Math421 - Chapter 2 Hyperbolic Functions 2 HYPERBOLIC...

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Unformatted text preview: Chapter 2 Hyperbolic Functions 2 HYPERBOLIC FUNCTIONS Objectives After studying this chapter you should • understand what is meant by a hyperbolic function; • be able to find derivatives and integrals of hyperbolic functions; • be able to find inverse hyperbolic functions and use them in calculus applications; • recognise logarithmic equivalents of inverse hyperbolic functions. 2.0 Introduction This chapter will introduce you to the hyperbolic functions which you may have noticed on your calculator with the abbreviation hyp. You will see some connections with trigonometric functions and will be able to find various integrals which cannot be found without the help of hyperbolic functions. The first systematic consideration of hyperbolic functions was done by the Swiss mathematician Johann Heinrich Lambert (1728-1777). 2.1 Definitions The hyperbolic cosine function, written cosh x, is defined for all real values of x by the relation cosh x = ( 1 x e + e− x 2 ) Similarly the hyperbolic sine function, sinh x, is defined by sinh x = ( 1 x e − e− x 2 ) The names of these two hyperbolic functions suggest that they have similar properties to the trigonometric functions and some of these will be investigated. 33 Chapter 2 Hyperbolic Functions Activity 1 Show that cosh x + sinh x = e x and simplify cosh x − sinh x. (a) By multiplying the expressions for ( cosh x + sinh x ) and (cosh x − sinh x ) together, show that cosh 2 x − sinh 2 x = 1 ( cosh x + sinh x )2 + (cosh x − sinh x )2 (b) By considering cosh 2 x + sinh 2 x = cosh 2x show that ( cosh x + sinh x )2 − (cosh x − sinh x )2 (c) By considering 2 sinh x cosh x = sinh 2x show that Activity 2 Use the definitions of sinh x and cosh x in terms of exponential functions to prove that (a) cosh 2x = 2 cosh 2 x − 1 (b) cosh 2x = 1 + 2 sinh 2 x Example Prove that cosh ( x − y ) = cosh x cosh y − sinh x sinh y Solution cosh x cosh y = = sinh x sinh y = = ( ) ( 1 x 1 e + e− x × e y + e− y 2 2 ) ( 1 x+y e + e x − y + e−( x − y) + e−( x + y) 4 ( ) ( 1 x 1 e − e− x × e y − e− y 2 2 ) ) ( 1 x+y e − e x − y − e−( x − y) + e−( x + y) 4 ) Subtracting gives cosh x cosh y − sinh x sinh y = 2 × = 34 ( ( 1 x−y e + e−( x − y) 4 ) ) 1 x−y e + e − ( x − y ) = cosh( x − y ) 2 Chapter 2 Hyperbolic Functions Exercise 2A 3. cosh ( x + y ) = cosh x cosh y + sinh x sinh y Prove the following identities. 1. (a) sinh ( − x ) = − sinh x (b) cosh ( − x ) = cosh x 2. (a) sinh ( x + y ) = sinh x cosh y + cosh x sinh y (b) sinh ( x − y ) = sinh x cosh y − cosh x sinh y 2.2 4. sinh A + sinh B = 2sinh A − B A + B cosh 2 2 5. cosh A − cosh B = 2sinh A − B A + B sinh 2 2 Osborn's rule You should have noticed from the previous exercise a similarity between the corresponding identities for trigonometric functions. In fact, trigonometric formulae can be converted into formulae for hyperbolic functions using Osborn's rule, which states that cos should be converted into cosh and sin into sinh, except when there is a product of two sines, when a sign change must be effected. cos 2x = 1 − 2 sin 2 x For example, can be