Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 47 CHAPTER 2 HYPERBOLIC AND INVERSE HYPERBOLIC FUNCTION CHAPTER 2 Hyperbolic Function Inverse Hyperbolic Function Definition & Identities Identities Derivative Integration Definition Derivative Integration Similar to chapter 1 : Inverse trigonometric function Proving question • Using identities • Using basic definition (if mention) Solving hyperbolic equation • Using identities (square @ double angle) • Using basic definition (no square & same angle)
MAT238/ MAT438 : Foundation of Applied Mathematics 48 2.1 Hyperbolic Function 2.1.1 Definition of Hyperbolic Function (the difference between Hyperbolic and Trigonometric Function Hyperbolic Function Trigonometric Function sinh x = 2 x x e e − − cosh x = 2 x x e e − + tanh x = cosh x sinh x = x x x x e e e e − − + − sec hx = cosh x 1 = x x e e − + 2 cos echx = sinh x 1 = x x e e − − 2 coth x = tanh x 1 = sinh x cosh x = x x x x e e e e − − − + sin x = c a cos x = c b tan x = cos x sin x = b a sec x = cos x 1 = b c cosecx = sin x 1 = a c cot x = tan x 1 = sin x cos x = a b 2.1.2 Identities of Hyperbolic Function (the difference between Hyperbolic and Trigonometric Function Hyperbolic Function Trigonometric Function Basic Identities 1 2 2 cosh x − sinh x = tanh x sec h x 2 2 1 − = coth x cos ech x 2 2 − 1 = Sum and Difference Angle Identities sinh(A B) = sinh A coshB cosh A sinhB cosh(A B) = cosh A coshB sinh A sinhB ( ) tanh A tanhB tanh A tanhB tanh A B = 1 Basic Identities 1 2 2 cos x + sin x = tan x sec x 2 2 1 + = cot x cos ec x 2 2 + 1 = Sum and Difference Angle Identities sin (A B) = sin A cosB cos A sinB cos (A B) = cos A cosB sin A sinB ( ) tan A tanB tan A tanB tan A B 1 = c a b x ℎ 2 ℎ 2 − ℎ 2 ℎ 2 = 1 ℎ 2 ℎ 2 − 1 = ℎ 2 ℎ 2 ℎ 2 − ℎ 2 ℎ 2 = 1 ℎ 2 1 − ℎ 2 = ℎ 2
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 49 Hyperbolic Function Trigonometric Function Double-Angle Identities sinh 2A = 2sinh Acosh A i) cosh A cosh A sinh A 2 2 2 = + ii) 2 2 1 2 cosh A = cosh A − iii) 2 2 1 2 cosh A = sinh A + tanh A tanh A tanh A 2 1 2 2 + = Double-Angle Identities sin 2A = 2sin Acos A i) cos A cos A sin A 2 2 2 = − ii) 2 2 1 2 cos A = cos A − iii) cos A sin A 2 2 = 1 − 2 tan A tan A tan A 2 1 2 2 − = 2.1.3 Proving Hyperbolic Identity Example 1/ OCT 2004/ MAT238/ Q2a (3 marks) Show that coth2x − csch2x = 1 Solution: Types of Question involving Definitions and Identities of Hyperbolic Functions I) Proving Hyperbolic Identity II) Solving Hyperbolic Equation ℎ 2 − ℎ 2 = ℎ 2 ℎ 2 − 1 ℎ 2 = ℎ 2 − 1 ℎ 2 = ℎ 2 ℎ 2 = 1 # (ℎ)
MAT238/ MAT438 : Foundation of Applied Mathematics 50 Example 2/ APR 2006/ MAT238/ Q1c (4 marks) Verify the following hyperbolic identity = +1 − cosh x coth x csc hx sinh x Solution: Example 3/ SEP 2013/ MAT238/ Q2a (5 marks) Use the definitions of hyperbolic function to prove 1 2 2 cosh x − sinh x = . Solution: ℎ ℎ − ℎ = ℎ ℎ ℎ − 1 ℎ = ℎ ( ℎ − 1 ℎ ) = ℎ ÷ ℎ − 1 ℎ = ℎ ∙ ℎ ℎ − 1 = ℎ 2 ℎ − 1 = ℎ 2 − 1 ℎ − 1 = (ℎ + 1)(ℎ − 1) ℎ − 1 = ℎ + 1 # () ℎ 2 − ℎ 2 = ( + − 2 ) 2 − ( − − 2 ) 2 = ( + − ) 2 − ( − − ) 2 4 = ( 2 + 2 + −2 ) − ( 2− 2 + −2 ) 4 = 2 + 2 + −2− 2 + 2 − −2 4 = 4 4 = 1 # ()
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 51 Example 4/ NOV 2005/ MAT238/ Q2a (5 marks) Prove the following identity from the basic definitions of the hyperbolic functions. coth x cos ech x 2 2 − 1 = Solution: ℎ 2 − 1 = ( + − ) 2 ( − −) 2 − ( − − ) 2 ( − −) 2 = ( + − ) 2 − ( − − ) 2 ( − −) 2 = ( 2 + 2 + −2 ) − ( 2− 2 + −2 ) ( − −) 2 = 2 + 2 + −2− 2 + 2 − −2 ( − −) 2 = 4 ( − −) 2 = 2 2 ( − −) 2 = ( 2 − − ) 2 = ℎ 2 # ()
MAT238/ MAT438 : Foundation of Applied Mathematics 52 Example 5/ OCT 2012/ MAT238/ Q2a (5 marks) Use the definitions of hyperbolic function to show that cosh x cosh y − sinh x sinh y = cosh(x − y ) Solution: ℎ ℎ − ℎ ℎ = ( + − 2 )( + − 2 ) − ( − − 2 ) ( − − 2 ) = ( + − )( + −) 4 − ( − − )( − −) 4 = ( + − )( + −) − ( − − )( − −) 4 = ( + − + − + − −) − ( − − − − + − −) 4 = + − + − + − − − + − + − − − − 4 = − + − + − + − 4 = − + −+ + − + −+ 4 = 2 − + 2 −+ 4 = 2 − + 2 −(−) 4 = 2( − + −(−) ) 4 = − + −(−) 2 = ℎ( − ) # (ℎ)
