R, y R}. The range is {z : z R}. (b) y f(x, z) 3 x + z; The domain is {(x, y) : x R, z R}. The range is {y : y R}. (c) z f(x, y) 1 x y. The domain is {(x, y) : x R, y R, x + y 1}. The range is {z : z R, z 1}.. Find the limit, or show that the limit does not exist: 3

5 (a) lim (x,y) (,) x y x + y ; Let y m x, then: x y lim (x,y) (,) x + y lim x m x x x + m x, 1 m 1 + m. But this is different for every value m we choose, so the limit DNE. (b) lim (x,y) (,) 6 x 3 x + y ; In polar coordinates: lim (x,y) (,) (c) lim (x,y,z) (,1,) x y 3 4 z x + y 3 + z. Substituting values, we find: 6 x 3 x + y lim r 6 r 3 cos 3 θ r cos θ + r sin θ, 6 r 3 cos 3 θ lim r r x y 3 4 z lim (x,y,z) (,1,) x + y 3 + z lim r 6 r cos 3 θ Show: sin(x + y) (a) lim (x,y) (,) x + y In polar coordinates: 1; sin(x + y) sin(r(cos θ + sin θ)) lim lim, cos θ + sin θ sin(θ + π/4) (x,y) (,) x + y r r(cos θ + sin θ) sin(r sin(θ + π/4)) lim r r sin(rα), Recall : lim 1. sin(θ + π/4) r rα (b) lim (x,y) (,) sin x + sin y x + y 1. [Hint: sin α + sin β sin ( ) ( ) α+β cos α β.] 4

6 Using the hint, we obtain: sin x + sin y lim (x,y) (,) x + y 4. Find the value of k such that f(x, y) continuous everywhere? Why? Taking the limit, we find: lim (x,y) (,) lim (x,y) (,) lim (x,y) (,) sin ( x+y ) ( cos x y ), Recall : cos() 1 x + y sin ( x+y ) 1 1, from part (a). x+y x 4 x + y, (x, y) (, ) k, (x, y) (, ) x 4 x + y lim r r 4 cos 4 θ r cos θ + r sin θ, lim r r cos 4 θ. is continuous at (, ). Is f(x, y) Therefore, k. It is then continuous everywhere, since f(x, y) is otherwise a rational function and there are no other singularities. 5

7 Lab Partial Derivatives Part I Problems 1. Find the differentials of the following functions and write the tangent plane approximation at the given point (x, y ). Use this to estimate the function as required below: (a) f(x, y) x y + tan(x y) + 1 x, (x, y ) (, ). Estimate f( 1.95,.); The differential is: df f x dx + f y dy, ( x + y sec (x y) 1x ) dx + ( 4 y + x sec (x y) ) dy. Evaluating the differential at the point (x, y ) (, ), we find: Equivalently, we find the tangent plane: Estimating the function at ( 1.95,.): f f f f x (x, y ) x + f y (x, y ) y, f 7 ( 4 1 ) (x + ) + ( )(y ). 4 T (x, y) 7 17 (x + ) y. 4 f( 1.95,.) T ( 1.95,.) 7 17 (.5) + (.), Compare : f( 1.95,.) (b) f(x, y, z) ln(x y) + e y z + sin(x z), (x, y, z ) (, 4, π). Estimate f( 1.9, 4.1, π). The differential is: df f x dx + f y dy + f z dz, ( ) 1 + z cos(x z) dx + x ( ) 1 y + z ey z dy + (y e y z + x cos(x z)) dz. 6

8 Evaluating the differential at the point (x, y ) (, 4, π), we find: f f f f x (x, y, z ) x + f y (x, y, z ) y + f z (x, y, z ) z, ( f ln(8) e 4 π π + 1 ) ( (x ) + π e 4 π + 1 ) (y 4) + ( 4 e 4 π + ) (z π). 4 Equivalently, we find the tangent plane: ( T (x, y, z) ln(8) + e 4 π + π + 1 Estimating the function at (1.9, 4.1, π): ) (x ) + f(1.9, 4.1, π) T (1.9, 4.1, π) ln(8) + e 4 π Compare : f(1.9, 4.1, π) ( π e 4 π + 1 ) (y 4) + ( 4 e 4 π + ) (z π). 4 ( π + 1 ) ( (1.9 ) + π e 4 π + 1 ) (4.1 4), 4. The ideal gas law relates pressure P, volume V, and temperature T of an ideal gas by P V kt. where k is a constant. Show that P V T V T P 1. 7

9 We find: Exercises Therefore, combining all the results, we find: P V V T 1. If z f(x, y) arctan(x y), verify that f xy f yx ; We find: f x P V kt V, V T k P P V P T V T, T P V k. T P kt V V T V k 1. y 1 + (x y), f xy 1 + (x y) y x y (1 + (x y) ) 1 (1 + (x y) ), while: f y x 1 + (x y), f yx 1 + (x y) x y x (1 + (x y) ) 1 (1 + (x y) ). Therefore, they are equal.. Find the differential of f(x, y) x y + x + y. Find the equation of the tangent plane to the graph of z f(x, y) at (x, y ) (3, 1) and use it to estimate f(.9,.9). The differential is: df f x dx + f y dy, ( y + 1 x + y ) ( dx + x y + Evaluating the differential at the point (x, y ) (3, 1), we find: Equivalently, we find the tangent plane: ) 1 dy. x + y f f f f x (x, y ) x + f y (x, y ) y, ( f ) ( (x 3) ) (y 1). 4 4 T (x, y) (x 3) + (y 1). 4 8

10 Estimating the function at (.9,.9): f(.9,.9) T (.9,.9) (.1) (.1), 4.5. Compare : f(.9,.9) Find the differential of f(x, y, z) xy cos z. Find the equation of the tangent plane to the graph of w f(x, y, z) at (x, y, z ) (4, 5, 4π) and use it to estimate f(3.99, 4.98, 4.3π). Part II Problems The differential is: df f x dx + f y dy + f z dz, ( y ) cos z dx + ( xy cos z ) dy + ( xy sin z ) dz. x Evaluating the differential at the point (x, y ) (4, 5, 4π), we find: f f f f x (x, y, z ) x + f y (x, y, z ) y + f z (x, y, z ) z, ( ) 5 f 5 (x 4) + () (y 5) + () (z 4π). 4 Equivalently, we find the tangent plane: Estimating the function at (3.99, 4.98, 4.3π): T (x, y, z) (x 4) + (y 5). 4 f(3.99, 4.98, 4.3π) T (3.99, 4.98, 4.3π) (.1) + (.), Compare : f(3.99, 4.98, 4.3π) Let z cos(x + y), x u v, and y u v. Find z u and Since: z u z x x u + z y we have: z u u { z x x u + z y y u ( z x x u + z y x y u }, ) x u + z x x u + z u v ; y u, ( z y y u + z x y ) x y u u + z y y u, 9

11 and: z u v { z x v x u + z } y, y u ( ) z x x v + z y x x y v u + z x x u v + ( z y y v + z x y ) x y v u + z y y u v. Therefore: z ( cos(x + y) u v cos(x + y)v) u v sin(x + y) v, u + ( 4 cos(x + y)v cos(x + y) u v) v, 4(u + 1) v cos(x + y) v sin(x + y), and: z u v ( cos(x + y)u cos(x + y)u ) u v sin(x + y) u, + ( 4 cos(x + y)u cos(x + y)u ) v sin(x + y),. Let z f(x, y) and y g(x). Show that: u v(u + 1)(u + ) cos(x + y) (u + 1) sin(x + y). d z dx z x + z x y g (x) + z y ( g (x) ) + z y g (x); 1

