Topic 4 Calculus

Topic 4: Calculus PART A Differentiation: Finding the derivative of a particular function means finding the rate of change of one variable with respect to the other. The rate of change can be defined as the average rate of change, or the instantaneous rate of change. Note:- The slope of an equation is also known as the rate of change of the dependent variable with respect to the independent variable. Using the examples below we will understand what exactly this means. a. b. To find the average rate of change from A to B for example in question (a) we can find the slope of the line using the two points. A(1,2) and B(3,3). Slope = 3−2 3−1 1 = ; Another way to look at it would be that we rise 1 up and run 2 to the right. 2 This gives us the rate of change from A to B. Similarly, in the (b) we can use the same logic to calculate the rate of change from A(-2,1) to B(3,3). It would be :- Slope:- 3−1 3−(−2) = 2 5 Finding the instantaneous rate of change, would mean finding the rate of change of ‘y’ with respect to ‘x’ at that particular x value. The example below illustrated the same. 73 2a. To find the instantaneous rate of change at 𝑥 = −1, we draw a tangent at that point, and find its slope. By choosing two suitable points (1,-1) and (3,7) we can find the slope. Slope:- 7−(−1) 3−1 8 = 2 = 4To find the instantaneous rate of change at 𝑥 = 2, we find the slope of the tangent using two suitable points (0,1) and (-2,-1). −1−1 Slope:- −2 =1 Derivatives can be found using the first principle’s method as well. The first principle’s method suggests that : 𝑓(𝑥 + ℎ) − 𝑓(𝑥) ℎ→0 ℎ 𝑓′(𝑥) = 𝑙𝑖𝑚 Let us take an example to prove the following. 𝐿𝑒𝑡 𝑓(𝑥) = 𝑥 2 (𝑥 + ℎ)2 − 𝑥 2 𝑓′(𝑥) = 𝑙𝑖𝑚 ℎ→0 ℎ 𝑥 2 + 2ℎ𝑥 − 𝑥 2 + ℎ2 ℎ→0 ℎ 2ℎ𝑥 + ℎ2 𝑓′(𝑥) = 𝑙𝑖𝑚 ℎ→0 ℎ ℎ (2𝑥 + ℎ) 𝑓′(𝑥) = 𝑙𝑖𝑚 ℎ→0 ℎ 𝑓′(𝑥) = 𝑙𝑖𝑚 2𝑥 + ℎ 𝑓′(𝑥) = 𝑙𝑖𝑚 ℎ→0 𝑓′(𝑥) = 2𝑥 + 0 = 2𝑥 Therefore, the derivative of 𝑥 2 is 2𝑥. For now let us use the concept of limits, which will be explained in detail at a later stage of the guide. Usually, in order to understand derivatives fully it is imperative to understand that the derivative of a function can be used to find the gradient of the tangent to any curve at that particular point. For example, find the gradient of the tangent to: a) 𝑓(𝑥) = 𝑥 3 at the point (2,8) The following is solved by first, finding the derivative of the function and then plugging in the value of ‘x’ to find the gradient of the tangent at x = 2. So, using the first principle’s method: 74 𝑓(𝑥) = 𝑥 3 (𝑥 + ℎ)3 − 𝑥 3 𝑓′(𝑥) = 𝑙𝑖𝑚 ℎ→0 ℎ 3 2 𝑥 + 3𝑥 ℎ + 3ℎ2 𝑥 + ℎ3 − 𝑥 3 𝑓′(𝑥) = 𝑙𝑖𝑚 ℎ→0 ℎ 2 ℎ(3𝑥 + 3𝑥ℎ + ℎ2 ) 𝑓′(𝑥) = 𝑙𝑖𝑚 ℎ→0 ℎ 2 𝑓′(𝑥) = 𝑙𝑖𝑚 3𝑥 + 3𝑥ℎ + ℎ2 ℎ→0 𝑓′(𝑥) = 3𝑥 2 + 3𝑥(0) + 0 = 3𝑥 2 Now that we know that 𝑓′(𝑥) = 3𝑥 2 ,we can substitute 2 in place of ‘x’ to get 𝑓′(2)= 3(4) = 12. The gradient of the tangent at x = 2 is 12. Q. Using the first principle’s method find the gradient of the tangent to : a) 𝑓(𝑥) = 𝑥 3 + 𝑥, at (1,2) 4 b) 𝑓(𝑥) = 𝑥 + 3, at (2,5) After we perform the same calculation over and over again for similar functions we can generalise a set of rules for differentiation. The Derivative of 𝑥 𝑛 : 𝑓′(𝑥) = 𝑛𝑥 𝑛−1 Below are some of the many common derivatives of different types of functions. 75 Q. Suppose 𝑦 = 𝑐 √1+𝑑𝑥 𝑑𝑦 1 where c and d are constants. When 𝑐 = 3, 𝑑 = 1and 𝑑𝑥 = − 8,find c and d. 𝑑 Q. Suppose 𝑓(𝑥) = 3(𝑐𝑥 − 𝑥 )3 . Given that 𝑓(1.5) = 3and 𝑓′(1.5) = 30, find c and d. However, it does not end here. To find the derivative for functions that are products of two functions or are the result of one function being divided by another, we have the ‘Product’ and ‘Quotient’ rule. Also, one of the most important parts of differentiation is the chain rule. The chain rule suggests that while differentiation a function that has several functions in composite, if 𝑑𝑦 𝑑𝑦 𝑑𝑢 𝑦 = 𝑔(𝑢), 𝑤ℎ𝑒𝑟𝑒 𝑢 = 𝑓(𝑥) 𝑡ℎ𝑒𝑛, 𝑑𝑥 = 𝑑𝑢 × 𝑑𝑥 ; the derivative of the outer function multiplied by the derivative of the inner function. The three rules are shown below. 76 Product Rule Example: Q. Find the derivative of : 𝑓(𝑥) = 𝑥 2 𝑠𝑖𝑛 𝑥 Let, 𝑢 = 𝑥2 𝑑𝑢 = 2𝑥 𝑑𝑥 ; ; 𝑣 = 𝑠𝑖𝑛 𝑥 𝑑𝑣 = 𝑐𝑜𝑠 𝑥 𝑑𝑥 Then, 𝑓′(𝑥) = (𝑥 2 × 𝑐𝑜𝑠 𝑥) + (2𝑥 × 𝑠𝑖𝑛 𝑥) = 𝑥 2 𝑐𝑜𝑠 𝑥 + 2𝑥𝑠𝑖𝑛 𝑥 Quotient Rule Example: 𝑑𝑦 Q. Use the quotient rule to find 𝑑𝑥 if 𝑦= 1 + 3𝑥 2−𝑥 Let, 𝑢 = 1 + 3𝑥 ; 𝑣 =2−𝑥 𝑑𝑢 𝑑𝑣 =3 ; = −1 𝑑𝑥 𝑑𝑥 (2 − 𝑥) × 3 − (1 + 3𝑥) × −1 6𝑥 + 7 𝑓′(𝑥) = = (2 − 𝑥)2 (2 − 𝑥)2 Chain Rule Example: Q. Differentiate with respect to x: 77 𝑓(𝑥) = 𝑠𝑖𝑛(𝑥 2 ) To solve such questions we need to make use of the chain rule. We first take the derivative of the outer function, which in this case in 𝑠𝑖𝑛 𝑥, and then multiply it by the derivative of the inner function 𝑥 2 . 𝑓′(𝑥) = 𝑐𝑜𝑠(𝑥 2 )(2𝑥) = 2𝑥 𝑐𝑜𝑠 (𝑥 2 ) Exercise: Q. Find the following:1. 2. 3. 4. 5. 6. 7. 8. 𝑑 ( 4𝑥 ) 𝑑𝑥 𝑥−5 𝑑 𝑥2 (− 𝑑𝑥 𝑠𝑖𝑛 𝑥 ) √𝑥 2 +3 𝑥 𝑥 𝑡𝑎𝑛 𝑥 𝑙𝑜𝑔(𝑐𝑜𝑠 2𝑥) 𝑐𝑜𝑠(𝑐𝑜𝑠 𝑥) 𝑥√𝑐𝑜𝑠𝑒𝑐 𝑥 𝜋 𝑥 The point P has y-coordinate 6, and lies on the graph of 𝑓(𝑥) = 𝑎𝑟𝑐𝑡𝑎𝑛 ( 3 ). Find the coordinates of P Find the gradient of tangent to 𝑦 = 𝑓(𝑥) at P. 9. 𝑥 2 𝑐𝑜𝑠𝑒𝑐 𝑥 10. 𝑙𝑛 (𝑥(𝑥 2 + 1)) Second and Higher Order Derivatives Taking the second order derivative means differentiating the first derivative, or differentiating the function twice. If we were to find the 𝑛𝑡ℎ derivative, it would mean differentiating the function 𝑛times. Exercise: 𝑑2 𝑦 𝑑𝑦 Q. If 𝑦 = 2𝑒 3𝑥 + 5𝑒 4𝑥 ,show that 𝑑𝑥 2 − 7 𝑑𝑥 + 12𝑦 = 0. 𝑑2 𝑦 Q. If 𝑦 = 𝑠𝑖𝑛(2𝑥 + 3),show that 𝑑𝑥 2 + 4𝑦 = 0. 𝑑4 𝑦 Q. If 𝑦 = 𝑠𝑖𝑛 𝑥,show that 𝑑𝑥 4 = 𝑦. Q. If 𝑦 = 2𝑠𝑖𝑛 𝑥 + 3𝑐𝑜𝑠 𝑥, show that 𝑦′′ + 𝑦 = 0. Q. Let 𝑓(𝑥) = 𝑒 −𝑥 (𝑥 + 2) .Find: I. 𝑓′(𝑥) II. 𝑓′′(𝑥) III. 𝑓′′′(𝑥) IV. 𝑓 (4) (𝑥) V. Hence form a general formula for 𝑓 (𝑛) (𝑥), 𝑛 ∈ 𝑍 + . 78 Implicit Differentiation The process of finding the derivative of a dependent variable in an implicit function by differentiating each term separately, by expressing the derivative of the dependent variable as a symbol, and by solving the resulting expression for the symbol. Example: 𝑑𝑦 Q. Find 𝑑𝑥 if a) 𝑥 2 + 𝑦 2 = 25 b) 𝑥 + 𝑐𝑜𝑠 𝑦 = 1 To solve this we need to carry out implicit differentiation in the following manner:𝑑𝑦 a) 2𝑥 + 2𝑦 𝑑𝑥 = 0 𝑑𝑦 Once we different each of the terms with respect to ‘x’, we must rewrite it to make 𝑑𝑥 the subject 𝑑𝑦 𝑥 of the formula. Therefore, for a) 𝑑𝑥 = − 𝑦. 𝑑𝑦 b) 1 − 𝑠𝑖𝑛 𝑦(𝑑𝑥 ) = 0 ; 𝑑𝑦 𝑑𝑥 1 = 𝑠𝑖𝑛 𝑦 Exercise: 𝑑𝑦 Q. Find 𝑑𝑥 if 1. 2. 3. 4. 𝑦 + 𝑥 𝑠𝑒𝑐 𝑦 = 𝑎𝑟𝑐𝑐𝑜𝑠 𝑥 𝑙𝑛(𝑥𝑦) = 𝑒 𝑦 𝑠𝑖𝑛 𝑦 + 𝑥𝑒 𝑦 = 2𝑦 The graph shows the relation (𝑥 2 + 𝑦 2 )2 = 𝑥 2 − 𝑦 2 . a) Find the axes intercepts. b) Find the coordinates of all points on the graph at which the tangent is horizontal. Q. An ellipse has the equation 𝑥 2 + 3𝑦 2 = 48. Find the coordinates of the points on the ellipse such that the normal at those points passes through (0,2). Properties of Curves 79 1. Tangents and Normals As mentioned before one of the key properties of differentiation is to find the equation of a tangent to a curve. The normal at that point is simply perpendicular to the tangent, and can be calculated since the slope of the normal is the negative reciprocal of the slope of the tangent. Example: Q. Find the equation of the tangent to: a) 𝑦 = 1 √3−2𝑥 at 𝑥 = − 3 To do this we must differentiate the equation with respect to ‘x’. 