# Video: Finding the Length of an Arc of a Given Curve

Work out the length of the arc π¦ = (π₯βΆ + 32)/(32π₯Β²) between π₯ = 1 and π₯ = 3. Give your answer as a fraction.

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### Video Transcript

Work out the length of the arc π¦ is equal to π₯ to the sixth power plus 32 all divided by 32π₯ squared between π₯ is equal to one and π₯ is equal to three. Give your answer as a fraction.

The question wants us to calculate the length of the arc given by the curve π¦ is equal to π₯ to the sixth power plus 32 all divided by 32π₯ squared between π₯ is equal to one and π₯ is equal to three. And it wants us to give our answer as a fraction. We recall, for a curve π¦ is equal to π of π₯, if our derivative function π prime is continuous on the closed interval from π to π. Then we can calculate the length of the arc π¦ is equal to π of π₯ from π₯ is equal to π to π₯ is equal to π by calculating the integral from π to π of the square root of one plus π prime of π₯ squared with respect to π₯.

The question wants us to calculate the length of the arc π¦ is equal to π₯ to the sixth power plus 32 all divided by 32π₯ squared between π₯ is equal to one and π₯ is equal to three. So weβll set π of π₯ to be equal to π₯ to the sixth power plus 32 all divided by 32π₯ squared, π equal to one, and π equal to three. Now, if we can show that our derivative function π prime is continuous on the closed interval from one to three, then the length of the arc π¦ is equal to π of π₯ from π₯ is equal to one to π₯ is equal to three is given by this integral.

So letβs check that our derivative function π prime of π₯ is continuous on the closed interval from one to three. We have that π of π₯ is equal to π₯ to the sixth power plus 32 all divided by 32π₯ squared. We could differentiate this using the quotient rule. However, itβs easier to divide both terms in our numerator by 32π₯ squared. π₯ to the sixth power divided by 32π₯ squared is equal to π₯ to the fourth power divided by 32. And 32 divided by 32π₯ squared is one divided by π₯ squared, which is equal to π₯ to the power of negative two.

Weβre now ready to calculate an expression for our derivative function, π prime of π₯. We recall for constants π and π the derivative of ππ₯ to the πth power is equal to π multiplied by π multiplied by π₯ to the power of π minus one. So to differentiate π₯ to the fourth power divided by 32, we multiply by four and reduce the exponent by one, giving us four π₯ cubed over 32, which simplifies to give us π₯ cubed divided by eight.

Similarly, to differentiate π₯ to the power of negative two, we multiply by the exponent of negative two and then reduce the exponent by one, giving us negative two multiplied by π₯ to the power of negative three. If we were to add the two terms of our function π prime of π₯, we would see that our function is a rational function. And we know that all rational functions are continuous on their domain. So we want to know the domain of our function π prime of π₯. Well, the only problem point would be when π₯ is equal to zero since then weβll be dividing by zero.

In particular, this means that our derivative function π prime of π₯ is continuous on the closed interval from one to three. So weβve shown that our prerequisite is true. Therefore, we can calculate the length of our arc, π¦ is equal to π of π₯ from π₯ is equal to one to π₯ is equal to three, as the integral from one to three of the square root of one plus π prime of π₯ squared with respect to π₯. Weβll start by substituting in what we calculated for our derivative function π prime of π₯. Next, weβll distribute the exponent over our parentheses.

Weβll start by squaring the first term. This gives us π₯ to the sixth power divided by 64. Then we get two lots of π₯ cubed over eight multiplied by negative two π₯ to the negative three. We get that π₯ cubed multiplied by π₯ to the negative three cancels. So we just have two multiplied by negative two over eight, which just simplifies to negative a half.

Finally, we want to square the final term in our parentheses. This gives us four multiplied by π₯ to the power of negative six. We can combine the terms of one and negative a half to just give us a half. At this point, there are several different ways of evaluating this integral. For example, we could use the substitution π’ is equal to π₯ to the power of negative six. However, weβre going to factor π₯ to the power of negative six divided by 64 from the expression inside of our square root. We have that π₯ to the power of 12 multiplied by π₯ to the power of negative six over 64 gives us our first term. Similarly, to get a half, we need to multiply π₯ to the power of negative six divided by 64 by 32π₯ to the sixth power. And 256 multiplied by π₯ to the power of negative six over 64 is equal to four π₯ to the power of negative six.

Now, first, this might not seem any simpler. Itβs the product of two functions. And we still canβt get rid of the square root sign inside of our integral. However, we can notice we can factor the expression inside of our parentheses. Itβs actually π₯ to the sixth power plus 16 all squared. Similarly, π₯ to the power of negative six divided by 64 is equal to π₯ to the power of negative three divided by eight all squared. So weβre actually taking the square root of the product of two squares. So we can simplify. Taking the square root of the product of our two squares gives us the integral from one to three of π₯ to the power of negative three divided by eight multiplied by π₯ to the sixth power plus 16 with respect to π₯.

So letβs clear some space and calculate this integral. Weβll start by distributing our coefficient over the parentheses. Distributing gives us the integral from one to three of π₯ cubed over eight plus two multiplied by π₯ to the power of negative three with respect to π₯. We recall for constants π and π, where π is not equal to negative one, to integrate π multiplied by π₯ to the πth power with respect to π₯, we add one to the exponent and divide by this new exponent. Then we add a constant of integration π. We can use this to evaluate our integral. We get π₯ to the fourth power over 32 minus π₯ to the power of negative two between the limits of one and three.

Finally, we can evaluate this at the limits of π₯ is equal to one and π₯ is equal to three to give us 81 over 32 minus one-ninth minus one over 32 plus one, which we can calculate to be equal to 61 divided by 18. Therefore, we have shown that the length of the arc π¦ is equal to π₯ to the sixth power plus 32 all divided by 32π₯ squared between π₯ is equal to one and π₯ is equal to three is equal to 61 divided by 18.