Well, everyone today will be doing problem 23 from chapter 16 Section one which states to compute the line integral over the curve See of the function X squared, divided by y to the 4/3 with respect to arc length where c is the curve dramatized by X equals C squared And why equals take you dor t ranges between one and two And so whenever we're computing ah, line in a girl with respect to our clank A good first step is to change our our client representation namely the differential d s to be in terms of tea or change it into DT. So to do that, recall that the derivative of the R clank D S D. T is equal to the magnitude the velocity vector v f t. This velocity vector is the derivative of the position vector o r of the privatization of our curve, See, And so to utilize this to get the correct differential, they can multiply the differential DT over with the magnitude of e to get a representation for D s. And that will be what we substitute into are integral above now to compute exactly what this magnitude is notice that in the statement of the problem, you're given what the position vector is. We're told that the X component of our position vector is T square and the lie component of our position vector is t cubed. And so to find the velocity vector, we just take the derivative of each component. So we get to t for the X component and three t squared for the y component. And now to compute the magnitude, which is what we need, recall that the magnitude could be computed by computing the the distance formula and the distance formula is the square root of the sums of each of the components squared. And so in this case, we square the X component first, which will give us for t squared. And then we'll come square the other term three t square, which gives us nine t to the fourth. And we can also now further simplify this by recognizing that each of these terms inside the square root share a factor of t squared. And so factoring out the T squared and then recall remembering that we need to take the square root of it to pull it out we get tea on the outside and we're left with four plus 90 squared on the inside. And so this gives us the complete substitution we'll make for the differential D s. So now making the substitution for are integral since everything on I'll turn in terms of d t o R where t is our variable are bounds Also need to be in terms of tea and this will be 1 to 2 because we're told that our parameter ization ranges between those t values and now tow change the function accordingly. We first need to make our substitution cz for X and y so that will be t squared squared which will give us t to the fourth in the numerator and now plugging. And why as well we'll have t cubed all to the 4/3 power which once we adjust for the two exponents, will also be left with t to the fourth in the denominator and then substituting in what we've done for D s. We're left with tee times the square root, a four plus 90 squared T t. And one thing we can initially recognizes these two t to the force will cancel to be one. And the reason we're allowed to do this is that tea is never zero. And that is because t ranges from 1 to 2 and are in a girl. And what we're left with now is two times the square root. And we should be able to recognize that this is a prime example to use u substitution. And this is because the different the derivative of the function embedded inside the square root has a derivative that is a scaler multiple of the remaining portion of the anagram, namely t, and so setting are U substitution up you will be the embedded function for plus 90 squared and so do you will be 18 t d t and in particular, 1/18 Do you equals t d t. So this leaves us now with the integral of 1/18 won over 18 times the square root of you, do you? And to get the proper bounds, remember that the bounds must also switch. Since we're now different thing with respect to you so plugging in our bounds of tea into our equation for you we get four plus nine times the one so 13 is our lower bound and four plus nine times four, which will give us a value of 40 for upper bound and so to complete the integral. Now, we we now complete the simple single variable derivative we're left with. And so this will give us won over 18 times the anti derivative of the square root of you, which is 2/3 you to the three halves from 13 to 40. I know this the constant simplifies to 1/27 so factoring that out and now plugging in the two bounds will have 40. The three has always plug in the upper bound than minus 13 to the three halves when we plug in the lower bound, and this will give us the desired answer for the problem.

