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Differentiation. Introduction. In this chapter you will learn how to differentiate equations that are given parametrically You will also learn to differentiate implicit functions, which aren’t necessarily written in the form ‘y = …’

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## Differentiation

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**Introduction**• In this chapter you will learn how to differentiate equations that are given parametrically • You will also learn to differentiate implicit functions, which aren’t necessarily written in the form ‘y = …’ • These will be combined with calculating normals and tangents as you have so far • We will also look at creating differential equations from information we are given**Differentiation**You can find the gradient of a curve given in parametric coordinates • When a curve is given in parametric equations: • Differentiate y and x with respect to the parameter t • We can then use the following calculation, a variation of the chain rule Rewrite as a multiplication Written differently Cancel dts 4A**Differentiation**You can find the gradient of a curve given in parametric coordinates Find the gradient at the point P where t = 2, on the curve given by parametric equations: Differentiate Differentiate Write in terms of t using dy/dt and dx/dt This is the gradient function for the curve, but in terms of t Sub in t = 2 Work out 4A**Differentiation**You can find the gradient of a curve given in parametric coordinates Find the equation of the normal at the point P where θ = π/6, on the curve with parametric equations: • We need to find the gradient at point P by differentiating and substituting θ in • We then need to find the gradient of the normal • We also need the coordinates of x and y at point P • Then we can use the formula y – y1 = m(x – x1) to obtain the equation of the line Differentiate Differentiate Gradient of the tangent at P: Sub in dx/dθand dy/dθ Gradient of the normal at P: Sub θ = π/6 in to obtain the gradient at P Work out the fraction 4A**Differentiation**You can find the gradient of a curve given in parametric coordinates Find the equation of the normal at the point P where θ = π/6, on the curve with parametric equations: Gradient of the normal at P: • We need to find the gradient at point P by differentiating and substituting θ in • We then need to find the gradient of the normal • We also need the coordinates of x and y at point P • Then we can use the formula y – y1 = m(x – x1) to obtain the equation of the line Sub in P = π/6 Sub in P = π/6 Work out Work out The coordinates of P are: 4A**Differentiation**You can find the gradient of a curve given in parametric coordinates Find the equation of the normal at the point P where θ = π/6, on the curve with parametric equations: Gradient of the normal at P: • We need to find the gradient at point P by differentiating and substituting θ in • We then need to find the gradient of the normal • We also need the coordinates of x and y at point P • Then we can use the formula y – y1 = m(x – x1) to obtain the equation of the line The coordinates of P are: Sub in y1, m and x1 Multiply by 2 Multiply by 5 (You could multiply by 10 in one step if confident) Multiply the bracket out Add 25√3 Divide by 2 4A**Differentiation**You can differentiate equations which are implicit, such as x2 + y2 = 8x, or cos(x + y) = siny Implicit effectively means all the terms are mixed up, not necessarily in the form ‘y = …’ This technique is useful as some equations are difficult to arrange into this form… Write dy/dx after differentiating the y term This is written differently The reason is that as the equation is not written as ‘y =…’, y is not a function of x Differentiate the y term as you would for an x term For example: This is what happens when you differentiate an equation which starts ‘y =…’ 4B**Differentiation**You can differentiate equations which are implicit, such as x2 + y2 = 8x, or cos(x + y) = siny Find dy/dx in terms of x and y for the following equation: Differentiate each part one at a time It would be very difficult to rearrange this into the form ‘y = …’ Isolate the terms with dy/dx in Factorise by taking out dy/dx Divide by (3y2 + 3) You now have a formula for the gradient, but in terms of x AND y, not just x 4B**Differentiation**You can differentiate equations which are implicit, such as x2 + y2 = 8x, or cos(x + y) = siny Find the value of dy/dxat the point (1,1) where: Differentiate each one at a time – remember the product rule for the first term!!!! Add 5, subtract 4y2 You can substitute x = 1 and y = 1 now (or rearrange first) Do not replace the terms in dy/dx Divide by 8 Keep your workings tidy – you may need to use the product rule on multiple terms as well as factorise – show everything you’re doing! 