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I Relate the Hyperbolic Functions to the Exponential Function I Sketch Graphs In th is ch apter you will learn how to: I relate the hyperbolic functions to the exponential function I sketch graphs of the hyperbolic functions I prove and use identities involving hyperbolic functions I use the definitions of the inverse hyperbolic functions and use the logarithmic forms. Cambridge International AS & A Level Further Mathematics: Further Pure Mathematics 2 ; '~ PREREQUISITE KNOWLEDGE Where it comes from What you should be able to do Check your skills AS & A Level Mathematics Use the trigonometric identities. 1 Prove: Pure Mathematics 2 & 3, cos A sin A cos(A + B) Chapter 3 ------ sinB cosB sinBcosB AS & A Level Mathematics Manipulate equations involving 2 Solve: e2x - e - 6 = 0 Pure Mathematics 2 & 3, the natural logarithm and Chapter 2 exponential functions. What are hyperbolic functions? In the Pure Mathematics 2 & 3 Coursebook, Chapter 4, you learned uses for trigonometric functions other than solving problems based on a triangle. In fact, trigonometric functions are also called circular functions, as they all have geometric meaning and derivations from the equation of the unit circle. There are similar relationships between the structure of circles, hyperbolas, parabolas and ellipses. These shapes are all called conic sections, because they are shapes produced when we slice a cone. Horizontal slices produce circles; oblique slices produce ellipses; slices parallel to the slope of the cone produce parabolas; steeper slices than this produce hyperbolas. circle ellipse parabola hyperbola Hyperbolic functions are a class of functions that are derived from the unit hyperbola, 2 2 x - y = 1. They are a significant tool in an engineer's toolkit. They also allow us to find highly elegant solutions to differential equations. 19.1 Exponential forms of hyperbolic functions In the introduction, we discussed the relationships between hyperbolic functions and trigonometric functions. Hyperbolics link to apparently vastly different function types, but as you read this chapter and do subsequent work on complex numbers, you will start to see the connections. We shall start by relating hyperbolic functions to exponentials. It is not clear from the ideas and development of hyperbolics whether the exponential definitions came first or the relationships with trigonometry came first. We shall use the exponential forms to develop the idea. Let us begin with the hyperbolic sine function: y = sinhx I. Chapter 19: Hyperbolic functions The first challenge is the pronunciation of the hyperbolic functions. The hyperbolic sine is ' ] pronounced 'shine' and its exponential definition is: . ex - e-x smhx = 2 Based on this, we can define the domain, find the range and draw the function. It is helpful to consider this graph as the average of y = e and y = -e-x. ------------- - 5 -4 - 3 - 2 -4 From this, we can see that the function is a one- one mapping. The domain is x E IR.. The range is f(x) E IR.. It is also an odd function. y = coshx The hyperbolic cosine is pronounced 'cosh' and its exponential definition is: ex+ e-x coshx=-- - 2 We consider this graph as the average of y =ex and y = e-x . --------------- ...... ------ .. -- - 5 -4 - 3 - 2 - 1 0 1 2 3 4 x ::-rom this, we can see that the function is not a one- one mapping. However, it can be .valuated with both vositive and negative values for x. the domain is x E IR.. The range is f( x) ~ 1. It is an even function. 'Or the function to be one- one, we need to restrict the domain to x ~ 0. This will be very mportant later. Cambridge International AS & A Level Further Mathematics: Further Pure Mathematics 2 J 0 DID YOU KNOW? The hyperbolic cosine is a very special curve and has many occurrences in nature. It is called a catenary, and is the shape of a hanging chain. EXPLORE 19.1 I Based on the exponential definitions of sinhx and coshx, find: ~(sinhx) and ~(cosh x) dx dx The exponential forms of sinh x and cosh x are shown in Key point 19.1. KEY POINT 19.1 The exponential forms of sinh x and cosh x: . ex - e- x smh x = , x E IR. , f(x) E IR. 2 ex + e-x I cosh x = , x E IR., f(x) ~ I 2 From this, we can deduce that ex = cosh x + sinh x. We can now derive further hyperbolic functions: Hyperbolic function Pronunciation Exponential form tanhx 'tanch' (preferred) ex - e-x tanhx= - - - 'than' (a long thhhhhh as in 'thank you') ex + e-x 'tank' Y ,~ Domain: x E lRI. 3 2 1 / Range: -1 < f(x) < 1 ..... ~ -4 - 3 - 2 0 1 2 3 4 x Y.