(b) Let S 1 : f(x, y, z) = (x a) 2 + (y b) 2 + (z c) 2 = 1, this is a level set in 3D, hence

Size: px

Start display at page:

Download "(b) Let S 1 : f(x, y, z) = (x a) 2 + (y b) 2 + (z c) 2 = 1, this is a level set in 3D, hence"

Priscilla Fitzgerald
5 years ago
Views:

1 Problem ( points) Find the vector eqution of the line tht joins points on the two lines L : r ( + t) i t j ( + t) k L : r t i + (t ) j ( + t) k nd is perpendiculr to both those lines. Find the set of ll points (, b, c) in -spce for which the two spheres ( ) + ( b) + ( c) nd + + intersect orthogonlit. (int: Their tngent plnes should be perpendiculr t ech point of intersection.) Let P nd Q be points on L nd L, then the vector PQ L [( + t) i t j ( + t) k] [s i + (s ) j ( + s) k] L ( + t s) i (t + s ) j ( + t s) k. P Q Problem ( points) Let : f(,, ) ( ) + ( b) + ( c), this is level set in, hence n f (( ), ( b), ( c)) is norml vector to, similrl, let be g(,, ) + +, this is level set in, hence n g (,, ). ince n n, hence n n, we hve ( ) + ( b) + ( c) + b + c. () Also, for two spheres to intersect, we hve to solve + + b + b + c + c () nd + + () so () (), we hve + b + b c + c. is perpendiculr to both lines if nd ( i j k) ( i + j k) From (), we hve + b + c. The set of points re sphere with center t the origin nd rdius. tht is if + t s + t + s + + t s t 7s + t s t s t 9s 7t s 7. This sstem hs solution t, s. We would epect to use s vector perpendiculr to both lines, but, s it hppens, if t s, becuse the two given lines intersection t (,, ). A nonero vector perpendiculr to both lines is i j k 5 i j + k. Thus the required line is r (,, ) + t(5,, ) -5 (5t + ) i (t + ) j + (t ) k

2 Problem ( points) Problem ( points) (i) If g(t) f(, ) with (t), (t), using the chin rule to find d g dt. Let f(, ) + if (, ) (, ) if (, ) (, ) (i) Clculte f (, ) nd f (, ) t ll points (, ) (include the point (, )) in the -plne. (ii) Compute f (, ) nd f (, ). Are the two mied prtils f nd f equl t the point (, ) (wh)? ( + ) ( + ) ( ) ( + ) f(h, ) f(, ) lim (,) h h f(, k) f(, ) lim. k (,) h g (t) d dt + d dt (Chin rule) g (t) d ( ) d + d ( ) d dt dt dt dt d ( ) d dt dt + d dt + d ( ) d dt dt + d dt (Product rule) [ ( ) d dt + ( ) d ] d dt dt + f d dt [ ( ) + d dt + ( ) d ] d dt dt + f d dt (Chin rule) ( ) ( ) d d d d f + f + f dt dt dt dt + f d dt + f d dt (i) Note lso tht for k k5 (,k) k k h (h,) h h (,k) lim (,) k k (h,) lim (,) h h ence (,). (,) (,) (,). The re not equl becuse Note f nd f is not continuous t (, ). ( + )[ + 5 ] ( + 5 )[( + ) ] ( + ) h (h ) (h,) h 8 k ( 5k ) + k 5 k (,k) k 8 lim h lim k (h,) (,k) hence f is not continuous t (, ), similrl for. 5

3 Problem ( points) Problem ( points) Use Lgrnge multipliers to find the point (, ) on the line + tht is closest to the point (, ). If the line chnges to +.9, estimte how this chnges will ffect the distnce between (, ) nd (, ). This problem is equivlent to minimie d f(, ) ( ) + ( ) subject to. Let g(, ). To find the criticl point, we hve f λ g ( ) λ( ) λ 5, 9 5 Therefore d 9/sqrt5. ( ) λ λ + λ λ λ 8 5 ( ) ( ) 9 nd M d min (, ) d (,) Find the re of the region inside r sin θ nd outside r. Find the volume of the solid tht is inside of the ellipsoid + + b bove the -plne, nd inside of the clinder +. r sin θ r r sin θ + i.e., + ( ) nd r is circle centre t (, ). These two circles intersect when A sin θ, i.e. θ or 5 sin θ r drdθ ( sin θ ) dθ +. r sin θ θ r Note tht, if g(, ) c, then Note tht M d, therefore dm dc λ M λ c. ( + + ) ( ) In polr coord. r r sin θ r sin θ V R da, where b In this cse c., d d λ c d d λ c b sin θ b ( r ) / r dr dθ ( r ) / sin θ dθ / R d 5 8. (.) b ( cos θ) dθ / The distnce decreses b.. 9 ( ) b. 7

