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© Boardworks Ltd 2006 1 of 50 These icons indicate that teacher’s notes or useful web addresses are available in the Notes Page. This icon indicates the slide contains activities created in Flash. These activities are not editable. For more detailed instructions, see the Getting Started presentation. © Boardworks Ltd 2006 1 of 50 A2-Level Maths: Core 4 for Edexcel C4.4 Differentiation

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© Boardworks Ltd 2006 2 of 50 Implicit differentiation Parametric differentiation Differentiating exponential functions Exponential growth and decay Connected rates of change Examination-style questions Contents © Boardworks Ltd 2006 2 of 50 Implicit differentiation

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© Boardworks Ltd 2006 3 of 50 Implicit differentiation When neither x nor y is given explicitly in terms of the other then the curve is said to be defined implicitly. For example: It is not always easy, or even possible, to rearrange an implicit function into an explicit form. When the equation of a curve is given in the form y = f ( x ) then the variable y is said to be given explicitly in terms of the variable x. For example: are explicit functions. and are implicit functions.and

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© Boardworks Ltd 2006 4 of 50 Implicit differentiation Differentiating every term in the equation with respect to x gives: For this reason we need to develop a technique to differentiate such functions in implicit form. Differentiate with respect to x. where is taken as an operator meaning ‘differentiate with respect to x’. The term in x and the constant can be differentiated directly to give:

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© Boardworks Ltd 2006 5 of 50 Implicit differentiation This would normally be done in a single step so: To differentiate the term in y with respect to x we have to use the chain rule. becomes We can now divide through by 2 y and rearrange to find :

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© Boardworks Ltd 2006 6 of 50 Implicit differentiation Differentiating term by term with respect to x gives: Differentiate with respect to x. Now rearrange to collect the terms in together: 3x23x2

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© Boardworks Ltd 2006 7 of 50 Implicit differentiation Differentiating term by term with respect to x : In some cases, we might also need to use the product rule. For example: Differentiate with respect to x. The first term is treated as a product to give: Using

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© Boardworks Ltd 2006 8 of 50 Implicit differentiation Once we have differentiated a curve that has been defined implicitly, we can find the equation of the tangent or the normal to the curve at a given point. Rearrange to collect the terms in together:

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© Boardworks Ltd 2006 9 of 50 Implicit differentiation Differentiating with respect to x gives: Find the equation of the tangent to the curve x 2 + y 2 – xy = 7 at the point (2, –1). Rearranging to collect the terms in together:

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© Boardworks Ltd 2006 10 of 50 Implicit differentiation At the point (2, –1), x = 2 and y = –1: The gradient of the tangent at (2, –1) is therefore. We can find the equation of the tangent using the coordinates: Using y – y 1 = m ( x – x 1 )

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© Boardworks Ltd 2006 11 of 50 Implicit differentiation Differentiating with respect to x gives: Find the equation of the normal to the curve 2 x 2 – ( y + 3) 2 = 1 at the point (5, 4).

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© Boardworks Ltd 2006 12 of 50 Implicit differentiation We need the gradient at the point (5, 4): The equation of the normal is given by: Using y – y 1 = m ( x – x 1 ) The gradient of the normal at (5, 4) is therefore.

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© Boardworks Ltd 2006 13 of 50 Implicit differentiation Parametric differentiation Differentiating exponential functions Exponential growth and decay Connected rates of change Examination-style questions Contents © Boardworks Ltd 2006 13 of 50 Parametric differentiation

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© Boardworks Ltd 2006 14 of 50 Parametric differentiation We can differentiate both of these equations with respect to the parameter t to give: can be found using the chain rule: Suppose we want to find for a curve that has been defined parametrically. For example: and

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© Boardworks Ltd 2006 15 of 50 Parametric differentiation Differentiating each equation with respect to t gives: In general, then, to differentiate a pair of parametric equations we can use the chain rule in the form: Find, in terms of t, for the curve defined by the parametric equations x = cos 2 t and y = sin t.

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© Boardworks Ltd 2006 16 of 50 Parametric differentiation This can be simplified further using the double angle formula for sin 2 t : = cosec t

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© Boardworks Ltd 2006 17 of 50 Parametric differentiation Differentiating with respect to x gives A curve is defined by the parametric equations x = (3 t + 2) 2 and y = 2 t 3 + 9. Find in terms of t and hence find the coordinates of the points where the gradient of the curve is –1.

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© Boardworks Ltd 2006 18 of 50 Parametric differentiation The gradient of the curve is –1 when: So the gradient of the curve is –1 when t = –1 and when t = –2. When t = –1: x = (–3 + 2) 2 = 1and y = 2(–1) 3 + 9= 7 When t = –2: x = (–6 + 2) 2 = 16and y = 2(–2) 3 + 9= –7 The gradient of the curve is –1 at (1, 7) and (16, –7).

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© Boardworks Ltd 2006 19 of 50 Implicit differentiation Parametric differentiation Differentiating exponential functions Exponential growth and decay Connected rates of change Examination-style questions Contents © Boardworks Ltd 2006 19 of 50 Differentiating exponential functions

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© Boardworks Ltd 2006 20 of 50 Differentiating exponential functions Suppose we want to differentiate a general exponential function of the form Take the natural log of both sides: Now differentiating with respect to x gives: y = a x where a is a constant.