3 −1 𝑦′ = (3 − 2𝑥)−2 × −2 = (3 − 2𝑥)−1.5 2 After this has been done, we must substitute -3 with ‘x’ to find the slope of tangent at x = -3. 1 𝑦′ = (3 − 2(−3))−1.5 = 27 We know the slope of the tangent, however we need the equation in the form y = mx + c. We now substitute 3 in the function to obtain its corresponding ‘y’ value. 𝑦= 1 √3−2(−3) 1 1 = 3, therefore (-3,3) lies on the curve. Now, we can obtain the equation of the tangent by doing:1 1 𝑦 − 3 = 27 (𝑥 + 3) Example: Q. Find the equation of another tangent to 𝑦 = 1 − 3𝑥 + 12𝑥 2 − 8𝑥 3 which is parallel to the tangent at (1,2). In order to solve this, we first differentiate with respect to ‘x’. 𝑑𝑦 = −3 + 24𝑥 − 24𝑥 2 𝑑𝑥 Slope of tangent = −3 + 24(1) − 24(1) = −3 −3 + 24𝑥 − 24𝑥 2 = − 3 24𝑥(1 − 𝑥) = 0 𝑥 = 1; 𝑥 = 0 𝑥 = 1, was already given to us, however the second answer we obtained is the key to answering the question. At x = 0, we need to find the corresponding ‘y’ value. 𝑦 = 1 − 3(0) + 12(0)2 − 8(0)3 = 1; (0,1) Now we can determine the equation of the tangent since we have the slope and a point on the tangent. Thus, the equation would be 𝑦 − 1 = −3(𝑥 − 0) 80 𝑦 = −3𝑥 + 1 Find a and b. To solve this question we first equate both the equations to get an equation in a and b. 𝑥 + 2 = 𝑎 √𝑥 + 𝑏𝑥 6 = 2𝑎 + 4𝑏 3 = 𝑎 + 2𝑏→ Eq 1 We know that the slope of the tangent to the curve at x = 4 is 1, since the coefficient of x in 𝑦 = 𝑥 + 2is 1. Therefore, we can find another equation in terms of a and b. 𝑦 = 𝑎√𝑥 + 𝑏𝑥 𝑑𝑦 𝑎 = +𝑏 𝑑𝑥 2√𝑥 Slope of the tangent at x = 4 is 1. 1= 𝑎 +𝑏 2(2) 𝑎 + 4𝑏 = 4→ Eq 2 Using these two simultaneous equations we can find the value of a and b. 𝑎 = 4 − 4𝑏 𝑎 = 3 − 2𝑏 4 − 4𝑏 = 3 − 2𝑏 1 1 1 = 2𝑏; 𝑏 = 2and therefore 𝑎 = 4 − 4(2) = 2; 𝑎 = 2 Exercise: Q. Consider the curve 𝑦 = 𝑎√1 − 𝑏𝑥 where a and b are constants. The tangent to this curve at the point where 𝑥 = −1is 3𝑥 + 𝑦 = 5.Find the values of a and b. Q. Find the exact area of the shaded triangle. 81 2. Increasing and Decreasing Functions If 𝑓(𝑥)is increasing then 𝑓′(𝑥) ≥ 0and a function 𝑓(𝑥)is decreasing when 𝑓′(𝑥) ≤ 0. Example: 3 2 Q. The graph of 𝑓(𝑥) = 𝑥 − 6𝑥 + 10. a. Find 𝑓′(𝑥), and draw its sign diagram. b. Find the intervals where 𝑓(𝑥)is increasing or decreasing. Note:- A sign diagram is very useful in determining the behaviour of a function. a. First we find the derivative of the function which is:𝑓′(𝑥) = 3𝑥 2 − 12𝑥 82 After we find the derivative we need to draw its sign diagram. To draw the sign diagram we must find the turning points. These are found by equating the derivative to 0. 3𝑥 2 − 12𝑥 = 0 3𝑥(𝑥 − 4) = 0 𝑥 = 0; 𝑥 = 4 We substitute a point smaller than 0 in the derivative and a value of ‘x’ greater than 4 in the derivative, and a value between 0 and 4. If the value is positive we put a +𝑣𝑒 sign and if the value is negative we put −𝑣𝑒 sign. For example the sign diagram for the above derivative is the following: b. To answer this we need to carefully observe the sign diagram. We know that if the derivative is positive, the function is increasing and if it is negative, it is decreasing. Therefore the function is increasing when x > 4 and x < 0, and the function is negative when 0 < x < 4. Exercise: Q. Find the intervals where 𝑓(𝑥) is increasing or decreasing:a) 𝑓(𝑥) = 𝑥 3 − 6𝑥 2 b) 𝑓(𝑥) = 2 √𝑥 c) 𝑓(𝑥) = √𝑥 𝑙𝑛 𝑥 4𝑥 Q. Consider 𝑓(𝑥) = 𝑥 2 +1 a) Show that 𝑓′(𝑥) = −4(𝑥+1)(𝑥−1) (𝑥 2 +1)2 and draw its sign diagram b) Hence find intervals where 𝑦 = 𝑓(𝑥)is increasing or decreasing. 3. Stationary Points Although you did not know what stationary points were, in the previous sums we have been finding them all along. A stationary point of a function is a point where 𝑓′(𝑥) = 0. It could be a local maximum or a local minimum or a point of inflection. Note:- At a stationary point the tangent is horizontal (gradient = 0). The diagram below explains what these three terms mean and the respective sign diagram for each of them. 83 Example: Q. The graph of 𝑓(𝑥) = 𝑥 3 + 6𝑥 2 − 15𝑥 − 40is shown. P and Q are stationary points. a) Classify points P and Q. b) Find 𝑓′(𝑥) c) Find the coordinates P and Q A. From our knowledge from functions we can classify point P as a maximum and point Q as a minimum. We can derive that 𝑓′(𝑥) = 3𝑥 2 + 12𝑥 − 15. Since stationary points are points at which the derivative is 0, to find the coordinates of points P and Q we equate the derivative to 0. 𝑓′(𝑥) = 3𝑥 2 + 12𝑥 − 15 = 0. This gives us 𝑥 = −5 ; 𝑥 = 1. To find the corresponding y values we simply substitute the x values in the 𝑓(𝑥). Q. Consider the function 𝑔(𝑥) = −2𝑥 3 + 6𝑥 2 + 18𝑥 − 7. a) Find 𝑔′(𝑥),and draw its sign diagram. b) Find intervals where the function is increasing and decreasing c) Find and classify any stationary points d) Sketch the graph 𝑦 = 𝑔(𝑥)showing the features you have found e) Describe the behaviour of the function as 𝑥 → ∞and as 𝑥 → −∞ 84 We begin by differentiating the function: 𝑔′(𝑥) = −6𝑥 2 + 12𝑥 + 18. To draw the sign diagram we equate the derivative to 0. −6𝑥 2 + 12𝑥 + 18 = 0 𝑥 = 3; 𝑥 = −1 Then we need to find the intervals where the function is increasing or decreasing. For this we need to see the sign diagram. Since the derivative is positive between -1 and 3, that is when the function is increasing. Since the derivative is negative when x > 3 and x < -1 it is decreasing during that time. Stationary points are when the derivative is equal to 0. Thus, the x-coordinate of the stationary points have already been determined: 𝑥 = 3; 𝑥 = −1. To find the corresponding ‘y’ values we have the substitute them in 𝑦 = 𝑔(𝑥). To classify them we need to observe the sign diagram again. If at x = a the sign goes from negative to positive there is a local minimum at that point, and if at x = a the sign goes from positive to negative there is a local maximum at that point. However, if the sign is the same it is a point of inflection. Therefore, since the sign changes from negative to positive at x = - 1, there is a local minimum at that point, and because the sign changes from positive to negative at x = 3, there is a local maximum at that point. Using the properties of the graph that we obtained we can easily sketch the above graph. Lastly, as 𝑥 → ∞we see that the graph is going in the downward direction where 𝑦 → −∞.and as 𝑥 → −∞in the second quadrant we see that the graph is moving in the upward direction where 𝑦 → +∞. 85 Exercise: Q. For each of the functions find and classify any stationary points. Sketch the function, showing all the important features. a) 𝑦 = 𝑥 4 − 6𝑥 2 + 8𝑥 − 3 b) 𝑦 = 𝑥 2 𝑒 𝑥 Q. The cubic polynomial 𝑄(𝑥) = 𝑎𝑥 3 + 𝑏𝑥 2 + 𝑐𝑥 + 𝑑 touches the line with the equation 𝑦 = 9𝑥 + 2at the point (0,2) and has a stationary point at (-1,-7). Find 𝑄(𝑥). Q. Find the greatest and least value of a) 𝑥 3 − 12𝑥 − 2 𝑓𝑜𝑟 − 3 ≤ 𝑥 ≤ 5 Q. Prove that 𝑙𝑛 𝑥 𝑥 1 ≤ 𝑒 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 > 0. Q. Consider the relation 𝑦 2 + 𝑥𝑒 𝑥 = 𝑦. 