online learning Pearson BTEC Higher Nationals in Electrical and Electronic Engineering (QCF) Unit 1: Analytical Methods for Engineers (core)

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1 online learning Pearson BTEC Higher Nationals in Electrical and Electronic Engineering (QCF) Unit 1: Analytical Methods for Engineers (core) Unit Workbook 3 in a series of 4 for this unit Learning Outcome Calculus Page 1 of 41 1 of 41

2 Contents INTRODUCTION... 3 GUIDANCE Calculus The Concept of the Limit, Continuity and the Derivative Derivatives of Standard Functions Notion of the Derivative and Rates of Change Differentiating a Function of a Function The Product Rule The Quotient Rule Integral Calculus Definite Integrals and Areas under Curves Further Differentiation Higher Derivatives Logarithmic Differentiation Differentiation of Inverse Trigonometric Functions Differentiation of Inverse Hyperbolic Functions Further Integration Integration by Parts Integration by Substitution Integration using Partial Fractions Solution of Engineering Problems with Calculus Charging RC Circuit Energising Inductor Page 2 of 41 2 of 41

3 INTRODUCTION This Workbook guides you through the learning outcomes related to: Calculus: the concept of the limit and continuity; definition of the derivative; derivatives of standard functions; notion of the derivative and rates of change; differentiation of functions using the product, quotient and function of a function rules; integral calculus as the calculation of area and the inverse of differentiation; the indefinite integral and the constant of integration; standard integrals and the application of algebraic and trigonometric functions for their solution; the definite integral and area under curves Further differentiation: second order and higher derivatives; logarithmic differentiation; differentiation of inverse trigonometric functions; differential coefficients of inverse hyperbolic functions Further integration: integration by parts; integration by substitution; integration using partial fractions Applications of the calculus: e.g. maxima and minima, points of inflexion, rates of change of temperature, distance and time, electrical capacitance, rms values, electrical circuit analysis, AC theory, electromagnetic fields, velocity and acceleration problems, complex stress and strain, engineering structures, simple harmonic motion, centroids, volumes of solids of revolution, second moments of area, moments of inertia, rules of Pappus, radius of gyration, thermonamic work and heat energy Engineering problems: e.g. stress and strain, torsion, motion, namic systems, oscillating systems, force systems, heat energy and thermonamic systems, fluid flow, AC theory, electrical signals, information systems, transmission systems, electrical machines, electronics GUIDANCE This document is prepared to break the unit material down into bite size chunks. You will see the learning outcomes above treated in their own sections. Therein you will encounter the following structures; Purpose Theory Example Explains why you need to stu the current section of material. Quite often learners are put off by material which does not initially seem to be relevant to a topic or profession. Once you understand the importance of new learning or theory you will embrace the concepts more readily. Conveys new material to you in a straightforward fashion. To support the treatments in this section you are strongly advised to follow the given hyperlinks, which may be useful documents or applications on the web. The examples/worked examples are presented in a knowledge-building order. Make sure you follow them all through. If you are feeling confident then you might like to treat an example as a question, in which case cover it up and have a go yourself. Many of the examples given resemble assignment questions which will come your way, so follow them through diligently. Page 3 of 41 3 of 41

4 Question Challenge Video Questions should not be avoided if you are determined to learn. Please do take the time to tackle each of the given questions, in the order in which they are presented. The order is important, as further knowledge and confidence is built upon previous knowledge and confidence. As an Online Learner it is important that the answers to questions are immediately available to you. Contact your Unit Tutor if you need help. You can really cement your new knowledge by undertaking the challenges. A challenge could be to download software and perform an exercise. An alternative challenge might involve a practical activity or other form of research. Videos on the web can be very useful supplements to your distance learning efforts. Wherever an online video(s) will help you then it will be hyperlinked at the appropriate point. Page 4 of 41 4 of 41

5 3.1 Calculus There are a couple of great webpages for checking your answers to Calculus problems Check Differentiation answers Check Integration answers The Concept of the Limit, Continuity and the Derivative Calculus deals with functions which continually vary and is based upon the concept of a limit and continuity. Let s refer to a diagram to understand the concept of a limit Here we have indicated a point P on part of a function (the red curve) with Cartesian co-ordinates (x 1, y 1 ). We require another point Q on the function to be a small increment away from P and will designate a small increment with the symbol δ (delta). What we then have is P = (x 1, y 1 ) Q = (x 1 + δx, y 1 + δy) Look now at the chord PQ (drawn as the straight line in blue). If we can determine the slope of this chord and then make it infinitesimally short we will end up with a tangent to the function. It is the slope of this tangent which forms the basis of Differential Calculus (normally called differentiation). By inspection, we see that the slope of the chord is given by slope of chord PQ = δy δx Page 5 of 41 5 of 41