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© Boardworks Ltd 2006 21 of 50 Differentiating exponential functions Since y = a x we can substitute a x for y. So in general: For example:

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© Boardworks Ltd 2006 22 of 50 Implicit differentiation Parametric differentiation Differentiating exponential functions Exponential growth and decay Connected rates of change Examination-style questions Contents © Boardworks Ltd 2006 22 of 50 Exponential growth and decay

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© Boardworks Ltd 2006 23 of 50 Exponential growth Exponential growth occurs when a quantity increases at a rate that is proportional to its size. In other words, when a quantity grows exponentially, the larger it becomes, the quicker it grows. Examples of quantities that grow exponentially include: the number of micro-organisms in a culture dish, investments with a fixed compound interest rate, population growth. For example, a bacteria colony starts with 100 bacteria and doubles every minute. The number of bacteria can be shown in a table.

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© Boardworks Ltd 2006 24 of 50 Exponential growth So if N is the number of bacteria after time t we have: 100 × 2 t …800400200100 Number of bacteria t …3210Time (mins) N = 100 × 2 t More generally, if b is the number of bacteria when t = 0, N = b 2 t In practice it is more usual to express exponential growth in terms of the constant e. We can write the number 2 in terms of e as e ln 2, and so the formula becomes N = be t ln 2 or N = be 0.693 t

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© Boardworks Ltd 2006 25 of 50 t 0 f(t)f(t) Exponential growth In general, exponential growth can be modelled by the function f ( t ) = Ae kt A is the original quantity (the quantity when t = 0), f ( t ) is the quantity after time t and where t is time, k is a positive constant (the growth rate) y = Ae kt An exponential growth curve has the following basic shape: A

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© Boardworks Ltd 2006 26 of 50 Exponential decay Exponential decay occurs when a quantity decreases at a rate that is proportional to its size. In other words, when a quantity decays exponentially, the smaller it becomes, the more slowly it decays. Examples of quantities that decay exponentially include: the number of atoms in a radioactive isotope, the value of a car as it depreciates, the rate at which an object cools (when the external temperature is constant).

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© Boardworks Ltd 2006 27 of 50 t 0 Exponential decay In general, exponential decay can be modelled by the function f ( t ) = Ae – kt A is the original quantity (the quantity when t = 0), f ( t ) is the quantity after time t, and where t is time, k is a positive constant (the decay rate). y = Ae – kt An exponential decay curve has the following basic shape. A f(t)f(t)

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© Boardworks Ltd 2006 28 of 50 Exponential decay Forensic scientists can predict a recently murdered victim’s time of death from the temperature of the body. They do this by applying Newton’s law of cooling: A body is discovered at 8.30 pm. The body’s temperature is recorded as 30°C and room temperature as 20°C. One hour later, the temperature of the body is 29°C. Assuming that the room temperature remains constant throughout, estimate the time of death. d = ae – kt where d is the temperature difference (between the body and its surroundings), a and k are constants and t is the time that has passed since the body started to cool. For example,

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© Boardworks Ltd 2006 29 of 50 Exponential decay Let t = 0 at 8.30 pm. d = 30 – 20 = 10 10 = ae – k (0) = ae 0 = a And using d = ae – kt : When t = 1, one hour later: d = 29 – 20 = 9 9 = ae – k (1) = ae – k Using d = ae – kt : But a = 10 so, 9 = 10 e – k e – k = 0.9

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© Boardworks Ltd 2006 30 of 50 Exponential decay We can solve e – k = 0.9 by taking natural logs on both sides: ln ( e – k ) = ln 0.9 Substitute these values of k and a into d = ae – kt : d = 10 e –0.105 t d = 36.9 – 20 = 16.9 If we assume that the murder victim had a normal body temperature of 36.9 °C when she died and that the room temperature was 20°C, then at the time of death: – k = ln 0.9 – k = –0.105 (to 3 s.f.) k = 0.105

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© Boardworks Ltd 2006 31 of 50 Exponential decay Substituting d = 16.9 into d = 10 e –0.105 t gives 16.9 = 10 e –0.105 t Solving this equation for t will give the time since the victim died. t = 0 at 8.30 pm, so t = –5 at 3.30 pm. e –0.105 t = 1.69 –0.105 t = ln 1.69 t = t = –5.00 (to 3 s.f.) The victim died at about 3.30 pm.

