Module MA1132 (Frolov), Advanced Calculus Tutorial Sheet 1. To be solved during the tutorial session Thursday/Friday, 21/22 January 2016

Transcription

1 Module MA113 (Frolov), Advanced Calculus Tuorial Shee 1 To be solved during he uorial session Thursday/Friday, 1/ January 16 A curve C in he xy-plane is represened by he equaion Ax + Bxy + Cy + Dx + Ey + F =. (1) In he x y -plane obained by roaing he xy-plane hrough an angle φ x = x cos φ + y sin φ, y = x sin φ + y cos φ, () he curve C is represened by a similar equaion where If he angle φ saisfies A x + B x y + C y + D x + E y + F =, (3) A = A cos (φ) + B sin(φ) cos(φ) + C sin (φ), B = A sin(φ) + B cos(φ) + C sin(φ), C = A sin (φ) B sin(φ) cos(φ) + C cos (φ), D = D cos(φ) + E sin(φ), E = E cos(φ) D sin(φ), F = F. co φ = A C B, () hen he curve C is represened by he equaion A x + C y + D x + E y + F =, B =. (6) () 1. Use Mahemaica, and he resul of he previous quesion o idenify he curve. Find a parameric represenaion and plo he curve in he xy-plane. The Mahemaica funcion ParamericPlo can be used o plo parameric curves in he xy-plane. (a) x xy + y x y =. Soluion: One finds ha φ = π, and in erms of x, y variables he equaion akes he form y x =. (7) Thus, he curve is a parabola, and is parameric represenaion is y =, x = / or in erms of x, y The parabola is shown below x = +, y = +. (8) 1

2 y x (b) 31x 1 3xy 3x + 1y 3 3y 8 =. Soluion: One finds ha φ = π 6, and in erms of x, y variables he equaion akes he form x + (y ) 9 Thus, he curve is an ellipse, and is parameric represenaion is or in erms of x, y x = 3 sin() The ellipse is shown below = 1. (9) x = cos, y = + 3 sin. (1) + 3 cos() + 1, y = 1 3(3 sin() + ) cos(). (11) y x -1

3 (c) 3x 7y xy 1 x + 7 y + 9 =. Soluion: One finds ha co(φ) = 3, and in erms of x, y variables he equaion akes he form y 9 (x ) = 1. (1) Thus, he curve is a hyperbola, and is parameric represenaion (one branch of i) is y = 3 cosh, x = + sinh. (13) or in erms of x, y x = sinh() + 3 cosh() + 8, y = The hyperbola is shown below 6 cosh() (sinh() + ). (1) y x A curve C is he inersecion of he cone wih a plane. z = x + y, (1) Idenify he curve, find a parameric represenaion and plo he curve in he xyz-space for he planes below. The Mahemaica funcion ParamericPlo3D can be used o plo parameric curves in he xyz-space. (a) z =.. Soluion: I is a circle of radius.: x =. cos, y =. sin, z =. 3. Consider he vecor-valued funcion (wih values in R 3 ) r() = ln( ) i j + k (16) 3

4 (a) Find he domain D(r) of he vecor-valued funcion r(). Soluion: The domain D(r) of r() is he inersecion of domains of is componen funcions. Since D(ln( )) = (, ), D() = (, ) and D( ) = (, ), one ges ha is he vecor funcion r() is defined for <. (b) Find he derivaive dr/d. Soluion: (c) Find he norm dr/d. Simplify he expressions obained. Soluion: The magniude or norm of his vecor is dr d = (1 ) + ( 1) + because <. ( ) = (1 D(r) = (, ) (17) dr d = (1, 1, ). (18) ) (d) Find he uni angen vecor T for all values of in D(r). Soluion: The uni angen vecor is T = dr ( d dr = ) +, +, + d ( ) (1 = + ) = 1, (19) (e) Find he vecor equaion of he line angen o he graph of r() a he poin P (, 1, 1 ) on he curve. Soluion: The poin P (, 1, 1 ) on he curve corresponds o = 1. We find () r = r( 1) = j + 1 k, v = dr d (1) = i j 1 k. (1)

5 Thus he angen line equaion is r = r + ( + 1) v = ( + 1) i j 1 ( + 1 ) k. () Noe ha he same line is also described by he following equaion which is obained from he one above by he shif of he parameer : 1 r = r + v = i ( 1) j 1 ( 1 ) k. (3)

6 MA113, 16, Tuorial Se 1 FS = FullSimplify; Assumpions$ = { Reals} { Reals} Problem 1 of TS1 Clear[Ap, Bp, Cp, Dp, Ep, Fp] Ap := Ac Cos[p] + Bc Cos[p] Sin[p] + Cc Sin[p] Bp := Bc Cos[ p] - Ac Sin[ p] + Cc Sin[ p] Cp := Cc Cos[p] - Bc Cos[p] Sin[p] + Ac Sin[p] Dp := Dc Cos[p] + Ec Sin[p] Ep := Ec Cos[p] - Dc Sin[p] Fp := Fc Par (a) Clear[x, y, xp, yp, f, ff, p] f = x^ - x y + y^ - ^(1 / ) x - ^(1 / ) y - x + x - y - x y + y Ac = 1 / D[f, {x, }] 1 Bc = D[D[f, x, y]] - Cc = 1 / D[f, {y, }] 1 Dc = D[f, x] /. {x - >, y } - Ec = D[f, y] /. {x - >, y } - Fc = f /. {x - >, y } Ac - Cc Bc