6 If we deliberately make δx approach zero (i.e. make it as short as possible) then we shall reach a limit, which may expressed mathematically as limit = δx 0 δy dx δx The term dx is written in Leibnitz notation and indicates the slope of the chord when the chord only touches the function at one single point. This is achieved by continual reduction of δx. What we can now say is that we are able to find the slope of any function by adopting this process. Hence, given a function f(x) we are able to differentiate that function, meaning find its slope at all points. We can write The slope at any point of a function f(x) is given by This process is called finding the derivative of a function. d(f(x)) dx Derivatives of Standard Functions What we don t want to be doing is to spend too much time drawing graphs of functions just to work out the derivative. Fortunately there are standard ways to determine the derivative of functions and some of the frequent ones which engineers meet are given in the table below. Function Ax n A sin (x) A cos (x) A e kx A log e (x) A sinh (x) A cosh (x) Derivative nax n 1 A cos (x) A sin (x) kae kx A x A cosh (x) A sinh (x) Let s look at some examples of using these standard derivatives Worked Example 1 Page 6 of 41 6 of 41

7 Differentiate the function y = 3x 4 with respect to x. We see that this function is similar to that in row 1 of our table. We can see that A = 3 and n = 4. We may then write dx = naxn 1 = (4)(3)x 4 1 = 12x 3 Worked Example 2 Differentiate the function y = 6x 5 with respect to x. We see that this function is similar to that in row 1 of our table. We can see that A = 6 and n = 5. We may then write dx = naxn 1 = ( 5)(6)x 5 1 = 30x 6 Worked Example 3 Differentiate the function v = 12 sin(t) with respect to t. Don t worry that y and x have disappeared here. We can use any letters we like. The letter v means voltage and t is time. All we want to do here then is to find dv dt. We see that the function is similar to that in row 2 of our table. We may write dv dt = A cos(t) = 12 cos (t) Worked Example 4 Differentiate the function i = 5 cos(t) with respect to t. Page 7 of 41 7 of 41

8 The letter i means current and t is time. All we want to do here then is to find di dt. We see that the function is similar to that in row 3 of our table. We may write di = A sin(t) = 5 sin (t) dt Worked Example 5 Differentiate the function y = 4e 2x with respect to x. From row 4 of our table we can write dx = kaekx = (2)(4)e 2x = 8e 2x Worked Example 6 Differentiate the function y = 16 log e (x) with respect to x. We refer to row 5 of our table and write Worked Example 7 dx = A x = 16 x Differentiate the function y = 9 sinh(x) with respect to x. We refer to row 6 of our table and write = A cosh(x) = 9 cosh (x) dx Worked Example 8 Differentiate the function y = cosh(x) with respect to x. Page 8 of 41 8 of 41

9 We refer to row 7 of our table and write = A sinh(x) = 14.5 sinh (x) dx Questions Q3.1 Differentiate the function y = 8x 5 with respect to x. Q3.2 Differentiate the function y = 7x 4 with respect to x. Q3.3 Differentiate the function v = 18 sin(t) with respect to t. Q3.4 Differentiate the function i = 11 cos(t) with respect to t. Q3.5 Differentiate the function y = 6e 3x with respect to x. Q3.6 Differentiate the function y = 20 log e (x) with respect to x. Q3.7 Differentiate the function y = 13 sinh(x) with respect to x. Q3.8 Differentiate the function y = cosh(x) with respect to x. ANSWERS Q3.1 40x 4 Q3.2 28x 5 Q cos(t) Q sin(t) Q3.5 18e 3x Q x Q cosh(x) Q sinh(x) Page 9 of 41 9 of 41

10 Video These videos will boost your knowledge of differentiation Notion of the Derivative and Rates of Change The notion of taking the derivative of a function and examining rate of change can be readily explained with reference to the diagram below Here we have plotted a section of a sine wave in red. Notice that we have used pink arrows to indicate the slope at various points on the sine wave. For example, at the start the sine wave is rising to a positive more rapidly than at any other point. The pink arrow indicates this positive growth by pointing upwards in the direction of the sine wave at that beginning point. The second arrow shows the slope of the sine wave at π 2 radians (or 90 0 if you like). Here it is at a plateau and has zero slope (meaning that it just moves horizontally, with no vertical movement at all). The third arrow indicates a negative slope (its moving down very steeply). The picture continues in a similar way for the remaining arrows. Here s the really interesting bit. When we record all of the slopes on our sine wave we get the dotted waveform in blue. What does that look like? Of course, it is a cosine wave. So we have just proved graphically that finding the slope (differentiating) at all points on a sine wave leads to a cosine wave. d{sin (t)} dt = cos (t) as given in our table of standard derivatives in section If we had loads of time to kill we could prove all of the other derivatives graphically. Page 10 of of 41