12 Firstly: dz dx z dx x dx + z dy y dx, z x + z dy y dx. Then: d z dx z dx x dx + z dy x y z x + z dy x y dx + z y 3. If y 3f(x + t) g(x t), find y tt and y xx ; dx + z dy dx x y dx dx + z dy dy y dx dx + z y ( ) dy + z d y dx y dx, z x + z x y g (x) + z y ( g (x) ) + z y g (x). d y dx, 4. Find z z and where z arctan z x + y. x y Differentiating implicitly: or: Again, differentiating implicitly: y t 3f (x + t) + g (x t), y tt 3f (x + t) g (x t), y x 3f (x + t) g (x t), y xx 3f (x + t) g (x t). F (x, y, z) x + y + arctan z z, F x + F z 1 + z x, ( z 1 z x z + 1 z z. ) z x, or: F (x, y, z) x + y + arctan z z, F y + F z 1 + z y, ( z 1 z y z + 1 z z. ) z y, Exercises 11

13 1. If z 3 x e x y, x t cos t, and y t sin t, find dz at t π/6; dt We find: dz dt z dx x dt + z dy y dt, 3(1 + x y)e x y (cos t t sin t) + 3 x e x y (sin t + t cos t), ( ) dz dt π 3 π ( ) ( e π 3/144 3/ π/1 + 3 π 3/144e π 3/144 1/ + π ) 3/ tπ/ (Conversion to polar coordinates) If T x 3 3 x y + y, x r cos θ, and y r sin θ, find T r and T θ ; We have: T r T x x r + T y y r, ( 6 x 3 y ) cos θ + ( y 3 x) sin θ, and: T θ T x x θ + T y y θ, ( 6 x 3 y ) r sin θ + ( y 3 x) r cos θ. 3. (Conversion to polar coordinates) If x r cos θ, y r sin θ, and z f(x, y), show that: ( ) z + x ( ) z y ( ) z + 1 r r ( ) z ; θ Since: and: z r z x x r + z y y r, z z cos θ + sin θ, x y 1 z r θ 1 z x r x θ + 1 z y r y θ, z z sin θ + cos θ, x y 1

14 summing the squares gives: ( ) z + 1 r r ( ) z ( z θ x ( z cos θ + z ) ( y sin θ + z z sin θ + x ( z ) y cos θ, ) sin θ, ) cos θ + z z cos θ sin θ + x x y y ( ) z + sin θ z ( ) z z sin θ cos θ + cos θ, x x y y ( ) z ( cos θ + sin θ ) ( ) z ( + cos θ + sin θ ), x y ( ) z ( ) z +. x y 4. If U 3f(x, y), x 5t s, and y t + s, express U in terms of the partial derivatives of f t s with respect to x and y; We have: U s 3 f x x s + 3 f y y s, 6 f x + 6 f y, U t s f x 6 x t 6 f y x y t + 6 f x y x t + f y 6 y t, 3 f x 1 f x y + 3 f y x + f 1 y, or: U t s 3 f x + 18 f y x + 1 f y. 5. Let z 3 + t z s + t 1. Find z z and at (t, s) (1, ); t s When t 1 and s, we have: z 3 + z, or z, ±i. We reject the imaginary solutions to have a real surface. Let F (s, t, z) z 3 + t z + 1 s t. Then: F t + F z z t, z t + ( 3 z + t ) z t, z t t z 3 z + t, z t (t,s,z)(1,,) 1. 13

15 As well: F s + F z z s, s + ( 3 z + t ) z s, z s z s. (t,s,z)(1,,) s 3 z + t, 6. Find the equation of the plane tangent to the surface y + x z + ln x y z + e at the point (x, y) (1, 1). When x 1 and y 1, we have: 1 + z + ln z + e, or z e. Let F (x, y, z) y + x z + ln x y z e. To find the tangent plane, we require the differential, which requires the partial derivatives: F x + F z z x, z + 1 ( x + x + 1 ) z z x, z + x 1 z x x + z 1, z x e + 1 (x,y,z)(1,1,e) 1 + e 1. And: z y F y + F z z y, y + 1 ( y + x + 1 ) z z y, y + y 1 x + z 1, z 5 y (x,y,z)(1,1,e) (1 + e 1 ). Then, the differential is: and the tangent plane is then: T (x, y) e ( ) ( ) e dz 1 + e 1 dx (1 + e 1 ) ( ) ( e e 1 (x 1) 5 (1 + e 1 ) dy, ) (y 1). 14

16 Lab 3 Directional Derivatives and Gradients Problems 1. Find equations for the tangent plane and the normal line to the surface x y + sin z at (1,, ); To form the tangent plane and the normal line, we require the vector n which is normal to plane (and therefore parallel to the line). If our function can be written as f(x, y, z) x y + sin z, as above, then this vector is in the direction of the gradient: f f(x, y, z) x, f y, f, z y, x, cos z, n f(x, y, z) (x,y,z)(1,,), 1, 1. Then the tangent plane is written as n (r r ), where r is a variable vector and r is the point in question:, 1, 1 ( x, y, z 1,, ). As well, the normal line is written as r(t) r + nt, where t is a parameter: x(t), y(t), z(t) 1,, +, 1, 1 t, 1 + t, + t, t.. Find the greatest rate of change of the function f(x, y) e x y at (x, y) (1, 1). In which direction does this occur? The gradient of f(x, y) is: f(x, y) f x, f, y f(x, y) (x,y)(1,1) 1, 1. e x y, 1e x y e x y 1, 1, Therefore, the greatest rate of change of the function f(x, y) is f 1 + 1, and this occurs in the direction of the gradient, that is: u 1 1, 1. 15

17 Exercises 1. Find the directional derivative of f(x, y, z) x y z + z x at the point P (1, 4, 3) in the direction of the point Q(, 1, 8) [note: you must first find the vector from P to Q]. What is the maximum rate of increase of f in any direction at P? The vector from P to Q is QP ( 1, 1 + 4, 8 3) (1, 3, 5) u , 3, 5. The gradient of f at P is: f f(x, y, z) x, f y, f, z x + z, z, x z y, f(x, y, z) (x,y,z)(1, 4,3) 11, 3, 1. Therefore, the directional derivative in the direction of the point Q is: D u f u f , 3, 5 11, 3, 1, The maximum rate of increase of f in any direction at P is: f 11, 3, 1, Find equations for the tangent plane and the normal line to the ellipsoid 4 x + 9 y + z 49 at the point P (1,, 3); To form the tangent plane and the normal line, we require the vector n which is normal to plane (and therefore parallel to the line). If our function can be written as f(x, y, z) 4 x + 9 y + z 49, 16

18 as above, then this vector is in the direction of the gradient: f(x, y, z) f x, f y, f z 8 x, 18 y, z, n f(x, y, z) (x,y,z)(1,,3) 8, 36, 6. Then the tangent plane is written as n (r r ), where r is a variable vector and r is the point in question: 8, 36, 6 ( x, y, z 1,, 3 )., As well, the normal line is written as r(t) r + nt, where t is a parameter: x(t), y(t), z(t) 1,, 3 + 8, 36, 6 t, t, 36 t, t. 3. The directional derivative of f(x, y, z) at a given point P is greatest in the direction of the vector,, 1, and takes the value in this direction. Find the directional derivative of f(x, y, z) at P in the direction of the vector 1, 1, [hint: for u a unit vector, then D u f u f f cos θ]; The maximal value of the directional derivative is given as. Therefore we know that f. We must then find the angle between the direction,, 1 and 1, 1,, or more precisely the cosine of 17