𝑑𝑦 a) Find 𝑑𝑥 . b) Find the equation of the normal to the equation at A. c) Find the exact coordinates of the stationary points. 4. Shape : we say that the curve is concave down Note:- At this point 𝑦′′ > 0. : we say that the curve is concave up Note:- At this point 𝑦′′ < 0. Example: 86 Q. For the given function determine the intervals on the function where it is:a) Increasing b) Decreasing c) Concave Up d) Concave Down −1 𝑓(𝑥) = √𝑥 To solve this we need to find the derivative of the function. 1 𝑓′(𝑥) = 𝑥 −1.5 2 Then, we equate the derivative to 0 to find the x-coordinate of the stationary points. 1 2 𝑥 −1.5 = 0; However when we equate it to zero we realise that this equation does not have an answer, there is no solution to it, therefore there is no stationary point. The domain of the function is 𝑥 > 0 and for all positive values of x since 𝑓′(𝑥) > 0, the function is always increasing, and therefore never decreasing. Then to answer about the shape of the function we find the second derivative of the function or 𝑓′′(𝑥). In this example −3 𝑓′′(𝑥) = 4𝑥 2 𝑥and for any positive value of x, 𝑓′′(𝑥)< 0. This means that the function is concave √ down. Exercise: Q. Consider 𝑓(𝑥) = 𝑙𝑛(2𝑥 − 1) − 3. a) Find the x-intercept b) Can f(0) be found ? State its significance c) Find the domain of the function d) Find the gradient of the tangent to the curve at x = 1 e) Find , and hence explain why the function is concave down for all x in the domain of f f) Graph the function showing the features you have found Q. Find intervals where the curve is concave up or concave down for 𝑓(𝑥) = 𝑥 2 +𝑥−3 𝑥+1 . 5. Inflection Points A point of inflection is a point at which the tangent to the curve crosses the curve. At a point of inflection 𝑓′′(𝑥) = 0. 87 Note:- There is a point of inflection at x = a, if 𝑓′′(𝑥) = 0and the sign of 𝑓′′(𝑥)changes at x = a. The point of inflection is a: ● Stationary inflection if 𝑓′(𝑥) = 0 ● Non-stationary inflection if 𝑓′(𝑥) ≠ 0 Example: Q. Describe the inflection points of the graph. 88 A. C is a non-stationary point of inflection because the tangent to the curve at C would not be horizontal and therefore would not be equal to 0. D is a stationary point of inflection because the tangent to the curve at D would be horizontal and therefore equal to 0. Q. Find and classify all points of inflection. a) 𝑦 = 𝑥 3 − 6𝑥 2 + 9𝑥 + 1 First we find the derivative, followed by the second derivative. 𝑦′ = 3𝑥 2 − 12𝑥 + 9 𝑦′′ = 6𝑥 − 12 Now, we equate the second derivative to 0 to obtain x = 2. Then we draw the sign diagram for the second derivative. For this we need to plug in one value greater than 2 and less than 2, in the second derivative and accordingly put the signs to get: Therefore, there is a point of inflection at x = 2, since at this point the double derivative is equal to 0 and there is a sign change in the sign diagram for the double derivative. To figure out whether it is a stationary or non-stationary inflection point we substitute 2 in the first derivative. 𝑦′ = 3𝑥 2 − 12𝑥 + 9 𝑦′ = 3(2)2 − 12(2) + 9 = −3. Since −3 < 0, there is a non-stationary point of inflection at x = 2. Exercise: Q. For 𝑓(𝑥) = 𝑥 4 − 4𝑥 2 + 3 a) Find and classify all the turning points b) Find and classify all points of inflection c) Find where the function is increasing or decreasing d) Find intervals where the function is concave up or down e) Sketch the function showing the features you have found Q. Consider the function 𝑦 = 𝑒 𝑥 − 3𝑒 −𝑥 a) Determine the axes intercepts b) Prove that the function is increasing for all ‘x’. 𝑑2 𝑦 c) Show that 𝑑𝑥 2 = 𝑦. d) Use the GDC to graph the function and its features you have found Q. Consider the surge function 𝑓(𝑡) = 𝐴𝑡𝑒 −𝑏𝑡 , 𝑡 ≥ 0,where A and b are positive constants. Prove that the function has: 89 a) A local max at t = 1/b b) A point of inflection at t = 2/b c) Sketch the function showing the features you have proved Differentiability and Continuity Continuity: At a point:A function f(x) is said to be continuous at a point x=p, if 𝑙𝑖𝑚−𝑓(𝑥) = 𝑙𝑖𝑚+𝑓(𝑥),that 𝑥→𝑝 𝑥→𝑝 is, the function is approaching the same y-value from both the left hand and right hand side of p. During an interval: The function g(x) is said to be continuous during an interval (c,d) given that c< d only if g(x) is continuous at all points during the interval. Standard examples of continuous functions: 1. All polynomial functions 2. Cosine and sine functions Essential and removable discontinuities: Functions have 3 types of discontinuities: 1. Hole 2. Jump 3. Break Of these 3, a ‘break’ and a ‘jump’ is essential while a ‘hole’ is removable. In this graph, there is a hole (aka a removable discontinuity) at x=5 and an essential discontinuity in the form of a break in the graph at x=6. We can remove the hole at x=5, by defining the value of f(x) as 4 when x=5 using a piecewise function. Differentiability: A function f(x) is said to be differentiable at a point x=a if 𝑅𝑖𝑔ℎ𝑡 ℎ𝑎𝑛𝑑 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 = 𝐿𝑒𝑓𝑡 𝐻𝑎𝑛𝑑 𝐷𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 90 𝑅𝑖𝑔ℎ𝑡 ℎ𝑎𝑛𝑑 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 = 𝑙𝑖𝑚 𝑓(𝑎+ +ℎ)−𝑓(𝑎+ ) ℎ ℎ→0 (where a+ indicates a is being approached from the right hand side) 𝐿𝑒𝑓𝑡 ℎ𝑎𝑛𝑑 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 = 𝑙𝑖𝑚 ℎ→0 𝑓(𝑎− +ℎ)−𝑓(𝑎− ) ℎ (where a- indicates a is being approached from the left hand side) Note:- All differentiable functions are continuous, but not all continuous functions are differentiable. The overall function is differentiable if a derivative can be found on all points on the graph of the function. Functions with cusps, sharp edges, or breaks are not differentiable. Examples: 𝑦 = |𝑥| 𝑎𝑡 𝑥 = 0. Note- This is merely an important concept. You will not be tested on this in an IB exam. Limits and L'Hopital's Rule When we discussed vertical asymptotes in Functions, we saw that the x-values of the function kept getting closer and closer to the vertical asymptote value (let it be noted by b), but were never quite equal to it. As the x-values keep approaching b, the function values also approach certain values. Limits help us determine what these values are. We say that the limit of f(x) is L as x approaches b. This can be written in the form given below. 𝑙𝑖𝑚 𝑓(𝑥) = 𝐿 𝑥→𝑏 Given that f(x) is as close to L as possible as we want x sufficiently close to b, from both sides, without actually letting x be b. Example: Find the limit: 𝑙𝑖𝑚 𝑥 2 −4𝑥+4 𝑥→2 𝑥 2 −2𝑥 0 Here, if we substitute x=2 in the expression, we get 0.There are 2 ways we can solve this. Firstly, we can graph the function and observe the behavior of the graph as the x-value approaches 2. Secondly, we can use L’Hopital’s Rule. Given that f(x) and g(x) are differentiable and g’(x) is not equal to zero on an interval that contains the point x=a. If 𝑙𝑖𝑚𝑓(𝑥) = 0and 𝑙𝑖𝑚𝑔(𝑥) = 0or, if as f(x)→ ±∞and g(x)→ ±∞, then 𝑥→𝑐 𝑙𝑖𝑚 𝑓(𝑥) 𝑥→𝑑 𝑔(𝑥) = 𝑙𝑖𝑚 𝑓′(𝑥) 𝑥→𝑐 , given that the limit on the right exists. 