MTH4101: Calculus II [SOLUTIONS]

Transcription

1 MTH40: Calculus II [SOLUTIONS] Duration: hours Date and time: 7th May 06, 4:30 6:30 Apart from this page, you are not permitted to read the contents of this question paper until instructed to do so by an invigilator. You should attempt ALL questions. Marks awarded are shown next to the questions. Calculators are not permitted in this examination. The unauthorised use of a calculator constitutes an examination offence. Complete all rough workings in the answer book and cross through any work that is not to be assessed. Possession of unauthorised material at any time when under examination conditions is an assessment offence and can lead to expulsion from QMUL. Check now to ensure you do not have any notes, mobile phones, smartwatches or unauthorised electronic devices on your person. If you do, raise your hand and give them to an invigilator immediately. It is also an offence to have any writing of any kind on your person, including on your body. If you are found to have hidden unauthorised material elsewhere, including toilets and cloakrooms it shall be treated as being found in your possession. Unauthorised material found on your mobile phone or other electronic device will be considered the same as being in possession of paper notes. A mobile phone that causes a disruption in the exam is also an assessment offence. Exam papers must not be removed from the examination room. Examiner(s): R. J. Harris, J. A. Valiente Kroon [Note: Some questions are modified versions of exercises found in the compulsory module textbook, Thomas Calculus (Twelfth Edition), as also used for online coursework via the MyMathLab system.] Turn Over

2 Page MTH40 (06) Question. (a) [Similar to example in lectures] We have cosn for alln > 0 so n cosn n for all n > 0. n Nowlim n /n lim n /n 0 and hence, by the Sandwich Theorem for sequences, cosn lim n n 0. (b) [Very easy, similar to coursework] Recognizing this as a geometric series, we have ( ) n 3 ( ) n n 3 n (c) [Similar to coursework] We have and so n ( /3) 4/3 3. f(x) sinx, f (x) cosx, f (x) sinx, f ), f ), f. 4 4 Hence the first three Taylor polynomials aboutx π/4 are obtained: P 0 (x) f, )( P (x) f +f x π ) ( x π ), 4 )( ) P (x) f +f 4 + ( x π 4 x π 4 ) 3 )( x π ) 4 + f 4 ( x π.

3 MTH40 (06) Page 3 (d) [Inverted version of coursework question] We have lim (x,y) (,0) x y 4 x y 4 ( x y +)( x y ) lim 3 x y (x,y) (,0) x y x y 4 x y + lim (x,y) (,0) x y (e) [Very easy, similar to coursework] By straightforward differentiation we have g x 3x y e x3, g y, yex3 g x 6xy e x3 +9x 4 y e x3, g y, ex3 g x y 6x ye x3 g y x. (f) [Similar to coursework but requiring understanding not just calculation of gradient] We have so f x e y, f y xe y, and f z z, f (,ln,/) f x (,ln,/)i+f y (,ln,/)j+f z (,ln,/)k i+j+k. Hence f(x, y, z) increases most rapidly in the direction u i+j+k i+ 3 j+ 3 k. Turn Over

4 Page 4 MTH40 (06) (g) [Similar to coursework] Integrating in the given order yields (x+y +z)dydxdz 0 0 ] y4 [xy + y +zy dxdz y0 (4x+8+4z)dxdz [ x +8x+4zx ] x x0 dz (8+6+8z)dz [ 4z +4z ] 48. (h) [Similar to example in lectures] By separation of variables and integration, we obtain +y dy e 3x dx so ln(+y) 3 e3x +C, 3 wherec is a constant of integration, and so ( ) y exp 3 e3x +C.

5 MTH40 (06) Page 5 Question. (a) [Bookwork] n a n diverges iflim n a n fails to exist or is different from zero. 4 [Give for contrapositive form: If n a n converges, then a n 0.] (b) [Unseen in precisely this form] Fora n ln(n+) n, we have lim a ln(n+) n lim n n n /(n+) lim n /( n) n lim n n+ / n lim n +/n 0. [using l Hôpital] Hence, in this case, thenth-term Test is inconclusive we cannot conclude from it whether the series converges or diverges. Turn Over

6 Page 6 MTH40 (06) Question 3. (a) [Similar to coursework] By the Chain Rule, z u z x x u + z y y u ( ) (3e y )()+(3xe y ) u and 3uv +3(u+v)uv u 6uv +3v 3. z v z x x v + z y y v ( ) (3e y )()+(3xe y ) v 3uv +3(u+v)uv v 9uv +6u v. [Give up to 4 for correct answers via direct method rather than Chain Rule.] (b) [Similar to coursework] The same answers would be recovered by expressingz in terms ofuand v directly and taking the corresponding partial derivatives.

7 MTH40 (06) Page 7 Question 4. [Similar to coursework and example in lectures] We want to find the locations of the extreme values off(x,y) 3x y+5 subject to the constraint g(x,y) x +y 4 0. Now so the condition is and hence we have f f x i+f y j 3i j, g g x i+g y j xi+yj, f λ g 3i j λxi+λyj 3 λx λ 3 x and λy y λ x 3. Substituting into the constraint equation g(x, y) 0 yields or x + ( x 3) 4 0 x 9 4 x Hence the extreme values are at the points(6/ 0, / 0) and ( 6/ 0,/ 0). Turn Over

8 Page 8 MTH40 (06) Question 5. (a) [Similar to coursework] From the second equation we have3v 3x+3y and adding this to the first one yields and hence The Jacobian is x u+3v x u 3v, (b) [Similar to coursework but hard] y v +x u v. (x, y) (u,v) x x u v y y u v 3 We have (x y) dxdy R. G G ( v) (x,y) (u, v) dudv v dudv. The regionghas boundaries obtained by transforming the boundaries ofr: x 6 u+3v 6, x 0 u+3v 0, y x v 0, y x+ v. HenceGis the shaded area in the sketch below. v v u+3v 6 ( 6, ) u+3v 0 (0,) (0,0) (6,0) u End of Paper.