MA 1116 Suggested Homework Problems from Davis & Snider 7 th edition

Transcription

1 MA 6 Suggested Homework Problems from Davis & Snider 7 th edition Sec. Page Problems Chapter. pg. 6 4, 8.3 pg. 8, 5, 8, 4.4 pg. 4, 6, 9,.5 pg , 3, 5.6 pg. 7 3, 5.7 pg. 3, 3, 7, 9, 7, 9.8 pg. 9, 4, 7, 9,, 5a.9 pg. 34, 5, 7,, a,, 4. pg. 38, 4, 6, 9,, 3. pg. 5, 3, 5, 8,, 3, 9,, 3.3 pg. 57, 3, 5, 7.4 pg Chapter. pg. 7, 3, 4, 5hi. pg. 85 3, 5, 8, 3.3 pg. 95, 5 7, 9, 5, 7.4 pg., 3, 4, 6, Chapter 3 3. pg., 6, 7, 8, 6, 7,, 4, 3 3. pg pg. 4, 3, 5, 7, 3.4 pg. 3,, 4, 7,, 3.5 pg. 35,, pg. 4, 3 5, pg. 5, 5, 6,, 3. pg. 69, 4, 6, 8, 3. pg , 8, 9,, 3 Chapter 4 4. pg. 9, 3, 6,, 4, pg. 4, pg.,, 6, 7, 9, 4.6 pg. 36, pg. 46, d, 5, 6,, 5, pg. 56 4, pg. 6 3, 4, 6, 9, 4, 5, 7 4. pg. 7 Chapter 5 5. pg

2 5. pg. 84,, 5, 8a 5.4 pg. 94 4, 5, 7, 9, 5.5 pg. 99 3, 7

3 . Problems: 4, 8 B C A F K G H D E Figure : Figure.6 for Problems -4 on section.. C in terms of E, D, F Note that Therefore. G in terms of C, D, E, K Note that and or Therefore or F + C + D E C E D F K + G F F + C + D E F E C D K + G E + C + D G E + C + D K 3. Given x + B F, solve for x 3

4 based on Figure Therefore x F B F B A x A 4. Given x + H D E, solve for x based on Figure Therefore x D E H D E G + H x G + H H G B A G C H F D E Figure : Six equilateral triangles 8. a. Note that G C but G H A and H B therefore G B A so C B A D A E B F C A B 8. b. Sum of all is zero (vector addition with last terminal point being the same as first initial point) 4

5 .3 Problems:, 5, 8, 4. If A 3 then 4A 4 A 4 3 Also A A 3 6 For s we have sa s A 3 s 3 6. The minimum value of s is zero, so that sa. 5. Is it possible that B γa? A αb If α then we can divide and so γ α. If α then A and one cannot have any nonzero vector as a multiple of zero. 8. Two vectors, one pointing up from plane and one is pointing downward. 4. Given the vectors A, B and C (see Figure 3) B B A / B 4 4 C 4 A A Figure 3: For problem 4 section. Let us sketch the vector B from the terminal point of the vector A. In order to get an isosceles triangle, we make this new vector the same length of A, i.e. the new vecor is B A. B Note that I used 8 degree-angle between the vectors A and B. C A + B A B vector of length A in the direction of B so that the angle is bisected 5

6 .4 Problems: 4, 6, 9, 4. 3i 4j 3 + ( 4) Given P (4, ), P (5, ) P P (5 4)i + ( )j i 3j 9. a. A αi + βj with A α + β and A makes an angle of 6 with the x axis. Clearly, sin(6 ) β A β, so 3 β and thus Therefore α β 3 4 A i + 3 j α β Figure 4: For Problem 9a of section.4 9. b. A αi + βj with A α + β and A makes an angle of 3 with the x axis. Clearly, sin( 3 ) β A β, so β 6

7 and thus Therefore α β 3 4 A 3 i j 4 3 α 3 β Figure 5: For Problem 9b of section.4 9. c. 3i + 4j 5. The vector 3 5 i + 4 j is in the same direction as the original and of unit 5 length (dividing by the length of the original vector) 9. d. A ( ) i + βj. The length of the vector is unity, implying β + or β 4 3 4, so β ± 3 and A i ± 3 j 9. e. A is a unit vector perpendicular to the line x + y, so A αi + βj with α + β. Therefore (see figure 6) α β ± and A ± (i + j) (Only two vectors!!) 7

8 x+y A 45 Figure 6: For Problem 9e of section.4. given the vector V whose endpoints are A and B, so that V i + 3j. Given the midpoint M (, ) we have to find the coordinates of the endpoints. Suppose A (x, y ) and B (x, y ) then x x, y y 3 and from midpoint formula x + x y + y, solve the set of two equations for the x coordinates, we get x and x 3. Solve the other set for the y coordinates and we have y and y 5. Therefore the endpoints A (,.5), B (3,.5) 8

9 .5 Problems: 4 6, 3, 5 4. B i + j k, so B sb implies s B. Therefore s 3 or s ± A 3i + 4j, so A and the unit vector A A 3 5 i j 6. A 3i + 4j, C 3i 4k We have C + D A, so DA-C 4j + 4k a. D b. The vector is parallel to yz plane since there is no component in x direction. 3. A i j + k therefore A Therefore cos α 3, cos β 3, cos γ 3 5. cos α describes a cone with vertex at the origin and base perpendicular to x axis. 9

10 .6 Problems: 3, 5 3. R (t ) 3i + 4j k, and R (t 3) 3i + 36j 7k, then the displacement is R R 3j 6k. 5. F 3k, F 6i, F 3 j, what is F 4, such that F + F + F 3 + F 4? F 4 F F F 3 6i j 3k. The magintude of F 4 is F lb

11 .7 Problems:, 3, 7, 9, 7, 9. A i + j + k, B 3i 4k A , and B and cos θ ( θ cos ) 5 3. Find the three angles of the triangle with vertices A (,, ), B (, 3, 5), and C (3, 4, 4). The sides are: AB i j 6k, AC i 3j 5k, and BC i j + bk. AB AC Therefore cos θ AB AC θ arccos 35 4 Since we need vectors emanating from the point B, we should use AB, cos θ AB BC AB BC ( + 6) θ arccos 6 4 θ 3 8 θ θ 7. Let the 4 sides be the vectors A, B, C, and D, so that A + B + C + D. Since the two opposing sides are equal and parallel, we have A C Therefore, when we substitute this in the previous equation, we get B + D and thus B D and so the other two are parallel and equal. 9. Show that AE EC, BE ED, see figure.

12 B B C A E C A D D Figure 7: For Problem 9 of section.7 A + B + C + D CA A + B Let us look at CE, and show that CE ( ) CA A + B and similarly BE BD. Now CE s ( ) A + B since it is in the direction of A + B. The point E is on the line connecting the points B and D and so EC C + t ( ) B + C, but C A, so Therefore so EC A + t ( B A ) EC A + t ( ) ( ) A B s A + B (s t) A ( s t) B Since A and B are not parallel, the scalars are zero s t, s + t Or so s s and therefore substituting s above CE ( ) A + B as we wanted to show. 7. The equation of a sphere r (x ) + (y 3) + (z 4)

13 where r is the vector from the center (,3,4) to the point (x,y,z) on the edge of the sphere, therefore r is the radius which is given as 3 and the equation 9. (x ) + (y 3) + (z 4) 9 x y z is a line (two equations!) x y is a plane and x z is a plane and we are taking the intersection of these two. 3

14 .8 Problems:, 4, 7, 9,, 5a. Line thru (,, ) parallel to 3i j + 7k or or in parametric form x 3 y z 7 x 3 y z 7 x 3t y t z 7t 4. Find the two unit vectors parallel to the line x 3 y + 4, z 9 The vector v in the direction of the line is v 3i + 4j, and v , therefore v the unit vectors are ± v ±3 5 i ± 4 5 j 7. Find equations of line thru (,, ) parallel to x 3 y + z +. 4 The vector v in the direction of the line is v i + 4j k and so the equations are or x y 4 x y 4 z z 9. Line thru (, 4, ) and (,, 7) and the line is or. Angle between lines v ( )i + ( 4)j + (7 + )k i j + 8k x y 4 z + 8 x y z 7 8 x y 3 z x 3 y z 4

