Question Video: Find an Integral to Represent the Arc Length of a Logarithmic Curve Mathematics • Higher Education

Write, but do not evaluate, an integral for the arc length of the curve 𝑦 = ln (2π‘₯⁡) + 7 between π‘₯ = 1 and π‘₯ = 4.

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Video Transcript

Write, but do not evaluate, an integral for the arc length of the curve 𝑦 is equal to the natural logarithm of two π‘₯ to the fifth power plus seven between π‘₯ is equal to one and π‘₯ is equal to four.

The question gives us a curve and it wants us to write an integral which gives us the arc length of this curve between the values of π‘₯ is equal to one and π‘₯ is equal to four. And we know how to find the arc length of certain curves. We know if 𝑓 prime is a continuous function on the closed interval from π‘Ž to 𝑏. Then the length of the curve 𝑦 is equal to 𝑓 of π‘₯ between the values of π‘₯ is equal to π‘Ž and π‘₯ is equal to 𝑏 is equal to the integral from π‘Ž to 𝑏 of the square root of one plus 𝑓 prime of π‘₯ squared with respect to π‘₯.

We want the arc length of the curve 𝑦 is equal to the natural logarithm of two π‘₯ to the fifth power plus seven between the values of π‘₯ is equal to one and π‘₯ is equal to four. So we’ll set our function 𝑓 of π‘₯ to be the natural logarithm of two π‘₯ to the fifth power plus seven. We’ll set π‘Ž equal to one and 𝑏 equal to four. Then, to use this arc length formula, we just need to show that 𝑓 prime is continuous on the closed interval from π‘Ž to 𝑏. So let’s find an expression for 𝑓 prime of π‘₯.

We can see if we try to differentiate 𝑓 of π‘₯, we need to differentiate a composite function. So we’ll do this by using the chain rule. If we set 𝑒 of π‘₯ to be our inner function two π‘₯ to the fifth power, then we can see that 𝑓 is a function of 𝑒 and 𝑒 is a function of π‘₯. Then by the chain rule, the derivative of 𝑓 of 𝑒 of π‘₯ is equal to 𝑓 prime evaluated at 𝑒 of π‘₯ times 𝑒 prime of π‘₯. Therefore, using the chain rule, we have 𝑓 prime of π‘₯ is equal to the derivative of the natural logarithm of 𝑒 with respect to 𝑒 times the derivative of two π‘₯ to the fifth power with respect to π‘₯.

And we can evaluate each of these derivatives separately. The derivative of the natural logarithm of 𝑒 with respect to 𝑒 is one over 𝑒. And the derivative of two π‘₯ to the fifth power with respect to π‘₯ is 10π‘₯ to the fourth power. Remember, this is an expression for 𝑓 prime of π‘₯. So we should give our answer in terms of π‘₯. We’ll do this by using our substitution 𝑒 is equal to two π‘₯ to the fifth power. Using this, we get 10π‘₯ to the fourth power divided by two π‘₯ to the fifth power, which simplifies to give us five over π‘₯. So we found that 𝑓 prime of π‘₯ is equal to five over π‘₯. This is a rational function. And we know that all rational functions are continuous on their domain. In fact, we can see the domain of 𝑓 prime of π‘₯ is all values of π‘₯ not equal to zero.

Remember, we wanted 𝑓 prime to be continuous on our closed interval from one to four. This doesn’t contain π‘₯ is equal to zero. So our function 𝑓 prime is continuous on this interval. Therefore, we’ve justified the use of our integral formula to find the length of our arc. Substituting in 𝑓 prime of π‘₯ as five over π‘₯, π‘Ž is equal to one, and 𝑏 is equal to four, we get the length of our arc is equal to the integral from one to four of the square root of one plus five over π‘₯ squared with respect to π‘₯. And the last thing we’ll do is simplify five over π‘₯ all squared to give us 25 over π‘₯ squared.

Therefore, we’ve shown we can find the arc length of the curve 𝑦 is equal to the natural logarithm of two π‘₯ to the fifth power plus seven between the values of π‘₯ is equal to one and π‘₯ is equal to four. By evaluating the integral from one to four of the square root of one plus 25 over π‘₯ squared with respect to π‘₯.

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