## Discussion

## Video Transcript

Well, everyone today will be doing problem 23 from chapter 16 Section one which states to compute the line integral over the curve See of the function X squared, divided by y to the 4/3 with respect to arc length where c is the curve dramatized by X equals C squared And why equals take you dor t ranges between one and two And so whenever we're computing ah, line in a girl with respect to our clank A good first step is to change our our client representation namely the differential d s to be in terms of tea or change it into DT. So to do that, recall that the derivative of the R clank D S D. T is equal to the magnitude the velocity vector v f t. This velocity vector is the derivative of the position vector o r of the privatization of our curve, See, And so to utilize this to get the correct differential, they can multiply the differential DT over with the magnitude of e to get a representation for D s. And that will be what we substitute into are integral above now to compute exactly what this magnitude is notice that in the statement of the problem, you're given what the position vector is. We're told that the X component of our position vector is T square and the lie component of our position vector is t cubed. And so to find the velocity vector, we just take the derivative of each component. So we get to t for the X component and three t squared for the y component. And now to compute the magnitude, which is what we need, recall that the magnitude could be computed by computing the the distance formula and the distance formula is the square root of the sums of each of the components squared. And so in this case, we square the X component first, which will give us for t squared. And then we'll come square the other term three t square, which gives us nine t to the fourth. And we can also now further simplify this by recognizing that each of these terms inside the square root share a factor of t squared. And so factoring out the T squared and then recall remembering that we need to take the square root of it to pull it out we get tea on the outside and we're left with four plus 90 squared on the inside. And so this gives us the complete substitution we'll make for the differential D s. So now making the substitution for are integral since everything on I'll turn in terms of d t o R where t is our variable are bounds Also need to be in terms of tea and this will be 1 to 2 because we're told that our parameter ization ranges between those t values and now tow change the function accordingly. We first need to make our substitution cz for X and y so that will be t squared squared which will give us t to the fourth in the numerator and now plugging. And why as well we'll have t cubed all to the 4/3 power which once we adjust for the two exponents, will also be left with t to the fourth in the denominator and then substituting in what we've done for D s. We're left with tee times the square root, a four plus 90 squared T t. And one thing we can initially recognizes these two t to the force will cancel to be one. And the reason we're allowed to do this is that tea is never zero. And that is because t ranges from 1 to 2 and are in a girl. And what we're left with now is two times the square root. And we should be able to recognize that this is a prime example to use u substitution. And this is because the different the derivative of the function embedded inside the square root has a derivative that is a scaler multiple of the remaining portion of the anagram, namely t, and so setting are U substitution up you will be the embedded function for plus 90 squared and so do you will be 18 t d t and in particular, 1/18 Do you equals t d t. So this leaves us now with the integral of 1/18 won over 18 times the square root of you, do you? And to get the proper bounds, remember that the bounds must also switch. Since we're now different thing with respect to you so plugging in our bounds of tea into our equation for you we get four plus nine times the one so 13 is our lower bound and four plus nine times four, which will give us a value of 40 for upper bound and so to complete the integral. Now, we we now complete the simple single variable derivative we're left with. And so this will give us won over 18 times the anti derivative of the square root of you, which is 2/3 you to the three halves from 13 to 40. I know this the constant simplifies to 1/27 so factoring that out and now plugging in the two bounds will have 40. The three has always plug in the upper bound than minus 13 to the three halves when we plug in the lower bound, and this will give us the desired answer for the problem.

## Recommended Questions

Evaluate $\int _ { C } \frac { x ^ { 2 } } { y ^ { 4 / 3 } } d s ,$ where $C$ is the curve $x = t ^ { 2 } , y = t ^ { 3 } ,$ for $1 \leq t \leq 2$

Evaluate the line integral, where $ C $ is the given curve.

$ \displaystyle \int_C (x/y) \, ds $, $ C: x = t^3 $, $ y = t^4 $, $ 1 \leqslant t \leqslant 2 $

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$$\int_{C} y^{3} d s, \quad C : x=t^{3}, y=t, 0 \leqslant t \leqslant 2$$

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Evaluate the line integral, where $ C $ is the given curve.

$ \displaystyle \int_C y \, ds $, $ C: x = t^2 $, $ y = 2t $, $ 0 \leqslant t \leqslant 3 $

Evaluate each line integral using the given curve $C$.

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$$\int_{C} x y d s, \quad C : x=t^{2}, y=2 t, 0 \leqslant t \leqslant 1$$

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