# Calculus Problem Solver

Calculus problem solvers are becoming increasingly popular among students and parents. Students looking for easy and accurate solutions turn to calculus problem solvers to get the right answers. Calculus gets much simpler if you are confident with the basics. A lot of online Calculus sites and resources are available to make you confident in the subject. These sites range from online tutoring sites, sites with worksheets and problems and homework and assignment help sites. There are a lot of calculus resources also available, you just need to find one which works best for you.

## Calculus Word Problem Solver

A calculus problem solver is a great way to learn and solve calculus problems because it gives you the chance to move back and forth between the basics and more advanced levels, depending on your learning needs. A calculus problem solver will not just blindly give you solutions to a problem. It will in fact give a step-by-step breakup of how the problem should be solved. The calculus solver offers a lot of advantages over traditional learning sources as it ensures that students actually learn the subject and don't just copy the answers.
Let us see with the help of examples how to solve problems:

### Solved Examples

Question 1: A one-mile recetrack has two semicircular ends connected by straight lines. Express the area enclosed by the track as a function of its semicircular radius. Determine its domain.

Solution:
Step 1:
The enclosed area consist of a rectangle whose dimensions are 'x' and 2r and two semicircles of radius r, then its combine area is $\pi r^2$.

=> A = Area or rectangular part + Area of 2 semicircles

= 2r x + $\pi r^2$

=> A =
2r x + $\pi r^2$                                  ..............................(1)

Step 2:
The perimeter of recetrack is the length of the two straight sides added to the lengths of the 2 semicircular arcs.

=> 2x + 2$\pi$ r = 1

=> 2x = 1 - 2$\pi$ r

=> x = $\frac{1 - 2\pi r}{2}$

Step 3:
Put the value of 'x'  in  equation (1).

=> A(r) = 2r($\frac{1 - 2\pi r}{2}$) + $\pi r^2$

= r - 2$\pi r^2$ + $\pi r^2$

= r - $\pi r^2$

since 'r' cant be negative, r$\geqslant$ 0.

Step 4:
The perimeter of the track is fixed so the maximum value of radius (r) occurs when x = 0.

=> 2x + 2$\pi$r = 1

=> 2$\pi$r = 1,   ( because, assume x = 0)

=> r =
$\frac{1}{2\pi}$

Step 5:
The area function and its domain are
A(r) = r - $\pi r^2$
where
0$\leq$ r $\leq$ $\frac{1}{2\pi}$.

Question 2: Does f(x) = $x^{\frac{2}{3}}$ have a tangent line at x = 0.
Solution:
Step 1:
Given f(x) =  $x^{\frac{2}{3}}$

According to definition of derivative:

f '(x) = $\lim_{h\to0}$$\frac{f(x + h) - f(x)}{h} Step 2: Check x = 0 is the tangent line to the curve or not. => \lim_{x\to0}$$\frac{f(x) - f(0)}{x - 0}$

= $\lim_{x\to0}$ $\frac{x^\frac{2}{3} - 0}{x - 0}$

= $\lim_{x\to0}$ $\frac{x^\frac{2}{3}}{x}$

= $\lim_{x\to0}$$\frac{1}{x^\frac{1}{3}}$

But the limit does not exist.

LHL = $\lim_{x\to0^+}$
$\frac{1}{x^\frac{1}{3}}$ = + $\infty$

RHL = $\lim_{x\to0^-}$
$\frac{1}{x^\frac{1}{3}}$  = - $\infty$

=> Curve have a tangent line at x = 0.