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 53 Tutorial 2.1 : Proving Hyperbolic Identity 1. Use the definition of hyperbolic functions to prove the identity ℎ 2 (2) − ℎ 2 (2) = 1 2. Use the definition of hyperbolic functions to prove the identity ℎ 2 () − 1 = ℎ 2 () 3. Use the definition of hyperbolic functions to prove the identity 1 − ℎ 2 () = ℎ 2 () 4. Use the definition of hyperbolic functions to prove the identity 1 + ℎ 2 () = ℎ 2 () 5. Use the definition of hyperbolic functions to verify that 2ℎ 2 (2) + 1 = ℎ(4) 6. Use the definition of hyperbolic functions to show that ℎ 2 (3) − ℎ 2 (3) = 1 7. Use the definition of hyperbolic functions to show that ℎ 2 (2) + ℎ 2 (2) = 1 8. Use the definition of hyperbolic functions to verify that ℎ(4) = ℎ 2 (2) + ℎ 2 (2)
MAT238/ MAT438 : Foundation of Applied Mathematics 54 2.1.4 Solving Hyperbolic Equation 2.1.4.1 using Hyperbolic identity (with terms containing square @ double angle) Example 1/ APR 2006/ MAT238/ Q1b (6 marks) Solve the following hyperbolic equation 9 6 17 2 cosh x − sinh x = Ans : x = −0.6251, x = 1.0986 Solution: − = = + 9ℎ 2 − 6ℎ = 17 9(ℎ 2 + 1 ) − 6ℎ = 17 9( 2 + 1 ) − 6 = 17 9 2 + 9 − 6 = 17 9 2 − 6 − 8 = 0 (3 + 2)(3 − 4) = 0 = − 2 3 = 4 3 ℎ = − 2 3 ℎ = 4 3 = ℎ −1 (− 2 3 ) = ℎ −1 ( 4 3 ) = −0.6251 = 1.0986
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 55 Example 2/ OCT 2007/ MAT238/ Q2a (5 marks) Solve the following hyperbolic equation 2 ℎ 2 + ℎ = 3 Ans : x = 0 Solution: 2 ℎ 2 + ℎ = 3 2(2 ℎ 2 − 1) + ℎ = 3 2(2 2 − 1) + = 3 4 2 − 2 + = 3 4 2 + − 5 = 0 ( − 1)(4 + 5) = 0 − 1 = 0 4 + 5 = 0 = 1 = − 5 4 ℎ = 1 ℎ = − 5 4 = ℎ −1 (1) = ℎ −1 (− 5 4 ) = 0 = −
MAT238/ MAT438 : Foundation of Applied Mathematics 56 Example 3/ DEC 2008/ MAT238/ Q2a (9 marks) Solve the following hyperbolic equation 2 ℎ 2 + 2 ℎ 2 2 = 2 [Hint : Use 2sinh cosh2 1 2 x = x − ] Ans : x = 0 Solution: 2 ℎ 2 + 2 ℎ 2 2 = 2 ℎ 2 − 1 + 2 ℎ 2 2 = 2 − 1 + 2 2 = 2 2 2 + − 3 = 0 ( − 1)(2 + 3) = 0 − 1 = 0 = 1 ℎ 2 = 1 2 = ℎ −1 (1) = 1 2 ℎ −1 (1) = 0 = − 2 + 3 = 0 = − 3 2 ℎ 2 = − 3 2 2 = ℎ −1 (− 3 2 ) = 1 2 ℎ −1 (− 3 2 )
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 57 Example 4/ MAR 2012/ MAT238/ Q1c (5 marks) Using the identities of hyperbolic function, solve the following equation. sec h x sinh x 2 2 3 − 3 = Ans : x = 0 Solution: 3 − 3 = 3 ( ) − 3 = − 3 ( 1 ) − 3 = − 1 (× ) ∶ 3 − 3 = 2 − 2 + 2 − 3 = 0 ( − 1)( + 3) = 0 − 1 = 0 + 3 = 0 = 1 = −3 = 1 = −3 = √1 = 1 = √−3 = ℎ −1 (1) = 0 − = = − =
MAT238/ MAT438 : Foundation of Applied Mathematics 58 Example 5/ OCT 2012/ MAT238/ Q1b (6 marks) Solve the following hyperbolic equation. 2 ( 1) 10 10 ( 1) 2 cosh x − + = sinh x − Ans : x = 2.4436, x = 2.8184 Solution: 2 ( − ) + 10 = 10 ℎ( − 1) 2( + ( − )) + 10 = 10 ℎ( − 1) 2(1 + 2 ) + 10 = 10 2 + 2 2 + 10 = 10 2 2 − 10 + 12 = 0 (÷ 2) 2 − 5 + 6 = 0 ( − 2)( − 3) = 0 − 2 = 0 − 3 = 0 = 2 = 3 ℎ( − 1) = 2 ℎ( − 1) = 3 − 1 = ℎ −1 (2) − 1 = ℎ −1 (3) = 1 + ℎ −1 (2) = 1 + ℎ −1 (3) = 2.4436 = 2.8184 − = = + → ( − ) = + ( − )
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 59 Example 6/ OCT 2009/ MAT238/ Q1b (7 marks) Use an appropriate identity to solve 3 1 2 sinh x − cosh x = Ans : x = 0.7954 Solution: 3ℎ 2 − ℎ = 1 3(ℎ 2 − 1) − ℎ = 1 3( 2 − 1) − = 1 3 2 − 3 − = 1 3 2 − − 4 = 0 3 2 − − 4 = 0 ( + 1)(3 − 4) = 0 = −1 = 4 3 ℎ = −1 ℎ = 4 3 = ℎ −1 (−1) = ℎ −1 ( 4 3 ) = 0.7954 − = = −