19 this angle. From linear algebra, we recall that: cos θ u v u v, 1, 1,,, 1 1, 1,,, 1,, , 3 6. Therefore, the directional derivative in the direction u 1, 1, is D u f Prove that the sphere x + y + z a and the cone x + y z are orthogonal at every point of intersection. Orthogonality requires their gradients to be orthogonal: n s f s x, y, z, n c f c x, y, z. Therefore, we require the dot product to be equal to : n s n c x, y, z x, y, z, 4 x + 4 y 4 z, but the values of z must be at once on the sphere, and as well on the cone. Since they are on the cone, z x + y, and therefore: n s n c 4 x + 4 y 4(x + y ). 18

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21 Lab 4 Maximum and Minimum Problems Problems 1. Find and classify all the critical points of the function f(x, y) x 4 1 x y + y 4 ; The critical points occur where the gradient (vector of partial derivatives) is : f(x, y) 4 x 3 1 y, 4 y 3 1 x,, 4 x 3 1 y, 4 y 3 1 x. The first equation implies y x 3 /3, and the second equation implies y (3 x) 1/3. Therefore, we find: x 3 /3 (3 x) 1/3, x 9 /7 3 x, x(x 8 /7 3), x(x 8 81). The real roots of this equation are x, ±81 1/8, ±3 1/. The critical points are therefore (x, y) (, ), ±(3 1/, 3 1/ ). The second derivative test requires the determinant: f xx (x, y) f xy (x, y) D(x, y) f yx (x, y) f yy (x, y), 1 x y, At the critical points, we find: 144(x y 1). Critical point D(x, y) f xx (x, y) Classification (, ) -144 N/A Saddle point (3 1/, 3 1/ ) Local minimum ( 3 1/, 3 1/ ) Local minimum

22 . Find the absolute minimum of the function f(x, y) x 4 + x y + y inside the circle x + y a for any a >. Conclude that this is the absolute minimum of the function on R ; The critical points occur where the gradient (vector of partial derivatives) is : f(x, y) 4 x x y, 4 y x y,, 4 x 3 + x y, y 3 + x y. The first equation implies either x or x + y, and the second equation implies either y or x + y. Therefore, the critical point is (x, y) (, ). The second derivative test requires the determinant: D(x, y) At the critical point, we find: f xx (x, y) f xy (x, y) f yx (x, y) f yy (x, y), 1 x + 4 y 8 x y 8 x y 1 y + 4 x, 144 x y + 48 x y x y 64 x y, 96 x y + 48 x y 4. Critical point D(x, y) f xx (x, y) Classification (, ) No info However, the value of the function at the critical point is f(, ) Notice, as well, that the function can also be written as f(x, y) x 4 + x y + y (x + y ) + 1. Therefore, on the circle x + y a, the function is f(x, ± a x ) a 4 + 1, so that for any a >, f(, ) is smaller than the value of f on the circle. 3. Find the absolute minimum and maximum of the function f(x, y) (x 4 x) cos y on the rectangular region 1 x 3, π 4 y π 4 ; 1

23 The critical points occur where the gradient (vector of partial derivatives) is : f(x, y) ( x 4) cos y, (x 4 x) sin y,, ( x 4) cos y, (x 4 x) sin y. The first equation implies either x or y ± π / [ π/4, π/4], and the second equation implies either x, x 4, or y. Therefore, the critical point is (x, y) (, ). The second derivative test requires the determinant: f xx (x, y) f xy (x, y) D(x, y) f yx (x, y) f yy (x, y), cos y ( x 4) sin y ( x 4) sin y (x 4 x) cos y, At the critical point, we find: At the critical point, f(, ) 4. (x 4 x) cos y ( x 4) sin y. Critical point D(x, y) f xx (x, y) Classification (, ) 8 Local minimum On the line x 1, we have f(1, y) 3 cos y, which has a local minimum at y and local maxima at y ±π/4. f(1, ) 3, f(1, ±π/4) 3/. On the line x 3, we have f(3, y) 3 cos y, which has a local minimum at y and local maxima at y ±π/4. f(3, ) 3, f(3, ±π/4) 3/. On the lines y ±π/4, we have f(x, ±π/4) (x 4 x)/. Completing the square, we find f(x, ±π/4) (x ) /. Due to the symmetry about the point x, the local minima are f(, ±π/4), and the local maxima are f( ± 1, ±π/4) 3/. Considering all points, the absolute minimum is f(, ) 4 and the absolute maximum is f( ± 1, ±π/4) 3/.

24 4. A flat circular plate has the shape of the unit disk x + y 1. The plate is heated so that the temperature at any point (x, y) is given by T (x, y) x + y x. Find the hottest and coldest points and temperatures on the plate. The critical points occur where the gradient (vector of partial derivatives) is : T (x, y) x 1, 4 y,, x 1, 4 y. The first equation implies x 1/ and the second equation implies y. Therefore, the critical point is (x, y) (1/, ). The second derivative test requires the determinant: T xx (x, y) T xy (x, y) D(x, y) T yx (x, y) T yy (x, y), 4, At the critical point, we find: At the critical point, T (1/, ) 1/4. 8. On the circle y ± 1 x, the temperature is: Critical point D(x, y) T xx (x, y) Classification (1/, ) 8 Local minimum T (x, ± 1 x ) x + (1 x ) x, x x, (x + 1/) + 9/4. Therefore, on the circle, the minimum temperature is T (1, ) and the maximum temperature is T ( 1/, ± 3/) 9/4. Therefore, the hottest points are at ( 1/, ± 3/) and the coldest point is at (1/, ). 3

25 Exercises 1. Classify the relative maxima and minima of f(x, y) x 3 + y 3 3 x 1 y + ; The critical points occur where the gradient (vector of partial derivatives) is : f(x, y) 3 x 3, 3 y 1,, 3 x 1, y 4. The first equation implies x ±1, and the second equation implies y ±. Therefore, the four critical points are (x, y) (±1, ±). The second derivative test requires the determinant: f xx (x, y) f xy (x, y) D(x, y) f yx (x, y) f yy (x, y), 6 x 6 y, At the critical points, we find: 36 x y. Critical point D(x, y) f xx (x, y) Classification (1, ) 7 6 Local minimum (1, ) -7 N/A Saddle point ( 1, ) -7 N/A Saddle point ( 1, ) 7-6 Local maximum. A rectangular box has four sides and a bottom but no top. The volume of the box is 18 m 3. Find the dimension of the box such that the surface area is a minimum; Let l, w, h be the length, width, and height of the box, respectively. Also, let V be the volume and S be the surface area. Then, we know that: V 18 l w h, and, S l w + l h + w h. 4

26 From the volume equation, we have h 18 l w. Therefore: S l w + (l + w) 18 l w, ( 1 l w + 16 w + 1 ). l Taking the gradient of S, we find: S, w 16 l, l 16 w. From the first equation, w and from the second w ± l l + is excluded on physical grounds. Solving for the length: l, l 16 l 3/, l 16 1/3 6, w 6, h l, where the negative width The second derivative test requires the determinant: S D(l, w) ll (l, w) S lw (l, w) S wl (l, w) S ww (l, w) 43 1 l , w 3 43 (l w) 3 1., At the critical point, we find: Critical point D(l, w) S ll (l, w) Classification (6, 6) 3 Local minimum 5

