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## 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.