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So I’ve been wanting to figure out a formula for an odd pattern I found… but to write a proof, I need to know one thing…

How do I find the distance between two points on a parabola?

Like, if I have y=x^2, how do I find out how much actual line is between say, (4,16) and (2,4)? If you use the normal line distance thing, it tells you how much space is between each point, but not how much of the actual parabola is between the points.

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So can you help me out?

Thanks.

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If you have a curve $y=f(x)$, the length of the curve between points $(x_1,f(x_1))$ and $(x_2,f(x_2))$ is given by

$$L(x_1,x_2) = \int_{x_1}^{x_2} \sqrt{(dx)^2 + (df(x))^2} = \int_{x_1}^{x_2}\sqrt{1+\left(\dfrac{df}{dx} \right)^2} dx$$

In your case, $f(x) = x^2$. Hence, $\dfrac{df}{dx} = 2x$. Hence, the length of curve between the points $(x_1,x_1^2)$ and $(x_2,x_2^2)$ is

$$L(x_1,x_2) = \int_{x_1}^{x_2} \sqrt{1+(2x)^2} dx = \int_{x_1}^{x_2} \sqrt{1+4x^2} dx$$

Can you evaluate this integral?

$$\int_{x_1}^{x_2} \sqrt{1+4x^2} dx = \left.\dfrac{2x \sqrt{4x^2+1} + \log(2x+\sqrt{1+4x^2})}4 \right \vert_{x_1}^{x_2}$$

**EDIT**

Added picture to highlight the different lengths.

Rectification (that’s what this is called) of the curve given by $f(x)$ is obtained by the integral

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

Hre you need to evaluate

$$\int_2^4\sqrt{1+4x^2}\,\mathrm dx.$$

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