Hyperbolic functions

FM Hyperbolic functions: KUS objectives BAT use the inverse hyperbolic functions Starter: Expand and simplify =ex+y ex x ey (e-x – 1)2 =e−2x – 2e−x + 1 =¼(e2x – 2 + e−2x) [½(ex – e-x)]2

If you take a rope/chain, fix the two ends, and let it hang under the force of gravity, it will naturally form a hyperbolic cosine curve. Hyperbolic functions Most curves that look parabolic are actually Catenaries, which is based in the hyperbolic cosine function. A good example of a Catenary would be the Gateway Arch in Saint Louis, Missouri.

tanh 𝑥 ≡ sinh 𝑥 cosh 𝑥 ≡ 𝑒 2𝑥 −1 𝑒 2𝑥 +1 Notes Hyperbolic functions have several properties in common with trigonometric functions, but they are defined in terms of exponential functions sinh 𝑥 ≡ 𝑒 𝑥 − 𝑒 −𝑥 2 cosh 𝑥 ≡ 𝑒 𝑥 + 𝑒 −𝑥 2 tanh 𝑥 ≡ sinh 𝑥 cosh 𝑥 ≡ 𝑒 2𝑥 −1 𝑒 2𝑥 +1 ‘shine x’ ‘coshine x’ There are corresponding reciprocal functions cosech 𝑥 ≡ 2 𝑒 𝑥 − 𝑒 −𝑥 sech 𝑥 ≡ 2 𝑒 𝑥 + 𝑒 −𝑥 coth 𝑥 ≡ 1 tanh 𝑥 ≡ 𝑒 2𝑥 +1 𝑒 2𝑥 −1 ‘cosheck x’ ‘sheck x’ These definitions can simply be stated but need to be memorised

WB A1 Find to 2dp the values of 𝑎) sinh 5 b) cosh ( ln 2 ) 𝑐) tanh 𝑥 2 tanh 𝑥 ≡ sinh 𝑥 cosh 𝑥 ≡ 𝑒 2𝑥 −1 𝑒 2𝑥 +1 a) sinh 5 ≡ 𝑒 5 − 𝑒 −5 2 = 74.20 b) cosh ( ln 2 ) ≡ 𝑒 ln 2 + 𝑒 −𝑙𝑛2 2 = 2+ 1 2 2 = 5 4 c) tanh 𝑥 2 ≡ 𝑒 2 𝑥 2 −1 𝑒 2 𝑥 2 +1 = 𝑒 𝑥 −1 𝑒 𝑥 +1

WB A2 Find the values of x for which 𝑎) sinh 𝑥 =5 𝑏) tanh 𝑥 = 15 17 cosh 𝑥 ≡ 𝑒 𝑥 + 𝑒 −𝑥 2 tanh 𝑥 ≡ sinh 𝑥 cosh 𝑥 ≡ 𝑒 2𝑥 −1 𝑒 2𝑥 +1 a) sinh 𝑥 ≡ 𝑒 𝑥 − 𝑒 −𝑥 2 =5 multiply through by 𝑒 𝑥 to get a quadratic Rearrange to 𝑒 𝑥 − 𝑒 −𝑥 −10=0 Use quadratic formula to solve 𝑒 2𝑥 −10 𝑒 𝑥 −1=0 𝑒 𝑥 = −10± (−10) 2 −4(−1) 2 =5+ 26 =2.31 b) tanh 𝑥 ≡ 𝑒 2𝑥 −1 𝑒 2𝑥 +1 = 15 17 17( 𝑒 2𝑥 −1) =15( 𝑒 2𝑥 +1) 𝑒 2𝑥 =32 2𝑥= ln 32 =4 ln 2 so 𝑥=2 ln 2