𝑥→𝑑 𝑔′(𝑥) Thus, we differentiate the numerator and denominator of the above expression separately and then substitute the value of 2. Note- A limit does not exist if the limit is ∞or if the value the function approaches as it approaches b from the left hand side is not equal to the value the function approaches as it approaches b from the right hand side. Sample Exercise 2: 91 𝑥 2 0 Find the limit:𝑙𝑖𝑚 1−𝑐𝑜𝑠(𝑥). Here, if we substitute zero in the expression, we get0. Hence, we apply 𝑥→0 2𝑥 L’Hopital’s Rule. However, even after applying the rule, we get 𝑙𝑖𝑚 𝑠𝑖𝑛(𝑥)which still gives us 𝑥→0 0 .Thus, we have to apply L’Hopital’s Rule again to solve this. 0 Exercise: 𝑥 2 +2𝑥 1. 𝑙𝑖𝑚 1−𝑐𝑜𝑠(𝑥) 𝑥→0 𝑥+𝑠𝑖𝑛𝑥 2. 𝑙𝑖𝑚 𝑥−𝑠𝑖𝑛𝑥 𝑥→0 3. 𝑙𝑖𝑚 𝑎𝑥 −𝑏 𝑥 𝑥→0 𝑠𝑖𝑛𝑥 1 4. 𝑙𝑖𝑚+( − 1 ) 𝑠𝑖𝑛𝑥 𝑥→0 𝑥 𝑙𝑛(𝑐𝑜𝑠(5𝑥)) 5. 𝑙𝑖𝑚+ 𝑙𝑛(𝑐𝑜𝑠(3𝑥)) 𝑥→0 Rates of Change One of the many applications of differentiation includes ‘Rates Of Change’. Here we examine how one variable changes with respect to another. Measuring the rate of something means how fast or slow it changes with respect to time. Example: Q. The future profit of a company is modelled as 𝑃(𝑥) = 2𝑥 2 − 12𝑥 + 118thousand dollars, where x is the time in years from now. 𝑑𝑃 a) Find 𝑑𝑥 at state its units 𝑑𝑃 b) Find 𝑑𝑥 at x = 8 and state its meaning. 𝑑𝑃 We have been given an equation that models profit of a company. The expression 𝑑𝑥 tells us how 𝑑𝑃 the profit of the company changes with respect to time. To find 𝑑𝑥 we simply differentiate P with respect to x. 𝑑𝑃 = 4𝑥 − 12 𝑑𝑥 𝑑𝑃 To find 𝑑𝑥 at x = 8, we substitute 8 in the derivative. 4(8) − 12 = 20; This means that the rate at which profit will be made in the 8th year will be 20,000 dollars per year. Example: 92 Q. The temperature of a liquid after being placed in a refrigerator is given by 𝑇 = 5 + 95𝑒 −𝑘𝑡 , where t is the time in minutes. a) Show that 𝑑𝑇 𝑑𝑡 = 𝑐(𝑇 − 5) This is an example of how the temperature of an object is changing with time, or one can say it shows the rate of change of T with t. First we differentiate the equation. 𝑑𝑇 = 0 + 95𝑒 −𝑘𝑡 (−𝑘) 𝑑𝑡 We can rewrite 𝑇 = 5 + 95𝑒 −𝑘𝑡 as 𝑇 − 5 = 95𝑒 −𝑘𝑡 Therefore, 𝑑𝑇 𝑑𝑡 = (−𝑘)(𝑇 − 5),where c = -k. Example: Q. Find exactly the rate of change in the area of triangle PQR as𝜃changes, at the time when 𝜃 = 45. . In this example, we have not been given any formula or equation. However, we know that the area of a triangle is 𝐴 = 2 𝑎𝑏 ⋅ 𝑠𝑖𝑛𝜃 . 1 We can substitute the two sides that we have been given. Thus, 𝐴 = 3(7) 𝑠𝑖𝑛𝐶 = 21𝑠𝑖𝑛𝜃. Now we can differentiate A with respect to t. 𝑑𝐴 𝑑𝑡 𝑑𝜃 = 21𝑐𝑜𝑠𝜃 ⋅ 𝑑𝑡 ; Now we simply substitute 45 in place of 𝜃to get 𝑑𝐴 𝑑𝑡 = 21 √2 𝑑𝜃 ⋅ 𝑑𝑡 , and that is how we can find the rate of change of the Area with respect to the time as the angle changes. Optimisation A very useful application of calculus is optimisation. It involves maximizing or minimizing a variable to make efficient or best use of it. For example, a company would aim to maximize profit and minimize costs and therefore can model equations and find ways to accomplish their goals. Let us solve a few examples together. Example: Q. The total cost of producing x chocolates per day is 𝑥2 4 + 8𝑥 + 20dollars and for this production 1 level each chocolate may be sold for (23 − 2 𝑥)dollars. How many chocolates should be produced per day to maximise the total profit? In this example we need to optimize the production level to find the best number of chocolates that should be produced per day for profits to be maximum. First of all, we need to frame an equation that gives us the profit. Let the profit 𝑃(𝑥). 93 Since profit = revenue - cost 1 𝑃(𝑥) = 𝑥(23 − 2 𝑥) − 𝑥2 4 + 8𝑥 + 20. To maximize/minimize something we need to equate the derivative to 0. So by differentiating we get 𝑃′(𝑥) = 31 − 31 = 3𝑥 3𝑥 2 . Now we must equate this to zero. ; 3𝑥 = 62; 𝑥 = 20.67. We can check if this is actually a maximum by equating the 2 double derivative to 0 and if it is a negative value it means that at that particular point there is a maximum. Example: Q. A cone has radius r cm and slant height s cm. Find the ratio s:r which maximises the volume of a cone. ℎ = √𝑠 2 − 𝑟 2 . We know that the volume of a cone is given by:1 1 𝑉 = 3 𝜋𝑟 2 ℎ; We can now replace ‘h’ with √𝑠 2 − 𝑟 2 . Therefore, 𝑉 = 3 𝜋𝑟 2 √𝑠 2 − 𝑟 2 . Using the product rule, 𝑑𝑉 𝑑𝑟 2 1 = 3 𝜋𝑟√𝑠 2 − 𝑟 2 − 3 𝜋𝑟 3 √𝑠2 −𝑟 2 . Now we need to maximise the volume and therefore we must equate the derivative of the volume with respect to the radius. 2(𝑠2 −𝑟 2 )−𝑟 2 √𝑠2 −𝑟 2 = 0; 2(𝑠 2 − 𝑟 2 ) − 𝑟 2 = 0; 2𝑠 2 = 3𝑟 2 ; 𝑠 𝑟 3 = √2 . Thus the ratio s: r is √3: √2. Note:- The main idea in optimisation is to get an equation of the variable you want to maximise or minimise and find its derivative. Once you have done this, you need to equate it to 0 and you are good to go! IB Questions: Q. André wants to get from point A located in the sea to point Y located on a straight stretch of beach. P is the point on the beach nearest to A such that AP = 2 km and PY = 2 km. He does this by swimming in a straight line to a point Q located on the beach and then running to Y. 94 When André swims he covers 1 km in 5√5 minutes. When he runs he covers 1 km in 5 minutes. (a) If PQ = x km, 0 ⩽ x ⩽ 2 , find an expression for the time T minutes taken by André to reach point Y. 5√5𝑥 𝑑𝑇 (b) Show that 𝑑𝑥 = (c) √𝑥 2 +4 − 5. 𝑑𝑇 (i) Solve 𝑑𝑥 = 0 (ii) Use the value of x found in part (c) (i) to determine the time, T minutes, taken for André to reach point Y. 𝑑2 𝑇 20 5 √ (iii) Show that 𝑑𝑥 2 = (𝑥 2+5) 1.5 and hence show that the time found in part (c) (ii) is a minimum. 95 Q. A packaging company makes boxes for chocolates. An example of a box is shown below. This box is closed and the top and bottom of the box are identical regular hexagons of side x cm. (a) Show that the area of each hexagon is 3√3𝑥2 2 𝑐𝑚2 . 3 (b) Given that the volume of the box is 90𝑐𝑚3 , show that when √20 the total surface area of the box is a minimum, justifying that this value gives a minimum. Q. Two non-intersecting circles C1 , containing points M and S , and C2 , containing points N and R, have centres P and Q where PQ = 50 . The line segments [MN] and [SR] are common tangents to the circles. The size of the reflex angle MPS is α, the size of the obtuse angle NQR is β , and the size of the angle MPQ is θ . The arc length MS is l1 and the arc length NR is l2 . This information is represented in the diagram below. The radius of C1 is x, where x ≥10 and the radius of C2 is 10. a) Explain why 𝑥 < 40 b) Show that 𝑐𝑜𝑠𝜃 = 𝑥−10 50 c) (i) Find an expression for MN in terms of x (ii) Find the value of x that maximises MN. 96 d) e) f) g) Find an expression in terms of x for 𝛼and 𝛽. Find an expression, b(x) , for the length of the perimeter in terms of x Find the maximum value of the length of the perimeter. Find the value of x that gives a perimeter length of 200. Related Rates of Change Essentially, in related rates, our two variables (let’s call them x and y) are changing with time. Thus, the problems in these types of questions is, in simple terms, determining the change in x with respect to time when we know the change in y with respect to time and a relation between x and y is given. Let us solve some sample questions together. Example: 1. p and q are variables related by the equation 𝑝𝑞 3 = 30. a. Differentiate this equation with respect to time t. b. When p=3, q is increasing at a rate of 1 unit per second. Find the rate of change of p. So, we differentiate this equation using the concepts of implicit differentiation and product rule. 