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.

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.