4B**Differentiation**You can differentiate equations which are implicit, such as x2 + y2 = 8x, or cos(x + y) = siny Find the value of dy/dxat the point (1,1) where: Give your answer in terms of e. Differentiate (again watch out for the product rule! Sub in x = 1 and y = 1 ln1 = 0 since e0 = 1, cancelling the term out Subtract dy/dx Factorise Divide by (e2 – 1) 4B**Differentiation**You can differentiate the general power function ax where a is a constant Differentiate where a is a constant This is important – as a is a constant it can be treated as a number Take natural logs Use the power law Differentiate each side ln a is a constant so can be thought of as just a number Multiply both sides by y y = ax as from the first line 4C**Differentiation**You can differentiate the general power function ax where a is a constant Differentiate Differentiate Differentiate where r is a constant Differentiate 4C**Differentiation**You can relate one rate of change to another. This is useful when a situation involves more than two variables. Given that the area of a circle A, is related to its radius r by the formula A = πr2, and the rate of change of its radius in cm is given by dr/dt = 5, find dA/dt when r = 3 • You are told a formula linking A and r • You are told the radius is increasing by 5 at that moment in time • You are asked to find how much the area is increasing when the radius is 3 This is quite logical – if the radius is increasing over time, the area must also be increasing, but at a different rate… Replace dr/dt and dA/dr We are trying to work out dA/dt We have been told dr/dt in the question You need to find a derivative that will give you the dA and cancel the dr out r = 3 Differentiate 4D**Differentiation**You can relate one rate of change to another. This is useful when a situation involves more than two variables. The volume of a hemisphere is related to its radius by the formula: The total surface area is given by the formula: Find the rate of increase of surface area when the rate of increase of volume: You need to use the information to set up a chain of derivatives that will leave dS/dt Rate of change of surface area over time We are told dV/dt in the question so we will use it We have a formula linking V and r so can work out dV/dr We have a formula linking S and r so can work out dS/dr This isn’t all correct though, as multiplying these will not leave dS/dt However, if we flip the middle derivative, the sequence will work! 4D**Differentiation**You can relate one rate of change to another. This is useful when a situation involves more than two variables. The volume of a hemisphere is related to its radius by the formula: The total surface area is given by the formula: Find the rate of increase of surface area when the rate of increase of volume: You need to use the information to set up a chain of derivatives that will leave dS/dt Sub in values Differentiate Differentiate You can write as one fraction Flip to obtain the derivative we want Simplify 4D**Differentiation**You can set up differential equations from information given in context • Differential equations can arise anywhere when variables change relative to one another • In Mechanics, speed may change over time due to acceleration • The rate of growth of bacteria may change when temperature is changed • As taxes change, the amount of money people spend will change • Setting these up involve the idea of proportion, a topic you met at GCSE… Radioactive particles decay at a rate proportional to the number of particles remaining. Write an equation for the rate of change of particles… Let N be the number of particles and t be time The rate of change of the number of particles The number of particles remaining multiplied by the proportional constant, k. Negative because the number of particles is decreasing 4E**Differentiation**You can set up differential equations from information given in context A population is growing at a rate proportional to its size at a given time. Write an equation for the rate of growth of the population Let P be the population and t be time The population multiplied by the proportional constant k Leave positive since the population is increasing The rate of change of the population 4E**Differentiation**You can set up differential equations from information given in context Newton’s law of cooling states that the rate of loss of temperature is proportional to the excess temperature the body has over its surroundings. Write an equation for this law… • Let the temperature of the body be θ degrees and time be t • The ‘excess’ temperature will be the difference between the objects temperature and its surroundings – ie) One subtract the other • (θ – θ0) where θ0 is the temperature of the surroundings The rate of change of the objects temperature The difference between temperatures (θ - θ0) multiplied by the proportional constant k Negative since it is a loss of temperature 4E**Differentiation**You can set up differential equations from information given in context The head of a snowman of radius r cm loses volume at a rate proportional to its surface area. Assuming the head is spherical, and that the volume V = 4/3πr3 and the Surface Area A = 4πr2, write down a differential equation for the change of radius of the snowman’s head… The first sentence tells us The rate of change of volume Is the Surface Area multiplied by the proportional constant k (negative as the volume is falling) Differentiate We want to know the rate of change of the radius r We know dV/dt We need a derivative that will leave dr/dt when cancelled Flip over 4E**Differentiation**You can set up differential equations from information given in context The head of a snowman of radius r cm loses volume at a rate proportional to its surface area. Assuming the head is spherical, and that the volume V = 4/3πr3 and the Surface Area A = 4πr2, write down a differential equation for the change of radius of the snowman’s head… The first sentence tells us The rate of change of volume Is the Surface Area multiplied by the proportional constant k (negative as the volume is falling) Fill in what we know Differentiate We know A from the question Flip over Simplify 4E**Summary**• You have learnt how to differentiate equations given parametrically • You can also differentiate implicit equations • You can differentiate the general power function ax • You can also set up differential equations based on information you are given

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