Unit 3 Outcome 2 Further Calculus

Reminder on Derivative Graphs The graph of y = g(x) is given below y =g(x) X 5 6 7 8 9 10 x 6 8 SPs occur at x = 6 & x = 8 g'(x) + 0 - 0 + New y-values

The graph of y = g '(x) is y = g '(x) X 6 8

Derivative Graph of y = sinx -/2 3/2 /2 5/2 SPs at x = -/2 , x = /2 , x = 3/2 & x = 5/2 x -/2  /2  3/2  5/2 f (x) 0 + 0 - 0 + 0 New y-values

Roller-coaster from 0 to 2 ie shape of y = cosx /2 3/2 -/2 2 5/2 Roller-coaster from 0 to 2 ie shape of y = cosx Hence If y = sinx then dy/dx = cosx

A similar argument shows that …. If y = cosx then dy/dx = -sinx If we keep differentiating the basic trig functions we get the following cycle f(x) f  (x) sinx cosx cosx -sinx -sinx -cosx -cosx sinx

Example If g(x) = cosx – sinx then find g  (/2) . ********** g  (x) = -sinx – cosx g  (/2) = -sin/2 - cos/2 = -1 - 0 = -1 Example Prove that y = 7x – 2cosx is always increasing! ********* dy/dx = 7 + 2sinx since -2 < 2sinx < 2 then 5 < dy/dx < 9 Gradient is always positive so function is always increasing.

Example Find the equation of the tangent to y = sinx – cosx at the point where x = /4 . ******** Point of contact If x = /4 then y = sin/4 – cos/4 = 1/2 - 1/2 = 0 (/4 ,0) Gradient dy/dx = cosx + sinx when x = /4 then dy/dx = cos/4 + sin/4 = 1/2 + 1/2 = 2/2 = 2 Line Using y – b = m(x – a) we get y - 0 = 2(x - /4) ie y = 2x - 1/4 2

Integrating sinx & cosx If y = sinx then dy/dx = cosx So  cosx dx = sinx + C If y = -cosx then dy/dx = sinx So  sinx dx = -cosx + C Example  (3cosx – 4sinx) dx = 3sinx + 4cosx + C

 3sinx dx Example = [-3cosx ] = (-3cos /3) - (-3cos0) = (-3 X ½ ) – (-3 X 1) = -11/2 + 3 = 11/2 Example Find the total shaded area. y = sinx  3/2

1st area =  sinx dx = [ -cosx ] = -cos - (- cos0) = -(-1) – ( -1) = -(-1) – ( -1) = 2 3/2 2nd area =  sinx dx = [ -cosx ] 3/2 = -cos 3/2 - (- cos )   = 0 - 1 = -1 (actual area = 1) So total area = 2 + 1 = 3units2 alternatively By symmetry, 2nd area = ½ of first = 1 etc.

Show that the shaded area = 22units2. Example Show that the shaded area = 22units2. y = cosx y = sinx Limits Curves meet when sinx = cosx sinx = cosx cosx cosx tanx = 1 Q1 & Q3 x = /4 or 5/4 tan-11 = /4

Between /4 and 5/4 the sin graph is higher So shaded area =  5/4 ( sinx – cosx )dx /4 = [ ] 5/4 -cosx - sinx /4 = ( -cos5/4 - sin5/4 ) - (-cos/4 - sin/4 ) = (1/2 + 1/2 ) - (-1/2 - 1/2 ) = 1/2 + 1/2 + 1/2 + 1/2 = 4/2 = 4 X 2 2 X 2 = 42 = 22units2 2

Distance,Speed,Acceleration! Speed = change in distance change in time Acceleration = change in speed change in time Distance/Time Functions Suppose a distance/time function is given by d = f(t) then distance: d = f(t) speed: s = f´(t) 2nd derivative acceleration: a = f´´(t)

This now means that if we have a “speed/time” function Ie S = f(t) then acceleration = f´(t) and distance =  f(t)dt NB: Integration gives area under curve. In Physics area under speed/time graph = distance .

Example The speed of an object(in m/s) is given by the formula f(t) = 2t2 – 5t (i) Find its speed after10 seconds. (ii) How far did it travel in the first 10 seconds? (iii) Find its rate of acceleration 10s into the journey. ********* (i) f(t) = 2t2 – 5t so f(10) = (2 X 102) – (5 X 10) Speed is 150m/s = 200 - 50 = 150

Distance travelled = 4162/3 m = 4162/3 (ii) d =  f(t) dt 10 =  2t2 – 5t dt 10 = [ 2/3t3 - 5/2t2 ] 10 = (2000/3 – 250) - 0 Distance travelled = 4162/3 m = 4162/3 (iii) f ´(t) = 4t - 5 so f ´(10) = (4 X 10) - 5 = 35 Acceleration = 35m/s/s

