Parametric equations 8A

Transcription

1 Parameric equaions 8A a so () y () Susiue () ino (): y ( ) y 5, So he domain of f() is 6. y, So he range of f() is y 7. d so () y () Susiue () ino (): y y, 0 So he domain of f() is. 5 so 5 () y () Susiue () ino (): y (5 ) y , So he domain of f() is. y, So he range of f() is y. c so () y () Susiue () ino (): y, 0 So he domain of f() is 0. y, 0 Range of f() is y. e y, 0 So he range of f() is y 0. so () y () Susiue () ino (): y y, So he domain of f() is 0. y, So he range of f() is y. Pearson Educaion Ld 07. Copying permied for purchasing insiuion only. This maerial is no copyrigh free.

2 f so () y () Susiue () ino (): y y, So he domain of f() is y,, So he range of f() is y 0. 0 a i ln (5 ) ln (5 ) ii e 5 So 5 e Susiue 5 e ino y (5 e ) 5 5 0e e 5 0 0e e y 5: ln(5 ), When =, ln 0 and as increases ln (5 ) decreases. So he range of he parameric funcion for is 0. Hence he Caresian equaion is y 0 0e e, 0 y 5, y 5 is a quadraic funcion wih minimum value 5 a = 0. So he range of he parameric funcion for y is y 5. Hence he range of f() is y 5. i ln ( ) e e Susiue e ino y e 5 e y 5 : ln( ), When =, ln 0 and as increases, ln ( ) increases. So he range of he parameric funcion for is 0. Hence he Caresian equaion is y, 0 e Pearson Educaion Ld 07. Copying permied for purchasing insiuion only. This maerial is no copyrigh free.

3 ii c i ii y, 5 When =, y and as increases, 5 decreases owards zero. So he range of he parameric funcion for y is 0 y Hence he range of f() is 0 y e So y e (e ) (Noe ha since y is a power of here is no need o susiue for.) e, for is 0. Hence he Caresian equaion is y, 0 ye, for y is y 0. Hence he range of f() is y 0. a so Susiue y ( ) ino y ( ) :, 0 5 for is 0 5. Hence he Caresian equaion is y, 0 5 y ( ), 0 5 When = 0, y = 0; when = 5, y = 0; and y ( ) is a quadraic funcion wih maimum value 8 a So he range of he parameric funcion 8 for y is 0 y 8 Hence he range of f() is 0 y Pearson Educaion Ld 07. Copying permied for purchasing insiuion only. This maerial is no copyrigh free.

4 a i ii iii Take he posiive roo since 0. Susiue ino y : y The Caresian equaion is y, 0 is a quadraic funcion wih minimum value a = 0. for is. Hence he domain of f() is. y, 0 y wih maimum value a = 0. So he range of he parameric funcion for y is y. Hence he range of f() is y. is a quadraic funcion i Susiue ino y ( )( ) : y 6 7 The Caresian equaion is y ( )( 7) ii, When =, = ; when =, =. for is. So he domain of f() is. y ( )( ), When =, y = 0; when =, y = 8; and ( )( ) is a quadraic funcion wih minimum value a = 0.5. for y is y 8. Hence he range of f() is y 8. Noe: Due o symmery, he minimum value of y occurs midway eween he roos = and =, i.e. a = 0.5. Pearson Educaion Ld 07. Copying permied for purchasing insiuion only. This maerial is no copyrigh free.

5 iii c i Susiue ino y : y The Caresian equaion is y ii, So he domain of f() is. d i d iii ( ) Susiue ( ) ino y : y ( ) ( ) The Caresian equaion is y ii, 0 When = 0, = and as increases increases. for is. So he domain of f() is. y, 0 for y is y 0. So he range of f() is y 0. y,, So he range of f() is y 0. iii e i ln ( ) e e Susiue e ino y : y e e The Caresian equaion is y e ii ln( ), When =, ln 0 and as decreases ln ( ) increases. So he domain of f() is 0. y, When =, y = and as decreases decreases. So he range of f() is y. Pearson Educaion Ld 07. Copying permied for purchasing insiuion only. This maerial is no copyrigh free. 5

6 e iii 5 a C : Susiue ino y : y So he Caresian equaion of C is y C: and y Susiue and ino y : y So he Caresian equaion of C is y Therefore C and C are segmens of he same line y Pearson Educaion Ld 07. Copying permied for purchasing insiuion only. This maerial is no copyrigh free. 6

7 5 For he lengh of each segmen find he domain and range of C and C. For C:, 5 When =, = 5; when = 5, =. for is 5, so he domain of C is 5. y, 5 When =, y = 8; when = 5, y = 7. for y is 8 y 7, so he range of C is 8 y 7. The endpoins of C have coordinaes (5, 8) and (, 7). lengh of C ( 5) (7 8) For C: 6 8 7, When =, = ; when =,. for is, so he domain of C is. y, When =, y = ; when =, y =. for y is y, so he range of C is y. The endpoins of C have coordinaes `, and (, ). 6 a lengh of C, 0 for is. (This is also he domain of he Caresian equaion y = f().) y, 0 When = 0, y = ; is a quadraic funcion wih maimum value a =. for y is y, y. (This is also he range of he Caresian equaion y = f().) Noe: To find he maimum poin of he quadraic y, dy eiher solve 0 d 0 y or complee he square y ( ) ( ) ( ) () () Pearson Educaion Ld 07. Copying permied for purchasing insiuion only. This maerial is no copyrigh free. 7

8 6 Susiue y y 6( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( 6 ) : This is a Caresian equaion in he form A( c) y wih ( ) A =, = 6 and c =. 8 a y, 5 When =, y and as increases, decreases 5 owards zero. for y is 0 y so he range of f() is 0 y Susiue ino y : 6 y 7 The Caresian equaion is 6 y 7 7 a ln ( ), e e Susiue e ino y e 5 e y 5 :, 0 When = 0, = 0; when =,. for is 0 so he domain of f() is 0. When =, ln 0 and as increases ln ( ) increases. for is 0, so he domain of f() is 0. Therefore he Caresian equaion is y, k where k 0. e dy d dy 0 when 0 d (as 0 ) Pearson Educaion Ld 07. Copying permied for purchasing insiuion only. This maerial is no copyrigh free. 8

9 8 c a d y 6 d d y When, 6 0 d So gives a minimum poin of he parameric funcion for y. The minimum value of y is 6 6 When = 0, y = 0; when =, y =. 6 for y is y. Therefore he range of f() is 6 f ( ). ( ) ( ) () y y () Susiue () ino (): ( y)( y ) ( y)( y) This is in he form ( a y)( y) wih a = and =. y, This is a quadraic funcion of, and (y symmery) he maimum value of y occurs a = 0, where y =. So is he maimum y-coordinae. Challenge a Squaring he parameric funcions gives y Add () and (): () () y ( ) ( ) ( ) ( ) ( ) ( ) So a Caresian equaion for curve C is y. y y 0 0 Curve C is he equaion of a circle wih cenre (0, 0) and radius. Pearson Educaion Ld 07. Copying permied for purchasing insiuion only. This maerial is no copyrigh free.