- 1 - 2 Odd function - 3 I. Chapter 19: Hyperbolic functions erbolic function Pronunciation Exponential form 2 'setch' sechx=--­ ex + e-x y Domain: x E !R 2 Range: 0 < f(x) ~ 1 -4 - 3 - 2 - 1 2 3 4 x - 1 Even function - 2 DID YO~ KNOW? The graph of y = sech2 x is a very important curve. It is used to model tsunamis and to model wavelets in wave- particle duality. We call this type of wave a soliton. It is a very special type of wave that does not lose energy as it travels. Its velocity is proportional to the amplitude of the wave. This is why tsunamis have such a destructive force. You might study more about solitons at university level. yperbolic function Pronunciation Exponential form 2 cosechx 'cosetch' cosech x = --- ex - e-x Domain: x-=I= 0 Range: f(x) -::fa 0 4 x Odd function coth x Domain: x -=I= 0 To help us remember the reciprocal functions of sech x, cosech x and coth x, we can apply the third letter rule Range: f(x) > 1, f(x) < -1 as with trigonometric reciprocal functions: -4 - 3 - 2 0 1 2 3 4 x 1 se chx= - - coshx Odd function 1 cosechx = -- sinhx 1 cothx= -- tanhx Cambridge International AS & A Level Further Mathematics: Further Pure Mathematics 2 .~ t' WORKED EXAMPLE 19.1 Sketch the graph of y = l3cosh(x + 1) - 41. Answer Given that f(x) = cosh x : Consider each transformation in turn. I I y = f(x + 1) I I I I I I I I I I I I I I -" 2 3 4 x I I I I y=3f(x +l) I I I I I I I I I I I I I I I I \ I \ I \ I \ I \ I \ I \ 2 I ''--!'' I -4 - 3 - 2 - 1 y=3f(x +l) -4 This is the graph of 13 cosh(x + 1) - 41. y = l3f(x + 1) - 41 -4 - 3 2 3 4 x i Chapter 19: Hyperbolic functions These functions can all be evaluated using a calculator, in the same way as for trigonometric functions. However, generally you will be required to write in exact form, unless told otherwise. We can use the previous definitions to solve equations involving hyperbolic forms. ~- WORKED EXAMPLE 19.2 & Leaving your answer in exact form, solve 4coshx - 3 sinh x = 4. Answer 4cosh x - 3 sinh x = 4 First, use the exponential definitions: . ex - e-x 4 (e' ~c x ) - 3(e' ~ e-x) = 4 smh x= --- 2 ex + e-x coshx= --- 2 4 e2-' + 4 - 3e 2x + 3 = 8 ex Now multiply up by 2 ex. e2x - 8 ex + 7 = 0 This is now a quadratic in terms of e and so can be solved. ( e' - 7)(e -' - 1) = 0 Leading to two solutions: x = ln 7, x = In 1 = 0 We are usually asked to give our solutions in exact form. It is possible to approach this problem using a similar technique as that for trigonometric equations using hyperbolic forms. However, this rarely gives such elegant solutions. We need to understand both inverse hyperbolic functions and their logarithmic forms. Solve 3 tanh2 x - 4 tanh x = 4. Answer 3tanh2x - 4tanhx - 4 = 0 First, recognise that this is a quadratic equation in terms of tanh x. To make it simpler, replace tanh x with T 3T2 -6T+ 2T-4=0 Solve the equation. 3T(T - 2) + 2(T - 2) = 0 (T - 2)(3T + 2) = 0 T = 2 2 Or T = - - 3 We are left then to solve: Remember: since the range of tanh x ) tanh x = -=- cannot exceed 1, T = 2 will not lead to any 3 solutions. x = tanb- 1 (-D ~ Cambridge International AS & A Level Further Mathematics: Further Pure Mathematics 2 · j Write all solutions in exact form unless otherwise stated. 1 Solve 19sinhx + 16coshx = 8. 2 Solve 29coshx = 11 sinhx + 27. 3 Solve 17 sinhx + 16 coshx = 8. 4 Solve 7 = 17 tanhx + 28 sechx. 5 Solve cosechx - 2cothx = 2. 6 Sketch the graph of y = 13 coshx - 51. 7 Sketch the graph of y = 4 - cosech(x - 2). 8 Sketch the graph of y = 12 + _!_coth(x + 1)1. 2 ~ 9 Solve 4coshx + sinhx = 8. Write your answers in the form b In a, where a and bare integers. ~ 10 Solve lOcoshx + 2sinhx = 11, giving your answers in the form ln a, where a is a rational number. ~ 11 Given that sech-1 x = In (1 +?),solve 4sechx - 3tanh2x =I. I. 19.2 Hyperbolic identities If we consider that hyperbolic functions are similar to trigonometric functions, it is reasonable to consider that hyperbolic identities exist in the same way. We can use the exponential forms to prove that they are true. Q __ I When forming a proof, WORKED EXAMPLE 19.4 it is very important to either start at the 2 2 left-hand side and Prove that cosh x - sinh x = 1. show it is the same as Answer the right-hand side, or 2 2 start at the right-hand cosh x - sinh x side to show it is the same as the left-hand F irst, convert to exponential fo rm and side.
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