Find the lengths of the curves in Exercises 19–22. ...

Question

Find the lengths of the curves in Exercises 19–22.

y = (5/12) x⁶/⁵ ― (5/8)x⁴/⁵ , 1 ≤ x ≤ 32

Accepted Answer

Welcome back everyone. In this problem we want to find the length l of the curve y equals fx over the given interval, for y equals 2/3 of x 3 halves minus 1/2 x122, where x is on the interval from 1 to 4. A says it's 316 units, B 31/3 units, C 31/4 units, and D 31/9 units. Now what do we know about the length of a curve over an interval? Well, recall that by the Arkland formula. The length l will be equal to the definite integral between a and b, the bones of y interval of the square root of 1 plus the derivative of y with respect to x squared, all with respect to x. Now, so far from our problem we know that our lower bound a is 1 and our upper bound B is 4. So if we can figure out the derivative of y with respect to x, then we should be able to substitute it in the arc length formula to solve for l. So let's go ahead and do that. So given Y, OK, is equal to 2/3 of x 3 halves minus 1/2 x 1/2, OK. Then the derivative of y with respect to x would be equal to 3 halves multiplied by 2/3. Of x of 3 halves minus 1 minus 1/22 multiplied by 122 of x 1/2 minus 1, and that would simplify the x 1/2, OK, minus a quarter of x 52. Now that we have the derivative of y with respect to x before we substitute it into our formula for L, let's first compute what's under our square root, that is 1 plus the derivative of y with respect to x all squared. OK, let's see if there's an opportunity here to simplify. Now if we think about it, this will be equal to one plus. X122 minus 25 of x 122, all squared. And when we expand that, it will be equal to 1 + 2 multiplied by x12 multiplied by a quarter of x 12, OK, plus 1/16 of x to the power of -1, and this would eventually simplify, OK, when we go ahead and multiply through to x plus 1/22 plus 1 divided by 16 x. Now notice here that this is a perfect square, so we could rewrite it then as x12, OK, plus a quarter of x 12 all squared. And now that we have that if we put it into our formula for the integral. Then this means that L. Will be equal to the definite integral between 1 and 4 of the square root of x to the 12 plus. A quarter of x to the negative12 are squared with respect to x. No, the square cancels the square root, so this is going to be our integral between 1 and 4 of x122 plus a quarter of x to the negative12 with respect to x. And now when we integrate by integrating each term. Then this will simplify too. X 3 halves divided by 3 halves or 2/3 of x 3 halves again. Plus a half of x to the power of 1/2 between the bones of 1 and 4, and all we need to do now is to substitute our upper and lower bones and then solve. So when we do that, that's going to be 2/3 of 43 halves, plus 1/22 of 4 1/2. Minus 2/3 of 13 halves, OK, plus 1/2 1 1/2. And that is going to be 2/3 of 8, OK, 2/3 of 8, which is uh 16/3, OK, plus 1/2 of 2, which is 1. So that's 16/3 + 1 minus 2/3 of 1, which is 2/3, OK, plus 1/2 1/2, which is 1, or sorry, which is 1/2. No, we can simplify that to be 19/3 minus 2/3 plus 1/2, which would be 7/6, and if we write both of these with the same denominator of 6, then that's going to be 38/6 minus 7/6, which equals 316 units. Thus, the length, OK, the length of the curve y equals X, Y equals fx over the given interval is 31/6 units. When we look back at our answer choices, that means A is the correct answer. Thanks a lot for watching everyone. I hope this video helped.

Find the lengths of the curves in Exercises 19–22. y = (5/12) x⁶/⁵ ― (5/8)x⁴/⁵ , 1 ≤ x ≤ 32

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Find the lengths of the curves in Exercises 19–22.
y = (5/12) x⁶/⁵ ― (5/8)x⁴/⁵ , 1 ≤ x ≤ 32