Learning Outcomes and Assessment Standards

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1 Lesson 5 CALCULUS (8) Rate of change Learning Outcomes and Assessment Standards Learning Outcome : Functions and Algebra Assessment standard 1..7(e) Solve practical problems involving optimisation and rates of change. Overview In this lesson ou will discuss displacement, velocit and acceleration and solve practical problems involving these concepts. look at other rates of change and solve practical problems. DVD Lesson From science s = ut + _ 1 at Where s is displacement, and u and a are constants _ ds dt = u + at This is velocit (instantaneous speed) V = _ ds dt V = u + at _ dv dt = a Which is acceleration Eample 1 This displacement of a bod is given b the formula s = 0t t where s is distance in metres and t is time in seconds? a) What is the height of the bod after 1 second? b) After how man seconds will the bod be at a height of 100 m? c) What is the velocit of the bod after 1 second? d) After how man seconds will the bod reach its maimum height, and what is the maimum height? e) What is the initial velocit of the bod? f) How long is the bod in the air? g) At what speed does the bod hit the ground? h) What is the acceleration of the bod? 6 Solutions a) s = 0(1) (1) = 8 metres b) 0t t = = t 0t = t 15t = (t 5)(t 10) t = 5 or t = 10 LC G1 Cal LWB.indb 6 008/09/0 0::05 PM

2 c) V = 0 t V = 0 = 6 m s 1 d) V = 0 0 t = 0 0 = t t = 7 _ 1 Now substitute to find the height s = 0t t s = 0(7,5) (7,5) s = 7,5 metres e) Velocit when t = 0 V = 0 m s 1 f) Make displacement equal zero. 0 = 0t t 0 = t(15 t) t = seconds g) Velocit at 15 seconds h) V = 0 t V = 0 (15) = 0 at 0 m s 1 dv_ dt = m.s Eample A particle moves so that its velocit at an time (t) in seconds is given b: V = t + a) Find an epression for the displacement (s) in terms of t. b) If the displacement is 6 metres after one second, what will the displacement be after 5 seconds? c) What is the acceleration of the particle? Solution: a) v(t) = t + Remember v(t) = s (t), and we used the power rule to differentiate with. So we reverse that, b undoing what we did when we differentiated. Remember: If ƒ() = a n ƒ () = n a n 1 So if ƒ () = a n ƒ() = _ an + 1 n LC G1 Cal LWB.indb 7 008/09/0 0::06 PM

3 Something else to remember is that the derivative of a constant is zero. So we need to make provision for this. So: s(t) = _ t t + c = t + t + c } We add one to the power and divide b the new power. b) s(1) = 6: (1) + (1) + c = 6 c = 6 = s(t) = t + t + and s(5) = 5 + (5) + = = metres c) a(t) = v (t) = m/s Eample A stone is thrown verticall upwards off the roof of a building and its displacement after t seconds is given b s = 0t 5t. t is time in seconds and s is distance in metres. a) What is the stone s initial velocit? b) What is the maimum height above the building reached b the stone? c) If the stone is thrown so that it misses the building on its wa down, and takes a total time of 6 seconds to reach the ground, how high is the building? Solution: a) V = _ ds dt V = 0 10t Initial velocit t = 0: v = 0 10(0) = 0 m/s b) V = t = 0 t = seconds Reaches maimum height after seconds s = 0() 5() = 0 m c) s = 0t 5t + c where c will be this etra distance that the stone will travel Displacement is 0 when t = 6 0 = c c = 60 m Building is 60 m high Rate means the derivative gradient Increase means gradient is positive Decrease means gradient is negative 8 LC G1 Cal LWB.indb 8 008/09/0 0::07 PM

4 Eample The manner in which the temperature, T C, at the centre of a smelting pot in a blast furnace increases with time is given b T = t (5 t) where t is time in minutes. a) Find the rate of increase in temperature when t = 5 minutes b) After how man minutes will the temperature be increasing at a rate of 6,8 C per minute. Solution: a) T = _ t (5 t) T = _ 9t _ 1 10 t + 15 dt_ dt = 9t _ 10 t b) After 5 minutes rate of increase is 9(5) _ (5) = 7,5 C per minute 10 dt_ dt = 6,8 9t _ 10 t = _ t t = 68 t 90t + 68 = 0 t 0t + 16 = 0 (t 18)(t 1) = 0 t = 18 or t = 1 At 18 minutes and at 1 minutes Eample 5 Water is leaking from a tank and the depth of the water in the tank ( mm) after t minutes is given b = 16 _ 1 8 t _ 1 t. a) At what rate is the depth of the water decreasing when t =? b) After how man minutes will the depth of the water be decreasing at a rate of 6 _ 7 8 m per minutes? Solution: a) Rate _ d dt _ d dt = _ 1 8 _ t = _ 1 8 At minutes Decreasing at a rate of _ 8 d b) _ dt = _ 55 8 _ 1 8 _ t = _ _ 8 = _ t 6t = 5 t = 9 t = minutes mm per minute 9 LC G1 Cal LWB.indb 9 008/09/0 0::08 PM

