Unit 2: Derivatives Instructions There are 16 questions in this Assessment Opportunity. Show all work for each question. 1. Find the derivative of the following. a. f(x) =13x* -7x3 +5x3+11x+75 b. f(x)=(x+2x7 +4)(x*-3) c. F( x) = 3x -6x +7 4x-1 d. f(x)=(4x -7x) e. 12 = y + x] f. f (x) =156 g. f(x) =(1-3x](x] -2)3 h. f( x)= 2x7 +1 3x +1Unit 2 Assessment Opportun 2. Evaluate f"(x) for f (x) - (3x] -7x+4) 3. Find the third derivative of f (x) =12x" -3x7+4x'+5x'+6 4. Find the equation of the tangent line to the curve y= x -2x+3 at the point (2,3). 5. Find the point on the parabola y= -x3 + 3x + 4 where the slope of the tangent line is 5. 6. Give one example of a rate that is decreasing and one example of a rate that is increasing. 7. When a vehicle is said to be travelling at -35 km/h, what does this infer? dV 8. If V represents the volume of an object, what does - represent?9. A construction worker accidently drops a hammer from a height of 90 m while working on the roof of a new apartment building. The height, s, in metres, of the hammer after t seconds can be modelled by the function s(t)=90-4.9t', t>0- a. Determine the velocity at 1 s and 4 s. b. When will the hammer hit the ground? c. Determine the impact velocity of the hammer. 10. The motion of a particle is described by the position function s(t)=213-15t'+33t +17, t> 0, where r is measured in seconds and s() in metres. a. When is the particle at rest? b. When is the velocity positive? c) Draw a diagram to illustrate the motion of the particle in the first 10 seconds. d) Find the total distance traveled in the first 10 seconds. 11. A snowball melts so that its surface area decreases at a rate of 0.5 cm /min. Find the rate at which the radius decreases when the radius is 4 cm. (51 = dar)12. Brent is driving west at 60 km/h and Aaron is driving south at 70 km/h. Both cars are approaching the intersection of the two roads. At what rate is the distance between the cars decreasing when Brent's car is 0.4 km and Aaron's is 0.3 km from the intersection? 13. A water tank at a filtration plant is built in the shape of an inverted cone with height 5.2 m and diameter 5 m at the top. Water is being pumped into the tank at a rate of 1.2 mymin. Find the rate at which the water level is rising when there is On m' of water. (4 marks) 14. For the function f (x) =ax thy -5x+9 , determine the values of a and b so that f(-1)=12 and f'(-1)=3. 15. Determine the equations of two lines that pass through the point (-1,-3) and are tangent to the graph of y= x* +1. 16. Two particles have positions at time t given by s, =4/ -7 and s, =50 -7. Find the velocities v, and v, at the instant the accelerations of the two particles are equal.Position function, sets = 243- 15 17 + 334 + 17, +30. sides wirit. I we get , fig = (272 - 72+ 4) 8 Unit 2 : Derivatives . If = U = 64 - 30 + + $ 5' ( ) : 8 (32 7- 72+4 ) . ( 42 -7 ) 1) -we have to differentiate each function wirit. x 1 - 0 . 8 " ( 2) . 56 ( 30 7 - 72 + 4 )6 (62- 7 ) (62-7 ) +8(37 - 72+4).6 & then we will get the first order derivative . = > 61 - 304 + 33 = 0 = 8 (30% - 72+ 4)6 17 (62-7) +6(37- 72+4)] => 2 47 - 10 1 + 11 2 0 5170 = 1824 - 723 + 5 2 7 + 11 2 + - 5" in) = 5273 - 217 2 + 10 2 + 11 10 4 100 - 4. 2 . 11 * 8 (30 - 72+ 4)6 (252 7-58 82+343 +187-427+24 2 . 2 = 8 (372 - 72 +4)6 [ 270 72 - 630 % + 367] (W FIND = (23 + 20 * + 4 ) (24-3 ) 10 #100 - 88 => f" C = (307 + 42 ) (24 - 3) + (23 + 202 + 4 ) 423 $ 12)- 12 210 -32 + 4x+ 87+6 =326 -927+475-122 + 426 + 8x#+1623 - + = 10 # 1/12 - 103253 - 5213 - film - 120 29 - . . 