MAT238/ MAT438 : Foundation of Applied Mathematics 60 2.1.4.2 using the definitions of hyperbolic function (with terms not containing square and same angle) Example 1/ APR 2007/ MAT238/ Q1d (7 marks) Solve the following equation by using the definition of hyperbolic functions. 3cosh x − 2sinh x − 3 = 0 Ans : x = 0, x = ln 5 Solution: Multiplying each term by 2 (to eliminate denominator 2) 3 ( + − 2 ) (2) − 2 ( − − 2 ) (2) − 3(2) = 0(2) 3( + − ) − 2( − − ) − 6 = 0 expand Multiplying each term by u (to eliminate denominator u) () + 5 () − 6() = 0() 2 + 5 − 6 = 0 2 − 6 + 5 = 0 ( − 1)( − 5) = 0 = 1 = 5 = 1 = 5 = (1) = (5) = 0 = 1.6094 3 + 3 −− 2 + 2 −− 6 = 0 + 5 −− 6 = 0 + 5 − 6 = 0 + 5 − 6 = 0 3 ℎ − 2 ℎ − 3 = 0 3 ( + − 2 ) − 2 ( − − 2 ) − 3 = 0
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 61 Example 2/ MAY 2011/ MAT238/ Q1b (7 marks) Use the definition of hyperbolic function to solve 4sinh2x = 2cosh 2x −1 Ans : = 0.13224 Solution: Simplify expand Multiplying each term by u (to eliminate denominator u) () − 3 () + 1() = 0() 2 − 3 + = 0 2 + − 3 = 0 = 1.30277 = −2.30277 2 = 1.30277 2 = −2.30277 2 = (1.30277) 2 = (−2.30277) = 0.13224 2 2− 2 −2 = 2 + −2− 1 2 2− 2− 2 −2− −2 + 1 = 0 2− 3 −2 + 1 = 0 2− 3 2 + 1 = 0 − 3 + 1 = 0 4 ℎ 2 = 2 ℎ 2 − 1 4 ( 2− −2 2 ) = 2 ( 2 + −2 2 ) − 1 2( 2− −2 ) = 2 + −2− 1
MAT238/ MAT438 : Foundation of Applied Mathematics 62 Example 3/ APR 2011/ MAT238/ Q1b (7 marks) Use the definition of hyperbolic function to solve 2 2 2 2 4 2 − + = − x cosh x sinh x e Ans : x = 0, x = 0.973 Solution: Multiplying each term by 2 (to eliminate denominator 2) expand simplify 2 ℎ 2 − ℎ 2 + 2 −2 = 4 2 ( 2 + −2 2 ) − ( 2− −2 2 ) + 2 −2 = 4 2 2 + 2 −2− 2 + −2 + 4 −2 = 8 2 2 + 2 −2 − 2 + −2 + 4 −2 } = 8 2 • 2 ( 2 + −2 2 ) − 2 • ( 2− −2 2 ) + 2 • 2 −2 = 2 • 4 2( 2 + −2 ) − ( 2− −2 ) + 4 −2 = 8 2 + 7 −2 = 8 2 + 7 2= 8 + 7 = 8 (× ) ∶ 2 + 7 = 8 2 − 8 + 7 = 0 ( − 1)( − 7) = 0 = 1 = 7 2 = 1 2 = 7 2 = (1) 2 = (7) 2 = 0 = 1 2 (7) = 0 = 0.973
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 63 Example 4/ OCT 2008/ MAT238/ Q2a (6 marks) Use the definition of hyperbolic function to solve y sinh x cosh x e 1 2 4 − = in terms of x. Ans: = 1 2 [1 − ( 2− −2 )] Solution: Take ln on both sides Take ln on both sides 4 ℎ ℎ = 1−2 4 ( − − 2 ) ( + − 2 ) = 1−2 4 • ( − − )( + − ) 4 = 1−2 ( − − )( + − ) = 1−2 ( ) 2 − ( − ) 2 = 1−2 2− −2 = 1−2 2− −2 = 2 2 = 2 − −2 ( 2) = ( 2 − −2 ) 2 = − ( 2− −2 ) 2 = 1 − ( 2− −2 ) = 1 2 [1 − ( 2− −2 )] 4 ℎ ℎ = 1−2 2 • 2 ℎ ℎ = 1−2 2 • ℎ 2 = 1−2 2 ( 2− −2 2 ) = 1−2 2− −2 = 1−2 2− −2 = 2 2 = 2 − −2 ( 2) = ( 2 − −2 ) 2 = − ( 2− −2 ) 2 = 1 − ( 2− −2 ) = 1 2 [1 − ( 2− −2 )] Double angle identity : ℎ 2 = 2 ℎ ℎ = 1 ( + )( − ) = − ( ) =
MAT238/ MAT438 : Foundation of Applied Mathematics 64 More Examples Example 5/ OCT 2006/ MAT238/ Q2c (8 marks) Show that cosh2 2cosh 1 2 x = x − . Hence, solve cosh2x + 4 = 5cosh x . Ans : x = 0, x = 0.9624 Solution: = 2 ℎ Hence, solve 2 − 1 = 2 ( + − 2 ) 2 − 1 = 2 ∙ ( + − ) 2 4 − 1 = ( + − ) 2 2 − 2 2 = ( + − ) 2 − 2 2 = ( ) 2 + 2 − + ( − ) 2 − 2 2 = 2 + 2 + −2− 2 2 = 2 + −2 2 = ℎ 2 = (ℎ) + 4 = 5 ℎ ( − ) + 4 = 5 ℎ (2 2 − 1) + 4 = 5 2 2 − 5 + 3 = 0 ( − 1)(2 − 3) = 0 − 1 = 0 = 1 ℎ = 1 = ℎ −1 (1) = 0 2 − 3 = 0 = 3 2 ℎ = 3 2 = ℎ −1 ( 3 2 ) = 0.9624 Related formula From ( + ) = + + ( + − ) 2 = ( ) 2 + 2 − + ( − ) 2 From • = + • − = − = 0 = 1 From ( ) = ( ) 2 = 2 and ( − ) 2 = −2
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 65 Example 6/ APR 2008/ MAT238/ Q2b (7 marks) Use the definition of hyperbolic functions to show that 1 tanh 2x sec h 2x 2 2 − = . Hence, solve 3 2 2 2 2 2 tanh x + sec h x = . Ans : x = 0.4407, x = −0.4407 Solution: Hence, solve = 1 − ℎ 2 2 = 1 − ( 2− −2 2 + −2 ) 2 = ( 2 + −2 ) 2 ( 2 + −2) 2 − ( 2− −2 ) 2 ( 2 + −2) 2 = ( 2 + −2 ) 2 − ( 2− −2 ) 2 ( 2 + −2) 2 = ( 4 + 2 + −4 ) − ( 4− 2 + −4 ) ( 2 + −2) 2 = 4 + 2 + −4− 4 + 2 − −4 ( 2 + −2) 2 = 4 ( 2 + −2) 2 = 2 2 ( 2 + −2) 2 = ( 2 2 + −2 ) 2 = ( 1 ℎ 2 ) 2 = ℎ 2 2 = (ℎ) 3 ℎ 2 2 + = 2 3 ℎ 2 2 + − = 2 3 2 + 1 − 2 = 2 2 2 = 1 2 = 1 2 = ±√ 1 2 = 1 √2 ℎ 2 = 1 √2 2 = ℎ −1 ( 1 √2 ) = 1 2 ℎ −1 ( 1 √2 ) = 0.4407 = − 1 √2 ℎ 2 = − 1 √2 2 = ℎ −1 (− 1 √2 ) = 1 2 ℎ −1 (− 1 √2 ) = −0.4407 Related formula From ( + ) = + + ( + − ) 2 = ( ) 2 + 2 − + ( − ) 2 From • = + • − = − = 0 = 1 From ( ) = ( ) 2 = 2 and ( − ) 2 = −2
