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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?

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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

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

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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. – Arkamis 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 $\frac{1}{4.4}\int_0^{3.7} u^{1/2}\ du.$ – Arkamis Feb 21 '13 at 16:42

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