Hyperbolic Functions 6D

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Hyperbolic Functions 6D a (sinh cosh b (cosh 5 5sinh 5 c (tanh sech (sinh cosh e f (coth cosech (sech sinh (cosh sinh cosh cosh tanh sech g (e sinh e sinh e + cosh e (cosh sinh h ( cosh cosh + sinh sinh cosh sinh cosh sinh i ( cosh cosh + sinh (cosh + sinh j k l (sinh cosh cosh cosh + sinh sinh cosh cosh + sinh sinh (ln cosh sinh cosh tanh Pearson Eucation Lt 8. Copying permitte for purchasing institution only. This material is not copyright free.

m (sinh cosh (cosh cosh sinh cosh sinh n ( ( o p (e cosh sinh e cosh cosh (cosech sinh sinh coth cosech y a coshn+ b sinhn Differentiate with respect to ansinhn+ nbcoshn an coshn+ bn sinhn n ( a coshn+ bsinh n n y To fin the stationary point of the curve y cosh sinh, we ifferentiate an set equal to. sinh cosh tanh artanh. + Using the formula artanh ln, we fin that + ln ln Pearson Eucation Lt 8. Copying permitte for purchasing institution only. This material is not copyright free.

Substituting this value for back into the given equation for y, we obtain y cosh ln sinh ln + +. So the coorinates of the stationary point are ln,. y cosh sinh sinh sinh+ cosh cosh 9 cosh sinh+ sinh cosh+ sinh cosh+ cosh sinh cosh sinh+ 6 sinh cosh (5cosh sinh+ sinh cosh 5 a Let y arcosh then coshy Differentiate with respect to sinh y y sinhy but coshy cosh y so b Let y arsinh( + then sinhy + coshy coshy but sinhy + sinh y+ so ( + + Pearson Eucation Lt 8. Copying permitte for purchasing institution only. This material is not copyright free.

5 c Let y artanh tanhy y sech y sech tanh 9 Let y arsech sechy cosh y coshy Differentiate with respect to coshy+ sinhy sinhy coshy coshy sinhy tanhy y ( sech y ( e Let y arcosh Let t y arcosht t t t ( Pearson Eucation Lt 8. Copying permitte for purchasing institution only. This material is not copyright free.

5 f y arcosh Let t y arcosht t t t 9 ( g y arcosh arcosh+ h y arsinh Lett y arsinht t t t + t t t + + + i y e arsinh e e arsinh+ + j y arsinharcosh arcosh+ arsinh + k y arcoshsech sech arcosh tanhsech sech arcosh tanh l y arcosh arcosh+ 9 arcosh+ 9 Pearson Eucation Lt 8. Copying permitte for purchasing institution only. This material is not copyright free. 5

6 a y arcosh coshy sinhy y y sinh cosh but coshy so b y artanh tanhy y sech tanh sech y y but tanh y so 7 e y artanh e Lett y artanht t e t t t e t t Then e e e e ( e e ( e e Pearson Eucation Lt 8. Copying permitte for purchasing institution only. This material is not copyright free. 6

8 y arsinh ( + + ( + 5 ( + ( ( ( ( + ( + ( + ( + + ( + ( + ( + ( + ( + + + + + 9 y (arcosh arcosh ( arcosh ( arcosh+ ( arcosh + ( + y artanh y ln ln 5 ln 5 69 5 Tangent is 69 ( y ln 5 5 5y 5 ln 5 69 56 ( Pearson Eucation Lt 8. Copying permitte for purchasing institution only. This material is not copyright free. 7

In orer to fin the normal to the curve y arcosh at the point, we first must fin the value of at that point., so at,. 5 The graient of the normal at this point is therefore is arcosh ( + 5 The y value at ln. So now we substitute our values for, y, m into y m+ c in orer to fin c. y m+ c ( 5 ( 5 ln + + c c ln + 5 + 5. 5 So we have y + 5+ ln( + 5. 5 a We ifferentiate an evaluate at until we have non-zero terms. f ( cosh f ( f '( sinh f '( f ''( cosh f ''( f '''( sinh f '''( ( ( f ( cosh f (. Now we use the stanar Maclaurin series epansion an obtain cosh + +!! + +... b Using the approimation, cosh. + +.67(6.p..!!.66667 cosh. error cosh. 6 8.7 %. Pearson Eucation Lt 8. Copying permitte for purchasing institution only. This material is not copyright free. 8