11 Challenge Determine a graphical proof (as above) for d{cos (t)} dt = sin (t) Differentiating a Function of a Function What do we mean by this term function of a function? Well, the motion of kicking a ball involves moving your leg, which in turn involves brain activity to control the kick. We can therefore say that kicking a ball is a function of a function of brain activity. So it is with much of mathematics. If you see a simple function like y = 2x you would agree that is a straight line graph with a slope of 2. It also a function, because we can plot functions can t we? What if we looked at something like y = sin (2x) Is that a function? Yes it is, and we can plot it. But, it is also a function of a function. In fact, it is a sine function of a 2x function. How are we then going to evaluate this function of a function? Let s analyse what we have Let u = 2x u = 2x 1 This looks like the first function in our table of standard derivatives. Let s differentiate in the way we did in Worked Example 1 du dx = (1)(2)x1 1 = 2x 0 = 2 1 = 2 OK, so where does that get us? If we now use this u in the original question we get Let s now differentiate that y = sin(2x) = sin (u) = cos (u) du The trick in function of a function problems is to combine those two derivatives by multiplying them together du dx du = = 2 cos (u) dx Those two du terms have cancelled each other out. Since we know what u is (it s 2x) we can now write the final answer = 2 cos (2x) dx Page 11 of of 41

12 Let s look at a couple of worked examples on the function of a function process Worked Example 9 Differentiate y = (x 4 + 3) 2 with respect to x. The best approach is to let the term in brackets equal u and proceed as before Let u = x du dx = (4)x4 1 + (0)x 0 1 = 4x 3 Notice above that the solitary 3 at the end disappeared when we differentiated it. Think about what the number 3 looks like on a graph. It s just a horizontal line passing through the y axis at the point y = 3 right? Sketching it you realise that the slope is zero everywhere on the line. So the golden rule for solitary numbers is that we ignore them when differentiating. Let s proceed with the remainder of the original problem y = (u) 2 du = (2)u2 1 = 2u 1 = 2u Combining our results gives dx = du dx du = 4x3 2u = 4x 3 2(x 4 + 3) = 8x 3 (x 4 + 3) = 8x x 3 Worked Example 10 Differentiate y = cos(4x 3 + 3x 2 6x + 8) with respect to x. Let u = 4x 3 + 3x 2 6x + 8 du dx = 12x2 6x 3 6 Also, y = cos (u) Page 12 of of 41

13 Combining derivatives = sin (u) du dx = du dx du = (12x2 6x 3 6)( sin(u)) = (12x 2 6x 3 6)sin (u) = (12x 2 6x 3 6)sin (4x 3 + 3x 2 6x + 8) Questions Q3.9 Differentiate y = (x 4 9) 3 with respect to x. Q3.10 Differentiate y = cos(7x 2 + 4x 1 5x + 2) with respect to x. ANSWERS Q3.9 12x 5 (x 4 9) 2 Q3.10 (14x 4x 2 5) sin(7x 2 + 4x 1 5x + 2) Video These videos give further guidance on differentiating a function of a function (also called the Chain Rule ) The Product Rule When we need to differentiate the product of two functions (i.e. two functions which are multiplied together) then we use the Product Rule. If we have two functions u and v, which are each themselves a function of x, then their product y = uv may be differentiated as follows dv du = u + v dx dx dx All we need to do in such cases is to find u, v and their differentials, then plug the findings into the above formula. Page 13 of of 41

14 Worked Example 11 Differentiate y = x 3 cos (5x) with respect to x. First let us imagine that the question looks like y = uv. The first function u can be represented by the x 3 part and the second function v can be represented by the cos (5x) part. That s your starting point, the rest is just differentiation and plugging results into the Product Rule formula Question u = x 3 v = cos(5x) du dx = 3x2 dv = 5 sin (5x) dx That last part involved function of a function, which you can now do. dv du = u + v dx dx dx = (x3 )( 5 sin (5x)) + (cos (5x))(3x 2 ) We virtually have the answer, just needing to ti up the terms a little dx = 5x3 sin(5x) + 3x 2 cos (5x) We could go further and take out an x 2 term from either side of the addition symbol but that would not produce a more concise answer, so we leave it as it is. Q3.11 Differentiate y = x 6 sin(3x) with respect to x. ANSWER Q3.11 3x 6 cos(3x) + 6x 5 sin(3x) Video Helpful videos on the Product Rule. Page 14 of of 41

