SUMMER KNOWHOW STUDY AND LEARNING CENTRE Differential Calculus
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Contents Limits..5 Gradients, Tangents and Derivatives.6 Differentiation from First Principles.8 Rules for Differentiation..10 Chain Rule.12 Product Rule 14 Quotient Rule..15 Maxima and Minima 16 Curve Sketching.18 Rates of Change.21 Small Changes and Approximation.. 23 3
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LIMITS The limit of a function describes the behaviour of a function as the variable approaches a particular value. Examples 1. Find the limit of the function x + 2 as x approaches 2 The behaviour of as x 2 is shown in the table: x 1.95 1.995 2 2.005 2.05 3.95 3.995 4 4.005 4.05 The table shows that as x approaches 2, approaches 4 im 2 4 2. Find (5 2 ) The 2xh and h terms will approach zero as h approaches zero. The limit can be found by substituting zero for h:. (5 2 ) 5x 2 + 2x.0 + 0 5x 2 3. Find It is not possible to find the limit by substituting h 0 But consider the behaviour of 0 It appears that im 1 Exercises Determine the limit of the following: 1) im 4 2 2) im 3) im 4) im h -0.5-0.3-0.2-0.1 0 0.01 0.1 0.2 0.3 0.5 f(h) 0.959 0.985 0.993 0.998? 0.99998 0.998 0.993 0.985 0.959 Answers 2 2) 3) x 2 4) 1 5
GRADIENTS, TANGENTS AND DERIVATIVES Gradient of a Curve The gradient at a point on a curve is the gradient of the tangent to the curve at that point. Special cases: horizontal and vertical lines A line parallel to the x-axis with equation of the form y k (k constant), has a gradient of zero. As a line becomes c oser to vertica it s gradient gets larger and larger. A line parallel to the y-axis with equation of the form x c ( c constant) has a gradient which is undefined. Consider the gradient of the curve at the point P (ie the gradient of the line AB). x This gradient cannot be calculated - only one point on the line is known. But the gradient of the line PQ can be calculated and this can be used to approximate the gradient of AB. 6
The gradient of PQ As the value of h decreases (i.e Q becomes closer to the point P), the approximation of the gradient is more accurate. The value of the gradient becomes most accurate as h approaches zero. The gradient formula for the curve y f(x) is defined as the derivative function im, h 0 The derivative function gives the slope of the tangent to the curve at any point x. Example If the derivative function of is, find the slope of the tangent to the curve at x 4 At x 4, (4) Exercises 1. If the derivative function for x 3 x is 3x 2-1, find the slope of the tangent to this curve at a) x 2 b) x 0 c) x -9 2. If the derivative function of sin(x) is cos(x) find the gradient of y sin(x) at a) x 0 b) x c) x 3.5 3. Determine im a) x 2 b) x 0 c) x -9 and hence find the slope of the tangent to the curve y x 2 at Answers 1. a) 11 b) -1 c) 242 2. a) 1 b) 0 c) -0.94 3. im 2x a) 4 b) 0 c) -18 7
DIFFERENTIATION FROM FIRST PRINCIPLES The process of finding the derivative function using the definition im, h 0 is called differentiating from first principles. Examples 1. Differentiate x 2 from first principles. im im, h 0 im im im im 2 2x If x 2 then 2x 2. Determine, from first principles, the gradient function for the curve 2x 2 x and calculate its value at x 3. im im, h 0 im im im im 4 2 1 4x - 1 The gradient function is 4x 1 At x 3, 4x - 1 4(3) 1 11 3. Use differentiation from first principles to find the gradient function of y im, h 0 8
im im im im im ( ) im ( ) - Exercises Find the derivative of the following, using differentiation from first principles. 1) 3x 2) 5x 2-4 3) 2x 2 - Answers 1) 3 2) 10x 3) 4x + 9
