Arc length of a parabola Let L(c) be the length of the parabol...

Question

Arc length of a parabola Let L(c) be the length of the parabola f(x) = x² from x = 0 to x = c, where c ≥ 0 is a constant.

a. Find an expression for L.

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says let LD be the arc length of the curve G of X, which is equal to 3x2, from X equal to 0 to X equal to D, where D is greater than or equal to 0. Find an explicit expression for L of D. So here we're asked to find the arc length of a curve, and so we call your formula for finding the arc length of a curve. So we'll have the arc length L of D. And this is going to be equal to the integral from 0 to D of the square root of the quantity of 1, plus the quantity of G of X. In quantity squared in quantity DX. So first step is to find G of X, and G of X is just going to be equal to the derivative with respect to X of G of X, which is equal to the derivative with respect to X of the quantity of 3 x squared in quantity, which when we evaluate this derivative is going to be equal to 6 X. So making the substitution back into our formula, we see that LD is going to be equal to the integral from 0 to D. Of the square root of the quantity of 1 plus the quantity of 6 x and quantity squared in quantity dx. Now to get this into a form that we can actually integrate, we need to make a substitution. So we'll set U equal to 6 X. So that means that DU is going to be equal to 6DX. However, we only have just DX in our integral, so we'll divide both sides of this expression by 6, and we'll get 1 divided by 6 multiplied by D U is equal to DX. So making these substitutions into our integral we see that LD is going to be equal to, and here we're going to factor out the 1 divided by 6 in front of the integral. So we'll have 1 divided by 6 multiplied by the integral, and here we have to change our limits of integration since we're changing variables. So when X is equal to 0 for the lower limit, we have U is equal to 6 multiplied by 0, which is 0, and for the upper limit, when X is equal to D, we see that U is equal to 6D. And then for the integram we'll have the square root of the quantity of 1 plus U squad in quantity D U. And so this is actually in a form that we can integrate. And so we can integrate this by using one of our standard formulas of integration. And so recall that the integral. Of the square root of the quantity of A2 plus U2 in quantity, D U is equal to. You divided by 2 multiplied by the square root of the quantity of a 2 plus u squad in quantity plus a 2 divided by 2 multiplied by the natural log of the absolute value of the quantity of u plus the square root of the quantity of a quad plus U2 in quantity plus C. So using this integration formula, we see that our integral is going to be equal to 1 divided by 6 multiplied by the quantity of u divided by 2 multiplied by the square root of the quantity of 1 plus u squad in quantity, plus 1 divided by 2 multiplied by the natural log of the absolute value of U plus the square root of the quantity of 1 plus u squared. In quantity And this expression is going to be evaluated from 0 to 6D. Then we just need to substitute in our limits of integration. And so this is this expression is going to be equal to 1 divided by 6, multiplied by the quantity, and for our upper limit, we'll have 6 D divided by 2. Multiplied by the square root of the quantity of 1 + 36 D squared. In quantity Plus 1/2 multiplied by the natural log of the absolute value. Of the quantity of 6D plus the square root of the quantity of 1 plus 36D squared in quantity. Then we'll have minus our lower limit, so we'll have the quantity of 0 divided by 2 multiplied by the square root of the quantity of 1 plus 0 squared in quantity, plus 1 divided by 2 multiplied by the natural log of the absolute value of 0, plus the square root of the quantity of 1 plus 0 squared. In quantity. And so now we just need to simplify this expression. So, in doing so, this is going to be equal to. Here, we'll go ahead and factor in the 1 divided by 6 from out front. So, in doing so, we will have D divided by 2 multiplied by the square roots. Of the quantity of 1 plus 36 multiplied by D squared in quantity, plus. 1 divided by 12 multiplied by the natural log of the quantity, and here we can actually drop the absolute value sign for this expression, since D is a positive quantity since it's greater than or equal to 0. So that means that everything in the argument of the natural log function inside that absolute value sign is a positive quantity. So we'll have the natural log of the quantity of 6D plus the square root. Of the quantity of 1 + 36 multiplied by D squared in quantity, then we'll have minus the quantity of 0. Plus 1 divided by 12 multiplied by the natural log of 1. In quantity And so simplifying the expression, we see that this is going to be equal to D divided by 2, multiplied by the square root of the quantity of 1 plus 36 D squared in quantity. Plus 1 divided by 12 multiplied by the natural log of the quantity of 60 plus the square root of the quantity of 1 plus 36D squared. In quantity and the rest of the terms are 0 since the natural law of 1 is 0. So this here is the answer that we are looking for for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

Arc length of a parabola Let L(c) be the length of the parabola f(x) = x² from x = 0 to x = c, where c ≥ 0 is a constant.
a. Find an expression for L.

Key Concepts

Arc Length Formula

Derivative of a Function

Definite Integral

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