5 0 Activit 1. A particle moves in such a wa that distance(s) in metres after t seconds will be given b s = 8t t. Find its position, velocit and acceleration after seconds.. A ball is thrown verticall upwards. Its height (h metres) at t seconds is given b s = 6t t a) Find at what height the ball is after seconds. b) After how man seconds will the ball again hit the ground? c) Calculate the velocit of the ball when it hits the ground. d) After how man seconds will the ball reach its maimum height? e) Calculate the maimum height of the ball. f) What is the acceleration of the ball?. If a particle moves so that V = t t + a) Find an epression for s b) Find s at t = 6 given that s = and t = 0 c) Find an epression for the acceleration of the particle.. A motorcclist takes off from a stop light and stops when he reaches a second one. His distance, ƒ(t) in metres, from the 1 st light t seconds after he takes off is given b the function ƒ(t) = 6t 0,t. Determine a) How fast he was travelling when t = 5? b) The time it took him to reach the nd light. c) The distance between the two stop lights. 5. The volume of water in a tank is given b the equation V = t t (V is volume in m, t = time in minutes). a) Find the rate at which the volume is increasing when t = minutes. b) At what time does the volume start decreasing? c) What is the maimum volume? 6. The sweetness, s, of a peach, t das after it began to ripen is given b s = 00 + (t 15) a) At what rate is s increasing when t = 0? b) The peach falls off the tree (ie it is full ripe) when s stops increasing. How man das does the peach sta on the tree? 7. The volume V(cm ) of water remaining in a leaking pail after t seconds is V = 000 0t + 0,t a) Find the average rate of change of volume from t = 0 to t = 0. b) How fast is the volume decreasing when t = 0? 8. A water tank with an inlet and an outlet is used to water a garden. The equation D = + _ 1 t _ 1 t gives the depth of water in metres where t is the time in hours that has elapsed since 09:00. a) What is the depth of the water at 11:00? b) At what rate does the depth of the water change at 1:00? c) At what time will the water be decreasing at a rate of 8 metres per hour? d) At what time will the inflow of water be the same as the outflow of water? LC G1 Cal LWB.indb 0 008/09/0 0::09 PM

6 ANSWERS AND ASSESSMENT Lesson 18 Activit 1 1. a) 1 b). a) 6 b) 6. a) 5 b) 5 c) 5 It is a straight line.. a) 1 m b) 0 seconds c) 8 m s 1 Activit 1. a) 0 b) c) d) e) _. a) b) c) 0 d) 1_ 16 e) f) 1_ 16 Lesson lim ( + ) = () + () = = 1. lim _ = 1 (1) (1) + 5 = _ 7 _. 1. lim _ + 5 = _ no limit _ lim 16 9 = lim _ _ lim 0 ( + ) = ( )( + ) = lim ( + ) = ( + ) = 6 _ lim _ 8 _ 16 = ( ) + ( 1_ ) = _ no limit 18. 1_ 19. no limit 0. no limit 1. a) does not eist b) 1 c) 1 d) no limit _ = e) no limit f) g) Lesson 0 1. a) 0 b) 0 c) 0 d) e) f) 1_ g) 18 7 _ h) ƒ() = 1 ƒ ()= 1_ i) ƒ() = 1 ƒ ()= 8 1 j) ƒ() = ƒ ()= =. s = 5t 1 t _ ds dt = 5t. m = n n + 1 n m = n + n dm_ = n n dn. ƒ() = ƒ () = d_ ( d + 1 ) _ 1 1 LC G1 Cal LWB.indb 1 008/09/0 0::09 PM