15 3.1.6 The Quotient Rule When we need to differentiate the quotient of two functions (i.e. two functions which are divided) then we use the Quotient Rule. If we have two functions u and v, which are each themselves a function of x, then their quotient y = u v may be differentiated as follows dx = v du dx u dv dx All we need to do in such cases is to find u, v and their differentials, then plug the findings into the above formula. v 2 Worked Example 12 Differentiate the following function with respect to x y = x 2 sin(2x) Let u = x 2 Let v = sin(2x) du dx = 2x dv dx = 2 cos(2x) Placing these results in the formula gives dx = v du dx u dv dx v 2 = (sin(2x))(2x) (x2 )(2cos(2x)) sin 2 (2x) = 2x[sin(2x) x cos (2x)] sin 2 (2x) Question Q3.12 Differentiate the following function with respect to x Page 15 of of 41

16 y = x 5 sin(4x) ANSWER Q3.12 5x4 sin(4x) 4x 5 cos(4x) sin 2 (4x) Integral Calculus Integration is the reverse process of differentiation. The process of integration actually involves summing parts together. Notice the bold s in the word summing. The symbol we use for integration is an elongated version of the letter s and Integration uses the symbol: You will see later on how we sum parts to find the areas under curves. For now let s look at this idea of integration being the reverse of differentiation. A bit more notation is needed to understand integration. When we integrate a function of x me must indicate the variable being integrated by placing a dx at the end of the expression f(x)dx This is known as an indefinite integral since there are no limits to the integration. More on this in a later section. Let s take a look at a simple differentiation problem and see if we can reverse the solution (i.e. do a bit of integration) y = 3x dx = 12x3 That was easy enough. Our current task is to take that answer and integrate it with respect to the variable x. Let s use our new notation for this problem 12x 3 dx Page 16 of of 41

17 We know that what comes out of this integration must involve the term 3x 4 so how do we go about that? For this type of problem what we do is add one to the power and divide by the new power 12x 3 dx = 12x4 4 = 3x 4 Quite good, but what about that 8? When we wish to integrate a function we have no idea whether their needs to be a constant. Just in case there is we tend to place a C at the end of our answer to accommodate any possible constant. To complete the answer to our integration problem then, we may write 12x 3 dx = 3x 4 + C Perfect! There are other types of integration problems which we shall face, each type with its own standard integral. The table below gives some of the standard integrals encountered by engineers Function f(x) ax n sin (ax) cos (ax) e ax Standard Integral f(x)dx ax n+1 n C 1 cos(ax) + C a 1 sin(ax) + C a 1 a eax + C Page 17 of of 41

18 1 x sinh (ax) cosh (ax) log e (x) + C 1 cosh(ax) + C a 1 sinh(ax) + C a Worked Example 13 Solve: (2x 3 + 5x 2 4x + 12)dx Each of the terms looks to be in the form ax n so we just need to apply the standard integral for that type of term = 2x x x x0+1 1 = x x3 2x x + C + C Question Q3.13 Solve: (6x 4 5x 2 + 2x 7)dx ANSWER Q3.13 2x 3 + 5x 1 + x 2 7x + C Worked Example 14 Solve: 5 sin(4x) dx Page 18 of of 41

19 From the table we see that the standard integral required is 1 cos(ax) + C a But we also have that 5 as a coefficient, so the answer must be multiplied by a cos(ax) + C = 5 cos(4x) + C 4 You may wonder here that maybe the constant of integration C should also have been multiplied by 5. Multiplying a constant by a constant still gives a constant so we can just represent it by C for now. For more advanced problems at level 5, and beyond, you will need to take more considered care with constants of integration as there may be more than one. Question Q3.14 Solve: 3 sin(9x) dx ANSWER Q cos(9x) + C 3 Worked Example 15 Solve: 10 cos(5x)dx Using the appropriate standard integral Question = 10 sin(5x) dx + C = 2 sin(5x) + C 5 Page 19 of of 41

20 Q3.15 Solve: 30 cos(6x)dx Q sin(6x) + C Worked Example 16 Solve: 24e 8x dx Using the appropriate standard integral Question = 24 8 e8x + C = 3e 8x + C Q3.16 Solve: 6e 2x dx ANSWER Q3.16 3e 2x + C Worked Example 17 Solve: 6 x dx Using the appropriate standard integral = 6 log e (x) + C Question Page 20 of of 41

