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Answer Key and Solutions Part C

# Answer Key and Solutions Part C - 12.Optimisation y 10 9 8...

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y x 1 2 3 -1 -2 -3 1 2 3 4 5 6 7 8 9 10 -1 -2 -3 -4 -5 -6 -7 -8 -9 A B O g(x) f(x) 12. Optimisation                       1. (a) y  intercept is  1 1 2 + a .    x  intercept is  a     (b)   area =  1 1 2 1 2 + a a .   (c)   at  a  = 1 area =  4 1 . 2. (a)   graph (b)   4p 2 -p 4 (c)    2 = p 3. 10.(a) ) 60 cos( 1800 60 ). 60 cos( 30 t t dt dV π π π π = =    üü (b) ) 60 sin( 108000 2 2 2 t dt V d π π - =    ü For  0 2 2 = dt V d     ,... 2 , , 0 60 0 ) 60 sin( π π π π = = t t    ü sec ,... 15 1 , 20 1 , 30 1 , 60 1 , 0 = t    ü For  π 1800 0 = = dt dV t For  π π π 1800 cos 1800 60 1 - = = = dt dV t For  π π π 1800 2 cos 1800 30 1 = = = dt dV t Maximum when  ,... 30 1 , 0 = t   Then  π 1800 = dt dV .   ü (c) 0 = dt dV  when  sec ,... 120 5 , 120 3 , 120 1 2 60 0 ) 60 cos( 0 ) 60 cos( 1800 = = = = t t t t π π π π π   üü    When  t  =  120 1  sec, there will be a maximum voltage.   ü Alternatively (1) : When  t  =  120 3  sec, there will be a minimum voltage.  ( ü ) Alternatively (2): This means that the voltage is at a maximum or minimum at these times (alternating max/min). ( ü ) NOTE : The following is not required. No marks to be allocated. (When  120 1  sec,  30 2 sin 30 ) 120 1 . 60 sin( 30 = = = π π V VET Calculus 146 21.Solutions

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Alternatively : When  t  =  120 3  sec,   30 2 3 sin 30 ) 120 3 . 60 sin( 30 - = = = π π V .) VET Calculus 147 21.Solutions
4. S (5,0)   ü P ( p ,-3 p )  or P ( x , -3 x )   ü D  =  25 ) 3 ( ) 5 ( 2 2 2 + - = - + - p p p p    ü 25 2 1 2 2 + - - = p p p dp dD    ü For  2 1 0 = = p dp dD    ü When  0 0 < = dp dD p  and  0 1 = dp dD p .   For  2 1 = p  a minimum is obtained.   ü Minimum distance  11 2 3 4 18 4 81 2 9 ) 2 1 4 ( 2 = + = + =  units.      [7  marks ]            = 4.97 (or 5) units.   ü Alternatively (1) : Determine nature of turning point from graphics calculator. Alternatively (2): Distance SP =  2 2 ) 0 3 ( ) 5 ( - - + - x x    =  25 2 + - x x At minimum when  2 1 = x    (from calculator) Then min SP =  25 2 1 4 1 + -  =  75 . 24    4.97 units.    ( üüüüü ) 13. Related Rates                         1. (a) –0.6.   (b)   decreasing at 0.12 units per second. 2. . min / 3 5 m 3. (a)(i)    ) 2 )( 1 ( 2 t t + +    (ii)    ) 1 )( 2 1 ( 1 h h + +    (b)    ) 1 )( 2 1 ( 2 1 h h C + + - 3. (a) 55 P 12 - dh dP (b) - 6   k Pa/sec. 5. 48 radians per hour 6. (a) 19.(a) y  = 2 x  + 1 dt dy  = 2  dt dx  = 2 x 4 = 8 units/sec.   ü (b) s 2  = ( x  – 5) 2  + ( y  – 3) 2     ü Then: 2 s dt ds = 2( x  – 5). dt dx  + 2( y  – 3). dt dy     ü VET Calculus 148 21.Solutions

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i.e. dt ds  =  2 1 ] ) 3 ( ) 5 [( ). 3 ( ). 5 ( 2 2 - + - - + - y x dt dy y dt dx x At (-1,-1): dt ds  =  2 1 ] ) 3 1 ( ) 5 1 [( ) 8 ).( 3 1 ( ) 4 ).( 5 1 ( 2 2 - - + - - - - + - -  =  52 56 -  units/sec (-7.8 u/s)    ü Alternatively: P( x ,2 x +1) s  =  2 2 ) 3 1 2 ( ) 5 ( - + + - x x   =  29 18 5 2 + - x x dt dx x x x dt ds ) 18 10 ( ) 29 18 5 ( 2 1 2 2 1 - + - = - When  x  = -1:  4 ). 18 10 ( ) 29 18 5 ( 2 1 - - + + - dt ds  =  52 56 -  units/sec (-7.8 u/s)   ( üüü ) (c) The negative implies that P is  getting closer  to Q.    ü   (d) dt ds = 0.
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Answer Key and Solutions Part C - 12.Optimisation y 10 9 8...

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