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. - - 5
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
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 + 8 + λ λ 8 5 ( ) ( ) 9 nd M d min 5 + 5 8 5..5 (, ) 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.5 - - 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. (.).. 8 5 5 b ( cos θ) dθ / The distnce decreses b.. 9 ( ) b. 7
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.5.5.5.5 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.75.5.5 /( + + ) 8 9
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.
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 + +.