converted, remembering that sin 2 x = sin x.sin x, into cosh 2x = 1 + 2 sinh 2 x . But sin 2 A = 2 sin A cos A simply converts to sinh 2 A = 2 sinh A cosh A because there is no product of sines. Activity 3 Given the following trigonometric formulae, use Osborn's rule to write down the corresponding hyperbolic function formulae. (a) sin A − sin B = 2 cos A + B A − B sin 2 2 (b) sin 3A = 3sin A − 4 sin 3 A (c) cos 2 θ + sin 2 θ = 1 2.3 Further functions Corresponding to the trigonometric functions tan x, cot x, sec x and cosecx we define tanh x = sinh x , cosh x coth x = 1 cosh x = , tanh x sinh x 35 Chapter 2 Hyperbolic Functions sech x = 1 cosh x and cosech x = 1 sinh x By implication when using Osborn's rule, where the function tanh x occurs, it must be regarded as involving sinh x. Therefore, to convert the formula sec2 x = 1 + tan2 x we must write sech 2 x = 1 − tanh 2 x . Activity 4 (a) Prove that tanh x = ex − e− x 2 and sech x = x − x , x −x e +e e +e and hence verify that sech 2 x = 1 − tanh 2 x . (b) Apply Osborn's rule to obtain a formula which corresponds to cosec 2 y = 1 + cot 2 y . Prove the result by converting cosech y and coth y into exponential functions. 2.4 Graphs of hyperbolic functions You could plot the graphs of cosh x and sinh x quite easily on a graphics calculator and obtain graphs as shown opposite. y cosh x 1 0 y The shape of the graph of y = cosh x is that of a particular chain supported at each end and hanging freely. It is often called a catenary (from the Latin word catena for chain or thread). 36 x sinh x x Chapter 2 Hyperbolic Functions Activity 5 (a) Superimpose the graphs of y = cosh x and y = sinh x on the screen of a graphics calculator. Do the curves ever intersect? (b) Use a graphics calculator to sketch the function f : x a tanh x with domain x ∈ R. What is the range of the function? (c) Try to predict what the graphs of y = sechx, y = cosech x and y = coth x will look like. Check your ideas by plotting the graphs on a graphics calculator. 2.5 Solving equations 3 and we wish to find the exact value of x. 4 Recall that cosh 2 x = 1 + sinh 2 x and cosh x is always positive, so 3 5 when sinh x = , cosh x = . 4 4 Suppose sinh x = From Activity 1, we have sinh x + cosh x = e x 3 5 + =2 4 4 x = ln 2 . ex = so and hence Alternatively, we can write sinh x = so sinh x = ( 1 x e − e− x 2 ) 3 means 4 ( ) 1 x 3 e − e− x = 2 4 ⇒ 2e x − 3 − 2e − x = 0 and multiplying by e x 2e 2 x − 3e x − 2 = 0 (e x )( ) − 2 2e x + 1 = 0 e x = 2 or e x = − 1 2 But e x is always positive so e x = 2 ⇒ x = ln 2. 37 Chapter 2 Hyperbolic Functions Activity 6 Find the values of x for which cosh x = 13 5 expressing your answers as natural logarithms. Example Solve the equation 2 cosh 2x + 10 sinh 2x = 5 giving your answer in terms of a natural logarithm. Solution cosh 2x = So ( ) ( 1 2x 1 e + e −2 x ; sinh 2x = e 2 x − e −2 x 2 2 ) e 2 x + e −2 x + 5e 2 x − 5e −2 x = 5 6e 2 x − 5 − 4e −2 x = 0 6e 4 x − 5e 2 x − 4 = 0 (3e 2x )( ) − 4 2e 2 x + 1 = 0 e2 x = 4 1 or e 2 x = − 3 2 The only real solution occurs when e 2 x > 0 2x = ln So 4 1 4 ⇒ x = ln 3 2 3 Exercise 2B 1. Given that sinh x = 5 , find the values of 12 2. Given that cosh x = 5 , determine the values of 4 (a) cosh x (b) tanh x (c) sech x (a) sinh x (d) coth x (e) sinh 2 x (f) cosh 2 x Use the formula for cosh ( 2 x + x ) to determine the value of cosh 3x. Determine the value of x as a natural logarithm. 