27 The dimensions of the box of volume 18 m3 with minimal surface area are (l, w, h) (6, 6, 3) m. 3. Find the absolute maximum and minimum of the function f (x, y) 1+4 x 5 y on the closed triangular region with vertices (, ), (, ), and (, 3). Since it is a plane, the absolute maxima and minima will be at one of the vertices. f (, ) 1, f (, ) 9 and f (, 3) 14. Therefore, the absolute maximum is 9 and the absolute minimum is 14. Next week, we will find the global minimum of the function: f (x, y) esin(5 x) + sin(6 ey ) + sin(7 sin(x)) + sin(sin(8 y)) sin(1(x + y)) + 6 x + y. 4

28 Lab 5 Lagrange Multipliers Problems 1. Say the temperature T at any point (x, y, z) in space is given by T 4 x y z. Find the hottest point on the sphere F x + y + z 1 ; We equate gradients: T λ F, 4 y z, x z, x y z λ x, y, z, giving us three equations for four unknowns. Adding the constraint F, we have four equations for four unknowns: 4 y z λ x, 4 x z λ y, 8 x y z λ z, x + y + z 1. Simplifying by substituting the constraint equation, we find: Again for λ: And: 4 y(1 x y ) λ x, 4 x(1 x y ) λ y, 8 x y λ. 4 y(1 x y ) 8 x y, 4 x(1 x y ) 8 x y. 1 x y x, 1 x y y. 7

29 Or: 1 3 x y, 1 x 3 y. Therefore, the values of x and y must solve the above system. By subtraction, we find x ±5 and y ±5. At these values, z 1, or z 5, and z ±5. The hottest point occurs when T at the highest value of T (±5, ±5, ±5 ). This occurs when the signs of the x and y values are the same. Therefore, the hottest points are located at (x, y, z) (5, 5, ±5 ) and ( 5, 5, ±5 ).. Show that the product of the angles of a triangle is largest when the triangle is equilateral; We seek to: Equating gradients, we find: The four equations we have are: maximize P θ 1 θ θ 3 subject to S θ 1 + θ + θ 3 18, The top three equations can be rearranged as: P λ S, θ θ 3, θ 1 θ 3, θ 1 θ λ 1, 1, 1. θ θ 3 λ, θ 1 θ 3 λ, θ 1 θ λ, θ 1 + θ + θ θ 3 (θ θ 1 ), θ 1 (θ 3 θ ), θ (θ 3 θ 1 ), θ i >, i 1,, 3. implying θ 1 θ θ 3 6 from the angular constraint. This is the equilateral triangle. 3. Let C denote the line of intersection of the planes 3 x + y + z 6 and x 4 y + z 8. Find the point on C that is closest to the origin; The point that is closest to the origin minimizes the distance, and also the distance squared, which is simpler to work with. Equating gradients: d (x, y, z) λp 1 + µp, x, y, z λ 3,, 1 + µ 1, 4,, 8

30 These three equations with the two planes give: x 3λ µ, y λ + 4µ, z λ µ, 3 x + y + z 6, x 4 y + z 8. This system of five equations and five unknowns has the solution (check) x , y 44 57, z 8 λ 19, µ The closest point is therefore: (x, y, z) 57 (116, 44, 8). 4. A container is constructed in the shape of a cylinder with a top and a bottom (e.g. a beer can). If the surface area of the container has a fixed value S, find the radius and height of the container that will maximize the enclosed volume. The volume of a cylinder is V πr h, while the surface area is S πrh + πr. Therefore, we wish to: maximize V πr h subject to S πrh + πr. 57, Exercises Equating gradients, we find: These two equations are: V λ S, π r h, r πλ h + r, r. π r h πλ(h + r), π r πλ r. The second equation gives r or λ r. Since r won t maximize anything, inserting the second possibility into the first equation above: π r h π r (h + r), h h + r, h r. Therefore, the height is twice the size of the radius. 1. Find the maximum value of f(x, y, z) x + y + z on the sphere x + y + z 5; Equating gradients: f λ F, 1, 1, 1 λ x, y, z, 9

31 giving us three equations for four unknowns. Adding the constraint F, we have four equations for four unknowns: 1 λ x, 1 λ y, 1 λ z, x + y + z 5. Simplifying by substituting into the constraint equation, we find: ( ) 1 ( ) 1 ( ) , λ λ λ 3 4 5λ, or λ ± 3/1. Choosing only the positive branch for the position (otherwise we actually find the minimum value of f), we find x y z 5/ 3. Therefore, the maximum value of f on the sphere is f(5/ 3, 5/ 3, 5/ 3) Let C be the curve of intersection of the surface y z 1 and the plane x y 1. Find the minimum value of f(x, y, z) x + y + z on C; Letting g 1 (x, y, z) y z 1 and g (x, y, z) x y 1, we equate gradients: f λ g 1 + µ g, x, y, z λ, y, z + µ 1, 1,. These three equations with the two surfaces gives five equations for five unknowns: x µ, y λ y + µ, z + λ z, y z 1, x y 1. From the last and first equations x 1 + y and therefore µ (1 + y). Inserting this into the second equation and simplifying: 4 y λ y, z + λz, y z 1. Therefore, from this middle equation, either z or λ 1. If z, then y ±1 from the bottom equation and x 1 ± 1, from above. Therefore, f(x, y, z) + (±1) + (1 ± 1) 3 ±. If λ 1, then y 1/3. However, if 1 < y < 1, then from the bottom equation z C and therefore the point is not on both curves. The minimum value is f(, 1, ) 1. 3

32 3. What is the volume of the largest rectangular box (with edges parallel to the axes) which can be inscribed in the ellipsoid x 36 + y 9 + z 16 1? Since x, y, and z are the coordinate lengths, the volumes of the rectangular boxes with edges parallel to the axes is V x y z 8 x y z. Therefore, equating gradients: With the ellipsoidal constraint: From the first equation λ 144 y z x. Therefore: V λ E, x 8 y z, x z, x y λ 18, y 9, z. 8 8 y z λx 18, 8 x z λy 9, 8 x y λz 8, x 36 + y 9 + z Simplifying the top two equations: 8 x z 88 y z, 9 x 144 y z 8 x y, 8 x x 36 + y 9 + z x 4 y, 4 x 9 z, x 36 + y 9 + z Solving the top two equations for y and z and inserting in the ellipsoidal equation, we have: y x 4, z 4 x 9, x 36 + x 36 + x Therefore, x 1, or x ± 3, y ± 3, and z ±4/ 3. Taking only the positive points, the largest box has the volume V ( 3, 3, 4/ 3) Find the point on the plane z 4 x + 9 y which is closest to the point (1, 1, ). 31

33 We seek to minimize the distance (squared) from the point to the plane: minimize d P (x 1) + (y 1) + (z ) subject to S 4 x + 9 y z. Equating gradients, we find: With the planar constraint: d P λ S, x 1, y 1, z λ 4, 9, 1. (x 1) 4λ, (y 1) 9λ, (z 1) + λ, 4 x + 9 y z. This system of four equations and four unknowns has the solution (check) x 5 49, y 5 49, z 55 and λ This week, we will find the global minimum of the function: f(x, y) e sin(5 x) + sin(6 e y ) + sin(7 sin(x)) + sin(sin(8 y)) sin(1(x + y)) + x + y. 4 This is problem 4 in The SIAM 1-Digit Challenge [1]. For x and y large, f(x, y) is dominated by the paraboloid x +y 4, since the values of the other terms lie in the intervals [1/e, e], [ 1, 1], [ 1, 1], [ sin 1, sin 1], and [ 1, 1], respectively. Therefore, we know the minimum will be reasonably near the origin. However, what is reasonable? We can break the problem down into three steps: 1. Find a bounded region that contains the minimum;. Identify the rough location of the lowest point in that region; 3. Zoom in closer to pinpoint the minimum to high precision. Part I We evaluate the function on the 61 values of x and y from.5 to.5 in steps of.1. The function can be as small as 3.4. Outside the circle of radius 1, the function is at least e sin > 3.3. Therefore, the minimum must be inside the circle of radius 1. Part II Many animals and insects interact with eachother, whether this may be for the purposes of protection from predators, or for cooperation in hunting, finding shelter, or otherwise []. Bees, ants, birds, and fish have been studied in particular because they are all relatively easy to study. For example, bee hives host around 5,-1, bees, which is few enough to be individually counted and monitored. Insects can communicate with one another by releasing pheromones into the environment. These pheromones can indicate the source and quality of food, or a threat. 49, 3