15 We need the angle between the vectors v 3i+4j+5k and v i j+ k in the direction of these lines. v , v Therefore cos θ 3 + 4( ) a. Given the lines We can also write them as follows R (5i + 4j + 5k)t + 7i + 6j + 8k R (6i + 4j + 6k)t + 8i + 6j + 9k x 7 5 x 8 6 y 6 4 y 6 4 z 8 5 z 9 6 Find the point(s) of intersection. Since both have y 4, we get 6 or 6x 4 5x 4 or Now substitute this into x 7 5 y 6 4 x 7 5 x 8 6 x to get 5 5 y 6 4 y Now we take x y and substitute in both equations z 8 5 z 9 6 The first one yields z 8 5 or z 3. This value of z satisfies the second equation. Therefore we have one point (,, 3) or 5

16 .9 Problems:, 5, 7,, a,, 4. The scalar product of (3i + 8j k) (5i + j + k) 3(5) + 8() () The angle between the two vectors i and 3i + 4j is cos θ (3) + (4) (5).6 θ Find the component of B 8i + j in the direction of A i + j k A B A 8() + () + ( ) A A and A B From the first we have that A is the zero vector, substituting in the second we get identity no matter what the vector B is.. a. B 6i 3j 6k, A i + j + k, then B B A A A A (i + j + k) i j k + + B B B (6i 3j 6k) ( i j k) 7i j 5k. Show that A + B A + B Let us square the left hand side A + B (A + B) (A + B) A A + A B + B A + B B A + A B + B Now square the right hand side So we need to show that ( A + B ) A + A B + B A B A B Now A B A B cos θ and since cos θ, we have the triangle inequality. 4. Given two lines, line l thru (5,, ) and (, 3, ) and line l thru (3, 8, ) and ( 3,, 7), then v 3i + 4j 3k v 6i + 8j 6k v v 3(6) + 4(8) + ( 3)( 6) Therefore the lines are not perpendicular. In fact v v so the lines are parallel 6

17 . Problems:, 4, 6, 9,, 3. Plane thru the origin and perpendicular to i 8j + k is x 8y + z We can divide the equation by to get x 4y + z. 4. Plane parallel to 3x + y z 8 thru (, 3, 3) is 3(x ) + y 3 (z 3) or 3x + y z 3 6. The distance from the point (3, 4, 7) to the plane x y z 4 Let R 3i + 4j + 7k the position vector to the point and n i j k and d 4 the constant on the right hand side in the equation of the plane, then the distance is R n d n Show that the normal to the plane x 8y + z 5 is parallel to the line x y z + 3 n i 8j + k v i + j + 3k n v Therefore the normal to the plane is perpendicular to the line.. Given the three point O(,, ), A(,, 3), B(,, ) and C(,, ) a. Find a vector perpendicular to the plane OAB Since OA and OB are in the plane, we take the cross product. Since we don t know yet how to find the cross product, we just get the equation of the plane and this will give us the normal. Using A and B, we get a(x ) + b(y ) + c(z ), a + b + 3c b + c since O is on the plane From the second b c, so a + b + 3b implies a 5b. Clearly we have a choice, we take b so that c and a 5. The equation of the plane is 5x + y + z 7

18 and the normal is 5i + j + k Note that a different choice of b will give a multiple of this vector. b. Find the distance from C to the plane 5() + () Plane thru the points (4,, ), (, 6, ), and (,, ), find equation of another plane thru (6,, 4) parallel to this one. a(x 4) + by + cz, since the point (4,, ) is on the plane a( 4) + 6b a( 4) + c So a 3c and b a/3, choose c gives a 3 and b, so the normal is 3i + j + k and the equations of the plane required is 3(x 6) + (y + ) + (z 4) or 3x + y + z

19 . Problems:, 3, 5, 8,, 3, 9,, 3. a. A 3i j + k, B i + j 4k A B b. i j k 3 4 i 4 j 3 4 +k A B i + 4j + 4k 3 i(4 ) j( )+k(3+) A B i j k 7 3 i 7 j k 3 i( 8) j( ) + k( 3) A B 8i + 3j k c. d. A B i j k 6 A B i i j k 6 i j 6 j + k + k i 6j + k k e. Given B A i j, find A B (B A) i + j 3. Find the area of the triangle with vertices (,, ), (, 5, 5) and (, 3, 5). Let A ( )i + (5 )j + (5 )k 4j + 3k and B ( )i + (3 )j + (5 )k i + j + 4k. The area of the triangle is half the area of the parallelogram A B A B i j k i j 3 3 +k 4 i( 6) j( 3)+k( 4) 6i+3j 4k and the area is

20 5. Given the vectors 3i + j and i j 5k, the vector perpendicular to both is the cross product i j k 3 5 i( 5) j( 5) + k( 5) 5i + 5j 5k To get a unit vector we divide by the magnitude, which is ± ( 5) ( 5) ± ± 75 ± 5() ±5 Therefore the unit vector is 5i + 5j 5k ±5 i + 3j k ± 8. Given A i + j, B 3i j + k and C 8i, show that Not equal. (A B) C A (B C) (A B) (A B) C (B C) A (B C) i j k 3 i j k 8 8 i j k 3 8 i j k 8 8 i j 8k 64j + 6k 8j + 8k 6i 6j + 6k. Find a unit vector in the plane of i + j and j + k perpendicular to i + j + k. We cross the first two vectors to find a vector perpendicular to the plane i j k 4i j + k When we cross this with the third vector we get a vector in the plane i j k 4 5i 6j + 8k

21 The unit vector is then 5i 6j + 8k ± ( 5) + ( 6) + 8 5i 6j + 8k ± 5i 6j + 8k ± Find the distance from the point (5, 7, 4) to the line thru (, 3, 8) and (3, 6, ), see figure. (5,7,4) (3,6,) A A perp B (,3,8) Figure 8: For Problem 3 of section. The distance is A 3i + 4j + 6k B i + 3j + 4k i j k A B i 6j + 5k 3 4 A B B A B and A B imply and A B cos θ A B sin θ The only way this is possible is that either vector is the zero vector (since the sine and cosine functions can vanish at the same point).. Given that A and B are parallel to yz plane, i.e. The norms are given and A a j + a 3 k, B b j + b 3 k A, B 4 A B

22 then the angle θ between them satisfies cos θ since they are not zero vectors. This implies that sin θ ± therefore A B (4)(±)n ±8n. Therefore A B 8 and since the vectors are in the yz plane, the normal n i, so A B ±8i 3. a. Do the lines intersect? Let s write the lines as x 3 y z x 5 y 3 z 4 x 3t, y t, z t x 5t, y 3t, z 4 + t Matching the x and y we have 3t 5t and t 3t for which there is no solution. Therefore the line do not intersect. b. The line perpendicular to both, will have a vector in the direction of the normal to the vectors in the direction of those lines, i.e. i j k i + 4j k Now this line should cross the original two lines, so or component-wise (3i + j + k)t (4k + (5i + 3j + k)t ) ( i + 4j k)t 3 3t 5t t 3 t 3t 4t 3 t 4 t t 3 solve the last equation in the above system for t 3 gives t 3 t + t + 4, substitute this in the first two equations 3t 5t ( t + t + 4) simplify t 3t 4( t + t + 4) t t 8 t t 6 and solve for t and t. So t 4 and t 64. We use the point t 4 ) to get the equation x t , y 4t +, z t +

23 Note that when t we get the point on the first line and when t 4 the other line. We can write the equations as we get a point on x 4 7 Another way: The normal line is t, y 8 + 4t, z 8 t x x t, y y + 4t, z z t Now for t t 3 this line intersects the first line and for t t 4 it intersects the second line, we write these 6 equations x t 3 3t y + 4t 3 t z t 3 t x t 4 5t y + 4t 4 3t z t 4 t + 4 Solving the system of 6 equations with 7 unknowns: x, y, z, t, t, t 3, t 4, we find a free parameter t 4 and all the other variables will be t 4, t 64, t t 4 x 3 + t 4, y t 4, z + t 4 c. To find the distance, we have already the normal to both i + 4j k and we just need two point, one on each line. We should take (,, ) on the first and (,, 4) on the second (i.e., the parameters are zero) and create the vector connecting these two points which is 4k. Now we take the dot product of these vectors and divide by the norm, i.e. distance 4k ( i + 4j k) ( ) ( ) 4 3