MAT238/ MAT438 : Foundation of Applied Mathematics 66 Tutorial 2.1a : Solving the hyperbolic equation using Identities of Hyperbolic Functions 1. Use the hyperbolic identity ℎ(2) = 2 ℎ 2 () − 1 to solve 4 ℎ(2) − 22 ℎ() + 19 = 0 2. Use the hyperbolic identity ℎ 2 () − 1 = ℎ 2 () to solve 2 ℎ 2 () + ℎ() − 8 = 0 3. Use a suitable hyperbolic identity to solve 3ℎ 2 () + ℎ 2 () = 2. 4. Use the hyperbolic identity ℎ 2 () − ℎ 2 () = 1 to solve 2ℎ 2 () = 5ℎ() − 4 5. Use the hyperbolic identity ℎ 2 () − ℎ 2 () = 1 to solve ℎ 2 () − ℎ() = 5 6. Solve the hyperbolic equation 2 ℎ 2 (2) − 5 = 5 ℎ(2) to solve [ ∶ ℎ 2 (2) − ℎ 2 (2) = 1] 7. Use a suitable hyperbolic equation to solve 3 ℎ 2 () + ℎ() − 3 = 0. Tutorial 2.1b : Solving the hyperbolic equation using Definition of Hyperbolic Functions (Exponential Function) Solve the hyperbolic equation : Answers (Tutorial 2.1b) 8. = 0, = −(7) 9. = (3) 10. = 0.2554 11. = 0.5493, = 0.4581 8. 3ℎ + 4ℎ = 4 9. 7ℎ − 5ℎ = 1 10. 13ℎ2 − 7ℎ2 + 1 = 0 11. 4ℎ2 + 13 −2 = 11 Answers (Tutorial 2.1a) 1. = 0.9624 , = 0.6931 2. = −0.481 , = 0.625 3. = 0.8814 , = −0.8814 4. = 1.317 5. = 1.7627 6. = 0.9092, = −0.2406 7. = 0, = −0.327
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 67 2.1.5 Derivatives of Hyperbolic Function (the difference between Hyperbolic and Trigonometric Function) Hyperbolic Function Trigonometric Function ( ) dx d sinh x = cosh x ( ) dx d cosh x = sinh x ( ) dx d tanh x = sech x 2 ( ) dx d sec hx = − sec hx tanh x ( ) dx d cos echx = − cos echx coth x ( ) dx d coth x = cos ech x 2 − ( ) cos x dx d sin x = ( ) sin x dx d cos x = − ( ) sec x dx d tan x 2 = ( ) sec x tan x dx d sec x = ( ) cosecx cot x dx d cosecx = − ( ) cosec x dx d cot x 2 = − Example 1/ MAY 2011/ MAT238/ Q2a (6 marks) Differentiate the following implicit function with respect to x. y cosh y tan x e 2 1 3 = − − . Solution: ℎ 2 = −1√ − 3 Differentiate implicitly with respect to x, ℎ 2 ∙ 2 = 1 1 + (√) 2 ∙ 1 2 − 1 2 − 3 ∙ 3 2 ℎ 2 = 1 1 + ∙ 1 2√ − 3 3 2 ℎ 2 + 3 3 = 1 2√(1 + ) (2 ℎ 2 + 3 3) = 1 2√(1 + ) = 1 2√(1 + )(2 ℎ 2 + 3 3) #
MAT238/ MAT438 : Foundation of Applied Mathematics 68 Example 2/ OCT 2007/ MAT238/ Q2b (8 marks) Use logarithmic differentiation to find the derivative ( ) cosh x x y sinh x 2 2 2 + = 2 Ans : = 1 ( 2 + 2 )[ℎ 2 ( ℎ 2 ) + ℎ 2 ℎ 2 ] − Solution: 2 ( + ) 2 + 2 = 2 ℎ 2 (ℎ 2) + 2 ℎ 2 ℎ 2 Dividing both sides by 2, + 2 + 2 = ℎ 2 (ℎ 2) + ℎ 2 ℎ 2 + = ( 2 + 2 )(ℎ 2 (ℎ 2) + ℎ 2 ℎ 2) = ( 2 + 2 )(ℎ 2 (ℎ 2) + ℎ 2 ℎ 2) − = ( 2 + 2 )(ℎ 2 (ℎ 2) + ℎ 2 ℎ 2) − # 2 + 2 = (ℎ 2) ℎ 2 Take ln on both sides ( 2 + 2 ) = (ℎ 2) ℎ 2 ( 2 + 2 ) = ℎ 2 (ℎ 2) Differentiate both sides implicitly with respect to x 1 2 + 2 ∙ (2 + 2 ) = ′ + ′ 2 + 2 2 + 2 = (ℎ 2) ∙ 2 ℎ 2 + ℎ 2 ∙ 1 ℎ 2 ∙ ℎ 2 ∙ 2
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 69 Example 3/ JUN 2008/ MAT238/ Q1b (5 marks) Find dx dy if tanh (ln y ) e sin ( x ) x 2 2 3 −1 = + . Ans : x(1 4x) sech (ln y) 6 y e x(1 4x) y 2 3x − − + Solution: ℎ( ) = 2 3 + −1 (2√) Differentiate both sides implicitly with respect to x ℎ 2 ( ) ∙ 1 ∙ = 2 3 ∙ 3 + 1 √1 − (2√) 2 ∙ 2 ∙ 1 2 − 1 2 ℎ 2 ( ) = 6 3 + 1 √1 − 4 2 ∙ 1 √ ℎ 2 ( ) = 6 3 + 1 √(1 − 4 2) Multiplying both sides by √(1 − 4 2) to eliminate denominator and √(1 − 4 2) √(1 − 4 2) ∙ ℎ 2 ( ) = 6 3 ∙ √(1 − 4 2) + 1 √(1 − 4 2) ∙ √(1 − 4 2) √(1 − 4 2) ℎ 2 ( ) = 6√(1 − 4 2) 3 + = 6√(1 − 4 2) 3 + √(1 − 4 2) ℎ 2( ) #
MAT238/ MAT438 : Foundation of Applied Mathematics 70 Example 4/ OCT 2008/ MAT238/ Q1b (7 marks) Differentiate xy ( x) e x x sinh 4 4 sin 2 3 − −1 = + with respect to x. Solution: Differentiate : ′ = ′ + ′ − − − − − − ① Where, By substituting A’, B, and C’ into ①, + = 12 ℎ 2 4 ℎ 4 − 4 − −1 2 + 8 − √1 − 4 2 Separating the terms containing not containing = 12 ℎ 2 4 ℎ 4 − 4 − −1 2 + 8 − √1 − 4 2 − write as a subject = 1 (12 ℎ 2 4 ℎ 4 − 4 − −1 2 + 8 − √1 − 4 2 − ) # = = = ′ = 1 ′ = Using product rule, ′ = ′+ ′ ′ = + = (ℎ 4 ) 3 Using Generalized Power rule, ′ = 3(ℎ 4 ) 2 ∙ ℎ 4 ∙ 4 ′ = 12 ℎ 2 4 ℎ 4 = 4 − −1 2 = 4 − = −1 2 ′ = 4 − ∙ −1 ′ = 1 √1− (2) 2 ∙ 2 ′ = −4 − ′ = 2 √1− 4 2 Using product rule, ′ = ′+ ′ ′ = −4 − −1 2 + 8 − √1− 4 2 = (ℎ 4 ) 3 + 4 − −1 2 = +
