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.

y = x^3/2 / 3 − x^1/2 on [4, 16]

Accepted Answer

Hello. In this video, we are going to be determining the arc length of the curve. Y is equal to X 5th divided by 2, divided by 5, plus 1 divided by X/ power on the boundary from 1 to 4. Now, in order to do this, let's go ahead and review what the arc length is defined as. The arc length of a function is defined. As the definite integral from A to B. Of the square root. Of one plus the functions derivatives squared. DX So what we want to go ahead and do is we want to go ahead and find the derivative of the given function. Now, if we were to rewrite the given function, we can rewrite the function to be F of X is equal to 15th multiplied by X 5 halves, plus X to the negative/ power. Now, what we're going to do is we're going to take the derivative of this function with respect to X. Now, the derivative with respect to X is going to give us F of X, and that is going to equal to 15th, multiplied by the derivative of X to 5 halves power, which will be 5 halves multiplied by X to the 3 halves power. Minus the derivative of X to the negative half power, which will be 1/2 x -32. So this is going to be the derivative of the given function. If we further simplify, this will be 1/2, X 3 halves, minus 1/2 x to the negative 3//s power. Now let's go ahead and square the function, because we need the derivative to be squared. By squaring both sides of this function, we get F of X quantity squared equal to the quantity 1/2 x 3 halves minus 1/2 x 3/2 quantity squared. And if we were to expand the right-hand side of this equation, we will be left with 1/4 X cubed minus 1/2 plus 1/4 X 3 power. So this is going to be the value of the squared function. And finally, in the square root, we need to add 1 to the squared function. So by adding 1, we get 1 plus F of X quantity squared, and that is going to equal to 1 plus 1/4 X cubed minus 1/2 plus 1/4 to multiply by X to the negative 3 power. And by combining like terms, this is going to simplify to be 1/4 X cubed plus 1/2 plus 1/4 X 3 power. And conveniently here, this is a factorable trinomial. This is going to factor to be 1/2 x to the 3 halves power, plus 1/2 x 3/2 power quantity squared. So this is going to be the simplified form of the integran that is going to be inside the square root. In the arc length function. So now if we take this quantity and plug this into the arc length function, The arc length will be defined as the definite integral of the square root. Of the quantity, 1/2 x 3 halves, plus 1/2 x 3/2 power quantity squared DX. Now for the definite integral, we are told that we want to work on the boundary from 1 to 4, so the boundaries are going to be from 1 to 4. Now, if we simplify the integrand, the square root will cancel the square quantity, so we are left with the definite integral from 1 to 4 of 12 X to the 3 halves power, plus 1/2 x to the negative 3//al power DX. Now what we will go ahead and do is take the antiderivative of each term. By taking the antiderivative, we are left with 15th multiplied by X to the 5/3 power. Minus X to the negative power, and this is going to be evaluated from 1 to 4. Now, by using the fundamental theorem part 2, this is going to simplify to be 1/5 multiplied by 4 5/3 power minus 45 power. Minus the value of the lower boundary, and if we were to plug in one to each of the terms, we will be left with 15 minus 1. All we need to do now is simplify. This will simplify to leave us with 32, divided by 5 minus 1/2, minus 8 divided by 10. -8 divided by 10. No. Wait, apologies, no. This will simplify to just be 1/50 minus 1. Then by further combining like terms, we will simplify this further to be 59 divided by 16, plus 8 divided by 10, and this is going to simplify even further to leave us with 67 divided by 10, which is going to be the arc length of the given curve, and the solution to the problem. And with that being said, the overall solution is going to be a. So I hope this video helps you in understanding how to approach this problem, and we will go ahead and see you all in the next video.

9–20. Arc length calculations Find the arc length of the following curves on the given interval.y = x^3/2 / 3 − x^1/2 on [4, 16]

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9–20. Arc length calculations Find the arc length of the following curves on the given interval.
y = x^3/2 / 3 − x^1/2 on [4, 16]