24 .3 Problems:, 3, 5, 7. a. Given A i, B 3j, C 5k then [A, B, C] 3 (3)(5)[i, j, k] 3 5 b. Using the last column to expand. 3 [A, B, C] ( 3) 3 3 c. [i j+k, i+j+k, i+3k] 3 +3 ( )+3(+) d. [k, i, j] [i, k, j] [i, j, k] 3. The volume is the triple [ AB, AC, AD] which is The vecors i + j, 3i + 4j, k give a volume of Plane parallel to i + j + k, i 3k and thru 93, 4, ). The normal is the cross product of the given vectors i j k i( 3) j( 7) + k( ) 3i + 7j k 3 and the equation is or 3(x 3) + 7(y 4) (z + ) 3x 7y + z 4

25 .4 Problems: a ω (ω R) (ω R)ω (ω ω)r 6. a. A B B A? No 6. b. (A B) C A (B C)? No 6. c. A B A C if and only if B C. Not necessarily 6. d. A B if and only if one is zero. Not necessarily 7. A B + (A B) A B A B sin θ+ A B cos θ A B ( A B sin θ) ( A B cos θ) A B ( sin θ + cos θ ) 5

26 . Problems:, 3, 4, 5hi. F(t) sin ti + cos tj + k a. F (t) cos ti sin tj b. F (t) parallel to xy plane since no component in z direction c. When is F (t) parallel to xz plane? This means sin t or t ±nπ, n,,,... d. F(t) sin t + cos t + +, so F(t) has a constant magnitude. e. F (t) sin t + cos t, so F has a constant magnitude. f. F (t) sin ti cos tj. 3. a. f(t) (3ti + 5t j) (ti sin tj) f (t) (3i + tj) (ti sin tj) + (3ti + 5t j) (i cos tj) f (t) 3t t sin t + 3t 5t cos t 6t t sin t 5t cos t b. f(t) ti + tj k, so f (t) (ti + tj k) (ti + tj k) Now differentiate f(t)f (t) (i + j) (ti + tj k) + (ti + tj k) (i + j) (4t + 4t + ) 6t f (t) f (t) 8t f(t) 8t 4t + 4t + 8t + 8t c. f(t) [(i + j k) (3t 4 i + tj)] k. We can do this either by first computing the function f(t) and then differentiating or use the rule to differentiate the product. Here we use the latter. Note that the only vector depending on t is the second one. First compute the cross product of the first vector with the derivative of the second vector i j k t 3 i j( 4t 3 ) + ( t 3 )k Since we want to dot this with k, the answer should be f (t) t 3 4. Prove d ( R dr ) R d R dt dt dt The left hand side is d dt R dr ( ) + R d R which is the right hand side. }{{ dt} dt 5. h. A 3i + j + 6k, B 3i + 4k, C i j + k then since A and B are constants d (A + Bt) B 3i + 4k dt 6

27 5. i. d dt (B Ct) d dt B B C Ct + B d dt (Ct) i j k 3 4 C 8i + 5j 6k B C 7

28 . Problems: 3, 5, 8, 3. Find unit tangent vector to: x a cos t, y b sin t, z T R a cos ti + b sin tj dr dt a sin ti + b cos tj dr dt ( a sin t) + (b cos t) dr dt dr dt a sin ti + b cos tj ( a sin t) + (b cos t) Now at t 3π/ we have sin(3π/), cos(3π/) and T a( )i + b()j (a) i. For the curve x sin t t cos t, y cos t + t sin t, z t, a. Find the arc length between (,, ) and ( π,, 4π ). First note that t π. Now and the arc length is b. Find T(t) ( ) dx + dt s dx dt dy dt cos t (cos t t sin t) t sin t sin t + (sin t + t cos t) t cos t ( ) dy + dt π dz dt t ( ) dz t sin t + t cos t +4t 5t dt t 5t dt 5 t π 5 4π 5π dr dt T(t) dr dt dr dt t sin ti + t cos tj + tk from part a. 8

29 T(t) dr dt 5t from part a. t sin ti + t cos tj + tk sin ti + cos tj + k 5t 5 b. Find T(π) T(π) sin πi + cos πj + k 5 j + k 5 3. For the curve x t, y sin t, z cos t, find the arc length between the points (,, ) π and (,, ). Clearly t π. dr dt i + cos tj sin tk π dr + 4π dt 4π + cos t + sin t 4π + π π + 4π + 4π s dt π + 4π π π Now to the unit tangent vector T(t) At the point (,, ) we have t, so T() T(t) dr dt dr dt i + cos tj sin tk π +4π i + j π +4π π π i + πj + 4π 5. a. For the curve x e t cos t, y e t sin t, z, find the arc length between t and t dx dt et cos t e t sin t Now ( ) dx + dt ( ) dy + dt dy dt et sin t + e t cos t dz dt ( ) dz ( e t cos t e t sin t ) ( + e t sin t + e t cos t ) dt 9

30 ( ) ( dx dy + dt dt ) + ( dz dt and the arc length is ) e t ( cos t sin t cos t + sin t ) +e t ( sin t + sin t cos t + cos t ) e t s e t dt e t (e ) b. To reparametrize, we need s(t) t et dt e t t and t ln ( s + ) Therefore x(s) y(s) ( ) ( ) s s + cos ln + ( ) ( ) s s + sin ln + z (e t ), so e t s(t) + c. Sketch. The curve starts at (, ) and moves counter clockwise to (, e π/ ) and down to ( e π, ) continue to (, e 3π/ ) and hits the x axis again at (e π, ). Spiraling outward. 8. Show that the curve x(t) t y(t) t, z(t) t 3 intersects the plane x+8y +z 6 at right angle. For some t we need dr to be perpendicular to the plane or parallel to dt i + 8j + k which is normal to the plane. Now dr dt i + 4tj + 3t k For t we get dr i + 8j + k dt The question is if the curve and the plane intersect at that point x(t ), y(t ) 8, z(t ) 8 and for these values the equation of the plane: +8(8)+(8) Therefore (, 8, 8) is the point of intersection and at that point the vectors are parallel. 3. No. For example x + y C, x, C > which is the right half circle. In this case x + y dy dy, so dx dx x which does NOT exist for y. y 3

31 .3 Problems:, 5 7, 9, 5, 7. Given x 3t cos t, y 3t sin t, z 4t a. To find the speed, we need the first time derivatives of x, y, z, ẋ 3 cos t 3t sin t, ẏ 3 sin t + 3t cos t, ż 4 v (3 cos t 3t sin t) + (3 sin t + 3t cos t) + 4 v 9 cos t 8t sin t cos t + 9t sin t + 9 sin t + 8t cos t sin t + 9t cos t + 6 v 9(cos t + sin t) + 9t (sin t + cos t) + 6 v 9 + 9t t ds dt 5 + 9t b. To find the acceleration we need the second derivative Therefore ẍ 3 sin t 3 sin t 3t cos t, ÿ 3 cos t + 3 cos t 3t sin t, z a ( 6 sin t 3t cos t)i + (6 cos t 3t sin t)j a ( 6 sin t 3t cos t) + (6 cos t 3t sin t) a 36 sin t + 36t sin t cos t + 9t cos t + 36 cos t 36t sin t cos t + 9t sin t c. T v v a t a T d s dt d ds dt dt d 5 + 9t dt a T + a N a t a N t a N 8t 5 + 9t 9t 5 + 9t 8t 5 + 9t t 8t 5 + 9t (3 cos t 3t sin t)i + (3 sin t + 3t cos t)j + 4k 5 + 9t d. Find the curvature κ v a N, κ a N v t 8t 5+9t 9t + 5 3

