Answer Key and Solutions Part C

Answer Key and Solutions Part C - y x 1 2 3-1-2-3 1 2 3 4 5...

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Unformatted text preview: 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 = = (b) ) 60 sin( 108000 2 2 2 t dt V d - = For 2 2 = dt V d ,... 2 , , 60 ) 60 sin( = = t t sec ,... 15 1 , 20 1 , 30 1 , 60 1 , = t For 1800 = = dt dV t For 1800 cos 1800 60 1- = = = dt dV t For 1800 2 cos 1800 30 1 = = = dt dV t Maximum when ,... 30 1 , = t Then 1800 = dt dV . (c) = dt dV when sec ,... 120 5 , 120 3 , 120 1 2 60 ) 60 cos( ) 60 cos( 1800 = = = = t t t t When t = 120 1 sec, there will be a maximum voltage. Alternatively (1) : When t = 120 3 sec, there will be a minimum voltage. ( ) Alternatively (2): This means that the voltage is at a maximum or minimum at these times (alternating max/min). ( ) NOTE : The following is not required. No marks to be allocated. (When t = 120 1 sec, 30 2 sin 30 ) 120 1 . 60 sin( 30 = = = V . VET Calculus 146 21.Solutions Alternatively : When t = 120 3 sec, 30 2 3 sin 30 ) 120 3 . 60 sin( 30- = = = V .) VET Calculus 147 21.Solutions 4. S (5,0) P ( p ,-3 p ) or P ( x , -3 x ) D = 25 ) 3 ( ) 5 ( 2 2 2 +- =- +- p p p p 25 2 1 2 2 +-- = p p p dp dD For 2 1 = = p dp dD When < = dp dD p and 1 = dp dD p . For 2 1 = p a minimum is obtained. Minimum distance 11 2 3 4 18 4 81 2 9 ) 2 1 4 ( 2 = + = + = units. [7 marks ] = 4.97 (or 5) units. Alternatively (1) : Determine nature of turning point from graphics calculator. Alternatively (2): Distance SP = 2 2 ) 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. ( ) 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. (b) s 2 = ( x 5) 2 + ( y 3) 2 Then: 2 s dt ds = 2( x 5)....
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This note was uploaded on 09/01/2010 for the course MATH MAT1300 taught by Professor Mcdonald during the Spring '10 term at Acadia.

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Answer Key and Solutions Part C - y x 1 2 3-1-2-3 1 2 3 4 5...

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