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 71 Example 5/ OCT 2009/ MAT238/ Q1c (7 marks) Differentiate ( ) y sinh x x tan y e 3 1 2 2 + = − with respect to x. Solution: ℎ 32 + (−1) = 2 (ℎ 2) 3 + (−1) = 2 Differentiate both sides implicitly with respect to x, 3(ℎ 2) 2 ℎ 2 ∙ 2 + ′ + ′ = 2 ∙ 2 6ℎ 22 ℎ 2 + (−1) ∙ 1 + ∙ 1 1 + 2 = 2 2 6ℎ 22 ℎ 2 + (−1) + 1 + 2 = 2 2 1 + 2 − 2 2 = −6ℎ 22 ℎ 2 − (−1) ( 1 + 2 − 2 2) = −6ℎ 22 ℎ 2 − (−1) = −6ℎ 22 ℎ 2 − (−1) 1 + 2 − 2 2 #
MAT238/ MAT438 : Foundation of Applied Mathematics 72 Example 6/ APR 2011/ MAT238/ Q1c (5 marks) Differentiate xy sin x y cosh 2 2 2 3 −1 + = with respect to x. Solution: 3 ℎ ( 2 ) + = 2−12 Differentiate both sides implicitly with respect to x, 3 ℎ ( 2 ) ∙ 1 2 + ′ + ′ = 2 ∙ 1 √1 − (2) 2 ∙ 2 3 2 ℎ ( 2 ) + + = 4 √1 − 4 2 Multiplying each term by (2√1 − 4 2) to eliminate denominator 2 and √1 − 4 2 (2√1 − 4 2) 3 ℎ ( 2 ) ∙ 1 2 + (2√1 − 4 2) + (2√1 − 4 2) = 4 √1 − 4 2 (2√1 − 4 2) 3√1 − 4 2 ℎ ( 2 ) + 2√1 − 4 2 + 2√1 − 4 2 = 8 3√1 − 4 2 ℎ ( 2 ) + 2√1 − 4 2 = 8 − 2√1 − 4 2 (3√1 − 4 2 ℎ ( 2 ) + 2√1 − 4 2) = 8 − 2√1 − 4 2 = 8 − 2√1 − 4 2 3√1 − 4 2 ℎ ( 2 ) + 2√1 − 4 2 #
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 73 Example 7/ APR 2006/ MAT238/ Q2b (5 marks) Differentiate the implicit function with respect to x. y = ln (x cosh y ) 2 Solution: 2 = ( ℎ ) Using the properties of logarithm, 2 = + (ℎ ) Differentiate implicitly with respect to x, 2 = 1 + 1 ℎ ∙ ℎ 2 = 1 + ℎ 2 − ℎ = 1 (2 − ℎ ) = 1 = 1 (2 − ℎ ) #
MAT238/ MAT438 : Foundation of Applied Mathematics 74 Example 8/ APR 2008/ MAT238/ Q2a (4 marks) Use the hyperbolic identity to show that ( cosh x sinh x) sinh x dx d 2 2 2 2 4 2 2 − = . Solution: = 2 ℎ 22 − ℎ 22 = 2 (ℎ 2) 2 − (ℎ 2) 2 Differentiate with respect to x, = 4 (ℎ 2)∙ ℎ 2 ∙ 2 − 2(ℎ 2)∙ ℎ 2 ∙ 2 = 8 ℎ 2 ℎ 2 − 4 ℎ 2 ℎ 2 = 4 ℎ 2 ℎ 2 = 2 ∙ 2 ℎ 2 ℎ 2 = 2 ℎ 4 ℎ, (2 ℎ 22 − ℎ 22) = 2 ℎ 4 # (ℎ)
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 75 Example 9/ OCT 2008/ MAT238/ Q2b (7 marks) Use a suitable hyperbolic identity to show that (cosh x sinh x) cosh x sinh x dx d 2 6 2 2 + = . Hence, find the value of x such that (cosh x sinh x) cosh x dx d + = 2 2 2 . Solution: = ℎ 2 + 2ℎ 2 = (ℎ ) 2 + 2(ℎ ) 2 Differentiate with respect to x, = 2 (ℎ )∙ ℎ + 4(ℎ )∙ ℎ = 2 ℎ ℎ + 4 ℎ ℎ = 6 ℎ ℎ (ℎ 2 + 2ℎ 2) = 6 ℎ ℎ # (ℎ) Hence, (ℎ 2 + 2ℎ 2) = ℎ 6 ℎ ℎ = ℎ 6 ℎ ℎ − ℎ = 0 ℎ (6 ℎ − 1) = 0 ℎ = 0 6 ℎ − 1 = 0 = ℎ −1 (0) 6 ℎ = 1 ℎ = 1 6 = ℎ −1 ( 1 6 ) = 0.1659
MAT238/ MAT438 : Foundation of Applied Mathematics 76 Example 10/ MAR 2014/ MAT238/ Q1b (6 marks) Find dx dy for x x e y x 3 tanh sin 2 1 2 2 = + − Solution : 2 = −1 2 ℎ 2 + 3 Differentiate implicitly with respect to x, 2 = ′ − ′ 2 + 3 2 = ℎ 2 ∙ 1 √1 − ( 2) 2 ∙ 2 ∙ 2 − −1 2 ∙ ℎ 2 2 ∙ 2 (ℎ 2) 2 + 3 2 = 2 2 ℎ 2 √1 − 4 − 2ℎ 2 2 −1 2 ℎ 2 2 + 3 = 2 2 ℎ 2 √1 − 4 − 2ℎ 2 2 −1 2 2 ℎ 2 2 + 3 2 #
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 77 TUTORIAL 2.2 : DIFFRERENTIATION OF HYPERBOLIC FUNCTIONS 1. Find dx dy if tanh (ln y ) e sin ( x ) x 2 2 3 −1 = + . ∶ 6 3√(1 − 4) + √(1 − 4) ℎ 2 ( ) 2. Differentiate xy ( x) e x x sinh 4 4 sin 2 3 − −1 = + with respect to x. ∶ 12 ℎ 2 4 ℎ 4 − 4 − −1 2 + 8 − √1 − 4 2 − 3. Differentiate ( ) y sinh x x tan y e 3 1 2 2 + = − with respect to x. ∶ = −6ℎ 22 ℎ 2 − (−1) 1 + 2 − 2 2 4. Differentiate xy sin x y cosh 2 2 2 3 −1 + = with respect to x. ∶ = 8 − 2√1 − 4 2 3√1 − 4 2 ℎ ( 2 ) + 2√1 − 4 2 5. Differentiate the implicit function with respect to x. y cosh y tan x e 2 1 3 = − − . ∶ = 1 2√(1 + )(2 ℎ 2 + 3 3) 6. Differentiate the implicit function with respect to x. y = ln (x cosh y ) 2 ∶ = 1 (2 − ℎ ) 7. Use the hyperbolic identity to show that ( cosh x sinh x) sinh x dx d 2 2 2 2 4 2 2 − = . 8. Find dx dy for x x e y x 3 tanh sin 2 1 2 2 = + − ∶ = 2 2 ℎ 2 √1 − 4 − 2ℎ 2 2 −1 2 2 ℎ 2 2 + 3 2