4 Problem 5 ( points) Problem ( points) Epress the integrl f(,, ) d d d s five other different itertions of the triple integrl of f(,, ). From to (surfces in ) From to (curves on the -plne) From to (points on the -is) The curve C is the intersection between the surfces nd. II I From these, we let t, t, then t. Therefore, the prmetric form of C is r(t) t i + t j + t k, t (since ). ence the integrl J cn be re-written s / () J f(,, ) d d d. / The projection of the curve C onto the plne should be independent of, i.e.,. / () J f(,, ) d d d / () f(,, ) d d d. Find the volume of the solid in the first octnt bounded b the coordinte plnes, nd the surfces + +, Evlute the integrl ( + ) nd ( + ). ( + + ) dd. Note tht + + is sphere, in sphericl coord. ρ ( + ) is cone, in sphericl coord. φ ( + ) is cone, in sphericl coord. φ V ρ sin φ dρ dφ dθ 8 sin φ dφ dθ ( ) dθ ( ). Note tht the region of integrtion is the entire first qudrnt of the -plne, therefore in polr coord. - - The projection of the curve C onto the plne should be independent of, i.e.,. () J I + II / f(,, ) d d d + f(,, ) d d d / II (5) f(,, ) d d d + f(,, ) d d d. / I R I lim R lim R lim R. ( + r r dr dθ ) ( + r ) R + R R /( + + ) 8 9

5 Problem 7 ( points) Let G be the region bounded inside b the sphere + + nd outside b the clinder +. The surfce of G consists of clindricl prt, nd sphericl prt,. Evlute the flu of F ( + ) i + ( ) j + ( e sin ) k Problem 7 ( points) (c) The flu of F out of G through the sphericl prt is F n d F n d F n d +. out of G through the whole surfce, the surfce, nd (c) the surfce. (d) Find the prmetric representtion of the surfce (r(θ, φ)), hence find the surfce re of b using the method of surfce integrl. n (d) Using the method of surfce integrl: the surfce is given b r(φ, θ) sin φ cos θ i + sin φ sin θ j + cos φ k, where φ [/, 5/], θ [, ]. Note tht r φ r θ sin φ cos θ i + sin φ sin θ j + sin φ cos φ k. r φ r θ sin φ d r θ rφ da θφ Aθφ F ( + ) i + ( ) j + ( e sin ) k divf + +. n 5/ / 8 [ cos φ] 5/ / 8. sin φ dφdθ The flu of F out of G through is F n d F dv G r. r r dr r ddrdθ On, n i + j, d dθd. The flu of F out of G through is F n d + dθ d ddθ.

6 Problem 8 ( points) A vector field is defined b F (,, ). Compute B F nd, from this nswer, compute B. Clculte eplicitl F dr, B n d, B n d C where the curve C is the circle of unit rdius in the -plne with centre the origin, the surfce is the disc of unit rdius in the -plne with centre the origin, nd the surfce is the hemisphere of unit rdius + + with. Eplin how our results illustrte (i) Green s Theorem; (ii) tokes Theorem; (iii) the ivergence Theorem. Problem 8 ( points) Therefore the flu cross is B n d. (,, ) (,, ) d ( + + ) da ( + + ) da ( + ) da since 8. r r dr dθ F (,, ) i j k B F (,, ) Prmetrie the closed curve C s so B () + () + ( ) +. C : r(t) (cos θ, sin θ, ), θ. C F dr B n d ( sin θ, cos θ, ) ( sin θ, cos θ, ) dθ. (,, ) (,, )r drdθ. (iii) Therefore, C F dr ( F) n d ( F) n d B n d B n d B n d B n d B n d (Green s Th) (i) B n d (tokes Th) (ii) ( B) dv illustrtes the ivergence Theorem; is closed hemisphere nd B. For the surfce : nd the norml vector to is n (,, ) / + + (,, ) nd, so + +.

Multiple Integrals. Review of Single Integrals. Planar Area. Volume of Solid of Revolution

Multiple Integrals. Review of Single Integrals. Planar Area. Volume of Solid of Revolution Multiple Integrls eview of Single Integrls eding Trim 7.1 eview Appliction of Integrls: Are 7. eview Appliction of Integrls: Volumes 7.3 eview Appliction of Integrls: Lengths of Curves Assignment web pge