# A2-Level Maths: Core 4 for Edexcel

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A2-Level Maths: Core 4 for Edexcel
C4.4 Differentiation 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. 1 of 50 © Boardworks Ltd 2006

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

Implicit differentiation
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: and are explicit functions. When neither x nor y is given explicitly in terms of the other then the curve is said to be defined implicitly. For example: and are implicit functions. It is not always easy, or even possible, to rearrange an implicit function into an explicit form.

Implicit differentiation
For this reason we need to develop a technique to differentiate such functions in implicit form. Differentiate with respect to x. Differentiating every term in the equation with respect to x gives: 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:

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

Implicit differentiation
Differentiate with respect to x. Differentiating term by term with respect to x gives: 3x2 Now rearrange to collect the terms in together:

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

Implicit differentiation
Rearrange to collect the terms in together: 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.

Implicit differentiation
Find the equation of the tangent to the curve x2 + y2 – xy = 7 at the point (2, –1). Differentiating with respect to x gives: Rearranging to collect the terms in together:

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 – y1 = m(x – x1)

Implicit differentiation
Find the equation of the normal to the curve 2x2 – (y + 3)2 = 1 at the point (5, 4). Differentiating with respect to x gives:

Implicit differentiation
We need the gradient at the point (5, 4): The gradient of the normal at (5, 4) is therefore . The equation of the normal is given by: Using y – y1 = m(x – x1)

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

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

Parametric differentiation
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 2t and y = sin t. Differentiating each equation with respect to t gives:

Parametric differentiation
This can be simplified further using the double angle formula for sin 2t: = cosec t

Parametric differentiation
A curve is defined by the parametric equations x = (3t + 2)2 and y = 2t3 + 9. Find in terms of t and hence find the coordinates of the points where the gradient of the curve is –1. Differentiating with respect to x gives

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 = 1 and y = 2(–1)3 + 9 = 7 When t = –2: x = (–6 + 2)2 = 16 and y = 2(–2)3 + 9 = –7  The gradient of the curve is –1 at (1, 7) and (16, –7).

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

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

Differentiating exponential functions
Since y = ax we can substitute ax for y. So in general: For example:

Exponential growth and decay
Implicit differentiation Parametric differentiation Differentiating exponential functions Exponential growth and decay Connected rates of change Examination-style questions Contents Note that the use of differential equations to model exponential growth and decay is covered in C4.6 Integration 2. 22 of 50 © Boardworks Ltd 2006

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. All of the models listed involve certain assumptions. For example, bacteria in a dish will only continue to grow as long as there is sufficient food and space. For example, a bacteria colony starts with 100 bacteria and doubles every minute. The number of bacteria can be shown in a table.

Exponential growth Number of bacteria t 3 2 1 Time (mins) 100 200 400 800 100 × 2t So if N is the number of bacteria after time t we have: N = 100 × 2t More generally, if b is the number of bacteria when t = 0, N = b2t In practice it is more usual to express exponential growth in terms of the constant e. The reason that exponential growth is written in terms of the exponential function to make it easier to calculate the rate of growth with respect to time. This will be explored in more detail in relation to differential equations in C4. We can write the number 2 in terms of e as eln 2, and so the formula becomes N = bet ln 2 or N = be 0.693t

Exponential growth In general, exponential growth can be modelled by the function f(t) = Aekt where t is time, A is the original quantity (the quantity when t = 0), f(t) is the quantity after time t and k is a positive constant (the growth rate) An exponential growth curve has the following basic shape: t f(t) y = Aekt A The dotted line shows values for negative t. Note that transformations of the exponential growth curve can also be used, particularly if the growth tends to a particular value.

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).

Exponential decay In general, exponential decay can be modelled by the function f(t) = Ae–kt where t is time, A is the original quantity (the quantity when t = 0), f(t) is the quantity after time t, and k is a positive constant (the decay rate). An exponential decay curve has the following basic shape. t f(t) A The exponential decay curve is a reflection of the exponential growth curve in the vertical axis. The dotted line shows values for negative t. y = Ae–kt

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: 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, 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.

