Hyperbolic Functions Flashcards

1
Q

sinh(x)

A

“shine(x)”

(e^x - e^-x) /2

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2
Q

cosh(x)

A

“cosh(x)”

(e^x + e^-x) /2

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3
Q

tanh(x)

A

“tanch(x)”

e^2x - 1)/(e^2x + 1

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4
Q

Reciprocal hyperbolics

A

“cosetch(x)”
“setch(x)”
“coth(x)”
1/the original like in trig

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5
Q

Graph of y = sinh(x)

A

Steep, flattens to a diagonal through (0,0) and then rises steeply again

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6
Q

Graph of y = cosh(x)

A

Parabolic curve with a base at (0,1)

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7
Q

Graph of y = tanh(x)

A

Flat before rising diagonally through (0,0) and flattening again
Asymptotes at y = 1 and y = -1

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8
Q

Graph of y = cosech(x)

A

Like a positive reciprocal graph, asymptotes at x,y = 0

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9
Q

Graph of y = sech(x)

A

Flat asymptotically to the x-axis before rising and flattening to a peak at (0,1) and falling at an increasing rate then flattening asymptotically to the x-axis
Asymptote at y = 0

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10
Q

Graph of y = coth(x)

A

Similar to cosech(x) but starting from y = 1 and -1

Asymptotes at x = 0, y = 1, y = -1

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11
Q

Proving inverse hyperbolic formulae

A
  1. Use the hyperbolic on both sides to write in terms of x
  2. Replace with the hyperbolic formula
  3. Form a hidden quadratic by multiplying by e^y
  4. ln both sides, only use the positive case
  5. Replace y with the inverse hyperbolic
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12
Q

Graph of y = arsinh(x)

A

Flat, rises diagonally through the origin and flattens again, similar to I/V characteristics of a filament lamp but slower flattening

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13
Q

Graph of y = arccosh(x)

A

Domain: x>= 1

Almost vertical at x = 1 before flattening

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14
Q

Graph of y = artanh(x)

A

Domain: -1 < x < 1

Almost vertically upwards at those asymptotes, flattens through the origin

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15
Q

Graph of y = arsech(x)

A

Domain x>0

Almost downwards at the asymptote x = 0, flattens and falls again until it reaches the x-axis

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16
Q

Graph of y = arcosech(x)

A

Domain x != 0

Like a positive reciprocal curve

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17
Q

Graph of y = arcoth(x)

A

Domain |x| > 1

Like a positive reciprocal curve but with asymptotes at x = +/- 1

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18
Q

Hyperbolic pythagorean identities

A

cosh^2x - sinh^2x = 1
sech^2x = 1 - tanh^2x
cosech^2x = coth^2x - 1

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19
Q

sinh(A+/-B)

A

sinhAcoshB +/- coshA/sinhB

20
Q

cosh(A+/-B)

A

coshAcoshB +/- sinhAsinhB

21
Q

tanh(A+/-B)

A

tanhA -/+ tanhB / 1 +/- tanhAtanhB

22
Q

Osborn’s rule

A

Replace sin with sinh and cos with cosh

Put a - in front of any multiplication of sinhx (including tanh^2) (sinh^4 cancels out)

23
Q

d/dx(coth(x))

A

-cosech^2(x)

24
Q

d/dx(sinh(x))

25
d/dx(cosh(x))
sinh(x)
26
d/dx(tanh(x))
sech^2(x)
27
Proving regular differentials
Put it into exponential form and differentiate
28
Proving inverse derivatives
1. x = sinh/cosh/tanh y 2. Do dx/dy 3. Use identities to substitute in terms of x 4. Find the reciprocal for dy/dx
29
∫sinh(x) dx
cosh(x) + c
30
∫cosh(x) dx
sinh(x) + c
31
∫tanh(x) dx
sech^2(x) + c
32
∫cosech^2(x) dx
-coth(x) + c
33
∫sech^2(x) dx
tanh(x) + c
34
∫sech(x)tanh(x) dx
-sech(x) + c
35
∫cosech(x)coth(x) dx
-cosech(x) + c
36
Multiple terms in the numerator
Split and use standard integrals/reverse chain
37
Proving ∫tanh(x)
use sinh(x)/cosh(x) and use reverse chain rule to ln|cosh(x)|
38
Small odd powers of cosh/sinh
Factor out (cosh/sinh)^power-1 and use identity
39
When to use the exponential definition
When there is an exponential term or no simpler way to integrate
40
∫sech(x) or ∫cosech(x) method
Use exponential form, multiply by e^x and use substitution with e^x
41
∫1/sqrt(a^2 + x^2) substitution
x = asinh(u)
42
∫1/sqrt(x^2 - a^2) substitution
x = acosh(u)
43
Completing the square
With a/quadratic or a/sqrt(quadratic), complete the square and use substitution with u as the thing that is squared
44
cosh^2(x) substitution
1/2 + 1/2 cosh(2x)
45
sinh^2(x) substitution
1/2 cosh(2x) - 1/2
46
Hyperbolics to R formula
Use cosh^2 - sinh^2 = 1 for R rather than +