32 5. R (cos t + sin t)i + (sin t cos t)j + tk a. Ṙ v ( sin t + cos t)i + (cos t + sin t)j + k v ( sin t + cos t) + (cos t + sin t) + 4 sin t cos t + + sin t cos t b. c. v 3 ds dt a (cos t + sin t)i + ( sin t + cos t)j+ T 3 ( sin t + cos t)i + 3 (cos t + sin t)j + 3 k d. Therefore a T, since ds dt is constant a + sin t cos t + sin t cos t κ Note that the curvature is constant. e. Compare to (.5) a N a N v 9 4 e i j e i + j ρ 4 9 a 6. x(t) 3t t 3, y(t) 3t, z(t) 3t + t 3 Then ẋ(t) 6t 3t, ẏ(t) 6t, ż(t) 3 + 3t The curvature is Now ẍ(t) 6 6t, ÿ(t) 6, z(t) 6t R R κ R R R 3 i j k 6t 3t 6t 3 + 3t 6 6t 6 6t 3

33 R R (36t 8 8t )i (36t 8t 3 8+8t 8t +8t 3 )j+(36t 8t 36t+36t )k Therefore R R 8(t )i 8(t + t )j + 8t k R R 8 (t ) + 8 (t + t ) + 8 t 4 R R 8 t 4 t + + t 4 + t 3 t + t t + + t 4 R R 8 3t 4 + t 3 3t t + R 36t 36t 3 + 9t t t + 9t 4 8t 4 36t 3 + 9t + 9 R 9(t 4 4t 3 + t + ) κ 8 3t 4 + t 3 3t t + (9(t 4 4t 3 + t + )) 3t4 + t 3 3t t + 3/ 3(t 4 4t 3 + t + ) 3/ 7. R(t) sin ti + cos tj + log(sec t)k a. Find ds ( ) ds dx + dy + dz (cos t dt) + ( sin t dt) + sec t tan t sec t dt ds cos t + sin t + tan tdt sec t dt sec t b. T v v c. N Now dt dt dt dt Note that tan t sec t cos ti sin tj + tan tk sec t ( ) dt cos t dt i sec t sin t, so dt dt ( ) sin t j + sec t ( ) tan t k sec t } cos{{ t sin } t i ( cos t sin t ) j + cos tk sin(t) cos(t) N dt dt dt dt sin(t)i cos(t)j + cos tk sin (t) + cos (t) + cos t sin(t)i + cos(t)j cos tk N + cos t 33

34 d. For the curvature use κ R R R 3 or use the acceleration κ a N v R(t) sin ti + cos tj + log(sec t)k R (t) v cos ti sin tj + tan tk R (t) a sin ti cos tj + sec tk v ds dt sec t, a T d s tan t sec t dt see part a a N a a T ( sin t) + ( cos t) + (sec t) (tan t sec t) a N sin t + cos t + sec 4 t tan t sec t sec t(sec t tan t) a N + sec t (sec t tan t) κ + sec t sec t +t +t + sec t 9. R(t) log(t + )i + (t tan t)j + tk a. R (t) v ( t + ti + ) j + k + t }{{ } b. Now v 4t ( + t ) + (t ) ( + t ) + ( ) v 4t + t 4 t + ( + t ) + 8 t4 + t + ( + t ) + 8 ( + t ) ( + t ) v ds dt 3, a T d s dt, a κ ds since dt a N v a v ( t ) i + + t constant is constant ( t ) j + t 34

35 or Therefore a (t + ) t(t) ( + t ) i + t(t + ) (t )t ( + t ) j a (t + 4t )i + (t 3 + t t 3 + t)j ( + t ) a ( t )i + 4tj ( + t ) a 4( t ) + (4t) ( + t ) t + 4t 4 ( + t ) 4 4( + t ) ( + t ) 4 4 ( + t ) κ a ( + t ) ( + t ) 3 9( + t ) 5. a. b. dr ds T d (T T), ds T T T T, unit vector d R c. dt T a T a T, d. T N e. dr dt T dr ds T ds dt v dt since ds is parallel to N which is perpendicular to T T v T v, f. dn ds B τ g. [T, N, B] right handed system h. d R ds i. db ds τn, d dr ds ds T dt ds Frenet formula κ N κ, 7. x(t) cos 3 t, y(t) sin 3 t, z(t) sin t, t π dr dt 3 cos t sin ti + 3 sin t cos tj + 4 sin t cos tk 35

36 dr dt ( 3 cos t sin t) + (3 sin t cos t) + (4 sin t cos t) dr dt 9 cos 4 t sin t + 9 sin 4 t cos t + 6 sin t cos t dr dt cos t sin t(9 cos t + 9 sin t + 6) T dr dt ds dt 3 5 cos ti sin tj k cos t sin t(9 + 6) 5 sin t cos t N dt dt dt dt 3 sin ti + 3 cos tj sin t cos t 3 sin ti + 3 cos tj ( 3 5 sin ti cos tj ) Substituting κ dt ds N sin ti + cos tj i j k 3 B T N cos t 3 sin tj sin t cos t ( 4 ) 5 sin t + k B i ( 4 ) 5 cos t j τ ± dt dt dt B 4 5 cos ti sin tj 3 5 k ( 3 5 cos t 3 5 sin t 3 ds from computing N 5 5 sin t cos t from computing T 3 5 sin t cos t τ ± db ds ( 4 sin 5 t) + ( 4 cos t) 5 5 sin t cos t ± db dt ds dt db dt 4 5 sin ti + 4 cos tj 5 Compare with (.45) to find that the sign is negative. 4 5 ± 5 sin t cos t ± 4 5 sin t cos t ) 36

37 .4 Problems:, 3, 4, 6,. r b( cos θ) implies dr dθ b sin θ or dθ dt 4 Now v dr dt a implies d θ dt dr dθ dθ dt d r dt r d r dt d dt Substituting u r + r dθ dt 4 ( ) dθ u r + dt u θ 4b sin θu r + 4b( cos θ)u θ r d θ dt (4b sin θ) 4b cos θ dθ dt 4 + dr dt 6b cos θ dθ dt u θ a [6b cos θ 6b( cos θ)] u r + 3b sin θu θ 3. r ( + sin θ) implies dr dθ cos θ a 6b( cos θ )u r + 3b sin θu θ dr and dt v dr dt u r + r dθ dt u θ e t cos θu r ( + sin θ)e t u θ 4. r + sin t implies dr dt cos t v dr dt u r + r dθ dt u θ cos tu r + ( + sin t) dθ dt u θ To find dθ dt we use the given speed ν cos t cos θ dθ dt e t cos θ ( ) dθ v cos t + ( + sin t) cos t dt Therefore ( ) dθ ( + sin t) cos t dt dθ dt cos t + sin t cos t dθ + sin t dt 37

38 For this integral we use a substitution τ + sin t, then dτ cos t dt and we have θ τ dτ ln(τ) + C ln + sin t + C But θ at t, so ln + sin + C and C ln Therefore θ ln + sin t ln ln + sin t ln( + sin t) At t π then θ ln + sin(π ) ln( 3) and r + sin(π ) 3, so the position at t π/ is (r, θ) (3, ln(3/)) 6. a d r dt r ( ) [ dθ u r + r d θ dt dt + dr dt ] dθ u θ dt a. Moving around circle at (, ) with constant nonzero angular velocity dθ dt c d θ dr. Since r is constant, we have and therefore dt dt ( ) [ ] ( ) a d r dθ dt r u r + r d θ dt dt + dr dθ dθ u θ r u r dt dt dt so b. Moving around circle at (, ) with constant nonzero angular acceleration d θ dt and as in part a dr dt so a d r dt r ( ) [ dθ u r + r d θ dt dt + dr dt ] dθ u θ r dt ( ) dθ u r + r d θ dt dt u θ c c. Moving along a straight line with constant speed, therefore dr dt dθ and dt. Therefore all terms are non zero. d. Walking from the center of a merry-go-round toward its outer edge. The acceleration depends on the details (see example.9). Constant radial speed dr dt cm sec Platform rotates with uniform angular velocity of 3 rev min a. radial acceleration a r d r dt r b. Coriolis acceleration dr dt ( dθ dt dθ 4π cm sec dt π rev 3 6 sec rev sec π rad ) sec π r (the sign means toward the center) 38

