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 = …’ 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

Teachings for Exercise 4A

You can find the gradient of a curve given in parametric coordinates 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 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑡 ÷ 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝑡 ÷ 𝑑𝑥 𝑑𝑡 Rewrite as a multiplication Written differently 𝑑𝑦 𝑑𝑡 × 𝑑𝑡 𝑑𝑥 = 𝑑𝑦 𝑑𝑡 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝑥 Cancel dts = 𝑑𝑦 𝑑𝑥 4A

You can find the gradient of a curve given in parametric coordinates 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑡 𝑑𝑥 𝑑𝑡 Differentiation You can find the gradient of a curve given in parametric coordinates 𝑥= 𝑡 3 +𝑡 Find the gradient at the point P where t = 2, on the curve given by parametric equations: 𝑦= 𝑡 2 +1 𝑥= 𝑡 3 +𝑡 𝑦= 𝑡 2 +1 Differentiate Differentiate 𝑑𝑥 𝑑𝑡 =3 𝑡 2 +1 𝑑𝑦 𝑑𝑡 =2𝑡 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑡 𝑑𝑥 𝑑𝑡 Write in terms of t using dy/dt and dx/dt  This is the gradient function for the curve, but in terms of t 𝑑𝑦 𝑑𝑥 = 2𝑡 3 𝑡 2 +1 Sub in t = 2 𝑑𝑦 𝑑𝑥 = 2(2) 3 (2) 2 +1 Work out 𝑑𝑦 𝑑𝑥 = 4 13 4A

You can find the gradient of a curve given in parametric coordinates 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑡 𝑑𝑥 𝑑𝑡 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: 𝑥=3𝑠𝑖𝑛𝜃 𝑦=5𝑐𝑜𝑠𝜃 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 𝑥=3𝑠𝑖𝑛𝜃 𝑦=5𝑐𝑜𝑠𝜃 Differentiate Differentiate 𝑑𝑥 𝑑𝜃 =3𝑐𝑜𝑠𝜃 𝑑𝑦 𝑑𝜃 =−5𝑠𝑖𝑛𝜃 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝜃 𝑑𝑥 𝑑𝜃 −5 3 3 Gradient of the tangent at P: Sub in dx/dθ and dy/dθ 3 3 5 𝑑𝑦 𝑑𝑥 = −5𝑠𝑖𝑛𝜃 3𝑐𝑜𝑠𝜃 Gradient of the normal at P: Sub θ = π/6 in to obtain the gradient at P 𝑑𝑦 𝑑𝑥 = −5𝑠𝑖𝑛 𝜋 6 3𝑐𝑜𝑠 𝜋 6 Work out the fraction 𝑑𝑦 𝑑𝑥 = −5 3 3 4A

You can find the gradient of a curve given in parametric coordinates 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑡 𝑑𝑥 𝑑𝑡 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: 𝑥=3𝑠𝑖𝑛𝜃 𝑦=5𝑐𝑜𝑠𝜃 Gradient of the normal at P: = 3 3 5 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 𝑥=3𝑠𝑖𝑛𝜃 𝑦=5𝑐𝑜𝑠𝜃 Sub in P = π/6 Sub in P = π/6 𝑥=3𝑠𝑖𝑛 𝜋 6 𝑦=5𝑐𝑜𝑠 𝜋 6 Work out Work out 𝑦= 5 3 2 𝑥= 3 2 3 2 , 5 3 2 The coordinates of P are: 4A

You can find the gradient of a curve given in parametric coordinates 𝑑𝑦 𝑑𝑥 = 𝑑𝑦 𝑑𝑡 𝑑𝑥 𝑑𝑡 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: 𝑥=3𝑠𝑖𝑛𝜃 𝑦=5𝑐𝑜𝑠𝜃 Gradient of the normal at P: = 3 3 5 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 3 2 , 5 3 2 The coordinates of P are: 𝑦 − 𝑦 1 =𝑚(𝑥 − 𝑥 1 ) Sub in y1, m and x1 − 5 3 2 3 3 5 𝑥 − 3 2 𝑦 = Multiply by 2 6 3 5 𝑥 − 3 2 2𝑦 − 5 3 = Multiply by 5 (You could multiply by 10 in one step if confident) 𝑥 − 3 2 10𝑦 − 25 3 = 6 3 Multiply the bracket out 10𝑦 − 25 3 = 6𝑥 3 − 9 3 Add 25√3 10𝑦 = 6𝑥 3 + 16 3 Divide by 2 5𝑦 =3𝑥 3 +8 3 4A

