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Indices

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Indices is closely related to Surds and Logarithms What's Next Click here to check out how to solve log equations Click here to check out how to rationalize surds Click here for Contents Page

Applications of differentiation Part VI

Displacement, Velocity and Acceleration Consider the motion of a particle which moves in a way such that its displacement s from a fixed point O at time t seconds is given by $ s = -t^2 + 5t + 6 $ The instantaneous velocity is given by the differentiation of displacement $ v = \frac{ds}{dt} = -2t + 5 $ The instantaneous acceleration is given by the differentiation of velocity $ a = \frac{dv}{dt} = -2 $ To obtain velocity function from accleration, we integrate the acceleration function To obtain the displacement function from velocity, we integrate the velocity function Example 1 The acceleration of a particle is given by the function $ a = 10 + 15t^2 $ where t is the time in seconds after leaving start point O. Find the velocity of the particle at t = 1 given that the initial velocity of the particle is 0m/s. To obtain velocity function from accleration, we integrate the acceleration function $\begin{aligned} v &= \int a \:dt \\ &= \int 10 + 15t^2 \: dt \\ &

Applications of DIfferentiation Part V

Connected Rates of Change Example 1 A curve has the equation $ y = (5x-2)(\sqrt{2x + 2}) $ a) Find $\frac{dy}{dx} $ in the form of $ \frac{ ax }{ \sqrt{2x + 2} } $ Using Product rule $\begin{aligned} \frac{dy}{dx} &= (5x-2)\frac{1}{2}( 2x + 2)^{-\frac{1}{2}}( 2 ) + 5\sqrt{2x + 2} \\ &= \frac{ 5x - 2 }{ ( 2x + 2)^{\frac{1}{2} } } + 5\sqrt{2x + 2} \\ &= \frac{ 5x - 2 + 2x + 2 }{ \sqrt{2x + 2} } \\ &= \frac{ 7x }{ \sqrt{2x + 2} } \\ \end{aligned} $ b) Hence find the rate of change of x when x = 7 when y is change at a rate of 4 units per second Given $ x = 7 \qquad \frac{dy}{dt} = 4 $ $\begin{aligned} \frac{dy}{dt} &= \frac{dy}{dx} \times \frac{dx}{dt} \\ 4 &= \frac{ 7x }{ \sqrt{2x + 2} } \times \frac{dx}{dt} \\ 4 &= \frac{ 49 }{ \sqrt{14 + 2} } \times \frac{dx}{dt} \\ \frac{dx}{dt} &= \frac{16}{49} \\ \end{aligned} $

Applications of DIfferentiation Part IV

Connected rates of change Example 1 Liquid is poured into a bucket at a rate of $ 20cm^{-3}s^{-1} $ The volume of the liquid in the bucket $ V cm^3 $ when the depth of the liquid is $ x $ is given by $ V = 0.1x^3 + x^2 + 100x $ a) Find the rate of increase in the depth of liquid when $ x = 10 $ b) Find the depth of liquid when the rate of increase in depth is $ 0. 08 cm s^{-1} $ $ \frac{dV}{dt} = \frac{dV}{dx} \times \frac{dx}{dt} $ a) Given $ \frac{dV}{dt} = 20 $ To find $\frac{dV}{dx} $ differentiate V with respect to x $\frac{dV}{dx} = 0.3x^2 + 2x + 100$ When x = 10, $ \frac{dV}{dx} = 30 + 20 + 100 = 150 $ $ \frac{dx}{dt} = \frac{\frac{dV}{dt}}{\frac{dV}{dx}} = \frac{20}{150} = \frac{2}{15} $ The rate of increase in the depth of liquid is $\frac{2}{15} \:cm\:s^{-1}$ b) $\begin{aligned} \frac{dV}{dt} &= \frac{dV}{dx} \times \frac{dx}{dt} \\ 10 &= \frac{dV}{dx} \times 0.08 \\ \frac{dV}{dx} &= 125 \\ 0.3x^2 + 2x + 100 &= 125 \\ 0.3x^2 + 2x - 25 &=