39 3. Problems:, 6, 7, 8, 6, 7,, 4, 3. a. f sin x + e xy + z b. f By symmetry R x +y +z f x cos x + yexy + f y + xexy + f z + + f (cos x + ye xy )i + xe xy j + k f x (x + y + z ) / (x) x + y + z f y y (x + y + z ) 3/ f z z (x + y + z ) 3/ x (x + y + z ) 3/ x f (x + y + z ) i y 3/ (x + y + z ) j z 3/ (x + y + z ) k R 3/ R 3 c. f R i j k The gradient is f k therefore f (xi + yj + zk) k z 6. Given f(x, y, z) is the distance from the z axis then the gradient of f is a unit vector directed from the z axis except at points on the z axis where it is undefined. Hint: Look at f(x, y, z) x + y 7. a. f x + y + z then f xi + yj + zk and at the point (3,, 4) we have f 6i + 8k The maximum value is f z h e (x +y ), so f z h e (x +y ) a. f xh e (x +y ) i + 4yh e (x +y ) j + k at (,, he 9 ) lava flows steepest descent is in the direction of the gradient and f (,,he 9 ) he 9 i + 8he 9 j + k and the direction is f f he 9 i + 8he 9 j + k 4h e h e

40 The projection on the xy plane is f he 9 i + 8he 9 j b. The curve x h e y z he 9 in the direction of f. 9 8h e 9 The projection on the xy plane is x h e y or 4(x ) y or y 4x 9 8h e 9 6. x + z 8 at (,, ) 5 y 5 5 x 5 Figure 9: For Problem 6 of section 3.. y is the vertical axis a. This is a cylinder whose cross section is a circle in the xz plane. A vector normal to the cylinder is always perpendicular to the y axis. b. f x + z therefore f xi + zk at the given point we get f 4i + 4k. This has no component in the y direction and therefore it is always perpendicular to y. 7. Plane tangent to f z xy 4 at (,, 4) f (,,4) ( yi xj + zk) (,,4) i j + 8k This vector is normal to the surface and also normal to the tangent plane. Therefore the equation of the plane is (x ) (y ) + 8(z 4) 4

41 or x y + 8z 8. Let T (x, y, z) x + y + 3z and let S be the isotimic surface T, i.e. S : x + y + 3z The normal to the surface is xi + 4yj + 6zk At the point (x, y, z) we want the normal i + j + k, therefore 4y x, 6z x and so x y 3z. ) and we have x 6 6 or x ±. Now Now plug this in S: x + ( ) ( x + 3 x 3 we compute y and z, y x ± 4. 6 and z x 3 ± 3 6 f (/, 3/, 3 6/) (xi + yj + zk) (/, 3/, 3 6/) i + 3j + 3 6k g (/, 3/, 3 6/) ((x )i + yj + zk) (/, 3/, 3 6/) i + 3j + 3 6k cos θ f g f g (6) (6) (6) θ cos 3 3. x + y + z 84 closest to x + y + 4z 77 First find the tangent plane and a point (x, y, z ) so that the normal is parallel to the given plane normal, i.e. to i + j + 4k. The gradient is f (x,y,z ) x + y + z α(i + j + 4k) Now we find x α, y α, z α and since this point is on the sphere we have ( α ) + α + (α) 84 or ( )α 84 α α ±4 and the points are (, 4, 8) and (, 4, 8). Note that the second point is the farthest away from the plane on the other side of the sphere. If you need the distance from (, 4, 8) to the plane we use distance + (4) + 4(8)

42 the farthest point + ( 4) + 4( 8) The difference between the two the diameter 3 x y Figure : For Problem 3 of section 3. 4

43 3. Problems: 4. F yi + xj The equations of the curves are: dx y dy x The solution is x dx y dy, integration yields: x y + c or x + y k, these are circles centered at the origin for k > We can create the vectors F(, ) j F(, ) i F(, ) j F(, ) i F(, ) i + j F(, ) i j F(, ) i j F(, ) i j x.5 y 5. Figure : For Problem of section 3. 43

44 . F x i + y j + k dx a. x dy y dz Solve the right equality y z + c or Now solve dx x dz to get x z + k or y(z + c) x(z + k) b. At the point (,, ) we have ( + k) and ( + c), therefore k 3, c 3 and the equations are x(z 3), y(z 3) 4 y x 4 Figure : For Problem b of section 3. z axis is horizontal 44

45 3. R xi + yj + zk describe the flow lines Integrate and similarly which are lines aways from the origin. dx x dy y dz z ln x ln y + c x cy y kz 4. The flow lines of the gradient field cross isotimic surfaces ortogonally. The distance between isotimic surfaces is constant and it is the normal to surfaces which are perpendicular to the gradients. 45

46 3.3 Problems:, 3, 5, 7,. F e xy i + sin(xy)j + cos (xz)k F divf ye xy +x cos(xy)+x( cos(xz) ( sin(xz)) ye xy +x cos(xy) x cos(xz) sin(xz) 3. F (3x y 3 z) 5. (φf) φ F + F φ The left hand side is F 6xy 3 zi + 9x y zj + 3x y 3 k divf 6y 3 z + 8x yz + 6yz(y + 3x ) x (φf ) + y (φf ) + x (φf 3) φ x F + F x φ + φ y F + F y φ + φ z F 3 + F 3 z φ φ x F + φ y F + φ ( z F F 3 +φ x + F y + F ) 3 z ( φ) F ( φ) F + φ ( F) 7. Nonconstant field with zero divergence, for example F f(y, z)i + g(x, z)j + h(x, y)k will have a zero divergence.. The divergence is zero since the arrows have the same length. 46

47 3.4 Problems:,, 4, 7,,. F xy i + xyj + xyk The curl is i j k F x y z xy xy xy i(x ) j(y ) + k(y xy) xi yj + y( x)k. F e xy i + sin(xy)j + cos(yz )k The curl is i j k F x y z i( z sin(yz ) ) j( ) + k(y cos(xy) xe xy ) e xy sin(xy) cos(yz ) 4. F (x + xz )i + xyj + yzk a. The divergence is: F z sin(yz )i + (y cos(xy) xe xy )k F ( + z ) + x + y + x + y + z b. The curl is i j k F x y z x + xz xy yz i(z ) j( xz) + k(y ) zi + xzj + yk 7. The flow lines of a velocity field F are straight lines, this does not mean the curl is zero.. F y i + z j + xk a. The curl is i j k F x y y z x z i( z) j( ) + k( y) zi j yk b. Find the tangent to the curve x cos(πt), y sin(πt), z t when t x y π sin(πt), t t π cos(πt), z t t π sin(πt)i + π cos(πt)j + tk T ( π sin(πt)) + (π cos(πt)) + (t) 47

48 T π sin(πt)i + π cos(πt)j + tk π + 4t At t T() π sin(π)i + π cos(π)j + k π + 4 πj + k π + 4 The component (scalar) of ( F) parallel to T() is given by ( F) ( F) T() Therefore (since y, z at t ) ( F) π( πj + k) π j + πk π 4 + 4π π π +. Find the curl of f( R )R a. i j k (f( R )R) x y z f( R )x f( R )y f( R )z Recalling that R x + y + z we have (f( R )R) i(z f y y f f ) j(z z x x f z Let s compute these partials Similarly f x f y f z ) + k(y f x x f y ) f R R x f R (x + y + z ) / (x) f R R y f R R z Now we evaluate the components of the curl, z f y y f z f R f R (x + y + z ) / (y) f R (x + y + z ) / (z) ( zy(x + y + z ) / yz(x + y + z ) /) The same for the other components, therefore (f( R )R). b. Use geometrical interpretation. 48

49 3.5 Problems:,, 4 8. f x + y then f(, 3, 4) (3) f x y + z 4. F x yi + zj (x + y z)k a. F xy xy b. F i f xyi + x j + k f (,3,4) ()(3)i + j + k i + 4j + k i j k F x y z x y z (x + y z) y z z (x + y z) j x z x y (x + y z) + k x x y F ( )i ( + )j + ( x )k i + j x k c. ( F) ( + xy), using part a, so ( F) yi + xj y z 5. ( F) vector scalar 6. ( F) vector vector 7. R xi + yj + zk, then R and R i j k x y z x y z i 8. Find ( f) where f xyz + e xz y z y z j x z x z +k x y x y and f (yz + ze xz ) i + xzj + (xy + xe xz ) k ( f) z e xz + + x e xz ( x + z ) e xz 49