34 PSO seeks to use simple algorithmic rules from the insect interactions, apply them to agents with random motion, and hope to derive meaningful global information of the problem the agents are trying to solve. For bees and ants that are harvesting food, they may look like []: 1. Wander to find food;. When I find food, harvest it and lay a trail of pheromones back to hive/hill; 3. If I find a trail, follow it, harvest food, and reinforce the shortest trail with pheromones to hive/hill. Bees and ants have become the prototype species for PSO and Ant-Colony Optimization (ACO). While they are similar, the ACO is very similar to the simple rules above, while the PSO takes advantage of the ability of bees to fly. This results in the Particle Swarm Optimization algorithm [3]: 1. Initialization of agents i 1,..., N, with random positions ~xi and speeds ~vi ; Loop over all agents. Evaluate the function f (~xi ); 3. Compare the value of the function f (~xi ) with the best value of the agent. Let this be the vector p~. If the current value is the best, replace it; 4. Identify the agent with the best value p~. Let this be the vector ~g representing the entire swarm s best value; 5. Update the agents positions and velocities as: ~vi ω~vi + φp U (, 1) (~ pi ~xi ) + φg U (, 1) (~ pg ~xi ), ~xi ~xi + ~vi. (5.1) 6. Terminate when the optimal value (to an error tolerance) is obtained. End loop 33

35 In (5.1), < ω < 1 is an inertial term, < φ p <, < φ g <, denotes component-wise multiplication, and U(, 1) denotes a uniformly distributed random variable on the interval [, 1]. Part III To get really high precision result, we need something that converges very quickly. Remember the Newton- Raphson method? Suppose we want to find the solution to: Using the linear approximation to f (x i ), we have: Setting the linear approximation to, we solve: min f(x) f (x). (5.) L(x) f (x i ) + (x x i )f (x i ). (5.3) f (x i ) + (x i+1 x i )f (x i ), (5.4) for x i+1 as: Suppose, now we have two variables: x i+1 x i f (x i ) f (x i ). (5.5) But setting the gradient to is actually a system of two equations: min f(x, y) f(x, y). (5.6) f x (x, y), (5.7) f y (x, y), (5.8) with two unknowns x and y. Fortunately, we can form the tangent plane approximation (the generalization of the linear approximation) to both equations: T fx (x, y) f x (x i, y i ) + (x x i )f xx (x i, y i ) + (y y i )f xy (x i, y i ), (5.9) T fy (x, y) f y (x i, y i ) + (x x i )f xy (x i, y i ) + (y y i )f yy (x i, y i ), (5.1) Setting both equations simultaneously to at the new iterate (x i+1, y i+1 ), we have: [ ] [ ] [ ] f xx (x i, y i ) f xy (x i, y i ) x i+1 x i f x (x i, y i ). (5.11) f xy (x i, y i ) f yy (x i, y i ) y i+1 y i f y (x i, y i ) This is a familiar system of two equations for two unknowns. We can solve this! [ x i+1 y i+1 ] [ x i y i ] [ f xx (x i, y i ) f xy (x i, y i ) f xy (x i, y i ) f yy (x i, y i ) ] 1 [ f x (x i, y i ) f y (x i, y i ) ]. (5.1) 34

36 Bibliography [1] F. Bornemann et al. Think Globally, Act Locally in The SIAM 1-Digit Challenge, SIAM, 4:77 1, 4. [] M. Beekman et al. Biological Foundations of Swarm Intelligence in Swarm Intelligence, Springer-Verlag, 1:3 41, 8. [3] J. Kennedy and R.C. Eberhart Particle swarm optimization, Proceedings of the IEEE international conference on neural networks IV, ,

37 Lab 6 Double Integrals Problems 1. Evaluate the following integrals: (a) 3 5 ( x y + x) dx dy; 3 5 ( x y + x) dx dy [x y + x ( 5 y + 5 ( 1 y + 1 [ y + y ] 3, ] 5 1 (9 + 3) 16. dy, ) 4 y dy, ) dy, (b) (c) 1 x D x y dy dx; 1 x x y dy dx 1 [ x y 1 x 3 [ x 4 8 e y da, where D {(x, y) x y, y 1}; dx, ] 1 ] x 1 8. dx, 36

38 (d) π π 1 D [x e y] y e y da dy 1 y 1 e y dx dy, y e y dy [ ] 1 e y e 1. sin y dy dx; x y In this case, D {(x, y) x y π, x π} {(x, y) x y, y π}. equivalence is shown graphically below. This pi The region D {(x,y) x y π, x π} or equivalently D {(x,y) x y, y π}. pi/3 D y pi/3 pi/3 pi/3 pi x Then: (e) 8 π π x sin y y dy dx π y π π sin y dx dy, y [ x sin y y ] y dy, sin y dy [ cos y] π. 1 x 1/3 y 4 dy dx. + 1 In this case, D {(x, y) x 1/3 y, x 8} {(x, y) x y 3, y }. This equivalence is shown graphically below. Then: 37

39 The region D {(x,y) x 1/3 y, x 8} or equivalently D {(x,y) x y 3, y } D 1. y x 8 x 1/3 1 y dy dx y 3 [ 1 y 4 dx dy, + 1 x y ] y 3 y 3 y dy dy, [ ] 1 4 ln(y4 + 1) 1 ln Exercises 1. Evaluate: (a) x y da, R [, a] [, b]; (b) R We find: R R x y da a b a a [ x y 3 x y dy dx, 3 ] b x b 3 dx, 3 [ x 3 b 3 ] a 9 dx, a3 b y da, R is the triangle with vertices (, ), (1, ), and (1, 1); 38

40 We find: R 1 y da 1 x y 1 y dy dx, 1 y dx dy, (1 y) 1 y dy, 1 1 y dy y 1 y dy, π [ ] (1 y ) 3/, (c) (d) D π x y da, D is the region in the first quadrant lying between the curves y x and x y ; The curves intersect at x y x 4, or x, 1. We find: D D x y da 1 x 1 1 e y3 da, D {(x, y) x y 1, x 1}. x [ x y 3 3 x y dy dx, ] x x dx, 1 (x 5/ x 7 ) dx, 3 [ ] 1 1 x 7/ x8, ( ) The region D {(x, y) x y 1, x 1} {(x, y) x y, y 1}. Therefore, we find: 1 y e y3 da e y3 dx dy, D 1 [ 1 3 ey3 y e y3 dy, ] 1 e The volume of the solid region that lies above the region D in the xy-plane and under the graph of z f(x, y) is given by the double integral V D f(x, y) da. Find the volume of the region enclosed by the planes z x + y, z 6, x, y, and z ; [Hint: sketch the region first.] The planes x y z mean the region is bound to one of the eight octants. With the planes z 6 x + y, we recognize it is the first octant. 39