𝑑𝑞 𝑑𝑝 This becomes: 𝑝3 𝑑𝑡 + 3𝑝2 𝑞. 𝑑𝑡 = 0. Then, we need to find the rate of change of q. We know that 𝑑𝑞 𝑑𝑡 = 1. We can easily find q as p=3, and it is given that 𝑝𝑞 3 = 30. Thus, we substitute the value of q in the equation we found and get the rate of change of p. Note- The rate of change of p will be negative and that means that p is decreasing as q is increasing, something that we can easily deduce if we rewrite the original relation. Example: 2. The length of a rectangle is decreasing at 2 cm per minute. However, the area remains constant at 300 cm2. Taking c as the length and d as the width of the triangle: a. Determine the relation between c and d b. Differentiate the relation obtained in a with respect to time c. At what rate is the width of a rectangle increasing at the moment when i. c=8 ii. The rectangle is a square Firstly, we know that the area of a rectangle is length into width. Thus, 𝑐𝑑 = 300.We then use implicit differentiation and the product rule to get the differentiated equation with respect to the time. Once we have achieved that, it is simple substitution. Example: 3. Two planes fly on parallel courses that are 9 km apart. Their air speeds are 300 ms-1 and 400 ms-1. How quickly is the distance between them changing at the moment when the faster jet is 3 km behind the slower one? 97 This is a slightly more complex question and we need to draw a diagram to understand the situation better. Firstly, we must understand that the relative speed of the two airplanes is 400300=100 ms-1. Thus, the distance that this covers in time t is given by d=100t. Thus, we can deduce that 𝑥 2 = (100𝑡)2 + 90002 . Once we have rewritten this by applying a square root on both sides of the equation, we must differentiate the equation in terms of t. We find the time by equating 100t to 9000 to get the time. Using this, we can easily find the rate of change of the distance between the two planes. Example: 4. A flashlight level with the ground is located 20 m from the foot of a building. The flashlight shines in the building’s direction. A 1.92 m tall person walks from the flashlight in the building’s direction at .54 ms-1. At what rate is the person’s shadow shortening at the point when they are a. 16 m b. 9 m from the building? 98 𝑥 20 = 1.92 20 − 0.54𝑡 We then rewrite the equation in terms of x and differentiate with respect to t. For the two parts, we find the time taken by first finding the distance travelled by subtracting distance from the 𝑑 building from the total distance. Then, we use 𝑡 = 𝑣 where we just found d and v is 0.54 ms-1. IB Questions: 1. [7 marks] A ladder of length 10 m on horizontal ground rests against a vertical wall. The bottom of the ladder is moved away from the wall at a constant speed of . Calculate the speed of descent of the top of the ladder when the bottom of the ladder is 4 m away from the wall. 2. [7 marks] A helicopter H is moving vertically upwards with a speed of 10 ms−1 . The helicopter is m directly above the point Q which is situated on level ground. The helicopter is observed from the point P which is also at ground level and PQ =40 m. This information is represented in the diagram below. When , h = 30 m ̂ is 0.16 radians per second; (a) show that the rate of change of H𝑃𝑄 99 (b) find the rate of change of PH. 3. [10 marks] Below is a sketch of a Ferris wheel, an amusement park device carrying passengers around the rim of the wheel. (a) The circular Ferris wheel has a radius of 10 metres and is revolving at a rate of 3 radians per minute. Determine how fast a passenger on the wheel is going vertically upwards when the passenger is at point A, 6 metres higher than the centre of the wheel, and is rising. (b) The operator of the Ferris wheel stands directly below the centre such that the bottom of the Ferris wheel is level with his eyeline. As he watches the passenger his line of sight makes an angle with the horizontal. Find the rate of change of at point A. 4. [8 marks] A lighthouse L is located offshore, 500 metres from the nearest point P on a long straight shoreline. The narrow beam of light from the lighthouse rotates at a constant rate of radians per minute, producing an illuminated spot S that moves along the shoreline. You may assume that the height of the lighthouse can be ignored and that the beam of light lies in the horizontal plane defined by sea level. When S is 2000 metres from P, 100 (a) show that the speed of S, correct to three significant figures, is metres per minute; (b) find the acceleration of S. 5. [6 marks] A stalactite has the shape of a circular cone. Its height is 200 mm and is increasing at a rate of 3 mm per century. Its base radius is 40 mm and is decreasing at a rate of 0.5 mm per century. Determine if its volume is increasing or decreasing, and the rate at which the volume is changing. 6. [6 marks] Paint is poured into a tray where it forms a circular pool with a uniform thickness of 0.5 cm. If the paint is poured at a constant rate of circle when the radius is 20 cm. , find the rate of increase of the radius of the 7a. [2 marks] A straight street of width 20 metres is bounded on its parallel sides by two vertical walls, one of height 13 metres, the other of height 8 metres. The intensity of light at point P at ground level on the street is proportional to the angle where , as shown in the diagram. Find an expression for in terms of x, where x is the distance of P from the base of the wall of height 8 m. 7b. [2 marks] 101 (i) Calculate the value of 𝜃 when x = 0. (ii) Calculate the value of 𝜃 when x = 20. 7c. [2 marks] Sketch the graph of 𝜃 , for 0 ≤ 𝑥 ≤ 20 7d. [6 marks] Show that : 7e. [3 marks] Using the result in part (d), or otherwise, determine the value of x corresponding to the maximum light intensity at P. Give your answer to four significant figures. 7f. [4 marks] The point P moves across the street with speed 0.5 m/s. Determine the rate of change of 𝜃with respect to time when P is at the midpoint of the street. 8a. [3 marks] A water trough which is 10 metres long has a uniform cross-section in the shape of a semicircle with radius 0.5 metres. It is partly filled with water as shown in the following diagram of the cross-section. The centre of the circle is O and the angle KOL is 𝜃 radians. Find an expression for the volume of water in the trough in terms of 𝜃. 8b. [4 marks] The volume of water is increasing at a constant rate of . 