7 = 1 6. s (t) = 0,t 0,8 7. = _ d d = 1 8. d_ [ t dt π ] = πt π 1 9. s = _ t 1_ + 1_ t 1 ds dt = 1_ t 1_ 1_ 6 t _ 10. ƒ() = ( 1)( + ) ( 1) ƒ () = D [ + 6 _ 5 ] = D ( 1_ ) 5 = 1 _ D 5 ( 1_ 6 _ + _ ) = _ 1 ( 5 _ 1 7_ 6 _ 1_ _ 5_ ) = 1_ _ d_ [ 1 + u du u 1 ] = u d u1. _ = 6 Gradient 6 d 1. = 1 = 16 = ( ; 18) 15. _ d d = 0 + = 0 ( + )( 1) = 0 = or = 1 ( ; 10_ 11 ) ( 1; _ 6 ) 16. = 6 = = 1 ( 1; 7) 17. The gradient of the tangent at = is = = 0 ( )( + 1) = 0 = or = ƒ () = when = 1 ƒ () = + + a 5 + a = a = 1 (1; ) is on the curve = = 1 = + a b b = 0. d (t) = t 00 m s 1 1. V(t) = t _ V (t) = _ t 1_ V (t) = 9 = 9_ 9_ units per ear LC G1 Cal LWB.indb 008/09/0 0::10 PM

8 Lesson 1 Activit 1 1. Since the tangent is parallel to the line = 11, we have that m T = 11 Thus: f () = + + But m T = f ( T ) 11 = = = 0 ( )( + ) = 0 = _ or = For T = _ : = T (_ = _ _ ) + (_ ) + (_ ) = T 7 1 ( _ : _ 55 7 ) = _ 55 7 For T = : T = ( ) + ( ) + ( ) + 5 = = 5 T ( ; 5) Eq tangent 1: T = m T ( T ) _ 55 7 = 11 ( _ ) 7 55 = ( 11 _ ) = = = ,5 Eq tangent : T = m T ( T ) ( 5) = 11( ( )) + 5 = 11 + = T is given as : For m T : f () = For T : T = f() f () = = 1 = () () + 7 m T = 1 = T = 5 Eq tgt: f( T ) = f ( T )( T ) 5 = 1( ) 5 = = +.. Find the value of where the cut. Then show that the product of the gradients at that point is 1. First find where the curves intersect = + 1_ = 1_ = 1 LC G1 Cal LWB.indb 008/09/0 0::11 PM

9 = ± 1_ For = For = + 1 d _ d = d d = at = 1_ m = 1 at = 1_ 1 m = 1 at = 1_ m 1 m 1 m = 1 = 1 at = 1_ m = 1 At both points the curves intersect at right angles. We sa the curves intersect orthogonall.. a) 6_ = 6 1 _ d d = Slope of tangent at = is 6_ = _ Desired point is ( ; ) Equation of normal + = _ ( + ) + 9 = + = 5 b) 6 = _ 5_ 18 = = 0 ( 9)( + ) = 0 = 6_ and = _ 5_ intersect The normal cuts the curve again at ( 9_ ; _ 5. h() = and T = 1 Thus: For m T h () = 9 + m T = h ( 1) m T = 9( 1) + ( 1) m T = 9 m T = 16 and m N = _ 1 since tangent is perpendicular to the normal at the point of tangenc 16 For T T = h( T ) T = ( 1) + ( 1) ( 1) + 5 T = T = 1 Thus Eq tgt: T = m T ( T ) Eq norm: T = mn ( T) 1 = 16( ( 1)) 1 = _ 1 ( ( 1)) 16 1 = = + 1 = = + 09 = _ 16 + _ To find where the tangent cuts the curve again: h() = (1) and = 16 () (1) () : 16 = = 0 LC G1 Cal LWB.indb 008/09/0 0::1 PM

10 The tangent touches the cubic and we therefore have a double root at = 1. Thus ( + 1) is a factor of this new equation : 1 8 = ( + + 1)( 8) Thus the tangent cuts the curve again at = _ 8 and f (_ 8 ) = (_ 8 ) + (_ 8 ) (_ 8 ) + 5 = _ _ = 9 = _ 11 9 Cuts again at point (_ 8 ; _ 11 9 ) Activit = = 0. = +. 9 = (1; ) or ( 1; ) Lesson Activit 1 1. _ d d = 6 increasing decreasing stationar point 6 0 (; 7) d. _ = Alwas decreasing, no stationar points d. ƒ () = ( + ) increasing decreasing stationar point and 0 0 ( ; ) (0; ). p () = 1 ( ) = ( )( + ) increasing decreasing stationar point and ( ; 17) (; ) d 5. _ d = 5 increasing everwhere ( ; ) stationar point (0; 0) 6 (a) = ; [ 5;9] LC G1 Cal LWB.indb 5 008/09/0 0::1 PM