Teachings for Exercise 4B

Differentiation 𝑑 𝑑𝑥 𝑦 𝑛 = 𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥 4B 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… 𝑑 𝑑𝑥 𝑦 𝑛 = 𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥 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 6𝑦 𝑑𝑦 𝑑𝑥 For example: 3 𝑦 2 8 𝑑𝑦 𝑑𝑥 8𝑦 This is what happens when you differentiate an equation which starts ‘y =…’ 𝑑𝑦 𝑑𝑥 𝑦 4B

Differentiation 4B 𝑑 𝑑𝑥 𝑦 𝑛 = 𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥 𝑑 𝑑𝑥 𝑦 𝑛 = 𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥 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: 𝑥 3 + 𝑥 + 𝑦 3 + 3𝑦 = 6 𝑥 3 + 𝑥 + 𝑦 3 + 3𝑦 = 6 Differentiate each part one at a time It would be very difficult to rearrange this into the form ‘y = …’ 3𝑦 2 𝑑𝑦 𝑑𝑥 3 𝑑𝑦 𝑑𝑥 3𝑥 2 + 1 + + = Isolate the terms with dy/dx in 3𝑦 2 𝑑𝑦 𝑑𝑥 3 𝑑𝑦 𝑑𝑥 + = − 3𝑥 2 − 1 Factorise by taking out dy/dx 𝑑𝑦 𝑑𝑥 3𝑦 2 + 3 = − 3𝑥 2 − 1 Divide by (3y2 + 3) 𝑑𝑦 𝑑𝑥 −3 𝑥 2 −1 3𝑦 2 + 3 = You now have a formula for the gradient, but in terms of x AND y, not just x 4B