CHAIN RULE Inner & Outer Functions - think back to composite functions. Suppose that f(x) = (3x – 1)4 . To evaluate f(2) we first find the value of 3x –1 when x = 2 ie 5. Then we find 54 ie 625. We can think of f(x) as having inner and outer parts. or f(x) = g(h(x)) g(…) = (….)4 - outer h(x) = (3x – 1) - inner

Similarly If f(x) = 2cos3x , then to find f(/3) firstly 3 X /3 =  then 2cos  = -2 h(x) = 3x - inner g(…) = 2cos(…) - outer f(x) = g(h(x))

Chain Rule for Differentiation Suppose that f(x) = (2x + 1)3 & we want to find f (x). *********** Breaking brackets (2x + 1)3 = (2x + 1)(2x + 1)2 = (2x + 1)(4x2 + 4x + 1) = 2x(4x2 + 4x + 1) + 1(4x2 + 4x + 1) = 8x3 + 8x2 + 2x + 4x2 + 4x + 1 = 8x3 + 12x2 + 6x + 1 So f (x) = 24x2 + 24x + 6 = 6(4x2 + 4x + 1) = 6(2x + 1)(2x + 1) = 6(2x + 1)2 This has taken a lot of work and would be really time consuming if we wanted the derivative of say (2x + 1)10 .

The Chain Rule offers a much quicker method and is defined as follows .. If f(x) = g(h(x)) Then f (x) = g (h(x)) X h (x) ie differentiate the outer part then multiply by derivative of inner part. Back to f(x) = (2x + 1)3 Outer = (…)3 & inner = 2x + 1 So f (x) = 3(2x + 1)2 X 2 = 6(2x + 1)2

Example outer = (…)20 g(x) = (5x – 9)20 20(…)19 g (x) = 20(5x – 9)19 X 5 inner = 5x - 9 = 100(5x – 9)19 Example 5 f(x) = (6x - 5) . Find f (5) ******** outer = (…) 1/2 f(x) = (6x - 5) = (6x – 5)1/2 ½(…)-½ f (x) = 1/2(6x – 5)- 1/2 X 6 inner = 6x - 5 = 3 6 (6x – 5)1/2 3 . So f (5) = = 3/5 = 3 25 (6x - 5)

Example V(t) = 2 = 2(4t – 3t2)-3 (4t – 3t2)3 so V  (t) = X (4 – 6t) = -6(4 – 6t) common factor -2 (4t – 3t2)4 = 12(3t – 2) (4t – 3t2)4

Chain Rule with Trig Expressions Example outer = 6sin(…) y = 6sin3 inner = 3 dy/d = 6cos3 X 3 = 18cos3 Example Find f´(/2) when f(x) = 3cos(2x - /4 ) outer = 3cos(…) inner = 2x - /4 f (x) = -3sin(2x - /4 ) X 2 = -6sin(2x - /4 ) So f (/2) = -6sin( - /4) = -6sin3/4 = -6sin135° . = -6 X 1/2 = -6/2 X 2/2 = -32

Example P() = 1 3cos2 show that P´() = 2tan 3cos2 *********** P() = 1 3cos2 = 1/3(cos)-2 outer = 1/3(….)-2 inner = cos P´() = -2/3(cos)-3 X -sin = 2sin . 3(cos)3 = 2sin . 3cos(cos)2 = 2tan . 3cos2

Applications of Chain Rule Example Find the equation of the tangent to the curve y = -2 (3x – 1)2 (x  1/3) at the point where x = 1. ********* Point Contact If x = 1 then y = -2 (3 – 1)2 = -2/4 = -1/2 (1, -1/2) Gradient y = -2(3x – 1)-2 outer = -2(…)-2 inner = 3x -1 dy/dx = 4(3x – 1) -3 X 3

dy/dx = 12(3x – 1) -3 = 12 (3x – 1)3 when x = 1, dy/dx = 12 23 = 3/2 ctd dy/dx = 12(3x – 1) -3 = 12 (3x – 1)3 when x = 1, dy/dx = 12 23 = 3/2 gradient Using y – b = m(x – a) we get y – (-1/2) = 3/2(x – 1) or y + 1/2 = 3/2x – 3/2 or y = 3/2x – 2 X2 or 2y = 3x - 4 or 3x - 2y = 4

Example Find the coordinates of the stationary points on the curve y = 2sinx – cos2x and determine their nature. ********** SPs occur when dy/dx = 0 2cosx + 2sin2x = 0 2cosx + 4sinxcosx = 0 2cosx(1 + 2sinx) = 0 2cosx = 0 or 1 + 2sinx = 0 1 + 2sinx = 0 cosx = 0 (graph) sinx = -1/2 Q3 or Q4 x = /2 or 3/2 sin-1(1/2) = /6 Q3: x =  + /6 = 7/6 Q4: x = 2 - /6 = 11/6