21 Q3.17 Solve: 9 x dx ANSWER Q log e (x) + C Worked Example 18 Solve: (8 sinh(4x) + 12 cosh (6x))dx Here we have a sum of two functions. That s not a problem, we just treat them as individual terms, find the appropriate standard integrals, and add the results of the integration = cosh(4x) + sinh(6x) + C = 2 cosh(4x) + 2 sinh(6x) + C 6 You could then possibly go on to simplify this answer = 2[cosh(4x) + sinh (6x)] + C Question Q3.18 Solve: (6 sinh(3x) + 9 cosh (4x))dx ANSWER Q cosh(3x) + 9 sinh(4x) + C 4 Perhaps you could try a more involved combination problem Question Page 21 of of 41

22 Q3.19 Solve: ( 3x sin(2x) + 8e 4x + 6 sinh (3x))dx ANSWER Q x 5 2 cos(2x) + 2e 4x + 2 cosh(3x) + C Definite Integrals and Areas under Curves In the previous section we mentioned that indefinite integrals do not have any limits imposed upon them. In the case where there are limits imposed on an integral we call this a definite integral. These definite integrals are useful for finding the area under a curve or function. Let s consider a nice simple function and see if we can work out the area beneath it between some given limits. Suppose we take the function y = 2x We wish to find the area under that function for extremes of x between 3 and 5. Here s the idea The notation we use for such a case of definite integration is 5 2x dx 3 Notice that the lower limit (3) of the variable x is placed at the lower end of the summation symbol and the upper limit (5) is placed at the upper end of the summation symbol. Another good thing to notice here is the use of that word summation. Looking at the graph for the function it is clear that the area under the curve between our two limits (shown shaded) is a summation of the area of the bit at the top, which looks like a triangle, and the lower bit, which looks like a rectangle. You are probably thinking that you can work out that area in your head. If you did it would evaluate as 4 square units for the triangle and 12 square units for the rectangle, giving a total shaded area of 16 square units. The actual summation for the integration is not done in chunks like that, it s done as a sum of areas moving horizontally, but the idea is the same. Page 22 of of 41

23 Let s see if we can perform our definite integration to arrive at the same answer. 5 2x dx 3 = [ 2x2 5 2 ] 3 5 = [ x2 ] 3 What we now do with our limits is place them into the variable in turn, separating each term with a minus sign 5 2x dx 3 = [ 2x2 5 2 ] 3 5 = [ x2 ] = = 25 9 = 16 square units 3 You will notice that the constant of integration C has not appeared here. It s not that it was forgotten, in fact it appeared twice, but that minus sign cancelled out those two C s. That function y = 2x was a straightforward one to visualise between horizontal limits. Let s take a look at one involving a sine wave Worked Example 19 Given the function y = sin (x) find the area under this function, bounded by limits of x between 0 and π. Let s go straight into this solution with our definite integral sin(x) dx = [ cos (x) ] 0 π π 0 = ( cos (π) ( cos(0)) = = 2 square units That one needed a lot of care with those minus signs. Let s see how it looks graphically Page 23 of of 41

24 Question Q3.20 Given the function y = cos (x) find the area under this function, bounded by limits of x between 0 and π 2. ANSWER Q square unit Page 24 of of 41

25 3.2 Further Differentiation Higher Derivatives So far we have differentiated functions only once. When we differentiate a function more than once we are finding higher derivatives of that function. At this point it is useful to introduce some shorthand notation Worked Example 20 dx f (x) Those three bars mean equivalent to. So, f (x) is a shorthand way of writing dx. Another piece of more succinct shorthand is to use y as the first derivative of a function y. We may use these notations interchangeably, therefore dx f (x) y When a function is differentiated for a second time we are saying d { dx } dx In other words, we are differentiating the contents of the curly brackets with respect to x. That doesn t look to good though, does it? The generally accepted ways to write this second derivative are d 2 y dx 2 f (x) y Each dash represents a differentiation step. Therefore, a function which is differentiated 3 times can be indicated in either of the following 3 ways d 3 y dx 3 f (x) y Let s take a look at an example of taking higher derivatives Given the function y = 4x 3 + 3e 2x + 5 sin(3x) find y. If you have worked through all of the previous examples and questions then you will have no problem at all in being able to tackle such a question. Let s get into the solution Firstly, we need to find y y = 12x 2 + 6e 2x + 15 cos (3x) Page 25 of of 41