38 (b) cosh 2 x (c) sinh 2 x Chapter 2 Hyperbolic Functions 3. In the case when tanh x = 1 1 , show that x = ln 3. 2 2 Hence write down the minimum value of 25cosh x − 24sinh x and find the value of x at which this occurs, giving your answer in terms of a natural logarithm. 4. Solve the following equations giving your answers in terms of natural logarithms. (a) 4 cosh x + sinh x = 4 8. Determine a condition on A and B for which the equation (b) 3sinh x − cosh x = 1 Acosh x + Bsinh x = 1 (c) 4 tanh x = 1 + sech x has at least one real solution. 5. Find the possible values of sinh x for which 12 cosh 2 x + 7sinh x = 24 9. Given that a, b, c are all positive, show that when a>b then a cosh x + bsinh x can be written in the form Rcosh ( x + α ). (You may find the identity cosh 2 x − sinh 2 x = 1 useful.) Hence determine a further condition for which the equation Hence find the possible values of x, leaving your answers as natural logarithms. a cosh x + bsinh x = c has real solutions. 6. Solve the equations (a) 3cosh 2 x + 5cosh x = 22 10. Use an appropriate iterative method to find the solution of the equation (b) 4 cosh 2 x − 2sinh x = 7 cosh x = 3x 7. Express 25cosh x − 24sinh x in the form Rcosh ( x − α ) giving the values of R and tanh α . 2.6 giving your answer correct to three significant figures. Calculus of hyperbolic functions Activity 7 (a) By writing cosh x = ( ) 1 x e + e − x , prove that 2 d ( cosh x ) = sinh x. dx (b) Use a similar method to find d (sinh x ). dx (c) Assuming the derivatives of sinh x and cosh x, use the sinh x quotient rule to prove that if y = tanh x = cosh x then dy = sech 2 x. dx Note: care must be taken that Osborn's rule is not used to obtain corresponding results from trigonometry in calculus. 39 Chapter 2 Hyperbolic Functions Activity 8 Use the quotient rule, or otherwise, to prove that (a) d (sech x ) = −sech x tanh x dx (b) d (cosech x ) = −cosech x coth x dx (c) d (coth x ) = −cosech 2 x dx Example Integrate each of the following with respect to x. (a) cosh 3x (b) sinh 2 x (c) x sinh x (d) e x cosh x Solution (a) ∫ cosh 3x d x = 3 sinh 3x + constant 1 (b) sinh 2 x d x can be found by using cosh 2x = 1 + 2 sinh 2 x giving 1 2 ∫ (cosh 2x − 1) d x = 1 1 sinh 2 x − x + constant 4 2 Alternatively, you could change to exponentials, giving sinh 2 x = ∫ sinh 2 xdx = ( 1 2x e − 2 + e −2 x 4 ) 1 2x 1 1 e − x − e −2 x + constant 8 2 8 Can you show this answer is identical to the one found earlier? (c) Using integration by parts, ∫ x sinh x d x = x cosh x − ∫ cosh x d x = x cosh x − sinh x + constant 40 Chapter 2 Hyperbolic Functions (d) Certainly this is found most easily by converting to exponentials, giving e x cosh x = ∫e x 1 2x 1 e + 2 2 cosh x d x = 1 2x 1 e + x + constant 4 2 Exercise 2C 1. Differentiate with respect to x 4. A curve has equation y = λ cosh x + sinh x (a) tanh 4 x (b) sech2 x (c) cosech ( 5x + 3) (d) sinh e x where λ is a constant. (e) cosh 3 2 x (f ) tanh (sin x ) (a) Sketch the curve for the cases λ = 0 and λ = 