RULES FOR DIFFERENTIATION Operational Rules The fo owing ru es for differentiation can be estab ished very easi y from first princip es If g (x) k f(x), where k is a constant then g'(x) k f '(x) If f(x) k where k is a constant then f '(x) 0 If f(x) g(x) + h(x) then f '(x) g '(x) + h '(x) Derivative of a power of x If y, then n Examples 1. If y, then 7 7 2. If y, then 2015 2015 3. If y, then [Rewrite the expression in index form before differentiating] 3. If y 2 2 then -2-2 And using some of the operational rules: 5. y + 7, 3 42 6. y + 10 4 + 10, -4 0 7. y - -, [divide through by ] + 10
Derivatives of some other functions Function Derivative og sin(x) cos(x) cos(x) -sin(x) tan(x) Examples 1. sin 3, cos 6 2. 5 3. 10 (cos ) 10 cos, 10 sin 5 2 Exercises 1. Differentiate the following a) b) c) d) e) 53 f) g) 5 h) i) j) 3 + 2x 2. Find the derivative of a) sin x - cos x b) 10 og c) tan x - d) 3sin(x) 2 + e) - Answers 1. a) b) c) -19 d) -4 e) 0 f) g) 30 h) 45 -i) - 5 j) 6x + 2 2. a) cos(x) + sin(x) b) - c) sec 2( x) - d) 3cos(x) - e) - 11
CHAIN RULE The chain ru e is used to differentiate a function which is the composition of two simpler functions If y g[u] where u h(x), then Examples 1) Differentiate y (2x - 1) 4 Let u 2x 1, then y u 4 2 and 4u 3 4u 3. 2 8u 3 8(2x 1) 3 [since u 2x 1] 2) Find the derivative of y y (5t 2t 1) [change to index form for easier differentiation] Let u 5t 2 + 2t + 1, then y du dt 10t + 2 and dy du -. (10t + 2) 1 (5t 2t 1). (10t + 2) [since u 5t 2 + 2t + 1] 3 (5t 2t 1) [after simplifying] 3) Differentiate y sin5x y sin5x Let y sin(u) where u 5x cos(u) and 5 Then cos(u). 5 5cos(5x) 12
4). If f(x) cos 3 x find f'(x) y cos 3 x [cos(x)] 3 Let y u 3 where u cos(x) 3u 2 and -sin(x) Then f '(x) 3u 2.(-sinx) 3cos 2 x.(-sinx) -3sinxcos 2 x 5) Differentiate (log e4x) 3 Let y u 3 where u log ev and v 4x [The chain rule can be extended to three or more functions!!] 3u 2, and 4 Then 3u 2.. 4 3(log ev) 2.. 4 3(log e4x) 2.. 4 (log e4x) 2 Exercise Find the derivatives of the following functions 1) y tan3x 2) f(x) log e 3) y sin ( 2 ) 4) y cos 2 x 5) f(x) 6) y 1 cos(5 ) Answers 1) y 3sec 2 3x 2) f (x) 3) y -2cos ( 2 ) 4) y -2 sinx cosx 5) f (x) e sinx cosx 6) y 13
PRODUCT RULE The product ru e is used when we want to differentiate the product of two functions If y u(x).v(x) then y u(x).v'(x) + u'(x).v(x) which is often abbreviated to y uv u v Examples 1. Find the derivative of (x + 3) 6 (2x 1) Let u (x+3) 6 and let v 2x - 1 u' 6(x+3) 5 v' 2 Then y' uv' + u'v (x + 3) 6.2 + 6(x + 3) 5 (2x 1) 2(x + 3) 6 + 6(x + 3) 5 (2x 1)...and this could (but does not have to be) simplified further.. 2(x + 3) 5 [(x + 3) + 3(2x 1)] [by factorizing] 2(x + 3) 5 (7x) 14x(x + 3) 5 2. Differentiate e x sin(2x) Let u e x and v sin2x u e x and v 2cos2x [using the chain rule] Then y' uv' + u'v e x.2cos2x + e x sin2x 2e x cos2x + e x sin2x Exercises 1. Use the product rule to differentiate the following a) y (x 2)(6x + 7) b) f(x) (2x 2 + 4)(x 5 + 4x 2 2) [simplify as far as possible] c) y ( - 1)(x 2 + 1) d) y (x 3 4x + )(3x 4 + 2) [simplify as far as possible] 2. Find the derivative of a) y e x tanx b) y og c) y sinx cosx d) y [Hint: x -1 ] Answers 1. (a) 12x 5 (b) (2x 2 + 4)(5x 4 + 8x) + 4x(x 5 + 4x 2 2) (c) 2x + (d) (x 3 4x + )(12x 3 ) + (3x 2 4 + )(3x 4 + 2) 2. (a) e x tanx + e x sec 2 x (b) x + 2x og (c) cos 2 x sin 2 x (d) 14
QUOTIENT RULE The quotient rule is used when we want to differentiate a function which is the quotient of two simpler functions: If then which is often abbreviated to y' Examples 1) If y, find u 1 + x u' 1 and v x 2 3 v' 2x y' and simp ify if possib e 2 3 2 2 2 ( 2 3) 2 2) Differentiate u x 2 u 2x and v og v y'.. after simplifying Exercise Find the derivatives of the following functions 1) 2) 3) y 4) Answers 1) 2) 3) 15 after simplifying 4) after simplifying