102 Calculate when . PART B Integration: Integration is the complete opposite of differentiation. If differentiating a function helps you find its slope at a particular point, integration helps you approximate the area and volume under a curve. Integration is also known as antidifferentiation. Introduction to Integration - Riemann Integral If 𝑓(𝑥) ≥ 0for all 𝑎 ≤ 𝑥 ≤ 𝑏, then 𝑏 ∫𝑎 𝑓(𝑥) 𝑑𝑥is defined as the shaded area A. This is known as the Riemann integral. We would say “the integral of f(x) from a to b respect to x”. To approximate the area under the curve we can several methods. One of the methods is calculating the area of lower and upper rectangles. with use The first image shows lower rectangles and how finding the sum of their areas can give an approximate value for the area under the graph. The similar is shown in the second image with upper rectangles. Example: Consider the region enclosed by 𝑦 = √1 + 𝑥 3 and the x-axis for 0 ≤ 𝑥 ≤ 2. a. Write an expression for the lower and upper rectangle sums using n subintervals. b. Find the lower and upper rectangle sums for n=50. 2 c. Hence estimate∫0 √1 + 𝑥 3 𝑑𝑥 . 103 The graph for the function is:- If we take 𝛥𝑥 as the width of each rectangle and 𝑛 as the number of rectangles then we can say that:𝛥𝑥 = 𝑏−𝑎 ; 𝑛 where the area needs to be calculated between x = a and x = b. 2−0 2 𝛥𝑥 = = 𝑛 𝑛 Therefore for 𝑛 = 50 𝛥𝑥 = 0.04 To make our lives easier, there is a formula that allows us to calculate upper and lower rectangle sums for increasing and decreasing functions. For an increasing function:𝐴𝐿 = ∑𝑛𝑘=1 𝑓(𝑎 + (𝑘 − 1)𝛥𝑥) ⋅ 𝛥𝑥 (sum of lower rectangles), 𝐴𝑈 = ∑𝑛𝑘=1 𝑓(𝑎 + 𝑘𝛥𝑥) ⋅ 𝛥𝑥(sum of upper rectangles), For a decreasing function:𝐴𝐿 = ∑𝑛𝑘=1 𝑓(𝑎 + 𝑘𝛥𝑥) ⋅ 𝛥𝑥(sum of lower rectangles), 𝐴𝑈 = ∑𝑛𝑘=1 𝑓(𝑎 + (𝑘 − 1)𝛥𝑥) ⋅ 𝛥𝑥 (sum of upper rectangles), Since, the above function is increasing we use the formulas for an increasing function. Using the formulas and substituting appropriately we get:𝐴𝐿 = 3.2016 𝐴𝑈 = 3.2816 Note:- It is helpful to remember that for an increasing function:𝐴𝐿 ≤ 𝑎𝑐𝑡𝑢𝑎𝑙 𝑎𝑟𝑒𝑎 ≤ 𝐴𝑈 , And for a decreasing function:- 104 𝐴𝑈 ≤ 𝑎𝑐𝑡𝑢𝑎𝑙 𝑎𝑟𝑒𝑎 ≤ 𝐴𝐿 Techniques of Integration 1. Rules of Integration To find the integral of a function we use the formula provided above. Example: Q. Find a) ∫ (𝑥2 + 3𝑥 − 2) 𝑑𝑥 Using the formula above:𝑥3 3𝑥2 + − 2𝑥 + 𝑐 3 2 Note:- ‘c’ is known as the constant of integration that must be added after integrating any indefinite integral. (𝑥2 + 3𝑥 − 2) 𝑑𝑥 = ∫ b) ∫ (2𝑥2 − 3𝑥 + 1) 𝑑𝑥 = c) ∫ (𝑥√𝑥 − 9) 𝑑𝑥 = 2𝑥2.5 5 2𝑥3 3 − 3𝑥2 2 +𝑥+𝑐 − 9𝑥 + 𝑐 However, not all functions are so simple. For different functions there are different integrals. Some of the standard integrals have been listed below. 105 Below are the properties of integrals:- Exercise: Q. Find a) ∫ (2𝑒𝑥 − 3𝑥) 𝑑𝑥 b) ∫ (3𝑠𝑖𝑛 𝑥 − 2) 𝑑𝑥 c) ∫ ( d) ∫ 4𝑥 𝑑𝑥 e) ∫ (𝑥 − 𝑥𝑙𝑛 2) f) ∫ ( 𝑥(𝑥−1) 3 3 + 𝑠𝑒𝑐 2 𝑥) 𝑑𝑥 1 4 √1−𝑥2 ) 𝑑𝑥 Note:- If in place of x there is a linear model the rules can be modified as the following:1. ∫ 1 (𝑎𝑥 + 𝑏)𝑛 𝑑𝑥 = 𝑎 ⋅ (𝑎𝑥+𝑏)𝑛+1 𝑛+1 + 𝑐, for 𝑛 ≠ −1and 𝑎 ≠ 0. 106 1 2. ∫ 𝑐𝑜𝑠(𝑎𝑥 + 𝑏) 𝑑𝑥 = 𝑎 ⋅ 𝑠𝑖𝑛(𝑎𝑥 + 𝑏) + 𝑐 3. ∫ − 𝑠𝑖𝑛(𝑎𝑥 + 𝑏) 𝑑𝑥 = 𝑎 ⋅ 𝑐𝑜𝑠(𝑎𝑥 + 𝑏) + 𝑐 4. ∫ 𝑠𝑒𝑐2 (𝑎𝑥 + 𝑏) 𝑑𝑥 = 𝑎 ⋅ 𝑡𝑎𝑛(𝑎𝑥 + 𝑏) + 𝑐 5. ∫ 1 𝑎𝑥+𝑏 6. ∫ 𝑒𝑎𝑥+𝑏 = 𝑎 𝑒𝑎𝑥+𝑏 + 𝑐 1 1 1 𝑑𝑥 = 𝑎 𝑙𝑛|𝑎𝑥 + 𝑏| + 𝑐for 𝑎 ≠ 0 1 Exercise: Q. Find a) ∫ b) ∫ 3 (2𝑥 + 5) 𝑑𝑥 4 √3−4𝑥 𝑑𝑥 c) ∫ (2𝑠𝑖𝑛(3𝑥) + 5𝑐𝑜𝑠(4𝑥)) 𝑑𝑥 d) ∫ 32𝑥−1 𝑑𝑥 e) ∫ 1 (2𝑥+5)𝑙𝑛 3 𝑑𝑥 Partial Fractions and integration Earlier, we had studied how to write a fraction as the sum of its partial fractions. The main reason to write it in such a form is so that it can be integrated easily. Example: ∫ 2𝑥−8 𝑥2 −4 𝑑𝑥; We cannot directly integrate this, nor is there any formula that allows us to do so. Therefore, we must rewrite it in such a way that we can individually integrate the fractions and add their integrals up to get the answer. 2𝑥−8 3 1 As studied previously, 𝑥 2 −4can be written as (𝑥+2) − (𝑥−2) . Now this form is familiar for us, and we can easily integrate it. ∫ 3 1 − 𝑑𝑥 = 3𝑙𝑛(𝑥 + 2) − 𝑙𝑛(𝑥 − 2) + 𝑐 (𝑥 + 2) (𝑥 − 2) (𝑥+2)3 This can be rewritten as ‘𝑙𝑛 | (𝑥−2) | + 𝑐.’ Exercise: a) ∫ b) ∫ 6𝑥2 +𝑥−19 (𝑥+3)(𝑥−1)2 2 4𝑥2 −1 𝑑𝑥 𝑑𝑥 107 20 c) ∫ 2𝑥2 −𝑥−3 𝑥−9 𝑥2 −2𝑥−3 d) ∫ 𝑑𝑥 𝑑𝑥 2. Substitution Method Let us take a simple example: ∫ 3𝑥2 (𝑥3 + 1)4 𝑑𝑥 𝑢𝑠𝑖𝑛𝑔 𝑢 = 𝑥3 + 1. The derivative of u can be found and it is as follows: 𝑑𝑢 𝑑𝑢 = 3𝑥 2 and we can rewrite this to 3𝑥 2 = 𝑑𝑥. 𝑑𝑥 Let us try utilising u in the integral: 𝑑𝑢 3𝑥2 × 𝑑𝑥 × 𝑢4 . However, dx can be written as 3𝑥 2. Thus, if we substitute this back into the ∫ 𝑢5 integral, 3x2 cancels out and we are left with ∫ 𝑢4 𝑑𝑢.We can easily integrate this to get 5 . To end the sum, we need to rewrite u in terms of x and since that is given to us, we can conclude that:∫ 3𝑥2 (𝑥3 + 1)4 𝑑𝑥=(𝑥 3 + 1)5 + 𝑐. 2 2𝑥𝑒𝑥 𝑑𝑥.We substitute u=x2 and follow the Let us try the same concept with the integral: ∫ 𝑑𝑢 exact same methodology as in the previous sum. 𝑑𝑥 = 2𝑥 and we rewrite this as Rewriting the integral in terms of u: ∫ 𝑐. 𝑒𝑢 𝑑𝑢 as 2x cancels out. Thus, ∫ 𝑑𝑢 = 𝑑𝑥. 2𝑥 2 2𝑥𝑒𝑥 𝑑𝑥 2 = 𝑒𝑥 + Exercise: Integrate by substitution 1. 3𝑥 2 √𝑥 3 +2 2. 2𝑥𝑠𝑖𝑛(𝑥 2 − 5) 3. 6𝑥 2 (2𝑥 3 −1)6 4. (2𝑥 − 1)𝑒 𝑥−𝑥 5. 6. 7. 8. 2 1−𝑥 3 𝑥 2 −3𝑥 𝑒 𝑥 +𝑒 −𝑥 𝑒 𝑥 −𝑒 −𝑥 𝑥3 (Hint, try using the denominator) ( Hint, try using the denominator and rewriting the equation to make x2 in terms of u) 1+𝑥 2 (𝑙𝑛𝑥)2 𝑥 9. 𝑠𝑖𝑛𝑥2𝑐𝑜𝑠𝑥 10. 𝑒 𝑡𝑎𝑛𝑥 . (𝑠𝑒𝑐𝑥)2 11. 𝑒 𝑐𝑜𝑡𝑥 . (𝑐𝑜𝑠𝑒𝑐 𝑥)2 12. Tan 3x 13. 𝑥 2 𝑙𝑛(𝑥 3 +5) 𝑥 3 +5 108 14. 𝑥√𝑥 − 6 (Hint, try using the term under the bracket and using a part of the hint of sum 7) 15. 3𝑥 2 √𝑥 − 2 Now, we will look at integrating using substitution for trigonometric functions. 1. If an integral contains √𝑎2 − 𝑥 2 ,try substituting 𝑥 = 𝑎𝑠𝑖𝑛𝜃 2. If an integral contains √𝑥 2 + 𝑎2 try substituting 𝑥 = 𝑎𝑡𝑎𝑛𝜃 3. If an integral contains √𝑥 2 − 𝑎2 try substituting 𝑥 = 𝑎𝑠𝑒𝑐𝜃 Example: √9 − 𝑥2 . Here, we use the substitution mentioned above of 𝑥 = 𝑎𝑠𝑖𝑛𝜃and a=3. Then, we ∫ differentiate x with respect to 𝜃 for further substitution. 𝑑𝑥 We get𝑑𝜃 = 3𝑐𝑜𝑠𝜃 Eventually, using all our information, we substitute 3𝑠𝑖𝑛𝜃into the integral. ∫ √9 − 9(𝑠𝑖𝑛𝜃)2 . 3𝑐𝑜𝑠𝜃 𝑑𝜃 When we simplify the root using (𝑠𝑖𝑛𝜃)2 + (𝑐𝑜𝑠𝜃)2 = 1and eventually get ∫ 3(𝑐𝑜𝑠𝜃)2 . In trigonometry, while we were learning the double angle formula, we learned that 𝑐𝑜𝑠2𝜃 = 9 2(𝑐𝑜𝑠𝜃)2 − 1. We can use this formula to simplify the integral and get ∫ (1 + 𝑐𝑜𝑠2𝜃) From 2 here, we can use the standard rules of integration to get our solution in terms of 𝜃. Then, we must refer back to our original substitution and the double angle formula again to get the final answer. Exercise: Integrate with respect to x: 1. 2. 3. 