MAT238/ MAT438 : Foundation of Applied Mathematics 78 2.1.6 Integration of Hyperbolic Function (the difference between Hyperbolic and Trigonometric Functions) Hyperbolic Function Trigonometric Function cosh x dx = sinh x + c sinh x dx = cosh x + c sec h x dx = tanh x + c 2 sec h x tanh x dx = − sec hx + c cos ec hx coth x dx = − cos ech x + c cos ech x dx = − coth x + c 2 cos x dx = sin x + c sin x dx = − cos x + c sec x dx = tan x + c 2 sec x tan x dx = sec x + c cosec x cot x dx = − cosec x + c cos ec x dx = − cot x + c 2 Example 1/ MAR 2005/ MAT238/ Q2b(iii) (5 marks) Evaluate 2sinh 4x cosh 4x dx . Solution : ∫ 2 ℎ 4 ℎ 4 = ∫ 2 ℎ 4 ℎ 4 ∙ 4 ℎ 4 = 1 2 ∫ ℎ 4 = 1 2 ∫ = 1 2 ( 2 2 ) + = 1 4 ℎ 24 + # = ℎ 4 = 4 ℎ 4 = 4 ℎ 4 ∫ 2 ℎ 4 ℎ 4 = ∫ 2 ℎ 4 ℎ 4 ∙ 4 ℎ 4 = 1 2 ∫ ℎ 4 = 1 2 ∫ = 1 2 ( 2 2 ) + = 1 4 ℎ 24 + # = ℎ 4 = 4 ℎ 4 = 4 ℎ 4
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 79 Example 2/ APR 2006/ MAT238/ Q1d (3½ marks) Evaluate ( ) ( ) 2 1 1 x csc h coth dx x x . Solution : Example 3/ Page 111/ MAR 2013/ MAT238/ Q2c (i) ( marks) Evaluate ( ) x cosh x dx 2 10 . Solution : ∫ ℎ ( 1 ) ℎ ( 1 ) 2 = ∫ ℎ ℎ 2 ∙ − 2 = − ∫ ℎ ℎ = −(− ℎ ) + = ℎ ( 1 ) + # = 1 = −1 = − −2 = − 1 2 = − 2 ∫ ℎ(10 2 ) = ∫ ℎ ∙ 20 = ∫ ℎ ∙ 20 = 1 20 ∫ ℎ = 1 20 ℎ + = 1 20 ℎ(10 2 ) + # = 10 2 = 20 = 20 ∫ 2 ℎ 4 ℎ 4 = ∫ ℎ 8 = 1 8 ℎ 8 + # From formula no in page 3, chapter2, ℎ 2 = 2 ℎ ℎ Therefore, (if = 4) ℎ 2(4) = 2 ℎ4 ℎ4 ℎ 8 = 2 ℎ4 ℎ4 In appendix (Final Exam paper, formula no. 8) ∫ ℎ = 1 ℎ +
MAT238/ MAT438 : Foundation of Applied Mathematics 80 Example 4/ MAY 2013/ MAT238/ Q2a (5 marks) Evaluate 3 2 3 ln ln tanh x sec h x dx . Solution : ∫ ℎ ℎ 3 = ∫ ℎ ℎ ∙ 1 ℎ 3 = ∫ ℎ ℎ 4 = ∫ ℎ 4 ∙ ℎ = ∫ 1 4 = ∫ −4 = −3 −3 + = − 1 3 3 + = − 1 3 ℎ 3 + Thus, ∫ ℎ ℎ 3 3 2 = [− 1 3 ℎ 3 ] 2 3 = (− 1 3 ℎ 3(3) ) − (− 1 3 ℎ 3(2) ) = (− 1 3 ℎ 3(3) ) − (− 1 3 ℎ 3(2) ) = 37 375 = 0.0987 # = ℎ = ℎ = ℎ
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 81 Example 5/ Page 119/ SEP 2013/ MAT238/ Q2c(i) Evaluate dt t sec h t tanh t . Solution : Example 6/ APR 2011/ MAT238/ Q2b (4 marks) Evaluate dx sinh x cosh x 3 − 2 2 . Solution : ∫ ℎ √ ℎ √ √ = ∫ ℎ ℎ √ ∙ 2√ = 2 ∫ ℎ ℎ = 2(−ℎ ) + = −2 ℎ √ + # = √ = 1 2 = 1 2 − 1 2 = 1 2√ = 2√ ∫ ℎ 2 3 − ℎ 2 = ∫ ℎ 2 ∙ −2ℎ 2 = ∫ 1 ∙ −2 = − 1 2 ∫ 1 = − 1 2 || + = − 1 2 |3 − ℎ 2| + # = 3 − ℎ 2 = −2ℎ 2 = −2ℎ 2
MAT238/ MAT438 : Foundation of Applied Mathematics 82 Example 7/ OCT 2012/ MAT238/ Q2c(i) (4 marks) Evaluate dx sec h x sec h x tanh x 25 − 5 5 5 2 . Solution : ∫ ℎ5 ℎ5 25 − ℎ 25 = ∫ ℎ5 ℎ5 (5) 2 − (ℎ5) 2 = ∫ ℎ5 ℎ5 2 − 2 ∙ −5ℎ5 ℎ5 = ∫ 1 2 − 2 ∙ −5 = − 1 5 ∫ 1 2 − 2 = − 1 5 ∙ 1 ℎ −1 ( ) + = − 1 5 ∙ 1 5 ℎ −1 ( ℎ5 5 ) + = − 1 25 ℎ −1 ( ℎ5 5 ) + # = 5 = ℎ5 = −5ℎ5 ℎ5 = −5ℎ5 ℎ5 ∫ ℎ5 ℎ5 25 − ℎ 25 = ∫ ℎ5 ℎ5 (5) 2 − (ℎ5) 2 = ∫ ℎ5 ℎ5 2 − 2 ∙ −5ℎ5 ℎ5 = ∫ 1 2 − 2 ∙ −5 = − 1 5 ∫ 1 2 − 2 = − 1 5 ∙ 1 2 | + − | + = − 1 5 ∙ 1 2(5) | ℎ5 + 5 ℎ5 − 5 | + = − 1 50 | ℎ5 + 5 ℎ5 − 5 | + #
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 83 Example 8/ MAY 2011/ MAT238/ Q2b(i) (5 marks) Evaluate ( ) dx tanh x sec h x + 2 0 2 4 2 2 . Solution : ∫ sech22 (4 + tanh2) = ∫ sech22 ∙ 2ℎ 22 = ∫ 1 ∙ 2 = 1 2 ∫ 1 = 1 2 || + = 1 2 |4 + ℎ2| + # Hence, ∫ sech22 (4 + tanh2) 2 0 = [ 1 2 |4 + ℎ2|] 0 2 = ( 1 2 |4 + ℎ(4)|) − ( 1 2 |4 + ℎ(0)|) = 0.1115 # = 4 + ℎ2 = 2ℎ 22 = 2ℎ 22
MAT238/ MAT438 : Foundation of Applied Mathematics 84 Example 9/ APR 2011/ MAT238/ Q2c (6 marks) Use the hyperbolic identity to prove that cosh x sinh x cosh x sinh x = + − 1 . Hence, evaluate dx cosh x − sinh x 1 0 4 . Solution : Hyperbolic identity : ℎ 2 − ℎ 2 = 1 Which is equal to 1, because the numerator exactly equals to the denominator We have proven that 1 ℎ − ℎ = ℎ + ℎ 1 ℎ − ℎ = 1 ℎ − ℎ ∙ + + = ℎ + ℎ (ℎ − ℎ)(ℎ + ℎ) = ℎ + ℎ ℎ 2 − ℎ 2 = ℎ + ℎ 1 = ℎ + ℎ # () Hence, ∫ 4 ℎ − ℎ 1 0 = 4 ∫ ( 1 ℎ − ℎ ) 1 0 = 4 ∫(ℎ + ℎ) 1 0 = 4[ℎ + cosℎ]0 1 = 4 ((ℎ(1) + cosℎ(1)) − (ℎ(0) + cosℎ(0))) = 4( − 1) = 6.873 #
