3 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:andare 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:andare implicit functions.It is not always easy, or even possible, to rearrange an implicit function into an explicit form.
4 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:
5 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:becomesWe can now divide through by 2y and rearrange to find :
6 Implicit differentiation Differentiate with respect to x.Differentiating term by term with respect to x gives:3x2Now rearrange to collect the terms in together:
7 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
8 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.
9 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:
10 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)
11 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:
12 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)
14 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:andcan be found using the chain rule:
15 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 parametricequations x = cos 2t and y = sin t.Differentiating each equation with respect to t gives:
16 Parametric differentiation This can be simplified further using the double angle formula for sin 2t:= cosec t
17 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
18 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= 7When 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).
20 Differentiating exponential functions Suppose we want to differentiate a general exponential function of the formy = axwhere a is a constant.Take the natural log of both sides:Now differentiating with respect to x gives:
21 Differentiating exponential functions Since y = ax we can substitute ax for y. So in general:For example:
23 Exponential growthExponential 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.
24 Exponential growthNumber of bacteriat…321Time (mins)100200400800…100 × 2tSo if N is the number of bacteria after time t we have:N = 100 × 2tMore generally, if b is the number of bacteria when t = 0,N = b2tIn 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 becomesN = bet ln 2orN = be 0.693t
25 Exponential growthIn general, exponential growth can be modelled by the functionf(t) = Aektwhere t is time,A is the original quantity (the quantity when t = 0),f(t) is the quantity after time t andk is a positive constant (the growth rate)An exponential growth curve has the following basic shape:tf(t)y = AektAThe 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.
26 Exponential decayExponential 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).
27 Exponential decayIn general, exponential decay can be modelled by the functionf(t) = Ae–ktwhere t is time,A is the original quantity (the quantity when t = 0),f(t) is the quantity after time t, andk is a positive constant (the decay rate).An exponential decay curve has the following basic shape.tf(t)AThe 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
28 Exponential decayForensic 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–ktwhere 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.
29 Exponential decay Let t = 0 at 8.30 pm. d = 30 – 20 = 10 And using d = ae–kt:10 = ae–k(0) = ae0 = aWhen t = 1, one hour later:d = 29 – 20 = 9Using d = ae–kt:9 = ae–k(1) = ae–kBut a = 10 so,9 = 10e–ke–k = 0.9
30 Exponential decayWe 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 = 0.105Substitute these values of k and a into d = ae–kt:d = 10e–0.105tIf 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
31 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 1.69t =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.
33 Rates of changeWhen 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
34 Rates of changeWe can find by differentiating the formula for the surfacearea of a sphere with respect to the radius.A = 4πr2 soSo, the rate of change of the surface area is connected to the rate of change of the radius byIf 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.
35 Connected rates of change We are given that = 15 and we want to find when r = 4.Using we have:= 0.149So, 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.
36 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
37 Connected rates of change Here = –30 and r = 5:We can now find using with r = 5:4= –12So, 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.
38 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.
40 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.xyOPQ
41 Examination-style question 1 Using gives:when t = 1:
42 Examination-style question 1 b) The gradient of the normal to the curve at the point P is –1.Also when t = 1,x = 2y = 3If 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.
43 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
44 Examination-style question 2 The mass, m grams, of a sample of radioactive iodine decays according to the formulawhere 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.
45 Examination-style question 2 a) When t = 0,m = 15e0= 15So, the initial mass of the sample is 15 g.mt15b) 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.).
46 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, when7.5 = 15e–0.083t0.5 = e–0.083tTake logs:ln 0.5 = ln e–0.083tln 0.5 = –0.083tPoint 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.
47 Examination-style question 2 d) The rate at which the sample is decaying is given byWhen t = 5Point 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.).
48 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.2rrIf 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
49 Examination-style question 3 Since the radius of the container is half its depth this can be written in term of x asThe liquid enters the container at a rate of 5 litres per minute so
50 Examination-style question 3 Using the chain ruleSoAt the instant when x = 2 cm the rate at which the liquid is entering the container is