𝑥2 (hint, try adding ±4to the numerator) 𝑥 2 +4 𝑥2 4−𝑥 2 3𝑙𝑛𝑥 𝑥(1+(𝑙𝑛𝑥)2 ) IB Questions: 𝑥 1. Let 𝑓(𝑥) = √1−𝑥 , 0 < 𝑥 < 1 1 −1 −3 a. Show that 𝑓’(𝑥) = 2 𝑥 2 (1 − 𝑥) 2 and deduce that f(x) is an increasing function b. Show that the curve y=f(x) has one point of inflection and find its coordinates c. Use the substitution 𝑥 = (𝑠𝑖𝑛𝜃)2 to prove that ∫ 𝑓(𝑥) 𝑑𝑥 = 𝑎𝑟𝑐𝑠𝑖𝑛√𝑥 − √𝑥 − 𝑥2 + 𝑐 2. Using the substitution𝑡 = 𝑡𝑎𝑛𝜃 , find the integral ∫ 3. Using the substitution 𝑢 = 1 + √𝑥, find ∫ 1 2 3(𝑐𝑜𝑠𝜃) +(𝑠𝑖𝑛𝜃)2 𝑑𝜃 √𝑥 𝑑𝑥 1+√𝑥 109 4. By using the substitution 𝑥 2 = 2𝑠𝑒𝑐𝜃, prove that ∫ 1 5. Use the substitution 𝑢 = 𝑥 2 to find ∫ 1 3 1 𝑥2 +𝑥2 1 𝑥√𝑥4 −4 1 2 = 4 𝑎𝑟𝑐𝑐𝑜𝑠(𝑥2 ) + 𝑐 𝑑𝑥. 3. Integration by parts Integration by parts is one of the techniques of performing integration. There are two types of integration by parts which will be explored in this topic. Example: Q. Find I. ∫ 𝑥𝑠𝑖𝑛𝑥 𝑑𝑥 To do this kind of integration we need to use a special technique. We need to separate this expression into two parts:(𝑥) ⋅ (𝑠𝑖𝑛 𝑥) 𝑑𝑥; The next step is to let one of these be ‘u’ and the other ‘dv’. Now the interesting part is how does one decide which of them should be ‘u’ and which should be ‘dv’. There is an acronym called “L I A T E” which helps you decide which of them should be ‘u’. L : Logarithmic function I : Inverse function A : Algebraic function T : Trigonometric function E : Exponential function Whichever function appears first is equated to ‘u’. For instance in the example above ‘x’ is an algebraic function and ‘sin x’ is a trigonometric function. Because A comes before T in the acronym, we let x = u, and solve the sum in the following manner:Let x = u dv = sin x x’ = 1 v = - cos x Now, using the formula we solve it. ∫ ∫ 𝑥𝑠𝑖𝑛𝑥 𝑑𝑥 = −𝑥𝑐𝑜𝑠 𝑥 + ∫ 𝑐𝑜𝑠 𝑥 𝑑𝑥 ∫ 𝑥 𝑠𝑖𝑛𝑥 𝑑𝑥 = −𝑥 𝑐𝑜𝑠 𝑥 + 𝑠𝑖𝑛 𝑥 + 𝑐 In this manner, various expressions can be integrated using this method. Let us understand a few more examples. Q. Find ∫ ln x dx 110 (1)(𝑙𝑛 𝑥) 𝑑𝑥 𝑑𝑣 = 1 We can rewrite this as ∫ 𝑙𝑒𝑡 𝑢 = 𝑙𝑛 𝑥 1 𝑢′ = 𝑥 𝑣=𝑥 Now, using the formula we get: (𝑙𝑛 𝑥) 𝑑𝑥 = 𝑥 𝑙𝑛 𝑥 − ∫ ∫ 1 𝑑𝑥 = 𝑥 𝑙𝑛 𝑥 − 𝑥 + 𝑐 Another type of integration sum can be of the repetitive type. This is an important part of calculus that has various applications. An example has been shown below. Q. Find ∫ 𝑒−2𝑥 𝑐𝑜𝑠 2𝑥 𝑑𝑥 𝐿𝑒𝑡 𝑢 = 𝑐𝑜𝑠 2𝑥 𝑑𝑣 = 𝑒 −2𝑥 𝑢′ = −2 𝑠𝑖𝑛(2𝑥) 𝑣= 𝑒 −2𝑥 −2 Using the formula we get:𝑒−2𝑥 𝑐𝑜𝑠 2𝑥 = ∫ −1 −2𝑥 𝑒 𝑐𝑜𝑠(2𝑥) − ∫ 2 We cannot stop here, we need to further integrate ∫ integration by parts again on∫ 𝑠𝑖𝑛(2𝑥)𝑒−2𝑥 𝑑𝑥 𝑠𝑖𝑛(2𝑥)𝑒−2𝑥 𝑑𝑥, so we perform 𝑠𝑖𝑛(2𝑥)𝑒−2𝑥 𝑑𝑥. To solve this we let 𝑢 = 𝑠𝑖𝑛 (2𝑥) and 𝑑𝑣 = 𝑒 −2𝑥 𝑢′ = 2𝑐𝑜𝑠(2𝑥) 𝑣= 𝑒 −2𝑥 −2 Using the formula we get:∫ 𝑠𝑖𝑛(2𝑥)𝑒−2𝑥 𝑑𝑥 = −1 −2𝑥 𝑒 𝑠𝑖𝑛(2𝑥) − ∫ 2 𝑒−2𝑥 𝑐𝑜𝑠 2𝑥 𝑑𝑥 Putting it all together we get:∫ ∫ 𝑒−2𝑥 𝑐𝑜𝑠 2𝑥 = 𝑒−2𝑥 𝑐𝑜𝑠 2𝑥 = −1 −2𝑥 𝑒 𝑐𝑜𝑠(2𝑥) − ∫ 2 −1 −1 −2𝑥 𝑒 𝑐𝑜𝑠(2𝑥) −( 2 𝑒 −2𝑥 𝑠𝑖𝑛(2𝑥) 2 −∫ 𝑠𝑖𝑛(2𝑥)𝑒−2𝑥 𝑑𝑥 𝑒−2𝑥𝑐𝑜𝑠 2𝑥 𝑑𝑥) However if we look closely we realise that we have ended up with the exact expression that we want to integrate.Therefore, we can call this expression I. 𝐼=∫ 𝑒−2𝑥𝑐𝑜𝑠 2𝑥 𝑑𝑥); If we do that we get:−1 −1 𝑒 −2𝑥 𝑐𝑜𝑠(2𝑥) −( 2 𝑒 −2𝑥 𝑠𝑖𝑛(2𝑥) − 𝐼 −1 −2𝑥 1 2𝐼 = 𝑒 𝑐𝑜𝑠(2𝑥) + 𝑒 −2𝑥 𝑠𝑖𝑛(2𝑥) 2 2 1 −2𝑥 2𝐼 = 2 𝑒 (𝑠𝑖𝑛(2𝑥) − 𝑐𝑜𝑠(2𝑥)) 𝐼= 2 1 𝐼 = 4 𝑒 −2𝑥 (𝑠𝑖𝑛(2𝑥) − 𝑐𝑜𝑠(2𝑥)) + 𝑐 111 IB Questions: 1. Find ∫ 𝑎𝑟𝑐𝑠𝑖𝑛 𝑥 𝑑𝑥 2. Find ∫ 𝑥2 𝑙𝑛 𝑥 𝑑𝑥 3. Using integration by parts find ∫ 4. Find∫ 𝑥(𝑠𝑒𝑐𝑥)2 𝑑𝑥 5. Find ∫ 𝑥3 𝑠𝑖𝑛 2𝑥 𝑑𝑥 𝑥𝑠𝑖𝑛𝑥 𝑑𝑥 6. Consider the functions f, g defined for 𝑥 ∈ 𝑅, given by 𝑓(𝑥) = 𝑒 −𝑥 𝑠𝑖𝑛𝑥 𝑎𝑛𝑑 𝑔(𝑥) = 𝑒 −𝑥 𝑐𝑜𝑠𝑥 a. Find f’(x) b. Find g’(x) c. Find ∫ 𝑒−𝑥 𝑠𝑖𝑛𝑥 𝑑𝑥 Definite Integrals 𝑏 ∫ 𝑓(𝑥) 𝑑𝑥 = 𝐹(𝑏) − 𝐹(𝑎) 𝑤ℎ𝑒𝑟𝑒 𝐹(𝑥) 𝑖𝑠 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑙 𝑜𝑓 𝑓(𝑥). 𝑎 Note- when using substitution in definite integrals, make sure to obtain the new values that you will be integrating with respect to. Example: 2 ∫1 2𝑥(𝑥 3 − 1)4 𝑑𝑥. We use the normal steps for substitution of taking u=𝑥 3 − 1and differentiating u with respect to x. However, we also must find the value of u at x=1 and at x=2, to make the calculation easier. - Area under a curve Example: When the graph of f(x) is positive and continuous; 𝑦 = 𝑙𝑛𝑥. Find the area of the region from x=1 to x=4. 112 𝑏 𝐴𝑟𝑒𝑎 = ∫ 𝑓(𝑥) 𝑑𝑥 𝑎 4 𝐴𝑟𝑒𝑎 = ∫ 𝑙𝑛 𝑥 𝑑𝑥 1 𝑑𝑥)41 = 4𝑙𝑛4 − 𝑙𝑛1 − (𝑥)41 = 4𝑙𝑛4 − 3 𝐴𝑟𝑒𝑎 = (𝑥𝑙𝑛𝑥 − ∫ Example: When the graph of f(x) is negative and continuous; 𝑏 𝐴𝑟𝑒𝑎 = − ∫ 9 𝑓(𝑥) 𝑑𝑥 𝑎 The area of the region bounded by 𝑓(𝑥) = − 𝑥, the line x=3 and x=k is 9ln2 units. Find k. 𝑘 𝑓(𝑥)𝑑𝑥 = 9𝑙𝑛2 −∫ 3 𝑘 9 − 𝑑𝑥 = 9𝑙𝑛2 𝑥 −∫ 3 Example: Area between the y-axis and the curve; 𝑦= 1 √𝑥 . Find the area of the region bounded by the y-axis and the curve from y=1 to y=3. 1 √𝑥 = . 𝑦 1 𝑥= 2 𝑦 2 1 𝑑𝑦 𝑦2 1 Example: Area between 2 functions In order to find the area enclosed between 2 functions, we must first find the points of intersection between the two functions and then we can get the area using the formula: 𝐴𝑟𝑒𝑎 = ∫ 𝑏 𝐴𝑟𝑒𝑎 = ∫ 𝑓(𝑥) − 𝑔(𝑥) 𝑑𝑥 𝑎 Find the area enclosed by 𝑦 = −𝑥 3 + 3𝑥 2 + 6𝑥 + 8and 𝑦 = 3𝑥 − 5. −𝑥 3 + 3𝑥 2 + 6𝑥 + 8 = 3𝑥 − 5. IB Questions: 113 1. The function f is defined on the domain [0, 3𝜋 2 ] by 𝑓(𝑥) = 𝑒 −𝑥 𝑐𝑜𝑠𝑥 a. State the two zeros of f . b. Sketch the graph of f . c. The region bounded by the graph, the x-axis and the y-axis is denoted by A and the region bounded by the graph and the x-axis is denoted by B . Show that the 𝜋 ratio of the area of A to the area of B is 𝑒 𝜋 (𝑒 2 +1) 𝑒 𝜋+ 1 𝑘 2. The diagram below shows the two curves 𝑦 = 𝑥and 𝑦 = 𝑥 where k>1 a. Find the area of region A in terms of k . b. Find the area of region B in terms of k . c. Find the ratio of the area of region A to the area of region B 3. Consider the graph of 𝑦 = 𝑥 + 𝑠𝑖𝑛(𝑥 − 3), −𝜋 ≤ 𝑥 ≤ 𝜋 a. Sketch the graph, clearly labelling the x and y intercepts with their values. b. Find the area of the region bounded by the graph and the x and y axes. 114 𝑥+1 4. The graph of the function 𝑓(𝑥) = 𝑥 2 +1 is shown below. The point (1, 1) is a point of inflexion. There are two other points of inflexion. a. Find f’(x) b. Hence find the x-coordinates of the points where the gradient of the graph of f is zero. 𝑝(𝑥) c. Find f’’(x) expressing your answer in the form (𝑥 2+1)3where p(x) is a polynomial of degree 3 d. Find the x-coordinates of the other two points of inflexion. 𝜋 e. Find the area of the shaded region. Express your answer in the form 𝑎 − 𝑙𝑛√𝑏, where a and b are integers. 5. Find the area enclosed by the curve 𝑦 = 𝑎𝑟𝑐𝑡𝑎𝑛𝑥, the x-axis and the line 𝑥 = √3. - Volume of revolution Rotating an equation around the y or x axis forms what is known as a solid of revolution. We can use integration to find the volume of this solid that forms between two intervals. The formula to find the volume of such a solid is given below. 115 If the curve is rotated around the y-axis we use the first equation. Whereas if the curve is rotated around the x-axis we use the second formula provided. Example: Q. Find the volume of revolution when 𝑦 = √𝑠𝑖𝑛 𝑥 for 0 ≤ 𝑥 ≤ 𝜋is rotated 2𝜋radians about the x-axis . The graph below is the graph of 𝑦 = √𝑠𝑖𝑛 𝑥 for 0 ≤ 𝑥 ≤ 𝜋. We need to use the formula to find the volume of the solid that forms when this curve rotates 𝜋 about the horizontal x-axis. Thus, 𝑉 = 𝜋 ∫0 (√𝑠𝑖𝑛 𝑥)2 𝑑𝑥. 𝑉 = 𝜋(−𝑐𝑜𝑠 𝑥)𝜋0 = 𝜋(−𝑐𝑜𝑠 𝜋 + 𝑐𝑜𝑠 0) = 2𝜋. Example: Q. Consider the part of the curve 4𝑥 + 𝑦 = 4shown in the diagram below. 2 2 The y-intercept is (0,2) and the x-intercept is (1,0). 𝑑𝑦 a) Find an expression for 𝑑𝑥 in terms of x and y. Here we carry out implicit differentiation that we learnt earlier. 𝑑𝑦 Therefore, 8𝑥 + 2𝑦 ⋅ (𝑑𝑥 ) = 0. 