50 3.6 Problems:, 3 5, 7. f x 5 yz 3 f x 5(4)(x3 yz 3 ) x 3 yz 3 3. F 3i + j x y 3 z 4 k f y f z 3()(x5 yz) 6x 5 yz f x 3 yz 3 + 6x 5 yz x 3 yz(z + 3x ) F x ()y3 z 4 k y 3 z 4 k F y 3()x yz 4 k 6x yz 4 k F x 4(3)x y 3 z k x y 3 z k F [ ( y 3 z 4 ) + ( 6x yz 4 ) + ( x y 3 z ) ] k yz (y z + 3x z + 6x y )k 4. a. f e z sin y Yes Yes f x (ez sin y) + y (ez sin y) + z (ez sin y) e z sin y + e z sin y b. f sin x sinh y + cos x cosh z f sin x sinh y cos x cosh z + sin x sinh y cos x cosh z c. f sin(px) sinh(qy) f p sin(px) sinh(qy) + q sin(px) sinh(qy) ( p + q ) sin(px) sinh(qy) In order for this to vanish we have to have p q. 5. a. f is a vector b. F is a scalar c. F is a vector d. ( f) is a scalar ( f) 5

51 e. ( f) is a vector f. f is meaningless g. F is a vector h. } {{ F} is a vector i. f scalar j. ( f scalar vector is meaningless ) is a vector 7. f x + y and R xi + yj + zk a. f 4xi + j b. R c. f 4 i j k d. (fr) x y z x(x + y) y(x + y) z(x + y) (fr) i(z ) j(4xz ) + k(4xy x) zi 4xzj + x(4y )k 5

52 3.8 Problems:, 5, 6,,. Prove that (φ φ ) φ φ + φ φ Starting on the left hand side we have (φ φ ) [( ) φ φ + x ( ) ] φ φ i +... x Similarly for the other components. Now collect the second terms for each component to get the first term on the right and similarly for the first terms giving the second term on the right. So Prove (φf) φ F + F ( φ) Again start with the left hand side (φf) ( ) φ F + φ F x x + ( ) φ F + φ F y y + ( ) φ F 3 + φ F 3 z z ( ) ( ) ( ) φ φ φ F (φf) F + F + F 3 +φ x y z x + F y + F 3 z F φ F (φf) F ( φ) + φ( F) 5. Why (F G) G ( F) + F ( G) is not valid? The right hand side is symmetric relative to F and G but the left hand side is not since F G G F. 6. Show that A R R i j k A R A A A 3 (A z ya 3 )i (A z A 3 x)j + (A y A x)k x y z Therefore A R R Now compute the divergence of the above (A z ya 3 )i (A z A 3 x)j + (A y A x)k x + y + z A R R (x + y + z ) / (x)(a z A 3 y) x + y + z + (x + y + z ) / (y)(a z A 3 x) x + y + z + (x + y + z ) / (z)(a y A x) x + y + z 5

53 A R R x(a z A 3 y) + y(a z A 3 x) z(a y A x) (x + y + z ) 3/ You can easily check the all terms cancel and we are left with zero.. a. Evaluate ( R A ) b. Evaluate ( R A ) ( R A ) xa + ya + za 3 R A ( R A ) i j k x y z R A R A R A 3 (ya 3 za )i (xa 3 za )j+(xa ya )k Therefore ( R A ) (R A) c. R ( R A ) R (R A) (R R)A R A d. (A R) 4 4(A R) 3 A e. ( R A) A R R f. R (A RA) R (A R) A (R A)A A g. (A R) R A A R by (3.3) h. (A R) (R )A (A )R + ( R) A ( A) R A is constant 3 (A R) (A )R + 3A A i A j A 3 k + 3A A i. (R R) (x + y + z ) 6. V (x + 4y)i + (y 3z)j + Czk F Since ( F), we get x (x + 4y) + y (y 3z) + z (Cz) The left hand side is + + C, therefore C 53

54 3. Problems:, 4, 6, 8,. Use (3.5)-(3.5) to derive e z k, e ρ e ρ e z grad z grad z k k, grad ρ x grad ρ + y x i + y x + y j grad ρ e θ Now differentiate the arc sine So grad θ grad θ y x sin x + y y x +y y y sin x + y y x +y grad θ x sin xi + yj x + y, e θ since z z x x +y i + y x +y i + y sin grad θ x y x +y x + y x + y x yi + xj x + y y x +y j y x +y j xi + yj x + y xy (x + y ) / x + y y x + y x + y y (x + y ) / x + y x + y y x (x + y ) x x + y x (x + y ) + y (x + y ) x + y y e θ i + x +y x +y x j x +y yi + xj x + y xi + yj + zk 4. Use (3.6)-(3.6) to derive e r x + y + z Since r x x + y + z we find r x + y + z i+ y x + y + z j+ z x + y + z k and So e r xi + yj + zk x + y + z r x + y + z x + y + z 54

55 x Now to e θ. Recall that θ arccos x + y. θ x ( x x +y ) x + y x x x +y x + y Now to the norm of θ, x x +y y x +y y x + y x + y x (x + y ) x + y y (x + y ) x + y θ y ( x x +y x x + y x x +y x y x +y ) x + y xy (x + y ) x + y θ z θ y (x + y ) + ( x) (x + y ) x + y But x + y r sin φ cos θ + r sin φ sin θ r sin φ and so θ r sin φ Therefore e θ r sin φ θ. z Now to e φ. Recall φ arccos x + y + z. Therefore 55

56 φ x z x +y +z x x +y +z x + y + z z xz x + y (x + y + z ) φ y z x +y +z y x +y +z x + y + z z yz x + y (x + y + z ) φ z z x +y +z x +y x +y +z x + y + z z x + y + z x + y + z z x + y + z (x + y + z ) x + y x + y (x + y + z ) z x +y +z x + y x + y + z We now compute the magnitude of φ. φ x z (x + y )(x + y + z ) + y z (x + y )(x + y + z ) + x + y (x + y + z ) x z + y z + x x + y + z x + y + y x x + y + z + y + z x + y + z (x + y )z + x x + y + z x + y + y r Therefore e φ r φ. 56

57 6. f in cylindrical coordinates Let F f f e ρ + f e θ + f e z ρ ρ θ z F ρ F θ F z Now f F ρ ρ (ρf ρ) + ρ θ F θ + z F z ( ρ f ) ρ ρ ρ In spherical coordinates: F f f f e r + e φ + r r φ r sin φ F r F φ Now f θ } {{ } F θ f F r r (r F r ) + r sin φ ( r f ) + r r r r sin φ e θ θ θ F θ + r sin φ φ (sin φf φ) ( r sin φ ( r f ) f + r r r r sin φ θ + r sin φ ) f + θ + ρ f θ + f z ( sin φ r sin φ φ r ( φ sin φ f φ xi + yj 8. a. Use Problem to write F in cylindrical coordinates. x + y xi + yj From problem, we have: e ρ x + y, e yi + xj θ x + y, e z k. Therefore the vector field F becomes: e ρ ρ and the divergence is F (ρf ρ ρ ρ) and the curl is and we have F(ρ, θ, z) e ρ ρ ρ e ρ therefore F ρ ρ and F θ F z F ρ b. Again use Problem to get: e ρ e ρ ρ e θ ρ θ e z z ρ xi + yj x + y, e θ ) f φ ) yi + xj x + y, e z k. Therefore yi + xj the vector field F becomes: F e θ x + y ρ therefore F ρ, F θ ρ, F z and the divergence is (note that the only non zero component is F θ ) 57

58 and the curl is. f(r, θ, φ) cos φ r F F θ ρ θ F ρ e ρ ρ θ ρ θ e z z f f r cos φ r 3 e r + r sin φ f θ e θ + r f φ sin φ r e φ cos φ r 3 e r sin φ r e 3 φ 58