MAXIMA AND MINIMA The maximum or minimum points of a function occur where the derivative is zero. We can therefore use calculus to solve problems that involve maximizing or minimizing functions. Examples 1) The distance s km, to the nearest km, of a ship from a lighthouse at any time, t hours, is given by the formula s 2 + 8t 2.5 t 2. When is the ship furthest from the lighthouse and what is its distance from the lighthouse? s 2 + 8t 2.5 t 2 8 5t The maximum distance will occur when 0: ie 8 5t 0 5t 8 t 1.6 When t 1.6: s 2 + 8(1.6) 2.5(1.6) 2 8.4 The ship is furthest from the lighthouse after 1.6 hours and the distance is 8.4 km. 2) Find two with the maximum product if the sum of the numbers is 10. Let the numbers be a and b and the product P [define the variables you are using] Then a + b 10 b 10 a Also P a b ie P a (10 a) P 10a a 2 and 10 2a The maximum value of P will occur where 0: If a 5 then b 10 - a 10 5 5 The numbers are both 5 ie 10 2a 0 10 2a a 5 16
Exercises 1. Find two positive numbers whose sum is 18 such that the sum of their squares is a minimum. 2. Find the minimum value of the function f(x) 5x 2-30x + 17 3. A ball is thrown vertically upward. The height, h(t) m above the ground is a function of time with the formula h(t) 15t 5t 2 + 1.6. Find the greatest height reached. 4. What is the maximum area that can be enclosed if a rectangle is created with a piece of wire 48 cm long? 5. The annual profit P made on a garment is related to the number n that are produced by the formula P(n) 300n 7200-0.2n 2. How many garments should be produced to maximize profit? Answers 1. The two numbers are both 9. 2. -28 3. 12.85m 4. 144cm2 750 17
CURVE SKETCHING To sketch a curve, find the maximum and minimum stationary points the intercepts on the axes A stationary point is a point on a graph of a function y f(x) where the tangent to the curve is horizontal. At a stationary point the derivative function y f '(x) 0. A maximum stationary point occurs at x a if f '(x) > 0 for x < a f '(x) 0 for x a f '(x) < 0 for x > a A minimum stationary point occurs at x a if f '(x) < 0 for x < a f '(x) 0 for x a f '(x) > 0 for x > a Example Find the turning point of the parabola defined by y x² + 4x + 5 f(x) x² + 4x + 5 f '(x) 2x + 4 At a stationary point f '(x) 0 2x + 4 0 2x -4 x -2 When x -2, y (-2)² + 4(-2) + 5 1 So there is a stationary point at (-2, 1). Sign Test A sign test can be used to check whether the stationary point is a minimum or maximum. 18
Check the slope of the tangent on each side of the stationary point: x -2.1-2 -1.9 f '(x) 0 + gradient \ / There is a minimum point at (-2,1) Example Sketch the graph of y x 3 x f(x) x 3 x f '(x) 3x² - 1 Stationary points: f '(x) 0 3x² - 1 0 3x² 1 x² x or x 0.58 When x 0.58, y -0.38 (0.58,-0.38) x -0.58, y 0.38 (-0.58, 0.38) Do sign tests to check whether stationary points are minima or maxima: x 0.5 0.58 0.6 f '(x) 0 + gradient \ / x -0.6-0.58-0.5 f '(x) + 0 gradient / \ There is a minimum point at (0.58, -38) and a maximum point at (-0.58, 0.38) x-intercepts: When y 0, x 3 x 0 x(x² - 1) 0 x(x- 1)(x + 1) 0 x-intercepts at x 0, x 1 and x -1 y-intercepts: When x 0, y 0 y x 3 - x 2 1.5 1 0.5 0-3 -2-1 0 1 2 3-0.5-1 -1.5-2 19
Exercise Sketch the graphs of the following functions showing all intercepts and turning points 1. y x² - 4x 2. y x 3 2x² + x 3. y 6 x - x² 4. y (x + 1) 4 Answers 1. 2. 6 5 4 3 2 1 0-2 -1 0 2 4 6-2 -3-4 -5 2 1 0-2 -1-1 0 1 2 3-2 -3-4 -5 3. 4. 10 5 0-5 0 5-5 -10 20 15 10 5 0-4 -3-2 -1 0 1 2-5 20