26 Easy enough. If we now take the derivative of what we just found we shall have y y = 24x + 12e 2x 45 sin (3x) Question Q3.21 Given the function y = 6x 2 + 2e 6x 4 cos(5x) find y. ANSWER Q e 6x cos(5x) Logarithmic Differentiation When functions get quite complex it is frequently convenient to take natural logarithms of all the terms before the process of differentiation takes place. Let s review the three basic laws of logarithms in their natural form log e (AB) = log e (A) + log e (B) log e ( A B ) = log e(a) log e (B) log e (A n ) = n log e (A) As well as using those log laws we also need to understand a couple of important formulae before we can use logarithmic differentiation. Consider the problem of differentiating y = log e (4x 3 + 3x 6) with respect to x. Our f(x) here is 4x 3 + 3x 6. You will, of course, realise that this is a function of a function type of problem Collect results: Let u = 4x 3 + 3x 6 Also, y = log e (u) du dx = 12x2 + 3 du = 1 u dx = du dx du = 12x2 + 3 = 12x2 + 3 u 4x 3 + 3x 6 That last result looks like f (x) f(x). So, our first important result is d dx [log e(f(x))] = f (x) f(x) Page 26 of of 41

27 On to our second important result then If we are given log e (y) and are asked to differentiate it with respect to x then that is not the type of problem which we have met before. All of our problems so far have involved explicit functions, where we would have been given something like y = f(x). That s not the case here. In fact, what we are currently given is called an implicit function. Here s how we go about differentiating log e (y) with respect to the implicit variable x Realising that: Let s just emphasise that last result Find d dx {log e(y)} Let u = log e (y) du = 1 y du dx = du dx d dx {log e(y)} = 1 y dx d dx {log e(y)} = 1 y dx We are now in a position to be able to perform logarithmic differentiation. Worked Example 21 There are plenty of problems where using the Product Rule, Quotient Rule etc. sends us to a dead end. This is usually where logarithmic differentiation comes to the fore. Consider differentiating the following function with respect to x y = x2 cosh (x) e 4x sin (x) Wow, that looks complicated. Below is a section of the function drawn in Graph (on the left) and the function along with the derivative (in red and blue respectively, on the right). Our job is to produce an equation for that blue curve. Page 27 of of 41

28 We start by taking natural logs of both sides y = x2 cosh (x) e 4x sin (x) log e (y) = log e [ x2 cosh (x) e 4x sin (x) ] Remembering those three log laws we looked at earlier, plus those two important results log e (y) = log e (x 2 ) + log e (cosh(x)) log e (e 4x ) log e (sin(x)) log e (y) = 2 log e (x) + log e (cosh(x)) 4x log e (sin(x)) d dx {log e(y)} = 1 y dx = 2 x + sinh(x) cos (x) 4 cosh(x) sin (x) Since we know that sinh (x) cosh (x) = tanh(x) and also that cos (x) = cot (x) sin (x) We may write d dx {log e(y)} = 1 y dx = 2 + tanh (x) 4 cot (x) x We multiply both sides by y to find our final answer dx = cosh (x) [x2 e 4x sin (x) ] [2 + tanh(x) 4 cot (x)] x Question Page 28 of of 41

29 Q3.22 Differentiate the following function with respect to x y = 3x sin(x) cos (x) ANSWER Q3.22 [3x sin(x) cos (x)] [ 1 + cot(x) tan (x)] x Differentiation of Inverse Trigonometric Functions Suppose we have the equation y = sin 1 (x) How do we then go about finding dx? Maybe we should write this in a different form Now can differentiate x = sin(y) dx = cos(y) flip both sides: dx = 1 cos (y) Next we shall remember one of our han trig. Identities cos 2 (y) + sin 2 (y) = 1 cos 2 (y) = 1 sin 2 (y) cos(y) = 1 sin 2 (y) Since we know x = sin(y) then if we square both sides we shall have x 2 = sin 2 (y) So we may re-write our transposed trig. identity as cos(y) = 1 x 2 Now we can put this latest result back into our differential dx = 1 cos (y) = 1 1 x 2 = That was a bit tricky, but we managed to find y. 1 (1 x 2 ) 0.5 = (1 x2 ) 0.5 Let s go on and see if we can find y y = (1 x 2 ) 0.5 Page 29 of of 41

30 This is where that shorthand notation for derivatives comes in really han Let u = 1 x 2 y = u 0.5 du = 0.5u 1.5 du dx = 2x y = du dx = 2x 0.5u 1.5 du y = xu 1.5 = x(1 x 2 ) 1.5 Question Q3.23 Find y for y = cos 1 (2x) ANSWER Q3.23 8x (1 4x 2 ) Differentiation of Inverse Hyperbolic Functions Consider the function y = sinh 1 (x) Our task here is to find y. We proceed in a similar manner to the previous problem x = sinh (y) dx = cosh(y) dx = 1 cosh (y) A useful identity is cosh 2 (y) sinh 2 (y) = 1 cosh 2 (y) = 1 + sinh 2 (y) cosh 2 (y) = 1 + x 2 cosh(y) = 1 + x 2 dx = 1 y = 1 + x = (1 + 2 x2 ) 0.5 Page 30 of of 41

