**what is a derivative?**

The derivative of a function
*ƒ* at a number *a*, denoted by *ƒ*'(*a*),
is

*ƒ*'(*a*)
= lim *ƒ*(*a* + *h*) - *ƒ*(*a*)

* h*->0*
h*

**how can
we interpret the meaning of a derivative?**

from dictionary.com, a derivative is:

a. The limiting
value of the ratio of the change in a function to the corresponding change in
its independent variable.

b. The instantaneous rate of change of a function with respect to its variable.

c. The slope of the tangent line to the graph of a function at a given point.

**the fundamental
theorem of calculus**

part 1:

If *ƒ*
is continuous on [*a*,*b*], then the function *g* defined
by

/

*g*(*x*)
= \ *ƒ*(*t*) *dt* *a*
<= *x* <= *b*

/

part 2:

If *ƒ*
is continuous on [*a*,*b*], then

/*b*

\ *ƒ*(*x*) *dx* = *F*(*b*) - *F*(*a*)

/*a*

where *F*
is any antiderivative of *ƒ*, that is, a function such that *F*'
= * ƒ*.

**how can
we interpret the meaning of an integral?**

from dictionary.com, an integral is:

A number computed by a limiting process in which the domain of a function, often an interval or planar region, is divided into arbitrarily small units, the value of the function at a point in each unit is multiplied by the linear or areal measurement of that unit, and all such products are summed.

**example problems solved
using differentiation and integration:**

1. differentiation

A water tank has the shape of an inverted cirular cone with base radius 2m and height 4m. If water is being pumped into the tank at a rate of 2 m^3/min, find the rate at which the water level is rising when the water is 3m deep.

let V, r, and h be the volume of the water, the radius of the surface, and the height at time t, where t is measured in minutes. given: dV/dt = 2 m^3/min, we are asked to find dh/dt when h = 3m. the quantities V and h are related by the equation V = 1/3 (pi*r^2*h), but it is useful to express V as a function of h alone. note that r/h = 2/4. r = h/2. the expression for V becomes V = (1/3)*pi*((h/2)^2)*h = (pi/12)*h^3. differentiate each side with respect to r: dV/dt = (pi/4)*h^2*(dh/dt), so dh/dt = (4/(pi*h^2))*(dV/dt). Substituting h = 3m and dV/dt = 2 m^3/min we have dh/dt = (4/(pi*3^2))*2 = 8/9pi ~= 0.28 m/min.

2. integration

Find the area under the cosine curve from 0 to b, where 0 <= b <= pi/2.

Since the antiderivative of cos(x) is sin(x), we have

b

sin(x)] = sin(b) - sin(0) = sin(b)

0

**what is a differential
equation?**

Many of the priciples, or laws, underlying the behavior of the natural world are statements or relations involving rates at which things happen. When expressed in mathematical terms the relations are equations and the rates are derivatives. Equations containing derivatives are differential equations. For example, the Bessel equation of order zero is

x^2*y'' + x*y' + x^2*y = 0

One solution for x > 0, called the Bessel function of the first kind of order zero, is

inf

J0(x) = 1 + sum( __(-1)^n * x^2n__ )

n=1 2^2n
* (n!)^2

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