1. (g) cosh 5x sinh 3x (h) ( ) ( coth 4 x ) 2. Integrate each of the following with respect to x. (b) cosh 2 3x (a) sinh 4 x 2 2 (c) x cosh 2 x (e) cosech2 x coth 3x (d) sech 7x (f ) tanh x (g) tanh 2 x (h) e 2 sinh 3x ( (i) x cosh x + 4 2 3 ) (k) cosh 2 x sinh 3x 4 ( j) sinh x (l) sech x (b) Determine the coordinates of the turning point of the curve in the case when λ = 4 . Is 3 this a maximum or minimum point? (c) Determine the range of values of λ for which the curve has no real turning points. 5. Find the area of the region bounded by the coordinate axes, the line x = ln 3 and the curve with equation y = cosh 2 x + sech2 x. 3. Find the equation of the tangent to the curve with equation y = 3cosh 2 x − sinh x at the point where x = ln 2. 2.7 Inverse hyperbolic functions The function f : x a sinh x ( x ∈ R ) is one-one, as can be seen from the graph in Section 2.4. This means that the inverse function f −1 exists. The inverse hyperbolic sine function is denoted by sinh −1 x. Its graph is obtained by reflecting the graph of sinh x in the line y = x. Recall that y sinh–1 x x d (sinh x ) = cosh x , so the gradient of the graph of dx y = sinh x is equal to 1 at the origin. Similarly, the graph of sinh −1 x has gradient 1 at the origin. 41 Chapter 2 Hyperbolic Functions Similarly, the function g: x a tanh x ( x ∈ R ) is one-one. y 1 You should have obtained its graph in Activity 5. The range of g is { y:−1 < y < 1} or the open interval ( −1, 1). y = tanh x 0 x –1 Activity 9 Sketch the graph of the inverse tanh function, tanh −1 x. y Its range is now R. What is its domain? y = cosh x The function h: x a cosh x ( x ∈ R ) is not a one-one function and so we cannot define an inverse function. However, if we change the domain to give the function f : x a cosh x with domain 1 0 x { x: x ∈ R, x ≥ 0} y then we do have a one-one function, as illustrated. So, provided we consider cosh x for x ≥ 0, we can define the inverse function f −1 : x a cosh −1 x with domain y = cosh–1 x { x: x ∈ R, x ≥ 1}. 1 This is called the principal value of cosh −1 x. 2.8 0 Logarithmic equivalents Activity 10 Let y = sinh −1 x so sinh y = x. Since cosh y is always positive, show that cosh y = (1 + x ) 2 By considering sinh y + cosh y, find an expression for e y in terms of x. Hence show that sinh −1 x = ln x + 42 (1 + x 2 ) 1 x Chapter 2 Hyperbolic Functions Activity 11 Let y = tanh −1 x so tanh y = x. Express tanh y in the terms of ey and hence show that e2 y = 1+ x . 1− x Deduce that tanh −1 x = 1 1+ x ln . 2 1− x (Do not forget that tanh −1 x is only defined for x < 1. ) Activity 12 Let y = cosh −1 x, where x ≥ 1, so cosh y = x. Use the graph in Section 2.7 to explain why y is positive and hence why sinh y is positive. Show that sinh y = ( x 2 − 1) . Hence show that cosh −1 x = ln x + ( x 2 − 1) . The full results are summarised below. sinh −1 x = ln x + tanh −1 x = ( x 2 + 1) 1 1+ x ln 2 1− x cosh −1 x = ln x + ( x 2 − 1) (all values of x) ( x < 1) ( x ≥ 1) 43 Chapter 2 Hyperbolic Functions Exercise 2D 3. Express sech −1x in logarithmic form for 0 < x < 1. 1. Express each of the following in logarithmic form. 3 (a) sinh −1 4 4. Find the value of x for which 1 (c) tanh −1 2 (b) cosh −1 2 sinh−1 x + cosh−1( x + 2 ) = 0. 