Calculus problems solver

All the algebra, functions, relations etc that we study in math can be termed as pre-calculus. The next huge topic in math is calculus. Calculus is distinct from pre-calculus. In calculus there is less of static properties and more of dynamic properties. We talk of one quantity tending to another in calculus.
Calculus is a wide topic that encompasses various subtopics. The most common of which are: Limits, continuity, differential calculus and integral calculus.

The origin of calculus goes back at least 2500 years to the ancient Greeks. They use to find help with calculus problems of how to find area of any polygon.  The method they used was like this: the polygon was divided into triangles and the area of the triangles was added to find the area of the entire polygon, which is very similar to what we do in integration.

In solving calculus problems like finding the area of closed region bounded by curves, it is not possible to split the region to triangles as the boundaries are curved. The Greek method was to inscribe polygons in the figure the circumscribe polygons about the figure and then let the number of sides of the polygon increase. The limiting position of these areas was the area under consideration. Eudoxus used this method to solve calculus problems pertaining to area of circle and to prove the famous formula of A = pi r^2.

The problem of trying to find the slope of the tangent line to a curve gave rise to the branch of calculus called differential calculus. In problems of finding rate of increase of quantities, velocity, acceleration etc, the idea of limits is useful.

Sample calculus problems:

1. The price of a box of pencils is as follows: 5x/(x-1). Where x is the number of boxes one buys. So that means the more then number boxes that a person buys, the lower is the per box cost. What would the price tend to if a person buys lesser and lesser number of boxes?
Solution: The above problem is same as solving the limit: lim(x->1)(5x/(x-1)). We can see that x tends to 1, the term tends to infinity. So if someone wants to buy just one box, the cost would be pretty high.

2. Radius of a circle increases at a constant rate of 4 cm/s. Find the rate of change of area of circle when radius is 6cm.
Solution: This is application of derivatives.
A = pir^2
dA/dt = 2pir.dr/dt
dr/dt=4 and r=6
so, dA/dt = 2*pi*6*4 = 48pi (cm)^2/s