Hyperbolic Functions Flashcards
sinh(x)
“shine(x)”
(e^x - e^-x) /2
cosh(x)
“cosh(x)”
(e^x + e^-x) /2
tanh(x)
“tanch(x)”
e^2x - 1)/(e^2x + 1
Reciprocal hyperbolics
“cosetch(x)”
“setch(x)”
“coth(x)”
1/the original like in trig
Graph of y = sinh(x)
Steep, flattens to a diagonal through (0,0) and then rises steeply again
Graph of y = cosh(x)
Parabolic curve with a base at (0,1)
Graph of y = tanh(x)
Flat before rising diagonally through (0,0) and flattening again
Asymptotes at y = 1 and y = -1
Graph of y = cosech(x)
Like a positive reciprocal graph, asymptotes at x,y = 0
Graph of y = sech(x)
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
Graph of y = coth(x)
Similar to cosech(x) but starting from y = 1 and -1
Asymptotes at x = 0, y = 1, y = -1
Proving inverse hyperbolic formulae
- Use the hyperbolic on both sides to write in terms of x
- Replace with the hyperbolic formula
- Form a hidden quadratic by multiplying by e^y
- ln both sides, only use the positive case
- Replace y with the inverse hyperbolic
Graph of y = arsinh(x)
Flat, rises diagonally through the origin and flattens again, similar to I/V characteristics of a filament lamp but slower flattening
Graph of y = arccosh(x)
Domain: x>= 1
Almost vertical at x = 1 before flattening
Graph of y = artanh(x)
Domain: -1 < x < 1
Almost vertically upwards at those asymptotes, flattens through the origin
Graph of y = arsech(x)
Domain x>0
Almost downwards at the asymptote x = 0, flattens and falls again until it reaches the x-axis
Graph of y = arcosech(x)
Domain x != 0
Like a positive reciprocal curve
Graph of y = arcoth(x)
Domain |x| > 1
Like a positive reciprocal curve but with asymptotes at x = +/- 1
Hyperbolic pythagorean identities
cosh^2x - sinh^2x = 1
sech^2x = 1 - tanh^2x
cosech^2x = coth^2x - 1
sinh(A+/-B)
sinhAcoshB +/- coshA/sinhB
cosh(A+/-B)
coshAcoshB +/- sinhAsinhB
tanh(A+/-B)
tanhA -/+ tanhB / 1 +/- tanhAtanhB
Osborn’s rule
Replace sin with sinh and cos with cosh
Put a - in front of any multiplication of sinhx (including tanh^2) (sinh^4 cancels out)
d/dx(coth(x))
-cosech^2(x)
d/dx(sinh(x))
cosh(x)
d/dx(cosh(x))
sinh(x)
d/dx(tanh(x))
sech^2(x)
Proving regular differentials
Put it into exponential form and differentiate
Proving inverse derivatives
- x = sinh/cosh/tanh y
- Do dx/dy
- Use identities to substitute in terms of x
- Find the reciprocal for dy/dx
∫sinh(x) dx
cosh(x) + c
∫cosh(x) dx
sinh(x) + c
∫tanh(x) dx
sech^2(x) + c
∫cosech^2(x) dx
-coth(x) + c
∫sech^2(x) dx
tanh(x) + c
∫sech(x)tanh(x) dx
-sech(x) + c
∫cosech(x)coth(x) dx
-cosech(x) + c
Multiple terms in the numerator
Split and use standard integrals/reverse chain
Proving ∫tanh(x)
use sinh(x)/cosh(x) and use reverse chain rule to ln|cosh(x)|
Small odd powers of cosh/sinh
Factor out (cosh/sinh)^power-1 and use identity
When to use the exponential definition
When there is an exponential term or no simpler way to integrate
∫sech(x) or ∫cosech(x) method
Use exponential form, multiply by e^x and use substitution with e^x
∫1/sqrt(a^2 + x^2) substitution
x = asinh(u)
∫1/sqrt(x^2 - a^2) substitution
x = acosh(u)
Completing the square
With a/quadratic or a/sqrt(quadratic), complete the square and use substitution with u as the thing that is squared
cosh^2(x) substitution
1/2 + 1/2 cosh(2x)
sinh^2(x) substitution
1/2 cosh(2x) - 1/2
Hyperbolics to R formula
Use cosh^2 - sinh^2 = 1 for R rather than +