a We ifferentiate an evaluate at until we have non-zero terms. f ( sinh f ( f '( cosh f '( f ''( sinh f ''( f '''( cosh f '''( ( ( f ( sinh f ( (5 (5 f ( cosh f (. Now we use the stanar Maclaurin series epansion an obtain 5 sinh + +! 5! 5 + +. 6 b Since for all integers n, ( n (n f ( sinh f (, ( n (n f ( cosh f (. We can conclue that only the o erivatives will contribute to the Maclaurin series epansion, each with a enominator of (n!. The first non-zero term occurs when n with this choice of superscript an so we can conclue n that the nth non-zero term is. (n! a We ifferentiate an evaluate at until we have non-zero terms. f ( tanh f ( f '( sech f '( f ''( tanh sech f ''( f '''( tanh sech sech f '''(. Now we use the stanar Maclaurin series epansion an obtain tanh!. b Using the approimation,.8 tanh.8.8.69(6.p..69 tanh.8 error tanh.8 5.%( s.f. Pearson Eucation Lt 8. Copying permitte for purchasing institution only. This material is not copyright free. 9

5 a We ifferentiate an evaluate at until we have three non-zero terms. f ( ar tanh f ( f '( f '( f ''( f ''( f '''( ( ( ( ( ( f ( + ( (5 (5 5 ( f '''( ( + f ( (5 + + f ( f (. Now we use the stanar Maclaurin series epansion an obtain 5 artanh + +! 5! 5 + +. 5 b By observation, the n th non-zero term in the series epansion appears to be n. n Pearson Eucation Lt 8. Copying permitte for purchasing institution only. This material is not copyright free.

5 c We ifferentiate an evaluate at until we have two non-zero terms. f( cosh artanh f( cosh f '( + sinh artanh f '( sinh cosh f ''( + + cosh artanh f ''( ( f '''( + 6sinh+ cosh 8 cosh ( ( cosh + + sinh artanh f '''( 5. Now we use the stanar Maclaurin series epansion an obtain 5 cosh artanh +! 5 + 6 A less teious way of oing this woul be to take the epansions of cosh an artanh then multiply together, omitting higher orer terms. cosh + + 5 artanh + + 5 cosh artanh + + +... 5 +. 6 6 We ifferentiate an evaluate at until we have three non-zero terms. f( sinh cosh f( f '( cosh cosh+ sinh sinh f '( f ''( 5sinh cosh+ cosh sinh f ''( f '''( cosh cosh+ sinh sinh f '''( ( ( f ( sinh cosh+ cosh sinh f ( (5 (5 f ( cosh cosh+ sinh sinh f (. Now we use the stanar Maclaurin series epansion an obtain 5 sinh cosh + +! 5! 5 + +. 6 Pearson Eucation Lt 8. Copying permitte for purchasing institution only. This material is not copyright free.

7 a We ifferentiate y cos cosh with respect to four times. y cos cosh cos sinh sin cosh sin sinh cos si h + sin cosh cos cosh y. ( n b We evaluate the ifferentials at until we have three non-zero terms. f ( cos cosh f ( f '( cos sinh sin cosh f '( f ''( sin sinh f ''( ( f '''( cos sinh + sin cosh f '''( ( ( f ( cos cosh f ( f (. Since we have the relation y, we can conclue that all ifferentials that are not of the n form, (where n is an integer is when evaluate at. So our thir non-zero term is n 8 8 6y ( y (8 6cos cosh f ( 6. Now we use the stanar Maclaurin series epansion an obtain 8 6 cos cosh +! 8! 8 + 6 5 Pearson Eucation Lt 8. Copying permitte for purchasing institution only. This material is not copyright free.

n 7 c From the previous question, we know that the n terms are the only non-zero contributions. ( k f ( Recalling that the Maclaurin series epansion is f ( k an consiering k! n n ( n n n n ( y n 8 n 8 n 8 n 8 n y, ( y along with all ifferentials that are not of the form evaluate at, we can write an epression. ( k f ( k f ( k! cos cosh k r r r r ( r! ( r r ( ( ( r! ( ( r r r! r y + r Challenge We ifferentiate an evaluate at until we have three non-zero terms. f ( sech f ( f '( tanh sech f '( f ''( tanh sech sech f '''( tanh sech tanh sech tanh sech 5 tanh sech f '''( tanh sech f ( 5sech 5 tanh sech ( 5 f ''( ( tanh sech + tan h sech f ( 5. Now we use the stanar Maclaurin series epansion an obtain 5 sinh cosh +!! 5 + k n n (where n is an integer is when Pearson Eucation Lt 8. Copying permitte for purchasing institution only. This material is not copyright free.