41 Then D {(x, y) y 6 x, x 6}, f (x, y) 6 x y, and: ZZ V Z 6 Z 6 x (6 x y) dy dx, f (x, y) da D Z 6 y (6 x)y 6 x dx, 6 (6 x) dx, 6 63 (6 x) Z 3. Find the volume of the region that lies below the surface z 1 x and above the region in the xy-plane given by D {(x, y) y x, x 1}. The region is visualized as: 4

42 Then: V D f(x, y) da 1 x 1 1 (1 x ) dy dx, [ y(1 x ) ] x dx, x(1 x ) dx [ (1 x ) 4 ]

43 Lab 7 Polar Coordinates, Centre of Mass, Surface Area Problems 1. Find the volume lying between the paraboloids z x + y and 3 z 4 x y ; In polar coordinates, the paraboloids are z r and 3 z 4 r. They intersect when 3 r 4 r, or r 1. Therefore: ( 4 x y ) π 1 ( ) 4 r V x y da r r dr dθ, D ( ) 4 r 4 r 3 π dr, 3 [ r r 4 ] 1 π π A thin plate lying in the xy-plane is bounded by the curves x + 4 y 1 and x 4 y and has a mass density ρ(x, y) x. Find the total mass of the plate; From the two curves, 16 y y 1, or 4 y 4 + y 3, or y 1± ±7 8 1, 3/4. For the real solutions, we find y 3/4 or y ± 3/. The region is visualized below. 4

44 The region D {(x,y) 4 y x (1 4 y ) 1/, 3 1/ / y +3 1/ /} y.5.5 D x The total mass is then: m D ρ(x, y) da 1 3/ 3/ 3/ 3/ 3/ 3/ 3/ 1 4 y x dx dy, 4 y [ ] x 1 4 y dy, 4 y ( 1 4 y 16 y 4) dy, ( 1 4 y 16 y 4) dy, by symmetry, [1 y 4 ] y3 3/ 16 y5, ( ) The centroid of an object is the centre of mass computed under the assumption that the mass density is ρ 1. Find the centroid of the region in the xy-plane that lies inside the cardioid r a(1 + cos θ). As θ π, 1 cos θ 1. Therefore, the radius of this cardioid is from r a, meaning r. This fact ensures that the cardioid does not have a little circle within itself that can happen when r < as in r a(1 + cos θ), for example. We find the mass: m π a(1+cos θ) π π [ r ] a(1+cos θ) 1 r dr dθ, dθ, a (1 + cos θ) dθ, π a 1 + cos θ + cos θ dθ. 43

45 π Now, cos m θ dθ [sin m θ] π when m, ±1, ±,... (check). Using cos 1+cos θ θ, then: Then for x: x 1 m 1 m 1 m 1 m a3 m a3 m m a π a π π a(1+cos θ) π a(1+cos θ) π π π π [ r 3 3 ] a(1+cos θ) cos θ 4 3 3πa dθ 4. 1 x r dr dθ, r cos θ dr dθ, cos θ dθ, a 3 (1 + cos θ) 3 cos θ dθ, 3 dθ, cos θ + 3 cos θ + cos 3 θ 3 cos θ + 3 cos θ + 3 cos 3 θ + cos 4 θ 3 cos θ dθ, By the same logic as above for the mass, cos θ integrates to and so does cos 3 θ (check). Therefore, again using cos 1+cos θ θ, we find: For ȳ: x a3 m a3 m a3 m a3 m a3 m π π π π π 3 cos θ + cos 4 θ 3 dθ, 6(1 + cos θ) + (1 + cos θ) dθ, cos θ cos θ + cos θ cos θ 1 dθ, cos 4θ 4 15πa3 1m 5πa3 4m 5a 6. ȳ 1 m 1 m 1 m 1 m π a(1+cos θ) π a(1+cos θ) π π [ r 3 3 dθ, ] a(1+cos θ) 1 y r dr dθ, r sin θ dr dθ, sin θ dθ, a 3 (1 + cos θ) 3 sin θ 3 dθ. dθ. dθ, 44

46 because it is an odd function about θ π, the middle of the integration interval. Exercises 1. Evaluate x + y da where R is the region in the xy-plane lying between the curves x + y 9 R and x + y 16;. Find the volume lying inside both the sphere x + y + z a and the cylinder x + y a x; 3. Evaluate I + Considering I instead of I, we have: I e x / dx; [Hint: consider I instead.] e x / e y / dx dy, e (x +y )/ dx dy. Therefore, in polar coordinates, we find: π I e r / r dr dθ, π [ e ] r / π. Therefore, since I >, we find I + I π. 45

47 Lab 8 Triple Integrals Part I Problems 1. Evaluate We find: 1 x y. Evaluate x E x y z dz dy dx; 1 x y x x dv where E x y z dz dy dx 1 x x 1 x 1 1 x [ x y 4 4 [ x y z ] y dy dx, x y 3 dy dx, ] x x dx, x 5 ( 4 1) dx, 4 [ x 6 ( 4 ] 1 1) { (x, y, z) R 3 : x } 4 y, y, z y ; 46

48 We find: E x dv 4 y y x dz dx dy, 4 y [ x z] y dx dy, 4 y x y dx dy, [ x y ] 4 y dy, y(4 y ) dy, [ (4 y ) 4 ] Find the volume of the tetrahedron enclosed by the coordinate planes and the plane x + y + z 4. The plane is z 4 x y and is therefore above the xy-plane in the first octant. Projecting the plane onto the xy-plane (z plane), we find x + y 4, or y 4 x. When y, x. Therefore, the region of the tetrahedron is E { (x, y, z) R 3 : z 4 x y, y 4 x, x }, and we find: 4 x 4 x y V dv dz dy dx, E 4 x (4 x y) dy dx, [(4 x)y y (4 x) dx, (4 x)3 [ 3 ] ] 4 x dx, Exercises 1. Le B be the rectangular box B { (x, y, z) R 3 : x a, y b, z c }. Evaluate z 3 ) dv ;. Evaluate E x + y + z 4; (3 + x y) dv, where E is the region lying above the xy-plane and below the sphere 3. Find the volume of the region E { (x, y, z) R 3 : x 5, y 3 x, y z x + }. Part II Problems B (x y + 47

49 1. Use cylindrical coordinates to find the volume of the region lying in the first octant, bounded above by the paraboloid z 4 x y and lying within the cylinder x + y x; In cylindrical coordinates, the two functions are z 4 r and r r cos θ or r cos θ. Therefore, the region in the problem is E { (r, θ, z) : z 4 r, r cos θ, θ π/ }, and we find: V E dv 4 π/ cos θ 4 r π/ cos θ π/ π/ π/ π/ r dz dr dθ, r(4 r ) dr dθ, [ (4 r ) ] cos θ dθ, ( 4 4 (4 4 cos θ) ) dθ, 4 ( sin 4 ) θ dθ, 4 (1 sin 4 θ) dθ. To integrate the sin 4 θ term, we expand using the angle doubling identity: π/ ( π/ ( ) ) 1 cos θ 4 (1 sin 4 θ) dθ 4 1 dθ, π/ ( cos θ + ) cos θ π/ dθ 4 4 π/ (4 1 + cos θ (1 + cos 4θ)/) dθ [ sin θ 5θ/ + ] sin 4θ π/ 5π 8 4. π/ ( 1 ) 1 cos θ + (1 + cos 4θ)/ dθ, 4 ) cos 4θ dθ, ( 5/ + cos θ. Find the volume of the region that lies inside the cone φ α and the sphere ρ a; The region can be characterized as E {(ρ, θ, φ) : ρ a, θ π, φ α}. Therefore, we find: α π a V dv ρ sin φ dρ dθ dφ, E α π [ ] ρ 3 a α sin φ dθ dφ π a3 sin φ dφ, Compute E π a3 3 [ cos φ]α πa3 (1 cos α). 3 z x + y + z dv where E is the region in problem. The region in problem is E {(ρ, θ, φ) : ρ a, θ π, φ α}. Since z ρ cos φ and 48