2. Given that y = sinh −1 x, find a quadratic equation 5. Solve the equation satisfied by e y . Hence obtain the logarithmic form of sinh −1 x. Explain why you discard one of the solutions. 2.9 Derivatives of inverse hyperbolic functions Let y = sinh−1 x so that x = sinh y. dx = cosh y but cosh2 y = sinh2 y + 1 = x 2 + 1 dy and cosh y is always positive. So dx = dy ( x2 + 1) In other words, dy = dx and therefore d (sinh−1 x ) = dx (x 1 2 (x 1 2 + 1) + 1) Activity 13 (a) Show that d (cosh−1 x ) = dx (b) Show that d 1 tanh−1 x = dx 1 − x2 ( 1 ( x2 − 1) ) An alternative way of showing that to use the logarithmic equivalents. 44 d (sinh−1 x ) = dx (x 1 2 + 1) is x −2 = ln 2. 2 tanh −1 x +1 Chapter 2 Hyperbolic Functions Since [ d x+ dx ( x2 + 1) ] = 1 + 1 .2x ( x2 + 1)− 2 x = 1+ ( x2 + 1) = [ we can now find the derivative of ln x + { ( x2 + 1) + x ( x2 + 1) ( x2 + 1) ] ( x2 + 1) + x . 1 2 ( x + 1) x + ( x2 + 1) ( x2 + 1) } = d ln x + dx 1 2 which cancels down to (x 1 2 + 1) d (sinh−1 x ) = dx So (x 1 2 + 1) You can use a similar approach to find the derivatives of cosh−1 x and tanh−1 x but the algebra is a little messy. Example Differentiate (a) cosh−1( 2x + 1) 1 (b) sinh−1 with respect to x ( x > 0 ) . x Solution (a) Use the function of a function or chain rule. [ 1 d cosh−1( 2x + 1) = 2. dx (b) {(2x + 1) − 1} 2 = ( 4x −1 1 1 d −1 1 = − 2. sinh x = x2 . x dx 1 + 1 2 x = 2 2 + 4x ) = (x 1 2 + x) x (1 + x2 ) −1 x (1 + x2 ) 45 Chapter 2 Hyperbolic Functions Exercise 2E 7. Differentiate sech −1x with respect to x, by first writing x = sech y. Differentiate each of the expressions in Questions 1 to 6 with respect to x. 1. cosh −1 ( 4 + 3x ) 2. sinh −1 8. Find an expression for the derivative of cosech−1x in terms of x. ( x) 3. tanh −1 ( 3x + 1) 9. Prove that d ( coth−1 x ) = ( x2−1 ) . −1 dx 4. x 2 sinh −1 ( 2 x ) 1 5. cosh −1 x ( x > 0) 6. sinh −1 ( cosh 2 x ) 2.10 Use of hyperbolic functions in integration Activity 14 Use the results from Section 2.9 to write down the values of ⌠ (a) ⌡ (x 1 2 + 1) dx and ⌠ (b) ⌡ (x 1 2 Activity 15 x Differentiate sinh−1 with respect to x. 3 ⌠ Hence find ⌡ (x 1 2 + 9) ⌠ What do you think ⌡ d x. (x 2 1 + 49) d x is equal to? Activity 16 Use the substitution x = 2 cosh u to show that ⌠ ⌡ 46 (x 1 2 − 4) x d x = cosh−1 + constant 2 − 1) dx Chapter 2 Hyperbolic Functions Activity 17 Prove, by using suitable substitutions that, where a is a constant, ⌠ (a) ⌡ x d x = sinh−1 + constant a (x + a ) ⌠ (b) ⌡ x d x = cosh−1 + constant a (x − a ) 1 2 2 1 2 2 Integrals of this type are found by means of a substitution involving hyperbolic functions. They may be a little more complicated than the ones above and it is sometimes necessary to complete the square. Activity 18 Express 4x 2 − 8x − 5 in the form A( x − B) + C, where A, B and C are constants. 