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

Question

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

y = x¹/² ― (1/3) x³/² , 1 ≤ x ≤ 4

Accepted Answer

Welcome back everyone. In this problem we want to find the length of the curve y equals fx over the given interval, for y equals 1/6 x cubed plus 1 divided by 2 x, where x is on the interval from 1 to 3. A says it's 10/3 units, B 14/3 units, C 8/3 units, and D 16/3 units. Now what do we know about the length of a curve over an interval that can help us to solve? Well, recall that by the arc length formula this length l will be equal to the definite integral between A and B, the bones of the interval of the square root of 1 plus the derivative of y with respect to x, all squared, all with respect to x, OK. No, in our problem, A is equal to 1 or lower bound, and our upper bound B is equal to 3. So if we can find the derivative of y with respect to x, we should be able to substitute it into the arc length formula to solve. So let's go ahead and try to do that. So given y equals 1/3 of x, 1/6 of x cubed plus 1 divided by 2 x, if we were to find the derivative of y with respect to x, we could differentiate both terms which the derivative of 1/6 of x cubed will be will give us 1/6 of 3 x 2 or 1/2 of x 2. And the derivative of 1 divided by 2 x would give us -12 multiplied by x to the -2. So since this is the derivative of Y with respect to X, then we can put it into our arc length formula. But before we do that, let's try to compute what's under our root here, that might make it easier for us to solve, OK. No. If we take the expression under a root 1 plus the derivative of y with respect to x all squared, let me just make some space here, OK. Then this will be equal to 1 + 1/2 x2 minus 1/2 x 2, OK, all squad. Now if we factor out or squared half here, really, then we could write this as 1 + 1/2 2 or 1 + 25 multiplied by x2 minus x -2 all squared, and when we apply our square here, then this will become 1 + 25 of x 22, OK, which would be x 4. -2 multiplied by x2 multiplied by x -2 plus x -22 or x -4. And no X 2 multiplied by x -2 equals 1. OK, so if we expand here, this is going to become 1 + 25 of x 4 minus, plus 25 multiplied by -2 or -52. OK. Plus, if we think about it here. A quarter of x to the 4. And that will be equal to 1/2 plus 0.25 x 4 plus 0.25 of x to the 4th. Now notice here that we can factor out 25%, OK, and when we do that, we'll end up with a 25% of. And it was a 1/5 of 2 + x 4 plus x 4, and this is the perfect square of a plus B squared, so we could rewrite this then as a quarter of x 2 plus x to the -2 all squared. So now that we know that this will be 1 plus the derivative of y with respect to xr 2, if we put it back into our arc length formula, then that means l will be equal to the integral between 1 and 3 of the square root of a quarter of x2 plus x to the -2 r 2 with respect to x. That would be equal to or integral between 1 and 3. OK, of a half of X2 + X to the -2 with respect to X. And now if we go ahead and integrate, we take out our constant 12 and our integral of x2 would be 1/3 of x cubed, while the integral of x -2 would be equal to x 1 divided by -1 or 1 divided by x, OK. So now all we need to do is to evaluate our expression between the bones of 1 and 3 and then multiply it by 1. Now this is going to be equal to 1/22. Of 3 cubed. Divided by 3 minus 1 divided by 3 minus 1 cubed divided by 3 minus 1 divided by 1, OK. I know when we simplify here we'll end up with a half of 27/3 minus 1/3 or 26/3. OK, 1/3 minus 1 or 2/3 or + 2/3. 26/3 plus 2/3 is 28/3, so this will be a 12 of 28/3 or 14/3 units. Thus, the length of y on our interval over the given interval is going to be 14/3 units. If we look back at our answer twice, that means b 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 = x¹/² ― (1/3) x³/² , 1 ≤ x ≤ 4

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Find the lengths of the curves in Exercises 19–22.
y = x¹/² ― (1/3) x³/² , 1 ≤ x ≤ 4