7 Tuorial1.nb Ac - Cc p = ArcCo Bc π Bp ff = Collec[Ap xp^ + Bp xp yp + Cp yp^ + Dp xp + Ep yp + Fp, {xp^, xp yp, yp^, xp, yp}, FS] - 8 xp + yp Clear[xp, yp] sub = Solve[ff, xp][[1]] xp yp yp = xp = xp /. sub x = xp Cos[p] - yp Sin[p] - + y = yp Cos[p] + xp Sin[p] +

8 Tuorial1.nb 3 ParamericPlo[{x, y}, {, - 1, 1}, AxesLabel {"x", "y"}] y x Par (b) Clear[x, y, xp, yp, f, ff, p] f = 31 x^ - 1 3^(1 / ) x y + 1 y^ - 3 x - 3 3^(1 / ) y x + 31 x y x y + 1 y Ac = 1 / D[f, {x, }] 31 Bc = D[D[f, x, y]] Cc = 1 / D[f, {y, }] 1 Dc = D[f, x] /. {x - >, y } - 3 Ec = D[f, y] /. {x - >, y } Fc = f /. {x - >, y } - 8

9 Tuorial1.nb Ac - Cc Bc Ac - Cc p = ArcCo Bc - π 6 Bp ff = Collec[Ap xp^ + Bp xp yp + Cp yp^ + Dp xp + Ep yp + Fp, {xp^, xp yp, yp^, xp, yp}, FS] xp - 6 yp + 16 yp ff xp + 16 (yp - ) / / FS xxp + 16 (yyp - ) 1 / / Expand xxp - yyp 9 + yyp 9 xp = (1 / 36)^(1 / ) Cos[] Cos[] yp = + (1 / 16)^(1 / ) Sin[] + 3 Sin[] ff / / FS x = xp Cos[p] - yp Sin[p] / / FS Cos[] + 3 Sin[] y = yp Cos[p] + xp Sin[p] / / FS - Cos[] Sin[]

10 Tuorial1.nb ParamericPlo[{x, y}, {,, Pi}, AxesLabel {"x", "y"}] y x - 1 Par (c) Clear[x, y, xp, yp, f, ff, p] f = 3 x^ - x y - 7 y^ + 7 ^(1 / ) y - 1 ^(1 / ) x x + 3 x + 7 y - x y - 7 y Ac = 1 / D[f, {x, }] 3 Bc = D[D[f, x, y]] - Cc = 1 / D[f, {y, }] - 7 Dc = D[f, x] /. {x - >, y } - 1 Ec = D[f, y] /. {x - >, y } 7 Fc = f /. {x - >, y } 9

11 6 Tuorial1.nb Ac - Cc Bc - 3 Ac - Cc p = ArcCo / / FS Bc - 1 ArcCo 3 Co[ p] - 3 Bp ff = Collec[Ap xp^ + Bp xp yp + Cp yp^ + Dp xp + Ep yp + Fp, {xp^, xp yp, yp^, xp, yp}, FS] 9-36 xp + xp - yp ff yp + (xp - ) / / FS ff (18) / / Expand - xp + xp - yp 9 ff (18) (yp) 9 (xp - ) + / / FS yp = 3 / ( + 1 / ) xp = + ( - 1 / ) ff / / FS yp = 3 Cosh[]; xp = + Sinh[]; ff / / FS x = xp Cos[p] - yp Sin[p] / / FS Cosh[] + Sinh[]

12 Tuorial1.nb 7 y = yp Cos[p] + xp Sin[p] / / FS 6 Cosh[] - + Sinh[] x1 = x /. { + I Pi} ; y1 = y /. { + I Pi} ; ParamericPlo[{{x, y}, {x1, y1}}, {, -, }, AxesLabel {"x", "y"}] y x Problem of TS1 Par (a) Clear[x, y, z, xp, yp, f, ff, p] f = x^ + y^ - z^ x + y - z z =.; x = z Cos[]; y = z Sin[]; f / / FS.

13 8 Tuorial1.nb Show[{ParamericPlo3D[{r Cos[], r Sin[], r}, {r, - 3, 3}, {, - Pi, Pi}, AxesLabel {"x", "y", "z"}, BoxRaios Auomaic, PloSyle Direcive[Opaciy[.], Speculariy[Whie, ]]], ParamericPlo3D[{x, y, z}, {,, Pi}, AxesLabel {"x", "y", "z"}, PloSyle {Blue, Thick}]}, PloRange All]