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 85 Example 10/ MAR 2012/ MAT238/ Q2c (6 marks) Using the definition of hyperbolic function, show that 2 2 1 2 cosh x = sinh x + . Hence, evaluate sinh x dx 2 . Solution : ∫ = + = 2ℎ 2 + 1 = 2 ( − − 2 ) 2 + 1 = 2 ∙ ( − − ) 2 2 2 + 1 = 2 ∙ ( 2− 2 + −2 ) 4 + 1 = ( 2−2 + −2 ) 2 + 2 2 = 2− 2+ −2 + 2 2 = 2 + −2 2 = ℎ2 = ℎ2 = 2ℎ 2 + 1 ℎ2 − 1 = 2ℎ 2 ℎ2 −1 2 = ℎ 2 ℎ 2 = 1 2 (ℎ2 − 1) Hence, ∫ ℎ 2 = ∫ 1 2 (ℎ2 − 1) = 1 2 ∫(ℎ2 − 1) = 1 2 ( 1 2 ℎ2 −) + = 1 4 ℎ2 − 1 2 + #
MAT238/ MAT438 : Foundation of Applied Mathematics 86 TUTORIAL 2.3 : INTEGRATION OF HYPERBOLIC FUNCTIONS 1. Evaluate ( ) ( ) 2 1 1 x csc h coth dx x x . ∶ ℎ ( 1 ) + 2. Evaluate ( ) x cosh x dx 2 10 . ∶ 1 20 ℎ + = 1 20 ℎ(10 2 ) + 3. Evaluate 3 2 3 ln ln tanh x sec h x dx . Ans : 0.0987 4. Evaluate dt t sec h t tanh t . ∶ −2 ℎ √ + 5. Evaluate dx sinh x cosh x 3 − 2 2 . ∶ − 1 2 |3 − ℎ 2| + 6. Evaluate dx sec h x sec h x tanh x 25 − 5 5 5 2 . ∶ − 1 25 ℎ −1 ( ℎ5 5 ) + − 1 50 | ℎ5 + 5 ℎ5 − 5 | + 7. Evaluate ( ) dx tanh x sec h x + 2 0 2 4 2 2 . ∶ 1 2 |4 + ℎ2| + , 0.1115 8. Use the hyperbolic identity to prove that cosh x sinh x cosh x sinh x = + − 1 . Hence, evaluate dx cosh x − sinh x 1 0 4 . ∶ 6.873 9. Using the definition of hyperbolic function, show that 2 2 1 2 cosh x = sinh x + . Hence, evaluate sinh x dx 2 . ∶ 1 4 ℎ2 − 1 2 +
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 87 2.2 Inverse Hyperbolic Function 2.2.1 Notation • If sinh x = A, then x = sinh−1A • If cosh x = A, then x = cosh−1A • If tanh x = A, then x = tanh−1A • If sech x = A, then x = sech−1A • If cosech x = A, then x = cosech−1A • If coth x = A, then x = coth−1A 2.2.2 Properties of Inverse Trigonometric Function • sinh(sinh−1x) = x, sinh−1 (sinh x) = x • cosh(cosh−1x) = x, cosh−1 (cosh x) = x • tanh(tanh−1x) = x, tanh−1 (tanh x) = x • sech(sech−1x) = x, sech−1 (sech x) = x • cosech(cosech−1x) = x, cosech−1 (cosech x) = x • coth(coth−1x) = x, coth−1 (coth x) = x 2.2.3 Definition of Inverse Hyperbolic Function = + + − 1 1 2 sinh x ln x x = + − − 1 1 2 cosh x ln x x − + = − x x tanh x ln 1 1 2 1 1 Types of Question involving Definitions of Hyperbolic Functions i) Proving inverse Hyperbolic identities ii) Evaluate (without using calculator) Useful formula : e lnx = x
MAT238/ MAT438 : Foundation of Applied Mathematics 88 2.2.4 Proving Inverse Hyperbolic Function (involving Definition of Inverse Hyperbolic Function) Example 1 Prove that = + + − 1 1 2 sinh x ln x x . Solution : ℎ −1 = = ℎ = − − 2 2 = − − 2 = − 1 2 = − 1 2() = () − 1 () 2 = 2 − 1 0 = 2 − 2 − 1 2 − 2 − 1 = 0 → = 1, = −2, = −1 using quadratic formula = − ± √ 2 − 4 2 = −(−2) ± √(−2) 2 − 4(1)(−1) 2(1) = 2 ± √4 2 + 4 2 = 2 ± √4( 2 + 1) 2 = 2 ± 2√ 2 + 1 2 = ± √ 2 + 1 → = + √ 2 + 1 , = ( + √ 2 + 1) → ℎ −1 = ( + √ 2 + 1) # ()
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 89 Example 2 Show that = + − − 1 1 2 cosh x ln x x . Solution : ℎ −1 = = ℎ = + − 2 2 = + − 2 = + 1 2 = + 1 2() = () + 1 () 2 = 2 + 1 0 = 2 − 2 + 1 2 − 2 + 1 = 0 → = 1, = −2, = 1 using quadratic formula = − ± √ 2 − 4 2 = −(−2) ± √(−2) 2 − 4(1)(1) 2(1) = 2 ± √4 2 − 4 2 = 2 ± √4( 2 − 1) 2 = 2 ± 2√ 2 − 1 2 = ± √ 2 − 1 → = + √ 2 − 1 , = ( + √ 2 − 1) → ℎ −1 = ( + √ 2 − 1) # (ℎ)
MAT238/ MAT438 : Foundation of Applied Mathematics 90 Example 3 Verify that ℎ −1 = 1 2 | 1+ 1− |. Solution : ℎ −1 = = ℎ = − − + − ( + −) = − − + − = − − − + − = − + 1 = − + 1 = − () + 1 () = () − () + 1 = 2 − 2 + 1 = 2 (1 − ) 2 = 1 + 1 − = ±√ 1 + 1 − → = √ 1 + 1 − , = √ 1 + 1 − = ( 1 + 1 − ) 1 2 = 1 2 ( 1 + 1 − ) → ℎ −1 = 1 2 | 1 + 1 − | # ()