RATES OF CHANGE If there is a relationship between two or more variables, for example, area and radius of a circle (A πr 2 ), or length of a side and volume of a cube ( V l 3 ) then there will also be a relationship between the rates at which they change. If y is a function of x ie y then is the rate of change of y with respect to x We can use differentiation to find the function that defines the rate of change between variables 2 π r (differentiating with respect to r) and 3 l 2 (differentiating with respect to l) The chain rule can be used to find rates of change with respect to time: So that A πr 2 2 π r and V l 3 3 l 2 Examples 1. A balloon has a small hole and its volume V (cm3) at time t (sec) is V 66 10t 0.01t 2, t > 0. Find the rate of change of volume after 10 seconds. V 66 10t 0.01t 2-10 0.02t When t 10, -10 (0.02)(10) - 10.2 cm3/sec 2. The pressure P, of a given mass of gas kept at constant temperature, and its volume V are connected by the equation PV 500. Find when V 20. PV 500 ie. P 21
ie. P 500V -1 Then -500V -2 When V 20: -500(20) -2-1.25 3. Water is running out of a conical funnel at the rate of 5cm 3 /s. The radius of the funnel is 10 cm and the height is 20 cm. How fast is the water level dropping when the water is 10 cm deep? Let h be the depth, r the radius and V be the volume of the water at time t Then -5 (since the rate is decreasing) 10 By similar triangles: 20 r r V πr 2 h [formula for volume of a cone] πh 3 [since r h 2 When h 10, -5 10 2 [ rearranging to find The water level is dropping at a rate of cm/s Exercises 1) The radius of a spherical balloon is increasing at a rate of 3 cm/min. At what rate is the volume increasing when the radius is 5cm? 2) If the displacement of an object from a starting point is given by s(t) sin(t) 2cos(t) find the velocity when t 1. Hint v(t) s (t) 3) The function n(t) 200t - 100 describes the spread of a virus where t is the number of days since the initial infection and n is the number of people infected. Find the rate at which n is increasing at the instant when t 4. 4) If y find when x 2, given 1. 5) A hollow right circular cone is held vertex downwards beneath a tap leaking at the rate of 2cm3/s. Find the rate of rise of water level when the depth is 6cm given that the height of the cone is 18 cm and its radius 12cm. Answers 1) 300π 942 cm3/min 2) 2.22 3) 175people/day 4) 5) cm/s 0.04 cm/s 22
SMALL CHANGES & APPROXIMATIONS Consider a function defined by y. If x is increased by a small amount x to x + x, then as x 0, and Therefore if x is small, y x or y x Examples 1. The side of a square is 5cm. How much will the area of the square increase when the side expands by 0.01cm? Let the area of the square be A and the length of a side be x cm. Then A x 2 and 2x x 2x x 2 5 0.01 0.1 The increase in area 0.1 cm 2 2. A 2% error is made in measuring the radius of a sphere. Find the percentage error in the volume. Let the radius be r and the volume be V, r 0.02r Then V πr 3 and 4πr 2 r 4πr 2 r 4πr 2 0.02r 0.08πr 3 23
The percentage error in the volume: % error 100 %. 100 %. 100 % Exercises 6 % The percentage error in the vo ume 6% 1. If the radius of a sphere is increased from 10cm to 10.1 cm what is the approximate increase in surface area? 2. The height of a cylinder is 10 cm and its radius is 4cm. Find the approximate increase in volume when the radius increases to 4.02 cm. 3. An error of 3% is made in measuring the radius of a sphere. Find the percentage error in volume. 4. The kinetic energy K of a body of mass m moving with speed v is given by K mv 2. If a body s speed is increased by 1.5% what is the approximate percentage change in the kinetic energy? Answers 1) 25.13cm 2 2) 5.03cm 3 3) 9% 4) 3% 24