𝑑𝑦 4𝑥 =− 𝑑𝑥 𝑦 b) A bowl is formed by rotating this curve through 2𝜋radians about the x-axis. Calculate the volume of this bowl. 116 1 Since it rotated about the x-axis we use the formula 𝑉 = 𝜋 ∫0 Therefore, 𝑉 = 8𝜋 3 1 𝑦 2 𝑑𝑥 = 𝜋 ∫0 (4 − 4𝑥 2 ) 𝑑𝑥 . Note:- If you are asked to find the volume between two curves the same formula applies except it can be modified to the following:𝑏 (𝑦𝑈 )2 − (𝑦𝐿 )2 𝑑𝑥 (if the curves are rotated about the ‘x’ axis) 𝑉 = 𝜋 ∫𝑎 Exercise: Q. The region enclosed between the curves 𝑦 = √𝑥𝑒 𝑥 and 𝑦 = 𝑒√𝑥is rotated through 2𝜋about the x-axis. Find the volume of the solid obtained. Improper Integrals ∞ The Math AA HL course only considers improper integrals that exist in the form ∫𝑎 𝑏 𝑓(𝑥) 𝑑𝑥. 𝑓(𝑥) 𝑑𝑥.For this improper integral to exist, f(x) must approach 0 as We rewrite this as 𝑙𝑖𝑚 ∫𝑎 𝑏→∞ x→ ∞. Sample Question 1: 𝑏 𝑙𝑖𝑚 ∫1 𝑏→∞ −1 −1 1 𝑥 −3 𝑑𝑥 = 𝑙𝑖𝑚 [2𝑥 2]𝑏1 = 𝑙𝑖𝑚 (2𝑏2) + 2. As b→ ∞, its reciprocal tends to zero. Thus the 𝑏→∞ 𝑏→∞ limit is 0.5. Exercise: 1. Find the total area between 𝑓(𝑥) = 𝑒 −𝑥 (𝑠𝑖𝑛𝑥)2 and the x-axis for 𝑥 ≥ 0. ∞ 1 2. Determine ∫𝑙𝑛√3 𝑒 𝑥 +𝑒 −𝑥 𝑑𝑥 using the substitution 𝑢 = 𝑒 𝑥 3. Find the total area between the function and the x-axis for 𝑥 ≥ 0 1 a. 𝑦 = 1+𝑥 b. 𝑦 = 5−𝑥 c. 𝑦 = 𝑥𝑒 −𝑥 d. 𝑦 = 𝑥𝑒 −𝑥 3 Kinematics Kinematics is the branch of mechanics concerned with the motion of objects without reference to the forces which cause the motion. As mentioned earlier there are multiple applications of calculus and some of the most fundamental physics concepts have its roots in calculus. This application mainly deals with displacement, velocity and acceleration. 117 Displacement is a vector whose length is the shortest distance from the initial position to the final position. Velocity can be defined as the rate of change of displacement and therefore in terms of mathematics can be called the derivative of displacement. Acceleration is the rate of change of velocity which is therefore the derivative of the velocity function. They are often denoted by the following:𝑠 = 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡 𝑣 = 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎 = 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑡 = 𝑡𝑖𝑚𝑒 Some important formulas include:- 1. 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 𝑠(𝑡2 )−𝑠(𝑡1 ) 𝑡2 −𝑡1 2. 𝑖𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 𝑣(𝑡) = 𝑠′(𝑡) 3. 𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡: ∆𝑠 = 𝑠2 − 𝑠1 or ∫𝑡𝑡12 𝑣(𝑡) 𝑑𝑡 𝑡 4. 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑 = ∫𝑡 2 1 5. 𝑎𝑣𝑒𝑟𝑎𝑔𝑒 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 = 6. 𝑎 = 𝑑𝑣 𝑑𝑡 = |𝑣(𝑡)| 𝑑𝑡 𝑣(𝑡2 )−𝑣(𝑡1 ) 𝑡2 −𝑡1 𝑑2𝑠 𝑑𝑡 2 7. 𝑠𝑝𝑒𝑒𝑑 = |𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦| Note:- If 𝑣(𝑡)and 𝑎(𝑡)have the same sign then the speed is increasing at time ‘t’. If 𝑣(𝑡)and 𝑎(𝑡)have opposite signs then the speed is decreasing at ‘t’. 𝑠(𝑡) ⇒ 𝑣(𝑡) ∶ 𝑦𝑜𝑢 ℎ𝑎𝑣𝑒 𝑡𝑜 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑒 𝑣(𝑡) ⇒ 𝑠(𝑡) ∶ 𝑦𝑜𝑢 ℎ𝑎𝑣𝑒 𝑡𝑜 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑒 𝑣(𝑡) ⇒ 𝑎(𝑡): 𝑦𝑜𝑢 ℎ𝑎𝑣𝑒 𝑡𝑜 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑡𝑒 𝑎(𝑡) ⇒ 𝑣(𝑡) ∶ 𝑦𝑜𝑢 ℎ𝑎𝑣𝑒 𝑡𝑜 𝑖𝑛𝑡𝑒𝑔𝑟𝑎𝑡𝑒 Example: Q. A mass on a spring oscillates with displacement 𝑠(𝑡) = 8𝑠𝑖𝑛(8𝜋𝑡) + 6 𝑐𝑚where 0 ≤ 𝑡 ≤ 0.25𝑠. a) At what time is the displacement 6cm ? b) Plot the graph of 𝑠(𝑡) for the given domain 118 c) Find where and when the mass changes direction We begin by substituting 6 in place of s(t) to get the time at which displacement is 6 cm. 6 = 8𝑠𝑖𝑛(8𝜋𝑡) + 6 8𝑠𝑖𝑛(8𝜋𝑡) = 0 𝑡 = 𝑠𝑖𝑛−1 (0) 8𝜋𝑡 = 0, 𝜋, 2𝜋. .. 1 1 𝑡 = 0, , 8 4 Now we must plot the graph of the given function. To find where the mass changes direction we need to observe the graph and mark its turning 1 3 points. The graph has a maximum at t = 16 𝑠and a minimum at t = 16 𝑠, and therefore changes direction at those points. IB Questions: 1. Particle A moves such that its velocity v ms-1, at time t seconds, is given by 𝑣(𝑡) = 𝑡 12+𝑡 4 , 𝑡 ≥ 0.Particle B moves such that its velocity v ms-1, is related to its displacement s m by the equation 𝑣(𝑠) = 𝑎𝑟𝑐𝑠𝑖𝑛(√𝑠) a. Sketch the graph of v(t). Indicate clearly the local maximum and write down its coordinates. 119 b. Use the substitution 𝑢 = 𝑡 2 to find ∫ 𝑡 𝑑𝑡. 12+𝑡4 c. Find the exact distance travelled by particle A between t=0 and 6 seconds. Give your answer in the form karctan(b), k, b ∈ 𝑅 d. Find the acceleration of particle B when s=0.1 m 2. A particle P moves in a straight line with displacement relative to origin given by 𝑠 = 2𝑠𝑖𝑛(𝜋𝑡) + 𝑠𝑖𝑛(2𝜋𝑡), 𝑡 ≥ 0,where t is the time in seconds and the displacement is measured in centimetres a. Write down the period of the function s b. Find expressions for the velocity, v and the acceleration a of P c. Determine all the solutions of the equation v=0, for 0 ≤ 𝑡 ≤ 4. 3. A particle moves in a straight line such that its velocity, v ms-1, at time t seconds, is given by: a. Find the value of t when the particle is instantaneously at rest. b. The particle returns to its initial position at t=T. Find the value of T 4. The particle P moves along the x-axis such that its velocity v ms-1 at time t seconds is given by 𝑣 = 𝑐𝑜𝑠(𝑡 2 ) a. Given that P is at the origin O at time t=0, calculate i. the displacement of P from O after 3 seconds ii. the total distance travelled by P in the first 3 seconds b. Find the time at which the total distance travelled by P is 1 m. 5. A particle moves in a straight line, its velocity v ms-1 at time t seconds is given by𝑣 = 9𝑡 − 3𝑡 2 , 0 ≤ 𝑡 ≤ 5.At t = 0, the displacement s of a particle from an origin O is 3m. a. Find the displacement of the particle when t=4. b. Sketch a displacement/time graph for the particle showing clearly where the curve meets the axes and the coordinates of the points where the displacement takes greatest and least values. c. For t>5, the displacement of the particle is given by 𝑠 = 𝑎 + 𝑏𝑐𝑜𝑠 2𝜋𝑡 5 such that s is continuous for all 0 ≤ 𝑡. Given further that s=16.5 when t=7.5, find a and b. d. For t>5, the displacement of the particle is given by 𝑠 = 𝑎 + 𝑏𝑐𝑜𝑠 2𝜋𝑡 5 such that s is continuous for all 0 ≤ 𝑡. Find the times t1 and t2 (0< ti< t2< 8) when the particle returns to its starting point. Maclaurin Series 120 A Maclaurin series is a power series that allows one to calculate an approximation of a function 𝑓(𝑥) for input values close to zero, given that one knows the values of the successive derivatives of the function at zero. Example: If 𝑓(𝑥) = 𝑒 𝑥 𝑠𝑖𝑛 𝑥, find the first three terms using the Maclaurin Series formula. 𝑓(𝑥) = 𝑒 𝑥 𝑠𝑖𝑛 𝑥 𝑓′(𝑥) = 𝑒 𝑥 (𝑠𝑖𝑛 𝑥 + 𝑐𝑜𝑠 𝑥) 𝑓′′(𝑥) = 𝑒 𝑥 (−𝑠𝑖𝑛 𝑥 + 𝑐𝑜𝑠 𝑥) + 𝑒 𝑥 (𝑠𝑖𝑛 𝑥 + 𝑐𝑜𝑠 𝑥) = 2𝑒 𝑥 𝑐𝑜𝑠 𝑥 𝑓′′′(𝑥) = 2𝑒 𝑥 (−𝑠𝑖𝑛𝑥) + 2𝑒 𝑥 𝑐𝑜𝑠 𝑥 = 2𝑒 𝑥 (𝑐𝑜𝑠 𝑥 − 𝑠𝑖𝑛 𝑥) Now, we substitute zero in each of the equations. 𝑓(0) = 0 𝑓′(0) = 1 𝑓′′(0) = 2 𝑓′′′(0) = 2 Using the formula we can say that: 121 𝑥2 𝑥3 𝑓(𝑥) = 𝑥 + + +. . . .. 2 3 Exercise: Q. The function f is defined on the domain] −𝜋 𝜋 2 , 2 [by f(x) = ln (1+sin x) −1 a) Show that 𝑓′′(𝑥) = (1+𝑠𝑖𝑛 𝑥) b) Find the Maclaurin series for 𝑓(𝑥)up to and including the term in 𝑥 4 . Q. Consider the functions f and g given by 𝑓(𝑥) = 𝑒 𝑥 +𝑒 −𝑥 2 and 𝑔(𝑥) = 𝑒 𝑥 −𝑒 −𝑥 2 . a) Show that 𝑓′(𝑥) = 𝑔(𝑥)and 𝑔′(𝑥) = 𝑓(𝑥) b) Find the first three non-zero terms in the Maclauric expansion of 𝑓(𝑥). 122