50 ρ x + y + z, we find: E z x + y + z dv α π a α π a π a5 5 α ρ cos φ ρ ρ sin φ dρ dθ dφ, ρ 4 cos φ sin φ dρ dθ dφ, cos φ sin φ dφ. Since sin φ sin φ cos φ: z α x + y + z dv π a5 sin φ dφ, E 5 [ ] π a5 cos φ α πa5 (1 cos α) Exercises 1. Compute E x dv where E is the interior of the unit sphere centered on the origin;. Consider the plane y z and the paraboloid z x + y. Convert these equations to cylindrical coordinates. Solve them simultaneously to determine the r and θ values where these two surfaces intersect. Find the volume of the region that lies between these two surfaces; 3. Find the volume of the region that lies between the paraboloids z 1 x y and z ( x + y 1 ). 49

51 Lab 9 Line Integrals Part I Problems 1. A thin wire has the shape of the right half of a unit circle and the mass density ρ(x, y) x e y. Find the mass of the wire; The thin wire with nonconstant mass density y x Parameterizing the circle as x(t) cos t and y(t) sin t, π t π, we find: π (dx ) ( ) dy m ρ(x, y) ds x e y + dt, dt dt. Evaluate C C π π π π π F dr where F(x, y) cos t e sin t sin t + cos t dt, cos t e sin t dt [ e sin t] π π e 1 + e 1 sinh 1. x, y (x + y ) 3/ and C is the curve r(t) et cos t, sin t, t 1; 5

52 The Force F along the path r. 1.5 y x We find: F(x, y) F(t) et cos t, e t sin t (e t ) 3/ e t cos t, sin t, dr r (t) dt e t cos t sin t, sin t + cos t. Therefore: 3. Evaluate We find: C C F dr e t cos t, sin t e t cos t sin t, sin t + cos t dt, e t ( cos t cos t sin t + sin t + sin t cos t ) dt, e t ( cos t + sin t ) dt, e t dt 1 e 1. F dr where F(x, y, z) x z, y + z, x where C is the curve r(t) e t, e t, e t, t 1; F(x, y) F(t) x z, y + z, x e 3 t, e t + e t, e t, dr r (t) dt e t, e t, e t. Therefore: C F dr 1 1 e4 1 4 e 3 t, e t + e t, e t e t, e t, e t dt, ( e 4 t e t e t + e 3 t) dt, + e e + (e3 1). 3 51

53 The Force F along the path r y 1.5 x 4. Evaluate C y 3 dx x 3 dy where C is the triangle with vertices (, ), (1, ), and (, 1), oriented counterclockwise. Is there a scalar field whose gradient is y 3, x 3? 1 The region D {(x,y) y 1 x, x 1} y x 5

54 Therefore we find: F(x, y) y 3, x 3, ( F dr C C 1 + C + C 3 ) F dr, r 1 (t) t,, r (t) 1 t, t, r 3 (t), 1 t, r 1(t) 1,, r (t) 1, 1, r 3(t), 1, 1 1 F dr 3, t 3 1, dt + t 3, (1 t) 3 1, 1 dt + C 1 [ t t 3 + (1 t) 3 4 dt 4 ] 1 (1 t) (1 t) 3, 3, 1 dt, The scalar field must satisfy f(x, y) x y 3 + g(y). Similarly f y (x, y) 3 x y + g (y) x 3, but this implies g (y) is also a function of x, which makes the existence of a scalar field impossible. Alternatively, we assert there cannot be a scalar field as the contour integral round a closed loop is not, making the vector field not conservative. Exercises 1. If F 3 x y, y + z, x, evaluate and final endpoint (1, 1, 1) and defined by: (a) the parametric equations x(t) t, y(t) t, z(t) t 3, t 1; C F dr where C is the curve with initial endpoint (,, ) (b) union of line segments joining (,, ) to (, 1, ) to (, 1, 1) to (1, 1, 1); (c) the straight line from (,, ) to (1, 1, 1); (d) the portion of the curve of intersection of the surfaces x z and z y between the two endpoints; t. Evaluate x y /5 ds where C is the curve r(t), t5/, t 1. Part II Problems 1. Evaluate C C arctan y dx + x 1 + y dy where C is any curve from (, 1) to (1, 1); arctan y, The scalar field f(x, y) x arctan y has gradient F(x, y) fundamental theorem for line integrals: arctan y dx +. Show that C C x dy f(1, 1) f(, 1), 1 + y arctan 1 π 4. x 1 + y ( 4 x x y ) dx + ( 6 x 3 y + 6 y 5) dy for any closed curve C.. Therefore, from the 53

55 Exercises We integrate the first part with respect to x to obtain f(x, y) x x 3 y + g(y) for some arbitrary g(y). Differentiating with respect to y, we find f y (x, y) 6 x 3 y + g (y) 6 x 3 y + 6 y 5, or g (y) 6 y 5, or g(y) y 6 + K for some arbitrary K. Therefore, f(x, y) x x 3 y + y 6 shows that F is a conservative vector field and any closed loop contour integral will evaluate to. 1. (a) Use the methods of Example 6 to check that F(x, y) 3 y + 5 y 6 x, 6 x y + 5 x is conservative; (b) Use the Fundamental Theorem for Line Integrals to compute: ( 3 y + 5 y 6 x ) dx + (6 x y + 5 x) dy, C along any path C joining initial endpoint (1, ) to final endpoint (, 1); (. Show that y + x y ) dx + ( x + x y ) dy for any closed curve C; C 3. Find a potential function for F(x, y, z) e x cos y + y z, x z e x sin y, x y. Use it to evaluate (e x cos y + y z) dx + (x z e x sin y) dy + x y dz where C is any curve from (,, ) to (, π, π 1). C 54

56 Lab 1 Green s Theorem Let C be a positively oriented, piecewise smooth, simple closed curve be the boundary of a region D. Say that P and Q have continuous first partial derivatives on an open region containing D. Green s theorem states that: ( Q P dx + Q dy x P ) da. (1.1) y Problems 1. Use Green s theorem to evaluate traversed counterclockwise; We find:. Evaluate C counterclockwise; We find: C C C D ( 3 x y ) dx+ ( 11 y 5 + x ) dy where C is the circle x +y 1, ( 3 x y ) dx + ( 11 y 5 + x ) dy ( ) da, D 4 da 4 π 1 4π. ( 6 x y 3 5 y ) dx + ( 9 x y 5 x ) dy where C is the ellipse x + 4 y 16, traversed C ( 6 x y 3 5 y ) dx + ( 9 x y 5 x ) dy (18 x y 5 18 x y + 5) da, D da. 3. Use Green s theorem to show that C D D ( x x y 3) dx+ ( y 6 x y ) dy where C is the square with vertices (±1, ±1). Note that this does not mean that F x x y 3, y 6 x y is conservative, and in fact it is not conservative. 55

