9–20. Arc length calculations Find the arc length of the follo...

Question

9–20. Arc length calculations Find the arc length of the following curves on the given interval.

x = y⁴/4 + 1/8y², for 1≤y≤2

Accepted Answer

Hi everyone, let's take a look at this practice problem. This problem says to find the arc length of the curve, Y is equal to X 6 divided by 6, plus 1 divided by the quantity of 16 multiplied by X to the fourth in quantity on the closed interval from 1 to 2. So here we're asked to find the arc length of a curve. So recall your formula for the arc length of a curve. That arc length is going to be L, and that's going to be equal to the integral from A to B. Of the square root of the quantity of 1 plus the quantity of the derivative of Y with respect to X in quantity squared in quantity DX. So our next step is to find the derivative of Y with respect to X. This is going to be equal to the derivative with respect to X of quantity of X 6 divided by 6 plus 1 divided by the quantity of 16 multiplied by X to the 4th in quantity. So taking this derivative, this is going to be equal to x 5. Minus 1 divided by the quantity of 4 multiplied by X to the 5th in quantity. So substituting this result into our formula for the arc length, we see that L is going to be equal to the integral, and our limits of integration here we're going to go from 1 to 2, since that is the interval over which we're looking for the arc length. And then for the ingrain we'll have the square root of the quantity of 1 plus the quantity of X 5th. -1 divided by the quantity of 4 multiplied by X to the 5th. In quantity squared and quantity DX. So now what we want to do is to multiply out that binomial underneath the radical. In doing so, this is going to be equal to the integral from 1 to 2 of the square roots of the quantity of 1 plus X 10th. -1 divided by 2. Plus 1 divided by the quantity of 16 multiplied by X to the 10th in quantity, then another in quantity, DX. And so, collecting like terms underneath the radical, we see that this is going to be equal to the integral from 1 to 2. Of the square root Of the quantity of 1 to 10th. Plus 1 divided by 2, plus 1 divided by the quantity of 16 multiplied by X 10th in quantity, then another in quantity DX. Now, the polynomial underneath the radical is a perfect square. So rewriting that, we will see that this is going to be equal to the integral from 1 to 2. Of the square root of the quantity. Of X to the 5th. Plus, 1 divided by the quantity of 4 multiplied by X to the 5th in quantity. And then another in quantity squared in quantity DX. So, taking the square root, we see that we have our arc length is going to be equal to the integral from 1 to 2. Of the quantity of X 5 plus 1 divided by the quantity of 4 multiplied by X 5th in quantity DX. So, integrating turn by turn, we see that this is going to be equal to. The quantity of X 6 divided by 6. Minus 1 divided by the quantity of 16 multiplied by X the 4th in quantity, and this needs to be evaluated from 1 to 2. So, substituting in our limits of integration, we see that L is going to be equal to. 2, raise to the power 6, divided by 6, minus 1 divided by the quantity of 16 multiplied by 2 raise to the power 4 in quantity. -1, raise to the power of 6 divided by 6. Plus 1 divided by the quantity of 16 multiplied by 1 to the 4th in quantity. And when we evaluate this expression, we see that this is going to be equal to. 2,703 divided by 256. 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.

9–20. Arc length calculations Find the arc length of the following curves on the given interval.x = y⁴/4 + 1/8y², for 1≤y≤2

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9–20. Arc length calculations Find the arc length of the following curves on the given interval.
x = y⁴/4 + 1/8y², for 1≤y≤2