Differentiation 4B 𝑑 𝑑𝑥 𝑦 𝑛 = 𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥 𝑑 𝑑𝑥 𝑦 𝑛 = 𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥 You can differentiate equations which are implicit, such as x2 + y2 = 8x, or cos(x + y) = siny Find the value of dy/dx at the point (1,1) where: 4𝑥𝑦 2 −5𝑦=10 4𝑥𝑦 2 −5𝑥=10 Differentiate each one at a time – remember the product rule for the first term!!!! 𝑑𝑦 𝑑𝑥 =𝑢 𝑑𝑣 𝑑𝑥 +𝑣 𝑑𝑢 𝑑𝑥 𝑢=4𝑥 𝑣= 𝑦 2 8𝑥𝑦 𝑑𝑦 𝑑𝑥 + 4𝑦 2 − 5 = 0 𝑑𝑦 𝑑𝑥 = 8𝑥𝑦 𝑑𝑦 𝑑𝑥 𝑑𝑢 𝑑𝑥 =4 𝑑𝑣 𝑑𝑥 =2𝑦 𝑑𝑦 𝑑𝑥 + 4 𝑦 2 Add 5, subtract 4y2 8𝑥𝑦 𝑑𝑦 𝑑𝑥 =5−4 𝑦 2 You can substitute x = 1 and y = 1 now (or rearrange first)  Do not replace the terms in dy/dx 8 𝑑𝑦 𝑑𝑥 =5 − 4 Divide by 8 𝑑𝑦 𝑑𝑥 = 1 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 4B 𝑑 𝑑𝑥 𝑦 𝑛 = 𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥 𝑑 𝑑𝑥 𝑦 𝑛 = 𝑛 𝑦 𝑛−1 𝑑𝑦 𝑑𝑥 You can differentiate equations which are implicit, such as x2 + y2 = 8x, or cos(x + y) = siny Find the value of dy/dx at the point (1,1) where: 𝑒 2𝑥 𝑙𝑛𝑦 = 𝑥+𝑦−2 Give your answer in terms of e. 𝑒 2𝑥 𝑙𝑛𝑦 = 𝑥+𝑦−2 𝑑𝑦 𝑑𝑥 =𝑢 𝑑𝑣 𝑑𝑥 +𝑣 𝑑𝑢 𝑑𝑥 𝑢= 𝑒 2𝑥 𝑣=𝑙𝑛𝑦 Differentiate (again watch out for the product rule! 𝑒 2𝑥 𝑦 𝑑𝑦 𝑑𝑥 + 2𝑙𝑛𝑦 𝑒 2𝑥 =1+ 𝑑𝑦 𝑑𝑥 𝑑𝑦 𝑑𝑥 = 𝑒 2𝑥 𝑦 𝑑𝑦 𝑑𝑥 𝑑𝑢 𝑑𝑥 =2 𝑒 2𝑥 𝑑𝑣 𝑑𝑥 = 1 𝑦 𝑑𝑦 𝑑𝑥 + 2𝑙𝑛𝑦 𝑒 2𝑥 Sub in x = 1 and y = 1 𝑒 2 1 𝑑𝑦 𝑑𝑥 =1+ 𝑑𝑦 𝑑𝑥 + 2𝑙𝑛1 𝑒 2 ln1 = 0 since e0 = 1, cancelling the term out 𝑒 2 𝑑𝑦 𝑑𝑥 =1+ 𝑑𝑦 𝑑𝑥 Subtract dy/dx 𝑒 2 𝑑𝑦 𝑑𝑥 − 𝑑𝑦 𝑑𝑥 =1 Factorise 𝑑𝑦 𝑑𝑥 ( 𝑒 2 −1)=1 Divide by (e2 – 1) 𝑑𝑦 𝑑𝑥 = 1 𝑒 2 −1 4B

Teachings for Exercise 4C

Differentiation You can differentiate the general power function ax where a is a constant Differentiate 𝑦= 𝑎 𝑥 where a is a constant This is important – as a is a constant it can be treated as a number 𝑦= 𝑎 𝑥 Take natural logs 𝑙𝑛𝑦= 𝑙𝑛𝑎 𝑥 Use the power law 𝑙𝑛𝑦=𝑥𝑙𝑛𝑎 Differentiate each side  ln a is a constant so can be thought of as just a number 1 𝑦 𝑑𝑦 𝑑𝑥 =𝑙𝑛𝑎 Multiply both sides by y 𝑑𝑦 𝑑𝑥 =𝑦𝑙𝑛𝑎 y = ax as from the first line 𝑑𝑦 𝑑𝑥 = 𝑎 𝑥 𝑙𝑛𝑎 4C

Differentiation 4C 𝑑𝑦 𝑑𝑥 = 𝑎 𝑥 𝑙𝑛𝑎 You can differentiate the general power function ax where a is a constant Differentiate 𝑦= 3 𝑥 𝑦= 3 𝑥 Differentiate 𝑑𝑦 𝑑𝑥 = 3 𝑥 𝑙𝑛3 Differentiate 𝑦= 6𝑟 𝑥 where r is a constant 𝑦= 6𝑟 𝑥 Differentiate 𝑑𝑦 𝑑𝑥 = 6𝑟 𝑥 𝑙𝑛6𝑟 4C

Teachings for Exercise 4D

Differentiation 4D 𝑑𝐴 𝑑𝑡 = 𝑑𝑟 𝑑𝑡 × 𝑑𝐴 𝑑𝑟 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… 𝑑𝐴 𝑑𝑡 = 𝑑𝑟 𝑑𝑡 × 𝑑𝐴 𝑑𝑟 𝑑𝐴 𝑑𝑡 = 𝑑𝑟 𝑑𝑡 × 𝑑𝐴 𝑑𝑟 Replace dr/dt and dA/dr 𝑑𝐴 𝑑𝑡 = 5 × 2𝜋𝑟 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 𝑑𝐴 𝑑𝑡 = 10𝜋𝑟 r = 3 𝐴=𝜋 𝑟 2 = 30𝜋 𝑑𝐴 𝑑𝑟 =2𝜋𝑟 Differentiate 4D

