Video: Arc Length of Planar Curves

Find the function 𝑠(π‘₯) which gives the length of the arc 𝑦 = √(π‘₯Β³) from (0, 0) to (π‘₯, √(π‘₯Β³)).

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Video Transcript

Find the function 𝑠 of π‘₯ which gives the length of the arc 𝑦 is equal to the square root of π‘₯ cubed from the point zero, zero to the point π‘₯, square root of π‘₯ cubed.

We recall that if the derivative 𝑓 prime is continuous on the closed interval from π‘Ž to 𝑏, then the integral from π‘Ž to 𝑏 of the square root of one plus the derivative 𝑓 prime squared with respect to π‘₯ is equal to the length of the arc 𝑦 is equal to 𝑓 of π‘₯ from the point π‘Ž, 𝑓 of π‘Ž to the point 𝑏, 𝑓 of 𝑏. So to utilize this formula for the length of an arc, we’re going to need to find the relevant values of π‘Ž and 𝑏. And we’re also going to need to show the arc derivative function 𝑓 prime is continuous on the closed interval π‘Ž to 𝑏.

Since the question is asking us to find the length of the arc 𝑦 is equal to the square root of π‘₯ cubed, we must have the function in our arc length formula 𝑓 of π‘₯ is equal to the square root of π‘₯ cubed. And since the question asked us to find the length of the arc from the point zero, zero to the point π‘₯, square root of π‘₯ cubed, if we compare these coordinatewise with the coordinates given to us in the formula, we can see that we must have π‘Ž is equal to zero and 𝑏 is equal to π‘₯. Now, we need to find the interval that our function 𝑓 prime of π‘₯ is continuous on.

We have from the question that we want to find the length of the arc 𝑦 equals the square root of π‘₯ cubed. So we must have that 𝑓 of π‘₯ is equal to the square root of π‘₯ cubed, which we can rewrite as π‘₯ to the power of three over two. So by the power rule for differentiation, we can differentiate π‘₯ to the three over two as three over two multiplied by π‘₯ to the power of a half, which we can rewrite as three root π‘₯ all divided by two. Now, we also know that the principal square root function is continuous for all π‘₯ greater than or equal to zero. We can use this to argue that for any constant π‘˜, π‘˜ multiplied by the square root of π‘₯ is continuous for all π‘₯ greater than or equal to zero.

Therefore, since our function 𝑓 prime of π‘₯ is equal to three over two multiplied by the square root of π‘₯, we can conclude that our function 𝑓 prime of π‘₯ is continuous for all π‘₯ greater than or equal to zero. Therefore, we have justified that we can use our arc length formula on the arc 𝑦 is equal to the square root of π‘₯ cubed from the zero, zero to the point π‘₯, square root of π‘₯ cubed. What this means is if we set the function 𝑠 of π‘₯ to be equal to the integral from zero to π‘₯ with respect to π‘₯ of the square root of one plus the square of our derivative function three root π‘₯ over two, then 𝑠 evaluated to π‘₯ will be equal to the length of the arc of 𝑦 equals the square root of π‘₯ cubed from the point zero, zero to the point π‘₯, square root of π‘₯ cubed.

We must now start evaluating this integral. We see that if we square the value inside of our parentheses, then we will get nine π‘₯ divided by four. At this point, the easiest way to evaluate this integral is by using a substitution. We will use the substitution 𝑒 is equal to one plus nine π‘₯ divided by four. This means that the derivative of 𝑒 with respect to π‘₯ is equal to nine over four, which is equivalent to saying four over nine d𝑒 is equal to dπ‘₯. Using this information, we can rewrite our integral as the integral of the square root of 𝑒 multiplied by four-ninths d𝑒, where we must be careful. Since we’re using integration by substitution, the limits of our integral will change.

To find the upper limit of our integral, we see that when π‘₯ is equal to π‘₯, we will have that 𝑒 is equal to one plus nine π‘₯ over four. So we can write in the upper limit of our integral as one plus nine π‘₯ over four. And we see that when π‘₯ is equal to zero, we have that 𝑒 is equal to one plus nine multiplied by zero divided by four which is equal to one. So we can write in the lower limit of our integral as one. At this point, we can see that we can take the constant four over nine outside of our integral, giving us four-ninths multiplied by the integral from one to one plus nine π‘₯ over four of the square root of 𝑒 with respect to 𝑒.

Since we can rewrite the square root of 𝑒 as 𝑒 to the power of a half, we can use our power rule for integration to find that the integral of the square root of 𝑒 is equal to two-thirds multiplied by 𝑒 to the power of three over two. At this point, we can take the constant of two-thirds outside of our evaluation, giving us eight divided by 27 multiplied by 𝑒 to the power of three over two evaluated at one plus nine π‘₯ over four and one. We then evaluate this at the limits of our integral, giving us eight over 27 multiplied by one plus nine π‘₯ over four to the power of three over two minus one to the power of three over two. We can then see that one to the power of three over two is just equal to one.

To add one and nine π‘₯ over four together, we will need the denominators of the fractions to be equal. So we can rewrite one as four divided by four, meaning that we can then add these two fractions together to give us four plus nine π‘₯ all divided by four. Next, we want to take out the shared factor of a quarter, giving us a new expression of eight divided by 27 multiplied by a quarter of four plus nine π‘₯ all to the power of three over two minus one.

Next, we want to distribute our exponent of three over two over each factor. We can then calculate that one-quarter raised to the power of three over two is equal to one-eighth. Finally, if we distribute our coefficient of eight over 27 over our parentheses, we get that our function 𝑠 of π‘₯ is equal to nine π‘₯ plus four to the power of three over two minus eight all divided by 27, which gives us the length of the arc 𝑦 is equal to the square root of π‘₯ cubed from the point zero, zero to the point π‘₯, square root of π‘₯ cubed.

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