Formulas - Math 115



Formulas

Geometry:

Circles: Area = (r2 Circumference = 2(r

Spheres: Volume = (r3 Surface Area = 4(r2

Cylinders: Volume = (Area of base) ( (Height)

Cones: Volume = (Area of base) ( (Height)

Trigonometric Identities:

sin2x + cos2x = 1 tan2x + 1 = sec2x ctn2x + 1 = csc2x

sin(x+y) = sinx cosy + cosx siny cos(x+y) = cosx cosy - sinx siny

sin 2x = 2 sinx cosx cos 2x = cos2x - sin2x

sinx siny = cosx cosy =

sinx cosy = sin2x = cos2x =

y = is that angle y such that x = and

sin-1(- x) = - sin-1 x cos-1(- x) = ( - cos-1x cos-1x = - sin-1x

Hyperbolic Functions:

sinh x = cosh x = tanh x = =

coth x = = sech x = = csch x = =

cosh2x - sinh2x = 1 tanh2x + sech2x = 1 coth2x = 1 + csch2x

sinh(x+y) = sinh x cosh y + cosh x sinh y cosh(x+y) = cosh x cosh y + sinh x sinh y

sinh 2x = 2 sinh x cosh x cosh 2x = cosh2x + sinh2x

sinh2x = cosh2x =

y = = is that y such that x =

Derivatives

= f’(x) =

Linear approximation (3 ways of writing the same thing)

f(x) ( f(xo) + f’(xo)(x-xo) f(x+(x) ( f(x) + f’(x) (x (y ( (x

Quadratic approximation

f(x) ( f(xo) + f’(xo)(x-xo) + (x-xo)2 f(x+(x) ( f(x) + f’(x) (x + ((x)2

(y ( (x + ((x)2

Mean Value Theorem (Slope of secant line = slope of tangent line at intermediate point. Same as linear approximation but uncertainty is where derivative is evaluated.)

f(x) = f(xo) + f’(c)(x-xo) f(x+(x) = f(x) + f’(c) (x ( (c)

L’Hospital’s Rule = provided f(x) and g(x) are either both 0 or both (

Summation Formulas

1 + 2 + 3 + ... + n = 12 + 22 + 32 + ... + n2 =

13 + 23 + 33 + ... + n3 = 1 + x + x2 + ... + xn =

Integrals

=

Area = L(x) = Length of cross section through x.

Volume = A(x) = Area of cross section through x.

Volume = ( For the solid created by rotating a region about the x axis. The region is bounded above by y = f(x), below by y = g(x), on the left by x = a and on the right by x = b.

Volume = 2( For the solid created by rotating the same region about the y axis.

Work = F(x) = Force on object at position x.

Table of Derivatives and Integrals

| tan x = sec2 x | = tan x |

| cot x = - csc2 x | = - cot x |

| sec x = tan x sec x | = sec x |

| csc x = - cot x csc x | = - csc x |

| ax = ax ln(a) | = |

| sin-1x = ) | = sin-1x |

| cos-1x = ) | = - cos-1x |

| tan-1x = | = tan-1x |

| cot-1x = | = - cot-1x |

| sec-1x = ) | = sec-1x |

| csc-1x = ) | = - csc-1x |

| | |

| sinh x = cosh x | = sinh x |

| cosh x = sinh x | = cosh x |

| tanh x = sech2 x | = tanh x |

| coth x = - csch2 x | = - coth x |

| sech x = - tanh x sech x | = - sech x |

| csch x = - coth x csch x | = - csch x |

| | |

| sinh-1x = ) | = sinh-1x |

| cosh-1x = ) | = cosh-1x |

| tanh-1x = | = tanh-1x |

| coth-1x = | = coth-1x |

| sech-1x = ) | = - sech-1x |

| csch-1x = ) | = - csch-1x |

More Integrals

= uv -

= ln( tan x + sec x)

= ln( csc x - cot x)