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© Boardworks Ltd 2006 32 of 50 Implicit differentiation Parametric differentiation Differentiating exponential functions Exponential growth and decay Connected rates of change Examination-style questions Contents © Boardworks Ltd 2006 32 of 50 Connected rates of change

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© Boardworks Ltd 2006 33 of 50 Rates of change When we talk about rates of change we are usually talking about the rate at which a variable changes with respect to time. Suppose, for example, that a spherical balloon is slowly being inflated. There are several changes we could measure, such as: We can connect these rates of change using the chain rule. The rate at which the radius r changes,The rate at which the surface area A changes,The rate at which the volume V changes, For example, the rate of change of the surface area is connected to the rate of change of the radius by

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© Boardworks Ltd 2006 34 of 50 Rates of change So, the rate of change of the surface area is connected to the rate of change of the radius by If the surface area of the balloon is increasing at a rate of 15 cm 2 s –1, find the rate at which the radius of the balloon is increasing at the moment when the radius is 4 cm. We can find by differentiating the formula for the surface area of a sphere with respect to the radius. A = 4 πr 2 so

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© Boardworks Ltd 2006 35 of 50 Connected rates of change We are given that = 15 and we want to find when r = 4. Using we have: = 0.149 So, the radius is increasing at a rate of 0.149 cm s –1 (to 3 s.f.) at the moment when the radius of the balloon is 4 cm.

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© Boardworks Ltd 2006 36 of 50 Connected rates of change So If the air in the fully-inflated balloon is released at a rate of 30 cm 3 s –1, find the rate at which the surface area is decreasing at the moment when the radius is 5 cm. Using the chain rule, the rate of change of the volume is connected to the rate of change of the radius by: We are given that = –30 and we want to find when r = 5.

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© Boardworks Ltd 2006 37 of 50 Connected rates of change Here = –30 and r = 5: We can now find using with r = 5: 4 = –12 So, the surface area is decreasing at a rate of 12 cm 2 s –1 at the moment when the radius of the balloon is 5 cm.

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© Boardworks Ltd 2006 38 of 50 Connected rates of change

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© Boardworks Ltd 2006 39 of 50 Implicit differentiation Parametric differentiation Differentiating exponential functions Exponential growth and decay Connected rates of change Examination-style questions Contents © Boardworks Ltd 2006 39 of 50 Connected rates of change

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© Boardworks Ltd 2006 40 of 50 Examination-style question 1 The diagram shows the curve defined by the parametric equations x = 2 t 2 and y = t 2 + 2 t. a)Find the gradient of the curve at the point P where t = 1. b)Find the equation of the normal at the point P. c)The normal to the curve at the point P cuts the curve again at the point Q. Find the coordinates of the point Q. x y O P Q

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© Boardworks Ltd 2006 41 of 50 Examination-style question 1 a) when t = 1: Using gives:

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© Boardworks Ltd 2006 42 of 50 Examination-style question 1 b) The gradient of the normal to the curve at the point P is –1. Also when t = 1, If the gradient of the normal is –1 and it passes through the point (2, 3) then its equation is given by: x = 2 y = 3 c) Substituting x = 2 t 2 and y = t 2 + 2 t into the equation for the normal gives:

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© Boardworks Ltd 2006 43 of 50 Examination-style question 1 The normal therefore cuts the curve when t = 1 and when t =. When t = : The coordinates of the point Q are therefore.

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© Boardworks Ltd 2006 44 of 50 Examination-style question 2 The mass, m grams, of a sample of radioactive iodine decays according to the formula where t is the number of days after it is first observed. a)What is the initial mass of the sample? b)Sketch the graph of m against t. c)What is the mass of the sample after 2 days? d)Calculate the time it takes for the sample to halve its original mass. e)Find the rate at which the sample is decaying when t = 5.

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© Boardworks Ltd 2006 45 of 50 Examination-style question 2 a) When t = 0, m = 15 e 0 So, the initial mass of the sample is 15 g. b) The graph of m = 15 e –0.083 t will be an exponential decay curve passing through the point (0, 15). = 15 c) When t = 2, m = 15 e –0.083 × 2 So, the mass of the sample after 2 days is 12.71 g (to 2 d.p.). = 12.71 (to 2 d.p.) m t 15

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© Boardworks Ltd 2006 46 of 50 Examination-style question 2 d) The initial mass is 15 g, so we need to find the time it takes for the sample to reach 7.5 g. 7.5 = 15 e –0.083 t 0.5 = e –0.083 t Take logs: ln 0.5 = ln e –0.083 t ln 0.5 = –0.083 t = 8.35 (to 2 d.p.) The sample takes 8.35 days (to 2 d.p.) to halve its mass. That is, when

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© Boardworks Ltd 2006 47 of 50 Examination-style question 2 d) The rate at which the sample is decaying is given by. When t = 5 Therefore, when t = 5, the sample is decaying at a rate of 0.82 grams per day (to 2 d.p.).

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© Boardworks Ltd 2006 48 of 50 Examination-style question 3 The height of a conical container is twice its radius, r cm, as shown in the diagram. 2r2r r Liquid is poured into the container at a rate of 5 litres per minute. If x cm is the depth of the liquid at time t minutes, write an expression for the rate at which the depth is increasing when x = 2 cm. (The volume of a cone of radius r and height h is given by ). The volume of the liquid in the container at time t is given by

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© Boardworks Ltd 2006 49 of 50 Examination-style question 3 The liquid enters the container at a rate of 5 litres per minute so Since the radius of the container is half its depth this can be written in term of x as

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© Boardworks Ltd 2006 50 of 50 Examination-style question 3 So Using the chain rule At the instant when x = 2 cm the rate at which the liquid is entering the container is

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