Exponential decay Let t = 0 at 8.30 pm. d = 30 – 20 = 10
And using d = ae–kt: 10 = ae–k(0) = ae0 = a When t = 1, one hour later: d = 29 – 20 = 9 Using d = ae–kt: 9 = ae–k(1) = ae–k But a = 10 so, 9 = 10e–k e–k = 0.9

Exponential decay We can solve e–k = 0.9 by taking natural logs on both sides: ln (e–k ) = ln 0.9 –k = ln 0.9 –k = – (to 3 s.f.) k = 0.105 Substitute these values of k and a into d = ae–kt: d = 10e–0.105t 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: d = 36.9 – 20 = 16.9

Exponential decay Substituting d = 16.9 into d = 10e–0.105t gives
Solving this equation for t will give the time since the victim died. e–0.105t = 1.69 –0.105t = ln 1.69 t = You may wish to discuss the limitations of the model. For example, it assumes that the temperature in the room has remained unchanged and that the victim had a normal body temperature when she died. t = –5.00 (to 3 s.f.) t = 0 at 8.30 pm, so t = –5 at 3.30 pm.  The victim died at about 3.30 pm.

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

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: 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, We can connect these rates of change using the chain rule. For example, the rate of change of the surface area is connected to the rate of change of the radius by

Rates of change We can find by differentiating the formula for the surface area of a sphere with respect to the radius. A = 4πr2 so 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 cm2 s–1, find the rate at which the radius of the balloon is increasing at the moment when the radius is 4 cm.

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 cm s–1 (to 3 s.f.) at the moment when the radius of the balloon is 4 cm.

Connected rates of change
If the air in the fully-inflated balloon is released at a rate of 30 cm3 s–1, find the rate at which the surface area is decreasing at the moment when the radius is 5 cm. We are given that = –30 and we want to find when r = 5. Using the chain rule, the rate of change of the volume is connected to the rate of change of the radius by: Stress that the rate of change of the volume is negative because the volume is decreasing. So

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 cm2 s–1 at the moment when the radius of the balloon is 5 cm.

Connected rates of change
This activity shows a cylindrical beaker that is being filled with water at a constant rate. This rate can be varied as can the radius of the cylinder. The rate of change of the volume of the water is related to the rate of change of the height of the water by dv/dt = dv/dh × dh/dt. But v = πr2h so dv/dh = πr2. This gives dv/dt = πr2 × dh/dt. Challenge students to use this relationship to find the rate at which the height is increasing given the radius of the cylinder and the rate at which it is filling with water.

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

Examination-style question 1
The diagram shows the curve defined by the parametric equations x = 2t2 and y = t2 + 2t. Find the gradient of the curve at the point P where t = 1. Find the equation of the normal at the point P. 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

Examination-style question 1
Using gives: when t = 1:

Examination-style question 1
b) The gradient of the normal to the curve at the point P is –1. Also when t = 1, x = 2 y = 3 If the gradient of the normal is –1 and it passes through the point (2, 3) then its equation is given by: c) Substituting x = 2t2 and y = t2 + 2t into the equation for the normal gives: Explain that the points where the normal intersects the curve can be found by substituting the parametric equations of the curve into the equation of the normal. Since this is the normal at the point where t = 1, (t – 1) must be a factor of the resulting quadratic.

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

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. What is the initial mass of the sample? Sketch the graph of m against t. What is the mass of the sample after 2 days? Calculate the time it takes for the sample to halve its original mass. Find the rate at which the sample is decaying when t = 5.

Examination-style question 2
a) When t = 0, m = 15e0 = 15 So, the initial mass of the sample is 15 g. m t 15 b) The graph of m = 15e–0.083t will be an exponential decay curve passing through the point (0, 15). For part b) we know that this is an example of exponential decay because the formula is of the form Ae–kt (where k is a positive constant). c) When t = 2, m = 15e–0.083 × 2 = (to 2 d.p.) So, the mass of the sample after 2 days is g (to 2 d.p.).

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. That is, when 7.5 = 15e–0.083t 0.5 = e–0.083t Take logs: ln 0.5 = ln e–0.083t ln 0.5 = –0.083t Point out that the time taken for a radioactive isotope to halve its mass is called its half life. = 8.35 (to 2 d.p.) The sample takes 8.35 days (to 2 d.p.) to halve its mass.

Examination-style question 2
d) The rate at which the sample is decaying is given by When t = 5 Point out that since the rate is negative the mass is decreasing over time. Therefore, when t = 5, the sample is decaying at a rate of 0.82 grams per day (to 2 d.p.).

Examination-style question 3
The height of a conical container is twice its radius, r cm, as shown in the diagram. Liquid is poured into the container at a rate of 5 litres per minute. 2r r 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

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

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

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