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 91 2.2.5 Evaluating Inverse Hyperbolic Functions (without using Calculator) Example 1/ MAR 2005/ MAT238/ Q1a (5 marks) Without using a calculator, find the exact value of ( ) − 3 4 1 2 1 tanh sinh ln . Solution: ℎ = − − 2 ℎ( 3) = 3 − − 3 2 = 3 − 1 3 2 = 3 − 1 3 2 = 4 3 Therefore, 2 ℎ −1 [ 1 4 ℎ( 3)] = 2 ℎ −1 ( 1 4 ∙ 4 3 ) = 2 ℎ −1 ( 1 3 ) = 2 ∙ 1 2 | 1 + 1 3 1 − 1 3 | = (2) # Note : Thus, 3 = 3 − 3 ≠ −3, − 3 = 1 3 = 1 3 = Since we solve without using calculator, so no need to write in decimal
MAT238/ MAT438 : Foundation of Applied Mathematics 92 Example 2/ NOV 2005/ MAT238/ Q3a (4 marks) Find the value of 2 ( 3) 1 sinh cosh ln − without using calculator. Solution: ℎ = + − 2 ℎ( 3) = 3 + − 3 2 = 3 + 1 3 2 = 3 + 1 3 2 = 5 3 Therefore, ℎ −1 [2 ℎ( 3)] = ℎ −1 (2 ∙ 5 3 ) = ℎ −1 ( 10 3 ) = ( 10 3 + √( 10 3 ) 2 + 1) = ( 10 + √109 3 ) # Since we solve without using calculator, so no need to write in decimal
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 93 Example 3/ APR 2007/ MAT238/ Q2a (5 marks) Without using a calculator, find the exact value of ( 2) 1 sinh tanh ln − . Solution: Since we solve without using calculator, so no need to write in decimal ℎ = − − + − Therefore, ℎ( 2) = 2 − − 2 2 + − 2 = 2 − ( 1 2 ) 2 + ( 1 2 ) = 2 − ( 1 2 ) 2 + ( 1 2 ) = 3 5 ℎ −1 = ( + √ 2 + 1), Thus, ℎ −1 [ℎ( 2)] = ℎ −1 [ 3 5 ] = ( 3 5 + √( 3 5 ) 2 + 1) = ( 3 + √34 5 ) #
MAT238/ MAT438 : Foundation of Applied Mathematics 94 Example 4/ OCT 2007/ MAT238/ Q1d (4 marks) Without using a calculator, find the value of ( 4) 1 cosh sinh ln − . Solution: Example 5/ DEC 2008/ MAT238/ Q2b (6 marks) By using the definition, find the exact value of ( 2) 1 tanh sinh ln − . [Hint : − + = − x x tanh x ln 1 1 2 1 1 ] Solution: ℎ = − − 2 ℎ( 4) = 4 − − 4 2 = 4 − 1 4 2 = 4− 1 4 2 = 15 8 Therefore, ℎ −1 [ℎ( 4)] = ℎ −1 ( 15 8 ) = ( ( 15 8 ) + √( 15 8 ) 2 −1 ) = ( 15 + √161 8 ) # = ℎ = − − 2 ℎ( 2) = 2 − − 2 2 = 2 − 1 2 2 = 2− 1 2 2 = 3 4 Therefore, ℎ −1 [ℎ( 2)] = ℎ −1 ( 3 4 ) = 1 2 | 1+ 3 4 1− 3 4 | = 1 2 (7) = (7) 1 2 = √7 # Note : Thus, 2 = 2 − 2 ≠ −2, − 2 = 1 2 = 1 2 = Since we solve without using calculator, so no need to write in decimal Since we solve without using calculator, so no need to write in decimal
Chapter 2 : Hyperbolic and Inverse Hyperbolic Functions 95 TUTORIAL 2.4 : Evaluate Inverse Hyperbolic Function (without using Calculator) Without using a calculator, find the value of 1. ℎ −1 [ℎ( 0.5)] ∶ −√7 2. 2 ℎ −1 [ 1 3 ℎ( 2)] ∶ ( 5 3 ) 3. ℎ −1 [8 ℎ( 4)] ∶ (17 + √290) 4. ℎ −1 [ℎ((0.5))] ∶ ( −3 + √34 5 ) 5. ℎ −1 [ℎ( 0.2)] ∶ ( −12 + √119 5 ) 6. ℎ −1 [ℎ( 0.2)] ∶ 1 2 ( 7 17) 2.2.6 Derivatives of Inverse Hyperbolic Function (comparison between Inverse Hyperbolic and Inverse Trigonometric Function Inverse Hyperbolic Function Inverse Trigonometric Function (sinh x) dx d −1 = 2 1 1 + x (cosh x) dx d −1 = 1 1 2 x − (tanh x) dx d −1 = 2 1 1 − x (sin x) dx d −1 = 2 1 1 − x (cos x) dx d −1 = 2 1 1 − x − (tan x) dx d −1 = 2 1 1 + x
MAT238/ MAT438 : Foundation of Applied Mathematics 96 Simple Examples Differentiate with respect to x. a) y = sinh−1 (3x) b) y = tanh−1 (2x 3 ) c) y = cosh−1 2 3x d) y = sinh−1 (e 5x ) e) y = tanh−1 (ln x3 ) Solution : APPENDIX (given in Final Examination) 15. = + − − C a x sin a x dx 1 2 2 16. C a x tan a 1 a x dx 1 2 2 = + + − 17. C a x a x a x dx = + + − 1 2 2 sec 1 18. C a x x a dx = + − − 1 2 2 cosh = x + x − a + C 2 2 ln 19. C a x x a dx + = + − 1 2 2 sinh = x + x + a + C 2 2 ln 20. + + + = − + = − − − C, if x a a x coth a 1 C, if x a a x tanh a 1 C x a x a ln 2a 1 a x dx 1 1 2 2 ) = ℎ −1 (3) = 1 √1 +(3) 2 ∙ 3 = 3 √1+ 9 2 # ) = ℎ −1 (2 3 ) = 1 1 − (2 3) 2 ∙ 6 2 = 6 2 √1 − 4 6 # ) = ℎ −1 ( 3 2 ) = 1 √( 3 2 ) 2 −1 ∙ 3 2 = 1 √ 9 2 4 − 1 ∙ 3 2 = 1 √ 9 2 − 4 4 ∙ 3 2 = 1 ( √9 2 − 4 2 ) ∙ 3 2 = 2 √9 2 − 4 ∙ 3 2 = 3 √9 2 − 4 # ) = ℎ −1 ( 5 ) = 1 √1+ ( 5) 2 ∙ 5 ∙ 5 = 5 5 √1+ 10 # ) = ℎ −1 ( 3 ) = 1 1− ( 3) 2 ∙ 1 3 ∙ 3 2 = 1 1− 2 3 ∙ 3 = 3 − 2 3 #