11 The graph is increasing from [ 5;0) and decreasing from (0;9]. At = 0 it is stationar. Mathematicall: _ d = and > 0 iff < 0. So if < 0 it will increase. It will then d decrease on the > 0 side. (b) = ; [ ;6] The graph is increasing from [ ; 0) and decreasing from (0; 6] whilst it is stationar at = 0. Mathematicall: _ d = and > 0 iff < 0. So if < 0 it will increase. It will then d decrease on the > 0 side. (c) = 6 ; [ ; ] The graph decreases on [ ; 0) and increases on (0;]. It is stationar at = _ 1. Mathematicall: _ d d = 1 and 1 > 0 iff >. So if < _ 1 it will decrease. It will then increase on the > _ 1 side. (d) = + 5 ; [ 7;9] It is increasing over its whole domain with no stationar points. Mathematicall: _ d d = + 5 and + 5 > 0. So is an increasing function. LC G1 Cal LWB.indb 6 008/09/0 0::1 PM

12 Activit ( 1; 6) (_ 11 ; _ 00 7 ) (0; ) - 7 LC G1 Cal LWB.indb 7 008/09/0 0::1 PM

13 (_ 1 ; _ ) LC G1 Cal LWB.indb 8 008/09/0 0::15 PM

14 LC G1 Cal LWB.indb 9 008/09/0 0::16 PM

15 Lesson a) i) The zeros of ƒ are 0; 6 and 6 (the -ais is a tangent to the curve). ƒ() = ( 6)( 6) = ( 1 + 6) = a = 1; b = 6; c = 0 ii) For point A ƒ () = = = 0 ( 6)( ) = 0 = 6 or = At A = an = = A(; ) b) i) Co-ordinates of T (; ƒ ()) ƒ () = = 7 T(; 7) ii) ƒ () = + 6 ƒ () = = 9 Equation of tangent + 7 = 9( ) LC G1 Cal LWB.indb /09/0 0::16 PM

16 = 9 5 iii) Equation of normal (m = 1_ 9 ) + 7 = 1_ 9 ( ) + 7 = _ 9 + 1_. = a( 1)( ). 18 = a()(1) a = 9 = 9( 1)( ) _ d d = + b + c at = 1 (1) + b(1) + c = 0 + b + c = 0 c = b (1; 0) is on the curve 0 = 1 + b + c c = 1 b c = b 0 = + b b = c = a) b) 6 b smmetr c) [1; 5] d) = 1 = 5 e) ƒ () 6. a) ƒ () = ƒ ( 1_ ) = _ 1_ = 1_ g () = 1 a g ( 1_ ) = 1 a_ 51 LC G1 Cal LWB.indb /09/0 0::17 PM

17 1 _ a = 1_ a = 1 a = a = 1 b) = = 0 ( 1)( 1) = 0 = 1 Lesson 1. Let the width of the rectangle be. HEDGE (10 ) Then the length wiill be (10 ) Then the area: Area = length breadth A() = (10 ) (We find an epression that relates what we want to work with.) = 10 This epression is a quadratic equation, which has a maimum turning point since a < 0. To determine the maimum: differentiate the epression and solve it equal to zero. Thus: A() = 0 to optimise 10 = 0 = 10 = 5 This is the dimensions of the rectangle for maimum A Then: Maimum Area: A(5) = 100(5) (5) = = m is the maimum area The Dimensions: Length = 70 m Breadth 5 m. a) V = ()(180 ) b) 0 mm. For a closed clinder of radius r and height h, we open it up to determine its surface area. h 5 For the circular part: There are two of them Surface area of the top and bottom: S 1 = πr For the walls: LC G1 Cal LWB.indb 5 008/09/0 0::19 PM

18 πr h Surface area : S = πrh Thus the total surface area: S(r; h) = S 1 + S = πr + πrh Our problem now is that our surface area is in terms of two variables, so we cannot et differentiate it in order to optimise the area. We need to get rid of either r or h, so that differentiation can take place as per normal. This is the reason wh the volume has been supplied. We use the volume, where h is the dominant dimension taking it into the third dimension, and reduce it to two dimensions b removing the height from the area equation. Thus: Volume = (Base Area) (Height) 000π = (πr ) (h) h = _ 000π πr = _ 000 r (a) Now: S(r; h) = πr + πrh S(r) = πr + πr (_ 000 r ) S(r) = πr + _ 000 r (b) = πr + 000πr 1 To optimise: S (r) = 0 πr _ 000 = 0 πr 000π = 0 r = r = 10 And now from (a): h = _ 000π πr = _ = _ = 0 Thus the dimensions for minimum surface area: height 0 cm Radius 10 cm For the value of this minimum surface area: From (b) it follows that S(r) = πr + _ 000 r S(10) = π(10) + _ 000π = 00π + 00π = 600π cm 10. Thus the minimum area is S = 600π cm.. We let the height of the can be h cm and the radius be r cm Then we can find an epression for the volume: Volume = Base area height 000 = πr h h = _ 000 πr r h Area = πr Area = πrh Area = πr Area of the top and bottom: A(r; h) = πr Area of the walls: A(r; h) = πrh The cost involved in production: Top : Walls = : 1 _ Top Walls = _ 1 Top =.Walls Cost = (Circular area) + Wall area C(r; h) = (πr ) + πrh C(r) = πr + πr (_ 000 πr ) C(r) = πr + _ r 5 LC G1 Cal LWB.indb 5 008/09/0 0::19 PM

