Today: 5.6 Hyperbolic functions

Transcription

1 Toay: 5.6 Hyerbolic functions Warm u: Let f() = (e ) an g() = (e + ) Verify the following ientities: () f 0 () =g() () g 0 () =f() (3) f() is an o function (i.e. f(-) = -f()) (4) g() is an even function (i.e. g(-) = g()) Ean an simlify the following eressions () g () f () () f()g(y)+f(y)g() (3) g()g(y)+f()f(y) (eress the last two as f(stu ) or g(stu ))

2 Define hyerbolic sine or sinch hyerbolic cosine or cosh sinh() = (e ) cosh() = (e + ) y=cosh() y=.5 - y=sinh() y=-.5 e - y=-.5 e - y=.5 - For >>0 (large), sinh() e cosh() e, an for <<0 (large but negative), sinh() cosh(). Define hyerbolic sine or sinch hyerbolic cosine or cosh sinh() = (e ) cosh() = (e + ) y=cosh() y=.5 - y=sinh() y=-.5 e - y=-.5 e - y=.5 - We showe: sinh( ) = sinh() cosh( ) = cosh() sinh() = cosh() cosh() =sinh() cosh() sinh () = sinh( + y) =sinh() cosh(y)+sinh(y) cosh() cosh( + y) = cosh() cosh(y) +sinh() sinh(y)

3 Comaring trig an hyerbolic functions sinh() = (e ) cosh() = (e + ) Trig Hyerbolic even/o: sin( ) = sin() sinh( ) = sin() cos( ) = cos() cosh( ) = cosh() aitive: sin( + y) =sin() cos(y) sinh( + y) =sinh() cosh(y) +sin(y) cos() +sinh(y) cosh() aitive: cos( + y) = cos() cos(y) cosh( + y) = cosh() cosh(y) sin()sin(y) + sinh()sinh(y) Pyth.: cos ()+ sin () = cosh () sinh () = ers: sin() = cos() sinh() = cos() cos() = sin() cosh() =+ sinh() Other hyerbolic functions For trig functions, we efine tan() = sin() cos(), sec() =/ cos(), csc() =/ sin(), cot() =/ tan(). For hyerbolic functions, similarly efine tanh() =sinh()/ cosh() = e + sech() =/ cosh(), csch() =/ sinh(), coth() =/ tanh(). ( tanch, sech, cosech, cotanch )

4 Grahs sinh() = (e ) cosh() = (e + ) y=cosh() y=.5 - y=sinh() y=-.5 e - y=-.5 e - y=.5 - tanh() = sinh() cosh() = e + Notice lim tanh() = lim!! + e = lim! e = +e Grahs Notice lim tanh() = lim!! lim tanh() = lim!! tanh() = sinh() cosh() = e e = lim! (u=-) ==== lim u! e u = lim u! e u + = e u e = +e e u e u + e u e u e u tanh() = sinh() cosh() qu. rule ==== cosh () sinh () cosh () = cosh () > 0

5 Grahs tanh() = sinh() cosh() = e + lim tanh() =, lim tanh() =,!! tanh() =sech () > 0, tanh(0) = ( )/( + ) = 0 y= y=tanh() y=- Inverse hyerbolic functions Define sinh () =y, sinh(y) = cosh () =y, cosh(y) = tanh () =y, tanh(y) = Solving for y =sinh (): =sinh(y) = (ey e y ), so that e y e y =0. Thus, multilying both sies by e y, 0=e y e y = u u, where u = e y. So e y = u =( ± 4 + 4)/ ey >0 === + +,sothat sinh () =y =ln( + + ). You try: Use cosh() = (e + ) an tanh() = e calculate a formulas for cosh () an tanh (). + to

6 Grahs y =sinh() = (e ) Grahs y =sinh () =ln( + + ) Domain: (, ) Range:(, )

7 Grahs y = cosh() = (e + ) y = cosh () =ln( ± ) Domain: al Grahs y = cosh () =ln( + ) Domain: al Range: 0 ale y

8 Grahs y = tanh() =( )/( + ) Grahs y = tanh () = ln(( + )/( )) Domain: (, ) Range:(, )

9 Derivatives an integrals Using sinh () =ln( + + ), cosh () =ln( + ) tanh () = ln(( + )/( )) = (ln( + ) ln( )) we have sinh () = +/ = +! =/ + cosh () = +/ + =! + + tanh () = =/ + + =/( ) = ++ ( + )( ) In summary: sinh() = (e ), cosh() = (e + ) tanh() = sinh() cosh(), sech() = cosh(), sinh() = cosh(), sinh () = tanh() =sech () +, etc. cosh() =sinh(), cosh () = tanh () =,

10 You try: See integration worksheet at htts://zaugherty.ccnysites.cuny.eu/teaching/m0s6/ (an 8 more... )

11 5.7: More on limits, ineterminate forms, an L Hosital s rule Consier the function F () = ln(). As!, both the numerator an the enominator aroach 0. Both aroach somewhat slowly, but oes one go faster than the other? Or oes it aroach some interesting ratio? Similar question for!, where both the numerator an enominator aroach. Ineterminate forms are ratios where the numerator an the enominator each either aroach 0, or each aroach ±. So far, we ve been able to calculate limits with ineterminate forms through algebraic tricks or substitution, or recognizing limits as erivatives. Past eamles of solving ineterminate forms. lim!. lim! e + 5e e = lim! e sin() 3. lim! Recall, f 0 f() f(a) (a) =lim.!a a Note, e sin() = = esin( ) = e 0 =. lim! So e sin() So similarly, sinceln() = 0, lim! 3+ 5 = = lim! 5 = 0+ 0 = = esin() = = cos()esin() = =( )e0 =. ln() = ln() = = =. =

12 L Hosital s rule L Hosital s rule relates the limit of the ratio of two functions to the limit of the ratio of their erivatives. Consier i erentiable functions f() an g() such that lim f() =0=lim g(),!a!a an g 0 () 6= 0for close to but not equal to a. Then f 0 f() (a) =lim!a f(a) a f() =lim!a a, an g 0 g() (a) =lim!a g(a) a =lim!a g() a. (If f or g are not efine at a, wecanworkarounthis:seeaenicinbook) So f 0 () lim!a g 0 () = f 0 (a) g 0 (a) =lim f()( a)!a ( a)g() =lim f()!a g(). L Hosital s rule Theorem Suose f an g are i erentiable functions an g 0 () 6= 0for close to but not equal to a. Suose that lim f() =0=limg() or lim f() =± =lim g().!a!a!a!a Then if the limit of f 0 ()/g 0 () as! a eists (or is ±), we have f() lim!a g() =lim f 0 ()!a g 0 (). The same hols for! ± an one-sie limits! a ±. ln() Eamle. Let s recheck lim!. ln() an i erentiable? X g 0 () =6= 0X, ln()! 0 an! 0 as! X lim! ln() =lim /! =X

13 You try For each of the following, verify that you can use L Hosital s rule to calculate the limit, an then o so. e sin() () lim! () lim! ln() (3) lim! Each of the following has some reason why you can t use L Hosital s rule. For each, what is the reason? () lim!0 () lim!0 + bc (3) lim! sin() cos() (Recall, bc is the floor function, an gives back the biggest integer less than or equal to, i.e.b.c =, b.c = 3, bc =,etc..)