(/2 ,3) & (3/2 ,-1) are maximum turning points. using y = 2sinx – cos2x (/2 ,3) x = /2 y = 2sin/2 – cos = 2 – (-1) = 3 (3/2 ,-1) x = 3/2 y = 2sin3/2 – cos3 = -2 – (-1) = -1 x = 7/6 y = 2sin7/6 – cos7/3 = -1 – 1/2 = -3/2 (7/6 , -3/2) x = 11/6 y = 2sin11/6 – cos11/3 = -1 – 1/2 = -3/2 (11/6 , -3/2) x  /2  7/6  3/2  11/6  2cosx + 0 - - - 0 + + + 1 + 2sinx + + + 0 - - - 0 + dy/dx + 0 - 0 + 0 - 0 + (/2 ,3) & (3/2 ,-1) are maximum turning points. (7/6 , -3/2) & (11/6 , -3/2) are minimum turning points.

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Integrating (ax + b)n NB: we do this by reversing the chain rule as follows Consider f(x) = (ax + b)n+1 a(n + 1) f´(x) = (n + 1)(ax + b)n X a a(n + 1) = (ax + b)n so it follows that  (ax + b)n dx = (ax + b)n+1 a(n + 1) + C

  30(5x – 2)2 dx Example = 30(5x – 2)3 + C = 2(5x – 2)3 + C 3 X 5 dt . (4t + 1)  6 =  6 = [ ] (4t + 1)1/2 6 (4t + 1)-1/2 dt 2 ½ X 4 2 2 = [ ] 6 ½ (4t + 1) 2 = ( ½ X 5) – ( ½ X 3) = 1

Example Find the shaded area ! ********* Limits (3x + 2)2 – 4 = 0 y = (3x + 2)2 - 4 Find the shaded area ! ********* Limits (3x + 2)2 – 4 = 0 (3x + 2)2 = 4 3x + 2 = -2 or 2 3x = -4 or 0 x = -4/3 or 0

Shaded area =  (3x + 2)2 – 4 dx = [ ] (3x + 2)3 – 4x 3 X 3 = [ ] Shaded area =  (3x + 2)2 – 4 dx -4/3 = [ ] (3x + 2)3 – 4x 3 X 3 -4/3 = [ ] 1/9(3x + 2)3 – 4x -4/3 = 8/9 - ( -8/9 + 16/3 ) = 16/9 - 16/3 = -35/9 Negative sign indicates area under X-axis. Actual area = 35/9 units2

Integration of sin(ax + b) & cos(ax + b) Consider If f(x) = 1/asin(ax + b) then f´(x) = 1/acos(ax + b) X a = cos(ax + b) If g(x) = -1/acos(ax + b) then g´(x) = 1/asin(ax + b) X a = sin(ax + b) It now follows that  cos(ax + b)dx = 1/asin(ax + b) + C (supplied)  sin(ax + b)dx = -1/acos(ax + b) + C

 4sin(2x + ) dx  cos(4 - ) d  -6sin(3 + /2) d Example = 1/4sin(4 - ) + C Example  -6sin(3 + /2) d = 1/3 X 6cos(3 + /2) + C = 2cos(3 + /2) + C Example   4sin(2x + ) dx = [ ]  ½ X –4cos(2x + ) /2 /2 = [ ]  –2cos(2x + ) /2 = (-2cos3) - (-2cos2) = 2 – (-2) = 4

Example By firstly rearranging the formula cos2 = 2cos2 - 1 Find  cos2 d /2 ***********  cos2 d /2 cos2 = 2cos2 - 1 /2 2cos2 - 1 = cos2 =  (1/2cos2 + 1/2) d 2cos2 = cos2 + 1 = [ ½ X 1/2sin2 + 1/2 ] /2 cos2 = 1/2cos2 + 1/2 = [1/4sin2 + 1/2 ] /2 = (/4 + 1/4sin) - (0 + sin0) = /4

Example Find the shaded area !! Curves cross when sin2x = cosx y = sin2x B y = cosx C Curves cross when sin2x = cosx 2sinxcosx – cosx = 0 cosx(2sinx – 1) = 0 cosx = 0 or sin x = 1/2 x = /2 or 3/2 x = /6 or 5/6 B A C

First area =  (sin2x – cosx) dx /2 /6 = [ ] /2 -1/2cos2x - sinx /6 = ( -1/2cos - sin/2) - (-1/2cos /3 - sin/6) = (1/2 – 1) - (-1/4 – ½) = ½ - 1+ ¼ + ½ = 1/4 The diagram has ½ turn symmetry about point B So the total area = 2 X ¼ = 1/2unit2