# How to Find The Distance Between Two Points On a Curve

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Many students have difficulty finding the distance between two points on a straight line, it is more challenging for them when they have to find the distance between two points along a curve.
This article, by the way of an example problem will show how to find this distance.

### Things You'll Need

• Paper and
• Pencil
• To find the distance between two points A(x1,y1) and B(x2,y2) on a straight line on the xy-plane, we use the Distance Formula, which is...
d(AB) = √[(x1-y1)^2+(x2-y2)^2]. We will now demonstrate how this formula works by an example problem. Please click on the image to see how this is done.

• Now we will find the distance between two points A and B on a curve defined by a function f(x) on a closed interval [a,b]. To find this distance we should use the formula s =The integral, between the lower limit, a, and the upper limit, b, of the integrand √(1 +[f'(x)]^2) in respect to variable of integration, dx. Please click on the image for a better view.

• The function that we will be using as an example problem, over the closed Interval, [1,3], is...
f(x)= (1/2)[(x+4)√[(x+4)^2-1]-ln[(x+4)+√[(x+4)^2-1]]]. the derivative of this function, is...
f'(x)=√[(x+4)^2-1], we will now square both sides of the function of the derivative. That is [f'(x)]^2 = [√[(x+4)^2-1]]^2, which gives us
[f'(x)]^2 = (x + 4)^2 - 1. We now substitute this expression into the arc length formula/Integral of, s. then Integrate.

Please click on the image for a better understanding.

• Then by substitution, we have the following:
s =The integral, between the lower limit, 1, and the upper limit, 3, of the integrand √(1 +[f'(x)]^2) = the integrand √(1 + (x + 4)^2 - 1).
which is equal to √((x + 4)^2). By performing the antiderivative on this Integrand, and By the Fundamental Theorem of Calculus, we get...
{[(x^2)/2] + 4x} in which we first substitute the upper limit, 3, and from this result, we Subtract the result of the substitution of the lower limit, 1. That is {[(3^2)/2] + 4(3)} - {[(1^2)/2] + 4(1)} which is equal to {[(9/2) + 12]} - {[(1/2) + 4]} = {(33/2) - (9/2)} which is equal to (24/2) = 12. So the Arclength/distance of the function/curve over the Interval [1,3], is, 12 units.

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