Challenging question

Question 1 Prove that $ \frac{\tan\theta}{1-\cot\theta} + \frac{\cot\theta}{1-\tan\theta} = 1 + \sec\theta\csc\theta $ From LHS (Do not need to show this in answer, I am doing this to explain why we need to multiply by $ \cos\theta - \sin\theta $ later on ) $\begin{aligned} \frac{\sin\theta}{\cos\theta} \left(\frac{\sin\theta}{\sin\theta-\cos\theta}\right) + \frac{\cos\theta}{\sin\theta} \left(\frac{\cos\theta}{\cos\theta - \sin\theta} \right) &= \frac{-\sin^3 \theta + \cos^3 \theta}{\sin\theta \cos\theta (\cos\theta - \sin\theta)} \\ \end{aligned} $ From RHS $\begin{aligned} 1 + \frac{1}{\sin\theta}\frac{1}{\cos\theta} &= \frac{\sin\theta \cos\theta + 1}{\sin\theta \cos\theta} \\ &= \frac{\sin\theta \cos\theta + 1}{\sin\theta \cos\theta} \left(\frac{\cos\theta - \sin\theta}{\cos\theta - \sin\theta} \right) \qquad \text{clue from LHS}\\ &= \frac{\sin\theta\cos^2\theta - \sin^2\theta\cos\theta + \cos\theta - \sin\theta}{\sin\theta \cos\theta (\cos\theta - \sin\theta

Applications of Differentiation Part III

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Stationary points Stationary points occur when $ \frac{dy}{dx} = 0 $ A stationary point can be a: maximum point: $\frac{d^2y}{dx^2} \quad \text{lesser than 0} $ minimum point: $\frac{d^2y}{dx^2} > 0 $ point of inflexion: $\frac{d^2y}{dx^2} = 0 $ Second derivative is inconclusive Example 1 Given $y =13-20x + 4x^2 +3x^3 $ Find the maximum and minimum values of y and state where those values occur. $\begin{aligned} \frac{ dy }{ dx } &= -20 +8x +9x^2 \\ \text{To find stationary values of x} \\ \frac{ dy }{ dx } &= 0 \\ -20 +8x + 9x^2 &=0 \\ ( 9x - 10) ( x + 2) &= 0 \\ x &= \frac{10}{9} \qquad x = -2 \\ \text{next we determine the second derivative} \\ \frac{d^2y}{dx^2} &= 18x + 8 \\ \text{When} \: x = \frac{10}{9} \qquad \frac{d^2y}{dx^2} = 18 \times \frac{10}{9} + 8 = 28 \\ \text{hence value of y is minimum when} \: x = \frac{10}{9} \\ \text{When} \: x = \frac{10}{9} \qquad y = 13-20(\frac{10}{9}) + 4(\frac{10}{9})^2 +3(\frac{10}{9}

Applications of Differentiation Part II

Tangents and Normal Example 4 The point A lies on the curve $ y = x^2 -x + c $ where c is a constant. The tangent to the curve at A is $ y = 3x - 1 $. Find the equation of the normal to the curve at A. Recall equation of straight line: $ y = mx +c \qquad \text{m is gradient} $ Since tangent to the the curve at A is $ y = 3x + 5 $ , gradient of tangent is 3 We can calculate the gradient using $\frac{dy}{dx} $ $\frac{dy}{dx} = 2x -1 $ To obtain x coordinates of A, we set $ \frac{dy}{dx} = 3 $ $2x -1 = 3 \qquad x = 2 $ To determine the equation of normal we need to know the gradient of the normal and a point on the normal Gradient of the normal is the negative reciprocal of gradient of tangent Gradient of normal = $-\frac{1}{3} $ A is a point on the tangent as well as the normal When $ x = 2 \qquad y = 3(2)-1 = 5 $ Sub $ x= 2 \quad y = 5 \quad m=-\frac{1}{3} $ into $ y = mx +c $ $\begin{aligned} 5 &= -\frac{1}{3}(2) + c \\ c &= 5\frac{2}{3} \\ \end{aligned}$ Equation

Applications of Differentiation Part I

Tangents and Normals Equation of a line is given by: $ y = mx + c $ where m is the gradient and c is the y intercept The gradient at a particular point on a curve can be found using $ \frac{dy}{dx} $ To find the equation of a line, you will need a point on the line and the gradient of the line Example 1 The equation of a curve is $ y = \frac{5}{1 + x^2} $ Find the equation of the tangent to the curve at the point where x = 2 . First we differentiate y wrt x To make our job easier, we rewrite y as $ y = 5( 1 + x^2)^{-1} $ $ \begin{aligned} \frac{dy}{dx} &= 5(-1)(1+x^2)^{-1-1}(2x) \qquad \text{applying chain rule} \\ &= (-10x)(1+x^2)^{-2} \\ &= \frac{-10x}{(1+x^2)^{2}}\\ \end{aligned} $ In order to find the equation of the tangent, we need to determine the gradient of the tangent. When x = 2 , $ \frac{dy}{dx} = \frac{-10(2)}{1+2^2}} = -4 $ Now that we have determined the gradient, we need to find a point on the tangent The point on the tangent is also the point at w