31 Now to find that second derivative y = (1 + x 2 ) 0.5 Let u = 1 + x 2 y = u 0.5 du dx = 2x du = 0.5u 1.5 y = du dx du = 2x 0.5u 1.5 = x(1 + x 2 ) 1.5 Question Q3.24 Find y for y = cosh 1 (2x) ANSWER Q3.24 8x (4x 2 1) 1.5 Page 31 of of 41

32 3.3 Further Integration Integration by Parts This technique provides us with a way to integrate the product of two simple functions. We shall develop a formula to use as a framework for the technique. The starting point for our formula lies with the Product Rule, which you mastered in Section Let s rearrange this formula d dv du (uv) = u + v dx dx dx u dv dx = d du (uv) v dx dx If we now integrate both sides with respect to x then we need to place a sign before each term and a dx at the end of each term, like so Some simplifications can be noticed here: u dv dx dx = d du (uv)dx v dx dx dx In the first integral the two dx terms cancel each other out In the second integral we are taking the integral of the differential of uv. Since integration is the reverse of differentiation then the integral of a differential drops away, just leaving uv In the third integral the two dx terms cancel each other out Performing these simplifications gives That is our formula for integration by parts. u dv = uv v du What we do with the formula is to look at the left hand side and see the integral of a product. That product is composed of u and dv. When using integration by parts you have a choice of which part of the product to call u and which to call dv. Fortunately, your choice is guided The u part will become a constant after taking multiple derivatives The dv part is readily integrated by using standard integrals. Page 32 of of 41

33 Worked Example 22 Determine: x sin(x) dx One part of our product is x and the other part is sin (x). We need to make a choice as to which one to make u and which to make dv. Our guidance tells us to make the x part equal to u since the derivative of x becomes 1 (i.e. a constant) Let u = x Let dv = sin (x) We now try to form the terms in our integration by parts formula We can now write If u = x du dx = 1 du = dx If dv = sin(x) v = cos (x) u dv = uv v du x sin(x) dx = x( cos(x)) ( cos(x))dx x sin(x) dx = x cos(x) + cos(x) dx x sin(x) dx = x cos(x) + sin(x) + C That was fairly straightforward. Always remember the constant of integration C at the end. Worked Example 23 Let s now try to integrate the product of an exponential and a cosine. Determine: e 2x cos(x) dx Page 33 of of 41

34 Our two products here are e 2x and cos (x). Neither of them will reduce to a constant when differentiated so it doesn t matter which one we call u and which we call dv. You will see here that we need to do integration by parts twice to arrive at our answer. Just as in the previous worked example we need to decide on assignments for u and dv, then find v and du. Let s do that Let u = e 2x Let dv = cos(x) du dx = 2e2x du = 2e 2x dx v = sin (x) To aid our working here (and save some ink) we normally assign the capital letter I to the integral in our question. I = e 2x cos(x) dx This needs tiing I = e 2x sin(x) sin(x) 2e 2x dx I = e 2x sin(x) 2 e 2x sin(x) dx We were rather hoping for an answer, but what we seem to have is another integration by parts problem embedded into our development. Don t worry, that s quite normal for this type of problem. All we do is attack that last part with integration by parts once more, using square brackets. Our desired result will then appear Expanding these terms gives I = e 2x sin(x) 2 [e 2x ( cos (x)) cos (x)2e 2x dx] I = e 2x sin(x) + 2e 2x cos(x) 4 e 2x cos(x) dx That last integral is the same as I, which is really han, so we may now write I = e 2x sin(x) + 2e 2x cos(x) 4I 5I = e 2x sin(x) + 2e 2x cos (x) If we divide both sides by 5 then we have our final answer Page 34 of of 41

35 Question I = e2x sin(x) + 2e 2x cos (x) 5 Q3.25 Determine: e 4x sin(2x)dx ANSWER Q3.25 4e4x sin(2x) 2e 4x cos(2x) Integration by Substitution Quite often we come across integrals which are a function of a linear algebraic function. To solve such problems we introduce the variable u, as seen earlier, and make a convenient substitution with the derivative of u. Let s look at an example Worked Example 24 Determine: (3x 2) 5 dx This looks like an integration of a function of a function problem. We know alrea how to find the derivative of function of a function problems, but now we learn how to find their integrals. As mentioned, we shall introduce u and make it equal to the contents of the brackets u = 3x 2 du dx = 3 We are going to find it useful to have an expression for dx du. This is not a problem, we just need to flip both sides Let s now express our original problem in terms of u dx du = 1 3 Page 35 of of 41