19 5 To optimise the cost: C (r) = 0 8πr _ = 0 r 8πr = πr = r = 10 _ π For the ratio: h : r = : 1 5. Let the one side of the paper be cm Then the other side will be _ cm 1 The printing dimensions will be: Thus: Printing area = (_ )( ) P() = _ = To optimise: P () = 0 _ _ = 0 = 896 = 8 _ 1 = 8 7 since > 0 _ Thus the length: 8 7 and the width _ = 8 _ Assuming this happens after the have travelled for hours, at point P: Bus A travelled a distance of: 75 km Bus B travelled a distance of: 100 km The distance that the are apart: PQ = (0 100) + (75) (Pthagoras) = = A 75 Q P 100 B This is a quadratic equation under the surd, and we onl have to find its turning point to find the minimum value: Thus: D() = To optimise: D () = = = _ 1 50 =,08 hours = hours min = hours (rounded to nearest hour) The actual distance apart: PQ = 15 65() 6 000() _ PQ = 0 1 = 19 km. 7. From Total Surface area: πrh + πr + πr = πrh + πr = 0π. 0 r h = _ r Volume = _ πr + πr h = _ πr + πr 0 π (_ r ) = _ πr + 10πr _ πr = 10πr + _ 9 6 πr = 10πr _ 5 6 πr h V (r) = 10π 5πr = 0 5πr = 10π r = r = Then Ma Volume = 10π() _ 5 6 π(8) = _ 0π LC G1 Cal LWB.indb 5 008/09/0 0::0 PM

20 8. = 1 and 6 = 1. = _ 1 and = _ 1 6 PQ = and QR = _ 1 _ = 6 _ 1 Area = ( 6 1 _ ) = 1 A () = 1 = 0 = = So PQ = and RQ = Thus Maimum Area = 16u 9. = 10 mm cm 11. = 0 = b 10 Lesson m; 1 m s 1 ; deceleration of m s. a) 56 m b) 9 seconds c) 6 m s 1 d) _ 1 seconds e) 81 m f) m s. a) s = _ 1 t _ 1 t + t + c b) = c c) a = t 1 s = _ 1 t _ 1 t + t + s = _ 1 (6) _ 1 (6) s = 60. a) 0 m s 1 b) 10 seconds c) 00 m 5. a) 6 m per minute b) 5 minutes c) 0 m 6. a) 675 u per da b) 15 das 7. a) Decreasing b 6 cm second ( 6 cm second) b) 8 cm second 8. a) m b) Decreasing at _ m per hour c) At 1:00 d) At 10:0 55 LC G1 Cal LWB.indb /09/0 0::1 PM

21 TIPS FOR THE TEACHER Lesson 18 This section counts for a lot of marks so it is important that the learners practise man eamples. Alwas stress that the derivative is the gradient at a point. Tr to bring in practical eamples. Be careful of the notation. Lesson 19 This section counts for a lot of marks so it is important that the learners practise man eamples. Alwas stress that the derivative is the gradient at a point. Tr to bring in practical eamples. Be careful of the notation. Lesson 0 This section counts for a lot of marks so it is important that the learners practise man eamples. Alwas stress that the derivative is the gradient at a point. Tr to bring in practical eamples. Be careful of the notation. Lesson 1 This section counts for a lot of marks so it is important that the learners practise man eamples. Alwas stress that the derivative is the gradient at a point. Tr to bring in practical eamples. Be careful of the notation. Lesson This section counts for a lot of marks so it is important that the learners practise man eamples. Alwas stress that the derivative is the gradient at a point. Tr to bring in practical eamples. Be careful of the notation. Lesson Keep stressing to the learners that the derivative is the gradient. It is a good idea to make them do rough sketches of graphs. 56 LC G1 Cal LWB.indb /09/0 0:: PM

22 Lesson Encourage the learners to draw pictures before the make equations. It is a good idea to make the first equation the equation the need to differentiate. Lesson 5 Stress the difference between increasing and decreasing functions. Rate is a gradient so whenever the see the word rate the must differentiate. 57 LC G1 Cal LWB.indb /09/0 0:: PM