Differentiation Part II

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Product rule Example 1 Example 2 Quotient rule Example 3 Example 4 What's Next? Now that you learnt the techniques of differentiation, it time to apply them click here for Applications of Differentiation Part I: Tangents and Normals click here for Applications of Differentiation Part III: Stationary Values click here for Applications of Differentiation Part IV: Connected Rates of Change

Differentiation Part I

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Basic Differentiation Differentiating term by term Example 1 Example 2 Example 3 Chain Rule Example 4 Example 5 What's Next? Click here for more techniques of differentiation: Product Rule and Quotient Rule Click here for Contents Page

Surds Part II

Example 1 A cone has a volume of $\ (\sqrt{5} + 3) \: \text{cm}^3 $. Its base area is $\ (\sqrt{5} +1)\: \text{cm} $ What is the height of the cone ? To find the height of the cone we divide its volume by its height $\begin {aligned} \frac{\sqrt{5} + 3}{\sqrt{5} +1} &= \frac{(\sqrt{5} + 3)(\sqrt{5} -1)}{(\sqrt{5} +1)(\sqrt{5} -1)}\\ &= \frac{(\sqrt{5})(\sqrt{5}) - \sqrt{5} + 3\sqrt{5} -3}{(\sqrt{5} +1)(\sqrt{5} -1)} \\ &= \frac{5+ 2\sqrt{5} - 3 }{5-1} \\ &= \frac{2+ 2\sqrt{5}}{4} \\ &= \frac{1 + \sqrt{5}}{2} \\ \end {aligned} $ Example 2 Solve the following equation: $\ y\sqrt{24} = y\sqrt{3} + \sqrt{6} $ To solve this, we make y the subject of the equation $\begin {aligned} y\sqrt{24} &= y\sqrt{3} + \sqrt{6} \\ y\sqrt{24} - y\sqrt{3} &=\sqrt{6}\\ y ( \sqrt{24} - \sqrt{3} ) &= \sqrt{6}\\ y &= \frac{ \sqrt{6}}{\sqrt{24} - \sqrt{3}} \\ y &= \frac{ \sqrt{3}\sqrt{2}}{\sqrt{8}\sqrt{3} - \sqrt{3}} \\ y&= \frac{ \sqrt{2}}{\sqrt{8} - 1} \\ \en

Surds Part I

Surds are terms that contain the square root sign e.g. $\sqrt{3} $ It is useful to know that $\sqrt{x} $ is the same as $\ x^{\frac{1}{2}} $ . Hence the rules used in the evaluation of surds is similar to the rules of indices. Commonly used rules Rule 1 $\ 3\sqrt{x} + \sqrt{x} = 4\sqrt{x} $ Note the similarity to to $\ 3y + y = 4y $ Rule 2 $\ \sqrt{x} \times \sqrt{x} = x $ Note the similarity to $\begin {aligned} x^{\frac{1}{2}} \times x^{\frac{1}{2}} &= x^{\frac{1}{2} + \frac{1}{2}} \\ &= x^1 \\ \end{aligned} $ Related to rule 2 is the rationalization of surds Rationalization of surds Example 1 Simplify $\ \frac{2}{\sqrt{3}} $ In order to rationalize this fraction, we need to get rid of the square root in the denominator. The way to get rid of the square root is to make use of rule 2 $\begin {aligned} \frac{2}{\sqrt{3}} &= \frac{2 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3} } \\ &= \frac{2\sqrt{3}}{3} \\ \end{aligned} $ Example 2 Simplify $\ \frac{2}{\s

Binomial Theorem Part II

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Example 1 ii) Example 2 Example 3  

Binomial Theorem Part I

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Introduction Another way is to use the Binomial Theorem Example 1 Example 2 The (r+1)th term The (r+1)th term is given in the formula list: $\ {n \choose r}a^{n-r}b^r $ Example 3   Example 4 Hence the term independent of x is 15