2 Example Evaluate in terms of natural logarithms 7 ⌠ ⌡ 4 ( 4x 2 1 − 8x − 5) dx Solution From Activity 18, the integral can be written as 7 ⌠ ⌡ 4 1 {4( x − 1) − 9} dx 2 You need to make use of the identity cosh2 A − 1 = sinh2 A because of the appearance of the denominator. Substitute 4( x − 1) = 9 cosh2 u in order to accomplish this. So 2( x − 1) = 3cosh u and 2 2 dx = 3sinh x du 47 Chapter 2 Hyperbolic Functions The denominator then becomes {9 cosh2 u − 9} = (9sinh2 u) = 3sinh u In order to deal with the limits, note that when [) and when ( (so u = ln[4 + 15 ]) x = 4, cosh u = 2 so u = ln 2 + 3 x = 7, cosh u = 4 The integral then becomes cosh −1 4 3 2 ∫ cosh −1 2 sinh u du = 3sinh u = cosh −1 4 ∫ 1 du 2 cosh −1 2 [ { ( ( ) 1 1 cosh−1 4 − cosh−1 2 = ln 2 + 3 − ln 4 + 15 2 2 Example Evaluate ∫ (x 1 2 −3 ) + 6x + 13 d x leaving your answer in terms of natural logarithms. Solution Completing the square, x2 + 6x + 13 = ( x + 3) + 4 2 sinh 2 A + 1 = cosh 2 A this time You will need the identity because of the + sign after completing the square. Now make the substitution x + 3 = 2 sinh θ dx = 2 cosh θ dθ When x = −3, sinh θ = 0 ⇒ θ = 0 giving ( When x = 1, sinh θ = 2 ⇒ θ = sinh−1 2 = ln 2 + 5 The integral transforms to sinh −1 2 ∫ (4 sinh 2 ) θ + 4 .2 cosh θ dθ 0 sinh−1 2 = ∫ 4 cosh θ dθ 2 0 48 ) )} Chapter 2 Hyperbolic Functions You can either convert this into exponentials or you can use the identity cosh 2 A = 2 cosh2 A − 1 sinh −1 2 ∫ (2 + 2 cosh 2θ )d θ giving 0 sinh −1 2 = [2θ + sinh 2θ ]0 sinh −1 2 = 2[θ + sinh θ cosh θ ]0 ( = 2 sinh −1 2 + 2 1 + 2 2 ( ) ) = 2 ln 2 + 5 + 4 5 Activity 19 Show that d (sec φ tan φ ) = 2 sec3 φ − sec φ dφ and deduce that ∫ sec 3φ dφ = 1 sec φ tan φ + ln (sec φ + tan φ ) { } 2 Hence use the substitution x + 3 = 2 tan φ in the integral of the previous example and verify that you obtain the same answer. 49 Chapter 2 Hyperbolic Functions Exercise 2F Evaluate the integrals in Questions 1 to 12. 15. Evaluate each of the following integrals. 3 2 ⌠ 1. ⌡0 1 ( x2 + 1) 1 ⌠ 3. ⌡0 ⌠ 2. ⌡3 dx 1 ( x2 + 2 x + 5) dx 6. ∫1 ( x2 + 2 x + 2 ) d x ⌠ 8. ⌡1 5 ⌠ 9. ⌡4 0 ⌠ 11. ⌡−1 ( x2 − 4 ) 2 1 ( x2 + x ) ∫ (3x 5 1 ( 2 x2 + 4 x + 7) dx 12. 2 4 + 6 x − 7) ∫ (x 3 dx (b) 2 ) + 6x − 7 d x 2 2 dx 2 ( x2 + 2 x ) ⌠ 10. ⌡1 dx ( x2 − 9) 1 ( x2 + 6 x + 8) 1 (x 2 1 16. Transform the integral 3 2 x +1 dx ∫ ( x − 9) d x 4 2 0 2 7. 1 1 ⌠ 4. ⌡−1 ∫ (4 + x ) d x 2 5. 4 ⌠ (a) ⌡2 dx dx 1 ⌠ dx ⌡1 x 4 + x2 1 . Hence, by u means of a further substitution, or otherwise, evaluate the integral. by means of the substitution x = 17. The point P has coordinates ( a cosh t, bsinh t ). Show that P lies on the hyperbola with equation x 2 y2 − = 1. a2 b2 −12 x + 8) d x Which branch does it lie on when a>0? 13. Use the substitution u = e to evaluate x 1 ∫0 sech x d x . Given that O is the origin, and A is the point (a, 0), prove that the region bounded by the lines OA and OP and the arc AP of the hyperbola has area −1 14. (a) Differentiate sinh x with respect to x. 1 abt. 2 By writing sinh −1 x as 1 × sinh −1 x, use 2 integration by parts to find ∫ sinh −1 d x. 1 (b) Use integration by parts to evaluate −1 ∫1 cosh x d x . 