2.2 Solving Hyperbolic Equation (using Identities & definition)

MOOC MAT438/ UiTM

 Definition and Identities of Hyperbolic Functions
 Proving the Hyperbolic Function using Identities and Definition
 Solving the Hyperbolic Equations using Identities and Definition
 Derivatives of Hyperbolic Functions
 Integration of Hyperbolic Functions

MOOC MAT438/ UiTM

Example 1
Solve the hyperbolic equation 6 ℎ2 + 7 ℎ = 8.

Use identities

Solution :

6 ℎ2 + 7 ℎ = 8 2 − 1 = 0 3 − 2 = 0
6 1 − ℎ2 + 7 ℎ = 8 2
1
= 2 = 3

6 1 − 2 + 7 = 8 Where = ℎ 1 2
6 − 6 2 + 7 = 8 ℎ = 2 ℎ = 3

0 = 2 − 7 + 6 2 = ℎ−1 1 = ℎ−1 2
6 2 − 7 + 2 = 0 2 3

2 − 1 3 − 2 = 0 = 0.5493 # = 0.8047 #

MOOC MAT438/ UiTM

Example 2
Solve the hyperbolic equation 2 ℎ 2 + ℎ = 7.

Solution : Use double-angle
identities

2 ℎ 2 + ℎ = 7 − 1 = 0 4 + 5 = 0
2 1 + 2 ℎ2 + ℎ = 7 = 1 5
2 1 + 2 2 + = 7
ℎ = 1 = − 4
2 + 4 2 + = 7 = ℎ−1 1
4 2 + − 5 = 0 Where = ℎ 5
= 0.8814 # ℎ = − 4

= ℎ−1 5
−4

− 1 4 + 5 = 0 = −1.0476 #

MOOC MAT438/ UiTM

Example 3
Solve the hyperbolic equation 3 ℎ2 − 1 + ℎ − 1 = 5.

Use identities

Solution :

3 ℎ2 − 1 + ℎ − 1 = 5 + 1 = 0 3 − 2 = 0
= −1
3 1 + ℎ2 − 1 + ℎ − 1 = 5 ℎ ( − 1) = −1 2
= 3
3 1 + 2 + = 5 Where = ℎ − 1
2
3 + 3 2 + = 5 ℎ ( − 1) = 3

3 2 + − 2 = 0 − 1 = ℎ−1 −1 − 1 = ℎ−1 2
3

+ 1 3 − 2 = 0 = 0.1186 # = 1.6251 #

MOOC MAT438/ UiTM

MOOC MAT438/ UiTM

 Definition and Identities of Hyperbolic Functions
 Proving the Hyperbolic Function using Identities and Definition
 Solving the Hyperbolic Equations using Identities and Definition
 Derivatives of Hyperbolic Functions
 Integration of Hyperbolic Functions

MOOC MAT438/ UiTM

Example 1
Solve the equation by using the definition of hyperbolic functions: 4 ℎ + ℎ − 4 = 0.

Use definition

Solution : 5 + 3 − − 8 = 0

4 ℎ + ℎ − 4 = 0 5 + 3 − 8 = 0


+ − − − 3 Where =
4 2 + 2 −4=0 5 + − 8 = 0

Multiplying each term by u (to eliminate denominator u)

Multiplying each term by 2 (to eliminate denominator 2) 3

+ − − − 5 + − 8 = 0
2 2
24 +2 − 2 4= 2 0 5 2 + 3 − 8 = 0

4 + − + − − − 8 = 0 5 2 − 8 + 3 = 0

4 + 4 − + − − − 8 = 0 − 1 5 − 3 = 0
MOOC MAT438/ UiTM

Example 1 (continue…)
Solve the equation by using the definition of hyperbolic functions: 4 ℎ + ℎ − 4 = 0.

Solution : 5 − 3 = 0
− 1 5 − 3 = 0 3
− 1 = 0
= 1 = 5

= 1 = 3
5
= 1
= 3
= 0 # 5
MOOC MAT438/ UiTM
= −0.5108 #

Example 2
Solve the equation by using the definition of hyperbolic functions: 2 ℎ 2 = 1 + 4 ℎ 2 .

Solution : Use definition − 2 + 3 − 1 = 0
2 ℎ 2 = 1 + 4 ℎ 2 2
We not use double-
angle identities 3 Where = 2
− + − 1 = 0
because the angles
are same!

2 + −2 2 2 − −2
2 2 =1+4 2
Multiplying each term by u (to eliminate denominator u)

2 + −2 = 1 + 2 2 − −2 3
2 + −2 = 1 + 2 2 − 2 −2 − + − 1 = 0
2 − 2 2 + −2 + 2 −2 − 1 = 0
− 2 + 3 − = 0
2 + − 3 = 0

− 2 + 3 −2 − 1 = 0

MOOC MAT438/ UiTM

Example 2 (continue…)
Solve the equation by using the definition of hyperbolic functions: 2 ℎ 2 = 1 + 4 ℎ 2 .

Solution : = −2.3028
2 + − 3 = 0

= 1.3028

2 = 1.3028 2 = −2.3028
2 = 1.3028 2 = −2.3028
( )
= 0.13225 #

MOOC MAT438/ UiTM

Example 3
Solve the equation by using the definition of hyperbolic functions: 2 ℎ 4 − ℎ 4 = 2 −4 .

Use definition 2 4 + 2 −4 − 4 + −4 − 4 −4 = 0

Solution :

2 ℎ 4 − ℎ 4 = 2 −4 4 − −4 = 0

4 + −4 4 − −4 = 2 −4 4 − 1 = 0
2 2 4
2 −
1
− = 0 Where = 4

Multiplying each term by 2 (to eliminate denominator 2) Multiplying each term by u (to eliminate denominator u)

4 + −4 −2 4 − −4 = 2 2 −4 1
22 2 2 − = 0

2 4 + −4 − 4 − −4 = 4 −4 2 − 1 = 0

+ 1 − 1 = 0

MOOC MAT438/ UiTM

Example 3 (continue…)
Solve the equation by using the definition of hyperbolic functions: 2 ℎ 4 − ℎ 4 = 2 −4 .

Solution : = 1
+ 1 − 1 = 0

= −1

4 = −1 4 = 1
4 = −1 4 = 1
( ) = 0 #

MOOC MAT438/ UiTM

MOOC MAT438/ UiTM