Logarithms Part IV

Changing the base of logarithms Changing from base a to base c $\ \log_a b = \frac{\log_c b}{log_c a} $ Changing from base 3 to base 10 $\ \log_3 5 = \frac{\log_{10} 5}{log_{10} 3} $ Solving log equations that involve changing of base Example 1 Solve the following equation $ \log_9 x + \log_3 x = 15 $ $ \begin{aligned} \log_9 x + \log_3 x &= 15 \\ \frac{\log_3 x}{\log_3 9} + \log_3 x &= 15 \qquad \text{changing base} \\ \frac{\log_3 x}{\log_3 3^2} + \log_3 x &= 15 \\ \frac{\log_3 x}{2\log_3 3} + \log_3 x &= 15 \qquad \text{applying power law}\\ \frac{\log_3 x}{2} + \log_3 x &= 10 \qquad \log_3 3 =15 \\ \log_3 x + 2\log_3 x &= 30 \\ \log_3 x + 2\log_3 x &= 30\log_3 3 \qquad 1 = \log_3 3 \\ \log_3 x + \log_3 x^2 &= \log_3 3^{30} \qquad \text{applying power law}\\ \log_3 x^3 &= \log_3 3^{30} \qquad \text{applying product law}\\ x^3 &= 3^{30} \qquad \text{comparing arguments} \\ x &= \sqrt[3] {3^{30}} \\ x &= 3^{10} \\ \end{ali

Logarithms Part III

Laws of logarithm Product Law $\ \log_a mn = \log_a m + \log_a n $ note that the logarithms are of the same base Quotient Law $\ \log_a {\frac{m}{n}} = \log_a m - \log_a n $ note that the logarithms are of the same base Power Law $\ \log_a m^n = n\log m $ Important: $\ \log_a xy^n \ne n\log_a xy $ The power n belongs to the base y only. The power n does not belong to base x $ \begin{aligned} \log_a xy^n &= \log_a x + \log_a y^n \quad \text{ applying power law} \\ \log_a xy^n &= \log_a x + n\log_a y \\ \end{aligned}$ Using the Laws of logarithms to solve equations Strategy Obtain a single logarithm function on the LHS as well as the RHS Ensure that logarithm functions on the LHS and RHS have the same base Compare the arguments to solve for x Example 1 $ \begin{aligned} \lg {(5y +2)} &= 1 +\lg y \\ \lg {(5y +2)} &= \lg 10 +\lg y \qquad \text{Note that} 1 = \lg10 \\ \lg {(5y +2)} &= \lg 10y \qquad \text{(applying product law)} \\ 5y + 2 &am

Logarithms Part II

Application of logarithms to real world problems Example 3 The value of a car decreases so that after a period of n year, the value of a car is $\ 50000e^{-0.1n} $. i) Find the value of a newly bought car ii) Find the value of the car after 10 years iii) The car is scraped when its value drops to 10000 dollars. Determine when the car will be scraped. Let v represent the value of a car after n years Note that v and n are variables i) When a car is newly bought, n = 0 $ \begin{aligned} v &= 50000e^{-0.1n} \\ v &= 50000e^{-0.1*0} \\ v &= 50000 \quad \textbf{note that e to the power of zero is 1}\\ \end{aligned}$ The value of a newly bought car is 50000 dollars ii) Value of a car after 10 years, n = 10 $ \begin{aligned} v &= 50000e^{-0.1n} \\ v &= 50000e^{-0.1*10} \\ v &= 18400 \quad \text{ (to 3 s.f.)}\\ \end{aligned}$ The value of the car after 10 years is 18400 dollars iii) Technique used: Taking log on both sides $ \begin{aligned} 10000 &= 5

Logarithms Part I

Intro to logarithms $ \begin{aligned} y &= a^x \\ log_{a} y &= log_{a} a^x \\ log_{a} y &= x\\ \end{aligned}$ y, $\ a^x $ is known as the argument a is known as the base x is known as the exponent or the power It is important to recognize that the base of the argument (a) equals to the base of the log (a) Example 1 $ \begin{aligned} y &= a^x \\ 3 &= 10^x\\ log_{10} 3 &= log_{10} 10^x \\ log_{10} 3 &= x\\ \end{aligned}$ log base 10 is commonly used as it can be evaluated using a calculator. In most cases, it is abbreviated as lg. $\ log_{10} $ is equivalent to $\ lg $. Another number that is commonly used is 2.718, commonly known as e. log base e (2.718) can also be evaluated using a calculator. $\ log_{e} $ is equivalent to $\ ln $. Example 2 $ \begin{aligned} 5 &= e^x\\ log_{e} 5 &= log_{e} e^x \\ log_{e} 5 &= x\\ x &= 1.61 \quad \text{to 3 s.f.} \end{aligned}$ The next example will show you how Taking log on