maths:calculus

# a brief summary of calculus

## Differential and integral calculus basic rules

 f(x) f'(x) Integral of f(x) x^n nx^(n-1) (xn+1)/(n+1) lnx x^(-1) ln(ax+b) a/(ax+b) e^(kx) ke^(kx) sin(kx) kcos(kx) cos(kx) ksin(kx) tan(kx) k(sec^2)(kx) Arsin(x/a) 1/SQR(a^2-x^2) Arcos(x/a) -1/SQR(a^2-x^2) Artan(x/a) a/(a^2+x^2) g(x)h(x) g'(x)h(x)+g(x)h'(x) g(x)/h(x) [g'(x)h(x)-g(x)h'(x)]/[h(x)]^2 goh(x) g'oh(x).h'(x)

• Rules:
• A function f, is differentiable at a point x=a, if the derivative f'(x) exists at x=a:
• ie. must be smooth & continuous;
• If a function is continuous at x=a, then the lim(x→a)f(x) and f(a) both exist and are equal;
• d(x^2-5x+6)/dx = d(x^2)/dx - d(5x)/dx + d(6)/dx = 2x - 5;
• d(uv)/dx = vd(u)/dx + ud(v)/dx;
• dy/dx = (dy/du) * (du/dx);
• dy/dx = 1/(dx/dy);
• if y=u/v, dy/dx = [(vdu/dx)-(udv/dx)]/v^2;
• examples
• the cooling rate of a cup of tea
• the rate of cooling is dependent upon the temperature difference of the tea and the room (ΔT)
• the difference in temperature changes over time = ΔT(t) = e-kt
• rate of change the difference in temperature changes over time = dΔT/dt = -kΔT

## Implicit Differentiation

• eg. x^5 + 2(x^2)(y^3) + y^5 = 0,
• d(x^5)/dx + d[2(x^2)(y^3)]/dx + d(y^5)/dx = 0,
• and, d(x^5)/dx = 5x^4,
• d[2(x^2)(y^3)]/dx = 2x^2d(y^3)/dx + y^3d(2x^2)/dx
• = 2x^2d(y^3)dy/(dy.dx) + 4xy^3
• = (2x^2)(3y^2)dy/dx + 4xy^3,
• d(y^5)/dx = d(y^5)dy/(dy.dx)
• = (5y^4)dy/dx,
• ⇒ 5x^4 + 6(x^2)(y^2)dy/dx + 4xy^3 + 5(y^4)dy/dx = 0,
• ⇒ dy/dx =[-x(5x^3 + 4y^3)]/[y^2(6x^2 + 5y^2)], y not=0

## Taylor Series

• these consist of polynomials used to approximate a different function and the higher the level of polynomial you use in the series to approximate, it may either “converge” to a closer approximation or, at some value for x, it may “diverge” from the approximation
• for example:
• the approximation for cos(x) near x = 0, is cos(x) approx equals 1 - x2/2
• the approximation for ex approx equals 1 + x + x2/2! + x3/3! + .. xn/n! and plugging in x = 1 will give you a convergent approximation of the value of e
• these Taylor polynomials are created by:
• taking the base Taylor polynomial form P(x) = c0 + c1x + c2x2 + …
• set x to be the value you are wanting to most closely approximate (usually 0 but you may use x-a instead of x in the polynomial)
• sequentially differentiate the base form and equate to the same degree of differentiation of the original function and these will give you each of the constant c terms each time you differentiate, note that in the ex, the differential is ex each time and for x = 0, the result is 1, hence the Taylor polynomial uses power of x divided by the power factorial (as the sequential differentials of a power of x until the power of x is zero is the factorial of the original power)

## Integral calculus:

• this is used to determine the area bounded by the x axis and a function such that any parts above the x axis are positive areas and those below it are negative areas.
• thus if f(x) = x(2-x) = 2x-x2 then the x-axis intercepts (ie. where f(x) = 0) ) are 0 and 2 and f(x) is positive between these intercepts and negative outside these. The integral of f(x) = x2 - x3/3 .  Thus the area above the x-axis = (02 - 03/3) - (22 - 23/3) = 4/3. The area below the x-axis between x = -1 and x = 0 equals (-12 - (-1)3/3) - (02 - 03/3) = -4/3. This just happens to be equal but opposite sign to the area from x = 0 to x =2 which was first calculated. If you determine the integral from x = -1 to x =2, the answer will then be zero as the two areas will cancel each other out.
• it can also be used to determine the area bounded by two functions f(x) and g(x) by first determining the x values of the points of intersection between f(x) and g(x) and then determining the integral from intersect 1 to intersect 2 of f(x) - g(x)

## Dimensional analysis in physics

maths/calculus.txt · Last modified: 2021/10/03 18:48 by gary1