= - sinn-1x cos x +

= cosn-1x sin x +

= tann-1x -

= - cotn-1x -

= tan x secn-2x +

= - cot x cscn-2x +

= after substituting u = sin x and using cos2x = 1 – sin2x

Similarly when sin is to an odd power

Use sin2x = and cos2x =

= after substituting u = tan x and using sec2x = 1 + tan2x

= after substituting u = sec x and using tan2x = sec2x - 1

= after tan2x = sec2x – 1. Now expand out and use reduction formula for

Integral involving Let x = a sin u

Integral involving Let x = a tan u

Integral involving Let x = a sec u

Integral involving Complete the square: = . Then substitute u = x + which reduces the integral to one of the three previous

Integral of a rational function . First, divide denominator into numerator to get a polynomial plus a rational function where the numerator has degree less than the denominator. Integrate the polynomial as usual. For the new rational function, factor the denominator into the product of linear and quadratic factors and split it up using partial fractions: = + + ( + + + . To find the numerators, cross multiply to get anxn + an-1xn-1 + ( + a1x + a0 = A1(x-r2)((x-s)2(x2+bx+c) + A2(x-r1)(x-r3)((x-s)2(x2+bx+c) + ( + B(x-r1)(x-r2)((x-s)(x2+bx+c) + C(x-r1)(x-r2)((x2+bx+c) + (Dx+E)(x-r1)(x-r2)((x-s)2. You will need to plug in additional values to find B, D and E. Example: = + ( 1 = A1(x-5) + A2(x-4). x = 4 ( A1 = -1. The resulting fractions with linear denominators are easy to integrate. Complete the square, if necessary, to integrate the ones with quadratic denominators.

= f(x) dx = f(x) dx = f(x) dx +

= if f(x) is unbounded as x ( a

Numerical Integration

Midpoint method

( 2((x) [ f(x1) + f(x3) + f(x5) + + f(xn-1) ] where (x = and xj = a + j((x) and n is even

Simpson’s method

( [ f(x0) + 4f(x1) + 2f(x2) + 4f(x3) + 2f(x4) + + 2f(xn-2) + 4f(xn-1) + f(xn) ] with (x, xj as above, n is even

Arc Length

L = length of curve y = f(x) where a ( x ( b =

Center of Mass

(, ) = center of mass of thin plate of uniform density occupying the region bounded above by y = f(x) and below y = g(x) and between x = a and x = b. Let A = area. Then

= =

Parametric Equations of Curves: x = f(t) and y = g(t)

To write as y = F(x): Solve x = f(t) for t in terms of x and substitute in y = g(t). Similarly to write as x = G(y)

Slopes: = Second derivatives: =

Areas: = where a = f(() and b = f(()

Arc Lengths: L = = length of the part corresponding to ( ( t ( (

Polar Coordinates r and (

Curve defined by a polar equation r = f((): x = f(() cos ( and y = f(() sin (

Slopes: =

Arc Lengths: L = = length of the part corresponding to ( ( t ( (

Areas: A = = area of region where g(() ( r ( f(() and ( ( ( ( (

Conic Sections

Ellipse: + = 1

a ( b ( foci at (u ( c, v)

c2 = a2 – b2

Hyperbola: - = 1

foci at (u ( c, v) c2 = a2 + b2

Sequences and Series

Geometric series: if -1 < x < 1

nth term test: ( 0 ( diverges

Absolute convergence: < ( ( converges

Integral test: f(x) positive & decreasing ( < ( ( < (

and, more generally < ( ( < (

p - series: Converges if p > 1 and diverges if p ( 1

Limit comparison test: Assume an > 0 & bn > 0 for all n

< ( & < ( ( < (

> 0 & = ( ( = (

Ratio test:

If an > 0 then < 1 ( < ( & > 1 ( = (

< 1 ( converges & > 1 ( diverges

Root test:

If an > 0 then < 1 ( < ( & > 1 ( = (

< 1 ( converges & > 1 ( diverges

Alternating series test: an decreasing with n & = 0 ( converges

Taylor series: f(x) =

where M = maximum of | f(n+1)(t) | for t between a and x

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