2 2.11 Miscellaneous Exercises .11 Miscellaneous Exercises 1. The sketch below shows the curve with equation y = 3cosh x − x sinh x, which cuts the y-axis at the point A. Prove that, at A, y takes a minimum value and state this value. dy = 0 at B, show that the dx x-coordinate of B is the positive root of the equation Given that x cosh x − 2sinh x = 0. y Show that this root lies between 1.8 and 2. B A 0 50 y = 3cosh x – x sinh x y = 3 cosh x − x sinh x x Find, by integration, the area of the finite region bounded by the coordinate axes, the curve with equation y = 3cosh x − x sinh x and the line x = 2, giving your answer in terms of e. (AEB) Chapter 2 Hyperbolic Functions 2. Starting from the definition 1 1 cosh θ = ( eθ + e−θ ) and sinh θ = ( eθ − e−θ ) 2 2 show that 7. Define cosech x and coth x in terms of exponential functions and from your definitions prove that coth 2 x ≡ 1 + cosech 2 x. Solve the equation sinh( A + B) = sinh Acosh B + cosh Asinh B. 3coth 2 x + 4cosech x = 23 There exist real numbers r and α such that 8. Solve the equation 5cosh x + 13sinh x ≡ r sinh ( x + α ). 3sech2 x + 4 tanh x + 1 = 0 3 Find r and show that α = ln . 2 Hence, or otherwise, (a) solve the equation 5cosh x + 13sinh x = 12sinh 2 (b) show that 9. Find the area of the region R bounded by the curve with equation y = cosh x, the line x = ln 2 and the coordinate axes. Find also the volume obtained when R is rotated completely about the x-axis. (AEB) 10. Prove that 1 15e −10 dx 1 ⌠ = ln ⌡1 5cosh x + 13sinh x 12 3e + 2 3. Given that x < 1, prove that ( ) 1 d tanh −1 x = . 1 − x2 dx Show by integrating the result above that 1 1+ x ln 2 1− x ∫ tanh −1 Given that 2 cosh y − 7sinh x = 3 and cosh y − 3sinh2 x = 2 , find the real values of x and y in logarithmic form. (AEB) 3 x dx. (a) 1+ t (b) cosh x = 1− t2 2 ∫ (x 3 (b) 1 2 1 2 ) + 6x + 5 d x 12. Use the definitions in terms of exponential functions to prove that 1 x 2 5. Solve the equations 1 3 (a) cosh ( ln x ) − sinh ln x = 1 2 4 (a) 1 − tanh 2 x = sech2 x 1 + tanh 2 x (b) 2sinh x − cosh x = 2 tanh d ( tanh x ) = 1 − tanh2 x dx Hence, use the substitution t = tanh x to find (b) 4sinh x + 3e x + 3 = 0 cosh x + sinh x = e ∫ ( x + 6 x + 5) d x 1 Hence, or otherwise, solve the equation 6. Show that and write down a similar expression for cosh −1 x. 1 x, prove the identities 2 2t (a) sinh x = 1 − t2 2 11. Evaluate the following integrals Use integration by parts to find 4. Given that t ≡ tanh (1 + x ) sinh −1 x = ln x + (AEB) tanh −1 x = (AEB) ∫ sech2 x d x x n n n−k k Deduce that cosh n x + sinh n x = cosh x sinh x k k=0 ∑ Obtain a similar expression for cosh n x − sinh n x. Hence prove that cosh 7x = 64 cosh 7 x −112 cosh5 x + 56 cosh3 x − 7cosh x (AEB) 13. Sketch the graph of the curve with equation y = tanh x and state the equations of its asymptotes. Use your sketch to show that the equation tanh x = 10 − 3x has just one root α . Show that α lies between 3 and 3 1 . 3 Taking 3 as a first approximation for α , use the Newton-Raphson method once to obtain a second approximation, giving your answer to four decimal places. 51 Chapter 2 Hyperbolic Functions (1 + x2 ) . 14. Prove that sinh −1 x = ln x + 19. Solve the equation 3sech 2 x + 4 tanh x + 1 = 0 . (AEB) (a) Given that exp( z ) ≡ ez , show that ( ) y = exp sinh −1 x satisfies the differential equation cosh y + sinh y 2 y = ln cosh y − sinh y dy d (1 + x2 ) d xy + x d x − y = 0 2 2 1 (b) Find the value of ∫ sinh −1 x d x, leaving your 0 answer in terms of a natural logarithm. 15. Sketch the graph of y = tanh −1 x. Determine the value of x, in terms of e, for 1 which tanh −1 x = . 