59 3. Problems: 3 5, 8, 9,, 3 3. Since x, y, z are orthogonal then any order of x, y, z will do the same for ρ, θ, z 4. One way to do this is to find the differentials for the transformation: u e x, u y, u 3 z then du du du 3 e x dx dy dz and since dv dx dy dz we have dv e x du du du 3 du du du 3. u Two other ways are possible. First to use h i u. Now u e x, u y, u 3 z i then u < e x,, >, u <,, >, and u 3 <,, >. The magnitudes of these vectors are u e x, u, u 3. The reciprocals are the scale factors. Second way is to first find the inverse transformation: and find the scale factors using Therefore and therefore x ln u, y u, z u 3 R h i u i. h (ln u i + u j + u 3 k) u u e x h (ln u i + u j + u 3 k) u h 3 (ln u i + u j + u 3 k) u 3 dv h h h 3 du du du 3 u du du du 3 e x du du du 3 5. Find the Laplacian of g u 3 + u 3 + u 3 3 where x u u, y u + u, z u 3 First we can find u i in terms of x, y, z compute g and its Laplacian in Cartesian coordinates and transform back. This will give us u (x + y), u (y x), u 3 z and g 8 (x + y)3 + 8 (y x)3 + z 3/ 59

60 g 3 4 (x + y) + 3 (y x) + 3 }{{ 4 } 4 (x + y) + 3 (y x) + 3 }{{ 4 } 4 z / g x g y g 3 (x + y) + 3 (y x) z / and in terms of u i g 3u + 3u + 3 4u 3 Another way is to use the scale factors h, h, h 3 u 3 and substitute in the equation for the Laplacian [ ( ) h h 3 f g + ( ) h h 3 f + ( )] h h f h h h 3 u h u u h u u 3 h 3 u 3 g [ ( ) u3 (3u 4u 3 u ) + ( u3 (3u u ) ) + ( ) ] (3u u 3 u 3) 3 8. x u u, y u u, z u 3 g 4u 3 [u u 3 + u u 3 + 3] 3u + 3u + 3 4u 3 a. e R (u u )i + u u j + u 3 k R u R u u i + u j 4u + 4u u i + u j u + u e R u R u u i + u j 4u + 4u e 3 R u 3 R u 3 k k u i + u j u + u Orthogonality: e e u u + u u and similarly for the rest e u + u e 3 and e e 3. The system is right handed because e e e 3 and so on. b. scale factors can be read from above h h u + u and h 3 [ c. g g + g + ( 4(u 4(u + u ) u u u + u )g ) ] 3 [ ] g g + g + g 4(u + u ) u u u 3 6

61 d. F u 3 e + u e + u e 3 then F 4(u + u ) u [ F u 4(u + u 3 ) F 4(u + u ) ( ) u + u u 3 + ( ) u + u u u (u + u ) / (u ) + u u + u e u + ( ) 4(u u + u )u } 3 {{} ] (u + u ) / (u ) u (u 3 + u ) (u + u ) 3/ u + u e e 3 u u 3 u 3 u + u u u + u u [ F u 4(u + u + u e ( ) u + u e ( u + u ) ) + e 3 u + u u u + u u 3 u + u u + u e + e + e 3 + u u u 3 u + u u + u (u + u ) 3/ 9. x u 3, y e u cos u, z e u sin u R a. sin u (e u )j + cos u (e u )k u R cos u (e u )j + sin u (e u )k u R i u 3 Since R i and the other have no component in the direction of i we get the u 3 orthogonality to R. Now to the orthogonality of R and R u 3 u u R R sin u cos u e u + sin u cos u e u u u b. h R sin u e u u + cos u e u e u, similarly h e u and h 3 c. (u + u + u 3) e u ( ) g + ( ) g u u u u 6 + u 3 ( ) e u g u 3

62 [ ] (u + u + u 3) e u + + e u 4e u + d. ( e u e 3 + u 3 e ) e u F e u. x (u v ), y uv, z z (u 3 e u )) + + ( e u e u )) u } u {{}} 3 {{} e u e e u e e 3 u u u 3 e u u 3 e u F e u [eu e ( e u ) e u e ( e u ) + e 3 (e u u 3 )] F [ ] e u e u e + e u e + e u u 3 e 3 F e u [ eu e + e u e + u 3 e 3 ] where < u <, v, < z < This is parabolic cylindrical coordinate system h u R u ui + vj u + v h v R v vi + uj v + u h z R k z 3. For the previous problem: dv h h h 3 du du du 3 (u + v )du dv dz 6

63 4. Problems:, 3, 6,, 4, 8. F x i + j + yzk and the curve C is x t, y t, z 3t, t Therefore dx dt, dy 4tdt, dz 3dt and the integral is C (x dx+dy+yzdz) 3. Integrate C [t +4t+t (3t)(3)]dt C (x dx + dy + yzdz) (xi yj + zk) dr from (,, ) to (,, 4) a. along line joining points x, y, z 4t, t Therefore dx, dy, dz 4dt and the integral is C (xi yj + zk) dr (t +4t+8t 3 )dt ( ) 3 t3 +t +8 4 t4 ( ) (4t)(4)dt 6 t 8 b. along x cos(πt), y sin(πt), z 4t, t Therefore dx π sin(πt)dt, dy π cos(πt)dt, z 4dt, C (xi yj + zk) dr [(cos(πt)( π sin(πt)) sin(πt)(π cos(πt)) + 4t(4)] dt The first two terms add up and we have (xi yj + zk) dr 4π sin(πt) cos(πt) +6t C dt sin (πt) + 8t substitute usin(πt) 6. F yi + xj + xyz k and the curve C is x x + y, z Note that we have a circle centered at (, ) with radius 3 in a plane parallel to xy coordinate plane. Let s parametrize the circle x 3 cos θ +, y 3 sin θ, z, θ π so dx 3 sin θdθ, dy 3 cos θdθ, dz The integral is π [ ] π 3 sin θ( 3 sin θ) + ( 3 cos θ + )( 3 cos θ) + dθ 3 sin θ + 3 cos θ + 3 cos θ) dθ 3 cos(θ) 63

64 . Evaluate the ellipse C 3 sin(θ) + 3 sin θ π [ (y + yz cos(xyz))dx + (x + xz cos(xyz))dy + (z + xy cos(xyz))dz ] along x cos θ, y 3 sin θ, z, θ π The last term vanishes since z. Now dx sin θdθ, dy 3 cos θdθ, dz. Substituting in the integral we have π 3 sin θ + 3 sin θ()(cos(6 } sin{{ θ cos θ} ) ( sin θ)dθ 3 sin θ π + 4 cos θ + cos θ(cos(6 } sin{{ θ cos θ} ) (3 cos θ)dθ 3 sin θ π ( 6 sin θ cos(3 sin(θ)) ) π ( dθ + 6 cos θ cos(3 sin(θ) ) dθ π ( + 6 sin θ + cos 3 θ ) dθ π (6 cos(θ) cos(3 sin(θ))) dθ + 6π Substitute u3 sin(θ) ] π [ 3θ + 3 ( ) sin(θ) π + 6 sin θ ( cos(θ))/ + [ sin θ 4 sin 3 θ ] π dθ + π ( sin θ) cos θdθ Substitute usin θ 4. F ω R where ω is constant. Therefore F is perpendicular to R and so F dr and F dr C 8. a. F x i + yj + k Find the flow lines through (,, ) y dx dy dz F F F 3 y dx x dy y solve this first dz x y C at (,, ) we have C, so Now solve the other one y x dy y dz 64

65 at (,, ) we have K, so Let s parametrize it ln y z + K ln y z x t, y t, z ln t b. At (e, e, ) then t e and x y e, z ln e which is on flow line. ( ) (e,e,) x ( ) (e,e,) x c. (,,) y i + yj + k dr dx + ydy + dz (,,) y The flow line is x t, y t, z ln t and so dx dt, dy dt, dz dt and the integral t becomes ( e t t + t + ) ( dt t t + ) e t + ln t e + e + e 65

66 4.3 Problems:, 4 7. a. F yi + xj φ x y φ y x φ z Solving the first equation we have φ yx+p(y, z). Substitute this φ in the second equation p(y, z) we have x + x which is not possible since p(y, z) is not a function of x and y therefore the field is not conservative. b. F yi + y(x )j φ x y φ y y(x ) φ z Solving the first equation we have φ yx + p(y, z). Substitute this φ in the second equation p(y, z) we have x + y(x ) which is not possible since p(y, z) is not a function of x and y therefore the field is not conservative. c. F yi + xj + x k φ x y φ y x φ x z Solving the first equation we have φ yx + p(y, z). Substitute this φ in the second equation p(y, z) we have x + x which yields p(y, z) g(z) and φ yx + g(z). Now substitute in y the third equation above x g (z) and therefore the field is not conservative. 66