Differentiation You can relate one rate of change to another. This is useful when a situation involves more than two variables. 𝑉= 2 3 𝜋 𝑟 3 The volume of a hemisphere is related to its radius by the formula: The total surface area is given by the formula: 𝑆=3𝜋 𝑟 2 𝑑𝑉 𝑑𝑡 =6 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 𝑑𝑆 𝑑𝑡 = 𝑑𝑉 𝑑𝑡 𝑑𝑟 𝑑𝑉 𝑑𝑉 𝑑𝑟 × 𝑑𝑆 𝑑𝑟 × 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. 𝑉= 2 3 𝜋 𝑟 3 The volume of a hemisphere is related to its radius by the formula: The total surface area is given by the formula: 𝑆=3𝜋 𝑟 2 𝑑𝑉 𝑑𝑡 =6 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 𝑑𝑆 𝑑𝑡 = 𝑑𝑉 𝑑𝑡 × 𝑑𝑟 𝑑𝑉 × 𝑑𝑆 𝑑𝑟 𝑉= 2 3 𝜋 𝑟 3 𝑆=3𝜋 𝑟 2 Sub in values Differentiate Differentiate 𝑑𝑆 𝑑𝑡 = × 1 2 𝜋𝑟 2 6 × 6𝜋𝑟 𝑑𝑉 𝑑𝑟 =2 𝜋𝑟 2 𝑑𝑆 𝑑𝑟 =6𝜋𝑟 You can write as one fraction Flip to obtain the derivative we want 𝑑𝑆 𝑑𝑡 = 36𝜋𝑟 2 𝜋𝑟 2 𝑑𝑟 𝑑𝑉 = 1 2 𝜋𝑟 2 Simplify 𝑑𝑆 𝑑𝑡 = 18 𝑟 4D

Teachings for Exercise 4E

Differentiation 𝑑𝑁 𝑑𝑡 = −𝑘𝑁 4E 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 𝑑𝑁 𝑑𝑡 = −𝑘𝑁 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 𝑑𝑃 𝑑𝑡 = 𝑘𝑃 4E 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 𝑑𝑃 𝑑𝑡 = 𝑘𝑃 The population multiplied by the proportional constant k  Leave positive since the population is increasing The rate of change of the population 4E

Differentiation 𝑑𝜃 𝑑𝑡 = −𝑘(𝜃− 𝜃 0 ) 4E 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 𝑑𝜃 𝑑𝑡 = −𝑘(𝜃− 𝜃 0 ) 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… 𝑑𝑉 𝑑𝑡 =−𝑘𝐴 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) 𝑑𝑟 𝑑𝑡 = 𝑑𝑉 𝑑𝑡 𝑑𝑟 𝑑𝑉 𝑉= 4 3 𝜋 𝑟 3 × Differentiate 𝑑𝑉 𝑑𝑟 =4𝜋 𝑟 2 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 𝑑𝑟 𝑑𝑉 = 1 4𝜋 𝑟 2 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… 𝑑𝑉 𝑑𝑡 =−𝑘𝐴 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) 𝑑𝑟 𝑑𝑡 = 𝑑𝑉 𝑑𝑡 𝑑𝑟 𝑑𝑉 𝑉= 4 3 𝜋 𝑟 3 × Fill in what we know Differentiate 𝑑𝑟 𝑑𝑡 = 1 4𝜋 𝑟 2 𝑑𝑉 𝑑𝑟 =4𝜋 𝑟 2 −𝑘𝐴 × We know A from the question Flip over 𝑑𝑟 𝑑𝑡 = 1 4𝜋 𝑟 2 𝑑𝑟 𝑑𝑉 = 1 4𝜋 𝑟 2 −𝑘(4𝜋 𝑟 2 ) × Simplify 𝑑𝑟 𝑑𝑡 = −𝑘 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