2 The point P is on the curve y = tanh −1 x where 1 . Find the equation of the tangent to the 2 curve at P. Determine where the tangent to the curve crosses the y-axis. y= 16. Evaluate the following integrals, giving your answers as multiples of π or in logarithmic form. 2 ⌠ (a) ⌡0 2 dx (3x2 − 6 x + 4 ) ⌠ (b) ⌡0 20. Define sinh y and cosh y in terms of exponential functions and show that dx (1 + 6 x − 3x2 ) 17. Find the value of x for which 3 4 + sinh −1 x = sinh −1 sinh 4 3 −1 18. Starting from the definitions of hyperbolic functions in terms of exponential functions, show that 1 By putting tanh y = , deduce that 3 1 1 tanh −1 = ln 2 3 2 (AEB) 21. Show that the function f ( x ) = (1 + x ) sinh ( 3x − 2 ) has a stationary value which occurs at the intersection of the curve y = tanh ( 3x − 2 ) and the straight line y + 3x + 3 = 0. Show that the stationary value occurs between –1 and 0. Use Newton-Raphson's method once, using an initial value of –1 to obtain an improved estimate of the x-value of the stationary point. sinh x , express the value cosh x of tanh x in terms of e x and e−x . 22. (a) Given that tanh x = (b) Given that tanh y = t, show that y = for − 1 < t < 1. (c) Given that y = tanh −1 (sin x ) show that dy = sec x and hence show that dx π 6 ∫ sec x tanh 0 2 (sin x ) d x = 1 3 − 2 + 1 ln 3 3 2 −1 cosh ( x − y ) = cosh x cosh y − sinh x sinh y and that (AEB) 23. Solve the simultaneous equations tanh −1 = cosh x − 3sinh y = 0 1 1+ x where −1 < x < 1. ln 2 1− x (a) Find the values of R and α such that 5cosh x − 4sinh x ≡ Rcosh ( x − α ) Hence write down the coordinates of the minimum point on the curve with equation y = 5cosh x − 4sinh x (b) Solve the equation 9sech y − 3tanh y = 7 , 2 leaving your answer in terms of natural logarithms. (AEB) 52 1 1+ t ln 2 1− t 2sinh x + 6 cosh y = 5 giving your answers in logarithmic form. (AEB) 24. Given that x = cosh y, show that the value of x is either ln x + ( x2 − 1) or ln x − Solve the equations (a) sinh2 θ − 5cosh θ + 7 = 0 (b) cosh ( z + ln 3) = 2 ( x2 − 1) Chapter 2 Hyperbolic Functions 25. Sketch the curve with equation y = sech x and determine the coordinates of its point of inflection. The region bounded by the curve, the coordinate axes and the line x = ln 3 is R. 28. Prove that 16sinh2 x cosh3 x ≡ cosh 5x + cosh 3x − 2 cosh x Hence, or otherwise, evaluate 1 ∫ 16sinh x cosh x d x, Calculate (a) the area of R (b) the volume generated when R is rotated through 2 π radians about the x-axis. 26. Solve the equation tanh x + 4sech x = 4. 27. Prove the identity 1 cosh x cos x − sinh x sin x ≡ (1 + cosh 2 x cos2 x ) 2 2 2 2 3 0 2 2 giving your answer in terms of e. 29. Show that the minimum value of sinh x + n cosh x is ( n2 −1) x= and that this occurs when 1 n − 1 ln 2 n + 1 Show also, by obtaining a quadratic equation in e x , that if k > ( n2 −1) then the equation sinh x + n cosh x = k has two real roots, giving your answers in terms of natural logarithms. (AEB) 53 Chapter 2 Hyperbolic Functions 54 ...
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This note was uploaded on 01/31/2012 for the course PHYS 418 taught by Professor Somalwar during the Spring '11 term at Rutgers.

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