8 " ( m ) = 10 80 7 8 - 2126+ 2024 + 10x -7 76 + 1205 + 1693 - 972 -12 2. (0 fem) = 3x 7 - 67+ 7 - : 4 = 6 403 or 5- 25: 3.37 sec or 1. 63 sec, 16- 12675+ 8023 410 .. f " ( 2) = 864027 - 24 - 63024 + 240 2. -16 8 SET upto two decimal . $ 20 = ( 42-1) ( 62 - 6 ) - (27" - 6249 ) 4 Places . 4 8 = 27- 20+3 ( 42 - 1) the particle is at rest of wood see or see sec . = $4 = 27 -2 , now * 230 = 4- 2= 2 = me slope of 2- 62 +6 - 127 42/49 - 28 () the velocity is positive when uso the line to (42 -1 )7 now , equation of the tangent line is : the cure . 127 - 62 . 2.4 6 +7 - 301 +33 7 0 . from above -we get, if cote [# , 1. 63) U(3.37, 00) ( 8-3) = m (2-2 ) ( ) SIN = (473 - 72) 7 8 - 3 = 2 ( 2 - 2) $8 , as parabolic 7 8 - 3 = 2 2-4 => 27 -8 -1 = 0. ". 8'em) = 7 ( 423 - 72) ( 127- 7 ) for that time interval , the velocity is * 7 (48 5 - 8923 -2823+ 98%) " y=-+32+9, given slope, on . do : 5 12 = yo tar when += 10s them 7= - 27 + 3 25 . 0 . 24 82 + 2 2 07 d4 = - 8 "= 333 m /d - 27 = 2 2 =-1 find = 156 = 1'2=0 [:: 156 is a constant function ] & 8 = - 1 - . : paint is = ( - 1 , - 8 ) . -4 = - 8 find = (1- 30 ) (2 5 - 2) 3 > s'(20 = 2 ( 1 - 32 ) 1 -3) (7 / 2) +31 27 - 2). 27 (1 - 37 ) 10 " Let , the velocity expression? & let ult ) = - 2t - 4, +30 = - 6 (1-32 ) (27 - 2)3 + 62 (27 - 2 ) (1- 37 ) 2 U = 2t - 4 in dy - acceleration du = a = - 2 co , = 11-370 (27 - 2) ( -6 (20- 292 + 62 (1-30 ) " ) rat - IF = acce decreasing . ( 27 - 327 - 2 + 67 ) 8-6 (24- 427 + 9) + 62 (1-62 + 8 x]] when + E ( 1. 63 , 3.37 ) then u CO . = a = 2 > 0 , inc ( 62 - 204 - 2) 8- 67 4 +2472-24 + 61- 36 " + 54x ) total distance in first 10 see, is = (62 - 207-2) (-624 + 5473 -127+67-24) s(t) at If V represents the volume of an object then dy fin = (27 +1 represents that the rate of change in volume = 1 10 ( 2 + 3 - 15 + 2 + 3 3 + + 17 ) de I - 35 km/ oh means the vehicle's velocity is - 35 km > f " ( x ) = 2 ( 20 2 + 1 ) x ( 3 2 3 +1 ) 42 - ( 27 71) . Bar ( 3 73 +1 ) = [234 - 15 +33 4 +17+] Pes hour . when it travels . of kim = 2 . 2 2 + 1 - 182 " - = [ : 104 - 5 103 + 33 10 +(17 x 10)] 373 41 - * 12 24 + 42 - ( 373 41 ) x - 62 4 - 5060 - 5000 + 1650+ 170 -373 +1 ( 3 7 3 + 1) 2 1820 m . 2 ( 272 41 ) (- 62" - 927 +40) (3 23 +1 ) 3 14: flo = 023 + bar - 52 + 9 a , 6 = 9 16 . $1 = 4t - => J' = 307 +267-5, differentiating twixt. both 151 = v 1 = 4-2t If' be the first order derivative now & d's , = Q2 = 10 - bt $ 1 -1) = - a + 6 + 5 + 9 ": accelerations are equal [ giveen] $ 5 '1-1) = 30 - 26 - 5 given that f ( -1 ) = 12 = > - 9 46 + 5+ 9 = 12 - a , = a z 5' - 1 = 3 #30 - 26 -5 => -2 = 10 - 6+ = 3 - a +6 = - 2 8 3 0 - 26 : 8 - ( w - 6+ = - 12 => + =2 => a - 6 = 2 _ ( 1) " at the instawl- + = 2 . now, solving is & ( i) we get, Then V. = 4 -2.2 & U 2 F 10. 2 - 3.3 2 10 x3 - 60 x 1 = 0 unit - 12 = 8 unit 3 1 - 36 = 6 1 5( t ) = 90 - 4. 9 4 +20. - b = - 2 + Differentiating both sides wix. t . I've get , . 6 = 2 & from (@ -a - 2 = 2 ( 0 ) as = v = velocity . - 9.8 + Hence : a = 4 =- 9.8 x 4 a = 4 8 6=2 . now , v 1 =- 9.8 m / 1 4 12 : 4 - 39.2 m / J . 1 4 = 27 + 1 equation of the tangent line , the hammer -will hit the ground meththen 8 - 4, =m ( 2 - 71), where the height , $ = 0 in is the tangent- of the cure 90 - 4. 9+ 2 = 0 . 4 2 72 + 1 4. 9 7 = 90 .: dy = 4. 28571 ..... now = 4:28, upto . two decimal places. -. + = 4:28 sec . 271 - 2071 +4 - 41 2 0 . " the impact velocity of the Rammer, Also, the lines passing through the paint ( - 1, - 2). with the help of thin -we -will get the required Vimp. " ? ah m result . * N 2 x 9 . 8 * 90 =1 17 64 = 42 m/8