Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The rectangular grounds at Hillingham University is going to have a new path built which takes the curved shape of $y=2.2x^2$, starting from the south-west corner - (taken as the origin in a graph) over to the point on the path where $x=3.7$. The constructors of this path want to know the path's length.

What is the exact length of this path?

share|cite|improve this question

The length of a curve is given by

$$\int_a^b \sqrt{1+(f'(x))^2}\ dx.$$

Let $f(x)= y$, $f'(x) = \frac{dy}{dx}$, $a=0$ and $b=3.7$, and compute the integral in a straightforward manner.

Where does this come from?

Consider estimating the length of a segment of the curve by forming a right triangle whose vertices are $(x,f(x))$, $(x+\Delta x, f(x))$ and $(x+\Delta x, f(x+\Delta x))$.

Let $\Delta y = f(x+\Delta x)-f(x)$.

Then, the length of the hypotenuse, $s$, is given by the Pythagorean Theorem $s = (\Delta x)^2 + (\Delta y)^2$.

Taking the limit as $\Delta x \to 0$, using the notation of infinitesimals, we see that $\Delta y \to dy$ and $\Delta x \to dx$.

To get the total length of the curve, we add up all these infinitesimal lengths:

$$\int_a^b ds.$$

But from the Pythagorean theorem we have that $ds^2 = dx^2+dy^2$, so $\frac{ds^2}{dx^2} = 1+\frac{dy^2}{dx^2} = 1+\left(f'(x)\right)^2$.

Finally, multiplying $dx^2$ back over and taking the square root, we have

$$ds = \sqrt{1+(f'(x))^2}\ dx.$$

The length of the curve is found by adding up all these infinitesimal hypotenuses, so

$$L(f(x)) = \int_a^b ds = \int_a^b \sqrt{1+(f'(x))^2}\ dx.$$

share|cite|improve this answer
is this a standard result? how do you know this? – Anona anon Feb 20 '13 at 17:37
also, why does the function have to be differentiated first? – Anona anon Feb 20 '13 at 17:39
This is a standard result that comes from basic calculus. I'll expand the answer. – Emily Feb 20 '13 at 17:39
would you integrate this using trig substitution ? if so how? thank you – Anona anon Feb 21 '13 at 15:41
Well, $f(x) = 2.2x^2$, so $f'(x) = 4.4x$, so you integrate $\left[1+4.4x\right]^{1/2}$ through basic techniques. Substitution is a way; alternatively, you can recognize that multiplying the integral by 4.4 inside, and 1/4.4 outside gives you $\frac{1}{4.4}\int_0^{3.7} u^{1/2}\ du.$ – Emily Feb 21 '13 at 16:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.