Graphs

WB3 sketch the graphs of sinh 𝑥 , cosh 𝑥 𝑎𝑛𝑑 tanh 𝑥

WB3a (cont) sketch the graphs of sinh 𝑥 , cosh 𝑥 𝑎𝑛𝑑 tanh 𝑥 tanh 𝑥 ≡ sinh 𝑥 cosh 𝑥 ≡ 𝑒 2𝑥 −1 𝑒 2𝑥 +1 the graph of sinh x is the ‘average’ of the graphs of 𝑦=𝑒 𝑥 and y=− 𝑒 −𝑥 the graph of sinh x is an odd function sinh (−𝑥) =− sinh 𝑥 Note that: As 𝑥→∞ ,− 𝑒 −𝑥 →0 so sinh x≈ 1 2 𝑒 𝑥 And As 𝑥→−∞ , 𝑒 𝑥 →0 so sinh x≈ − 1 2 𝑒 −𝑥 Domain 𝑥∈𝑅 Range 𝑦 ∈𝑅

WB3b (cont) sketch the graphs of sinh 𝑥 , cosh 𝑥 𝑎𝑛𝑑 tanh 𝑥 tanh 𝑥 ≡ sinh 𝑥 cosh 𝑥 ≡ 𝑒 2𝑥 −1 𝑒 2𝑥 +1 the graph of cosh x is the ‘average’ of the graphs of 𝑦=𝑒 𝑥 and y= 𝑒 −𝑥 the graph of cosh x is an even function cosh (−𝑥) = cosh 𝑥 Note that: As 𝑥→∞ , 𝑒 −𝑥 →0 so cosh x≈ 1 2 𝑒 𝑥 And As 𝑥→−∞ , 𝑒 𝑥 →0 so cosh x≈ 1 2 𝑒 −𝑥 Domain 𝑥∈𝑅 Range 𝑦 ≥1

WB3c (cont) sketch the graphs of sinh 𝑥 , cosh 𝑥 𝑎𝑛𝑑 tanh 𝑥 tanh 𝑥 ≡ sinh 𝑥 cosh 𝑥 ≡ 𝑒 2𝑥 −1 𝑒 2𝑥 +1 Consider the graphs of cosh x and sinh x to understand the graphs of tanh x tanh 𝑥 ≡ sinh 𝑥 cosh 𝑥 the graph of tanh x is an odd function tanh (−𝑥) = 𝑡𝑎𝑛h 𝑥 Note that: As 𝑥→∞ , tanh 𝑥 →1 And As 𝑥→−∞ , tanh 𝑥 →−1 Domain 𝑥∈𝑅 Range −1<𝑦 <1

Any comments about these answers? WB A4 Evaluate the following, leaving answers to 4 s f a) sinh 7 b) cosh(-5) c) tanh 0.2 Solve the hyperbolic equations, leaving answers correct to 3sf d) sinh x = -2 e) cosh x = 3 f) tanh x = 0.8 Try finding the exact solutions using the definitions for sinh, cosh and tanh. Compare your answers to those in question 2. Any further comments now? 548.3 74.21 0.1974 coshx is a many-one function and therefore two solutions are required. Calculators only give the principal value. Any comments about these answers? x = -1.44 x = 1.76 x = 1.10 Y = sinhx and tanhx are one-one functions and therefore there is just the one answer Y = coshx is a many-one function and therefore two solutions were/are required. Calculators give the principal value. 𝑦=𝑠𝑖𝑛ℎ𝑥 𝑦=𝑡𝑎𝑛ℎ𝑥 𝑦=𝑐𝑜𝑠ℎ𝑥

NOW DO EX 6A WB A5 sketch the graphs of cosech 𝑥 , 𝑠𝑒𝑐h 𝑥 𝑎𝑛𝑑 coth 𝑥 y=cosech 𝑥 y=sinh 𝑥 y=coth 𝑥 y=tanh 𝑥 y=sech 𝑥 y=cosh 𝑥 NOW DO EX 6A

One thing to improve is – KUS objectives BAT use the inverse hyperbolic functions self-assess One thing learned is – One thing to improve is –

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