36 Question u 5 dx We can t solve that in its current form. What we need to do is to change the variable. Here s the trick u 5 dx = u 5 dx du du Notice that those du terms potentially cancel each other out, so what we have written is perfectly fine. Another thing we notice here is that we actually know what dx du is. We worked it out before as 1/3. We can now write u 5 dx = u 5 dx du du = 1 3 u5 du = 1 3 u6 u6 + C = C If we now replace u with its original value of (3x 2) we shall arrive at our answer (3x 2) 5 dx = (3x 2) C Q3.26 Determine: (2x 6) 4 dx ANSWER Q3.26 (2x 6) C Integration using Partial Fractions Many functions must be expressed in terms of partial fractions before they can be integrated. Partial fractions were covered in Workbook 1. Let s take a look at an integration problem requiring the use of partial fractions. Worked Example 25 Determine: x 2 dx (x + 2)(x 1) 2 Page 36 of of 41

37 We need to break the problem up into smaller fractions. Partial fractions are the result. Let s use our knowledge from Workbook 1 to determine these partial fractions. x 2 (x + 2)(x 1) 2 A (x + 2) + B (x 1) + C (x 1) 2 Now multiply both sides by (x + 2)(x 1) 2. This will give us a one line expression Let x = 1: Let x = 2: (1) 2 = 3C C = 1 3 ( 2) 2 = 9A A = 4 9 x 2 = A(x 1) 2 + B(x + 2)(x 1) + C(x + 2) We now need to find B. One way to do that is to equate coefficients of x 2 [x 2 ]: 1 = A + B B = 1 A = = 5 9 Since we now know the value of each of the three constants we may express the initial problem as x 2 4 (x + 2)(x 1) 2 dx = ( 5 9 (x + 2) (x 1) + 3 (x 1) 2) dx The right hand side may be split into three integrals, as follows x 2 (x + 2)(x 1) 2 dx = (x + 2) dx (x 1) dx (x 1) 2 dx These three integrals on the right hand side may be evaluated as (x + 2)(x 1) 2 dx = 4 9 log e(x + 2) log e(x 1) 1 3x 3 + C x 2 Question Q3.27 Determine: x 2 dx (x 2)(x + 1) 2 Page 37 of of 41

38 ANSWER Q log e(x + 1) log e(x 2) + 1 3x+3 + C Page 38 of of 41

39 3.4 Solution of Engineering Problems with Calculus Charging RC Circuit For the circuit below, use Calculus to find the charge stored in the capacitor 2 seconds after the switch closes. Assume zero charge on the capacitor before this event. For a capacitor charging via a DC source voltage, through a series resistor, the formula for instantaneous current is i = E R e( t RC) Amps Since current is the defined as the rate of change of charge (q, carried by passing electrons) we may say i = dq dt If we integrate both sides with respect to t we will have So we may say that i dt = dq dt q = i dt dt = dq = q If we wish to find the amount of charge store stored within a certain time period (i.e. 2 seconds in this example) we write = E 1 R [ e{ RC }t ] 1 RC 2 q (0 2) = i dt = E R e( t RC = ERC R 0 2 ) dt [e( t RC ) 2 ] = EC [e ( t RC ) 2 ] 0 0 Page 39 of of 41

40 Since we know that R = 1MΩ and C = 1μF then RC = = = 10 0 = 1 we may write q (0 2) = EC [e ( t 1 ) 2 ] = EC[e t ] 2 0 = EC[e 2 e 0 ] = EC[ ] 0 We also know that the DC supply voltage (E) is 10V q (0 2) = [ ] = Coulombs = μc The plot below illustrates the integration we have just performed Current is on the vertical axis and time on the horizontal. When we multiply current by time we get charge, which is shown as the shaded area between 0 and 2 seconds. You may like to experiment with the Graph simulator to produce similar results Energising Inductor Consider the series RL circuit below If E = 10 V, R = 1MΩ and L = 200mH, use Calculus to determine the voltage across the inductor after 0. 1μs. Use the following formulae in your solution development Page 40 of of 41

41 i = E R (1 e Rt L) and v L = L di dt We start with v L = L di dt Putting in the values for E, R, t and L gives E d { = L R (1 e Rt L)} dt = LE R d (1 e Rt L ) = LE dt R (R L e Rt L) = Ee Rt L v L = 10e ( ) = volts Page 41 of of 41