57 We find: C ( x x y 3) dx + ( y 6 x y ) dy D D ( 6 y 3 x y ) da, (3 x y 6 y) da, 1 (3 x y 6 y) dy dx, [ x y 3 3 y ] 1 1 dx, [x(1 ( 1)) 3(1 1)] dx, x dx by symmetry. Exercises 1. Use Green s theorem to compute counterclockwise and F x y, 3 x y 3 : C F dr where C is the curve in each of the diagrams below traversed x +y y 1 y yx.6.4 yx 1/ y(x ) x (a) x (b). Evaluate C [ ( e x 11 x 3 ) ] + y dx+[ x + arcsin y] dy where C is the circle (x ) +(y +3) 64, traversed counterclockwise; 3. Show that for a region D enclosed by a positively oriented simple closed curve C, the area of D is given by: A(D) x dy y dx. C C 56

58 Lab 11 Surface Integrals Problems 1. Find the surface area of the portion of the paraboloid x + y 6 z that lies within the sphere x + y + z 16; The surface can be parameterized by r(u, v) u, v, u + v. Therefore: 6 i j k r u r v 1 u/3 u/3i v/3j + k 1 + u /9 + v /9. 1 v/3 A(S) S ds D π 4 π 4 r u r v da, 1 + u /9 + v /9 r dr dθ, 1 + r /9 r dr dθ, π [3(1 + r /9) 3/] 4 π. Let the mass of a thin sheet in R 3 be given by m density and ds is the differential surface element. Let: r 1 rρ(x, y, z) ds, m S x, ȳ, z 1 x, y, z ρ(x, y, z) ds, m S 1 x ρ(x, y, z) ds, y ρ(x, y, z) ds, m S S S ( ) 196π 9 9. ρ(x, y, z) ds, where ρ(x, y, z) is the mass S z ρ(x, y, z) ds, be the centre of mass of this thin sheet. Find the mass m and centre of mass r for a constant unit density upper hemisphere of radius a centred on the origin: z a x y ; 57

59 The surface can be parameterized by r(u, v) u, v, a u v. Therefore: i j k r u r v 1 u(a u v ) 1/ 1 v(a u v ) 1/ u(a u v ) 1/ i + v(a u v ) 1/ j + k 1 + u + v (a u v ) 1 a u v a u v + u + v a a u v. Then: m S ρ(x, y, z) ds D r u r v da, a a u v da, D a a u a a u a dv du. a u v To do the inner integral, let v a u sin θ, then: a u a u a π/ a u v dv a a π/ π/ Then, inserting this into the outer integral, we find: m a a π/ π/ Finding the components of r now: x 1 xρ(x, y, z) ds 1 m S m ȳ 1 m z 1 m S S π/ a a u cos θ a u (a u ) sin θ dθ, a u cos θ a u cos θ dθ, dθ πa. πa du πa. D a u a u v da, by symmetry about u, yρ(x, y, z) ds 1 a v m a u v da, D by symmetry about v, zρ(x, y, z) ds 1 a a u v m D a u v da, 1 a da a A(D) m m a πa πa a. To conclude, the mass of a constant unit density hemisphere is m πa and is located at r,, a. D 58

60 3. Find the flux of the vector field F x, 3 y, z outward through the surface comprised of the finite cylinder { (x, y, z) R 3 : x + y 4, z 3 } together with the two end caps { (x, y, z) R 3 : x + y 4, z } and { (x, y, z) R 3 : x + y 4, z 3 } ; Generally speaking, the flux of the vector field is given by: Φ F ds F (r u r v ) da. S In this case, our surface should be split into the three regions S 1 is the circumferential surface, and S and S 3 are the end caps. S 1 can be parameterized by r(u, v) cos u, sin u, v, where u [, π] and v [, 3]. Then: r u r v D i j k sin u cos u 1 cos u, sin u,. This vector points outward for every u [, π], therefore we are finding the outward flux. And: F ds 4 cos u, 6 sin u, v cos u, sin u, da, S 1 D π 3 π π π ( 8 cos u + 1 sin u ) dv du, ( sin u ) du, (8 + (1 cos u)) du, (1 cos u)) du 1π. S can be parameterized by r(u, v) v cos u, v sin u,, where u [, π] and v [, ]. Then: i j k r u r v v sin u v cos u,, v sin u v cos u,, v. cos u sin u 59

61 This vector points outward (downward), therefore we are finding the outward flux. And: F ds v cos u, 3 v sin u, 4,, v da, S D π 4 v dv du, π [ v ] 16π. S 3 can be parameterized by r(u, v) v cos u, v sin u, 3, where u [, π] and v [, ]. Then: i j k r u r v v sin u v cos u,, v sin u v cos u,, v. cos u sin u This vector points inward (downward), therefore we negate the cross product to find the outward flux. And: F ds S 3 v cos u, 3 v sin u, 6,, v da, D 3 In total, we have: Φ S π 6 v dv du, π [ 3 v ] 4π. ( F ds S 1 + S + S 3 ) F ds 14π. 4. Find the flux of the vector field F i + j + k upward through the surface S described parametrically by r u vi + u v j + v 3 k, u 1, v 1. From the surface, we find: r u r v i j k u v v u u v 3 v 3 v 4, 6 u v 3, 3 u v. 6

62 Then: Φ S F ds , 1, 1 3 v 4, 6 u v 3, 3 u v du dv, ( 6 v 4 6 u v u v ) du dv, [ 6 u v 4 3 u v 3 + u 3 v ] 1 dv, ( 6 v 4 3 v 3 + v ) dv, [ 6 v v4 4 + v3 3 ] Exercises 1. Evaluate S f(x, y, z) ds where S is the upper hemisphere of the sphere of radius a centred on the origin: z a x y and f(x, y, z) x ;. Evaluate f(x, y, z) ds where S is the portion of the plane x + 3 y + z 6 lying in the first octant and f(x, y, z) x + y; S 3. Find the flux of the vector field F(x, y, z) x, y, z upward through the hemisphere z a x y ; 4. Find the flux of the vector field F(x, y, z) x, y, z outward through the surface S of the parallelepiped bounded by the planes x ±1, y ±, and z ±3. (Note that S is comprised of six different planar faces.) 61

63 Lab 1 Stokes Theorem and the Divergence Theorem Part I Stokes theorem Let C be a simple, piecewise-smooth closed curve bounding an oriented, piecewise-smooth surface S. Then Stokes theorem relates the line integral of a vector field F over C to a surface integral over S: F dr ( F) ds, C where C is parameterized so that C and the normal to the surface S obey the right-hand rule. For example, if C is traversed counterclockwise about the vertical axis, then the normal points upward. Problems 1. Use Stokes theorem to evaluate paraboloid z 4 x y with the cylinder x + y 1; C S y dx z dy + 3 dz where C is the curve of intersection of the 6

64 We begin by computing the curl of F: F i j k x y z y z 3 1i + j k. Then, by intersecting the cylinder and the paraboloid, we find z 4 x y Since in Stokes theorem, we can take any surface with boundary C, the simplest seems to be z 3. This surface is r(u, v) v cos u, v sin u, 3 for D { (u, v) R 3 : u π, v 1 }. Since: we find: r u r v S i j k v sin u v cos u cos u sin u ( F) ds D π 1 i + j vk, 1,,,, v da, v dv du π. Since we do not know in which direction the contour C is taken, we do not know in which direction the surface normal should point. Therefore F dr ±π.. Use Stokes theorem to evaluate C C y dx + x dy + z dz where C is the curve of intersection of the plane z x with the cylinder x + y 1, oriented so that the curve runs counterclockwise about the vertical axis (as viewed from above). We begin by computing the curl of F: F i j k x y z y x z i + j 1k. 63