67 d. F zi + zj + (y )k φ x z φ y z φ y z Solving the first equation we have φ zx + p(y, z). Substitute this φ in the second equation p(y, z) we have x + z which yields p(y, z) yz + g(z) and φ yz + zx + g(z). Now y substitute in the third equation above y y + x + g (z) and therefore the field is not conservative. x e. F x + y i + x x + y j φ x x x + y φ y x x + y Solving the second equation we have φ tan y x y x equation we have + p (x) + y x is not conservative. 4. Evaluate F dr where F C x x + y which yields p (x) x+y x +y + p(x). Substitute this φ in the first and therefore the field yi + xj x + y and the curve C is given by x + y r Using polar coordinates x r cos θ, y r sin θ, so dx r sin θdθ and dy r cos θdθ and the integral is C F dr π [ r sin θ ( ) y 5. tan is a multiple valued function. x r ( r sin θdθ) + r cos θ ] π (r cos θdθ) r dθ π 67

68 6. F (y + z cos(xz))i + xj + x cos(xz)k φ x y + z cos(xz) φ y φ z x x cos(xz) Solving the second equation we have φ xy + p(x, z). Substitute this φ in the first p(x, z) p(x, z) equation we have y + z cos(xz) y + which yields z cos(xz) and x x p(x, z) sin(xz) + g(z) and φ xy + sin(xz) + g(z). Now substitute in the third equation above x cos(xz) + x cos(xz) + g (z) and therefore g(z) is a constant and the field is conservative with a potential φ xy + sin(xz). 7. F xyi + (x + z)j + yk φ x xy φ y x + z φ z y Solving the first equation we have φ x y + p(y, z). Substitute this φ in the second equation we have x + z x p(y, z) p(y, z) + which yields z and p(y, z) yz + g(z) and y y φ x y + yz + g(z). Now substitute in the third equation above y + y + g (z) and therefore g(z) is a constant and the field is conservative with a potential φ x y + yz. 68

69 4.4 Problems:,, 6, 7, 9,. a. F (xy + yz)i + (6x + xz)j + xyk. Is F? i j k F i(x x) j(y y) + k(x + z (x + z)) x y z xy + yz 6x + xz xy YES b. F ze xz i + xe xz k. Is F? i j k F x y z ze xz xe xz i( ) j(xze xz + e xz (xze xz + e xz )) + k( ) YES c. F sin xi + y j + e z k. Is F? i j k F x y z sin x y e z i( ) j( ) + k( ) YES d. F 3x yz i + x 3 z j + x 3 yzk. Is F? i j k F i(x 3 z x 3 z) j(3x yz 6x yz) + k(3x z 3x z ) x y z 3x yz x 3 z x 3 yz 69

70 NO e. F x x + y i + y x + y j + zk. Is F? This test is not applicable because of the singularity at x y but we can try to find the potential. φ x x x + y φ y y x + y Solving the first equation we have φ ln φ z z ( ) + x y Substitute this φ in the second equation we have which yields or and p(y, z) ln y + g(z) and + p(y, z). y x + y ( x y 3 p(y, z) ) + + x y y p(y, z) y φ ln p(y, z) y y x + y + x y(x + y ) ( ) + x y y + x y(x + y ) y + ln y + g(z). Now substitute in the third equation above z ( + + ) g (z) and therefore g(z) z and the field is conservative with a potential φ ln + x + ln y + z. There is a problem when y but not when x. Given F φ and G ψ, is F + G conservative? The field F + G is conservative with a potential φ + ψ y 7

71 6. F ( 6x e x y ) i ye x j + cos zk To check if the field is conservative, φ x 6x ex y Solving the first equation we have φ y φ z ye x cos z φ 3x y e x + p(y, z). Substitute this φ in the second equation we have which yields and p(y, z) g(z) and ye x ye x + p(y, z) y p(y, z) y φ 3x y e x + g(z). Now substitute in the third equation above cos z + + g (z) and therefore g(z) sin z and the field is conservative with a potential φ 3x y e x + sin z. b. R t i + (t )(t ) j + π t3 k, t x y z Since F is conservative, the integral F dr can be found by evaluating the potential C at the end points. At t, we have x, y, z π. At t, we have x, y ( )( ), z. The values at t is φ(,, π ) 4 and at t φ(,, ) 4, therefore the integral is 4 ( 4) 8 c. R (t ) i + (t)(3 t) j + π (t ) k, t 3 } {{} 4 y x z Since F is conservative, the integral F dr can be found by evaluating the potential C at the end points. At t, we have x, y, z. At t 3, we have x, y, z π. The values at t 3 is φ(,, π ) 4 and at t φ(,, ) 4, therefore the integral is 4 ( 4) 8 7

72 7. F ( + x)e x+y i + ( xe x+y + y ) j zk To check if the field is conservative, φ x φ y ( + x)ex+y xe x+y + y Solving the first equation we have φ e x+y + e y xe x dx } {{ } (x )e x φ z z Substitute this φ in the second equation we have which yields and p(y, z) y + g(z) and e x+y + (x )e x+y + p(y, z) xe x+y + p(y, z). xe x+y + y xe x+y + p(y, z) y y p(y, z) y φ xe x+y + y + g(z). Now substitute in the third equation above z + + g (z) and therefore g(z) z and the field is conservative with a potential φ xe x+y + y z. b. R ( t)e t i + x t y j + t z k, t and G ( + x)e x+y i + ( xe x+y + z ) j yk Note that G is not conservative, and it is messy to try and do it using brute force. But we can write G F+(z y)j+(z y)k. Since F is conservative, the integral F dr can be found by evaluating the potential at the end points. At t, we have x, y, z. At t, we have x, y, z. The values at t is φ(,, ) ( 4) 3 and at t φ(,, ) e, therefore the integral is 3 e. Now to the integral over the rest C Therefore [(z y)dy + (z y)dz] C [(t t)dt + (t t)(dt)] G dr 3 e + 3 e 7 C 6tdt 3t 3

73 9. F (xyz + z y + )i + (x z 4xy)j + (x y + xz )k To check if the field is conservative, φ x (xyz + z y + ) φ y φ z Solving the first equation we have (x z 4xy) (x y + xz ) φ x yz + (z y + )x + p(y, z). Substitute this φ in the second equation we have which yields and p(y, z) g(z) and x z 4xy x z 4xy + p(y, z) y p(y, z) y φ x yz + (z y + )x + g(z). Now substitute in the third equation above x y + xz x y + zx + g (z) and therefore g(z) z and the field is conservative with a potential φ x yz + (z y + )x z. x b. G (x + z ) i + z (x + z ) k To check if this field is conservative we should not use the fact that G. The reason is that this is true except on the y axis (since then x + z ) and that is not a star shaped domain.. F (5x 4 3x y )i x 3 yj To check if the field is conservative, φ x (5x4 3x y ) Solving the first equation we have φ y x 3 y φ 3x 5 x 3 y + p(y). Substitute this φ in the second equation we have x 3 y x 3 y + p (y) 73

74 which yields p (y) and p(y) c and φ 3x 5 x 3 y. Therefore the field is conservative. To evaluate (,) (,) F dr φ(, ) φ(, ) (3 4) 74

75 4.6 Problems:, 3 6. The surface S is given by x u, y uv, z v ds R u R v dudv R u R v i j k u v u v i R u ui + vj R v uj + vk v u j u v u k v i 4uvj+ u k v ds ds ( v i 4uvj + u k ) dudv 4()v 4 + 6u v + 4()u 4 dudv 8v 4 + 6u v + 8u 4 dudv 8(u + v ) dudv 3. x a cos u, y a sin u, z v R u R v ds (u + v ) dudv R u ds R u R v dudv a sin ui + a cos uj R v k i j k a sin u a cos u 4. z x + y so R xi + yj + (x + y )k a cos ui + a sin uj ds (a cos ui + a sin uj) dudv ds a cos u + a sin u dudv a dudv ds R x R y dxdy R x i + xk 75