Find the lengths of the curves in Exercises 1-12 . If you...

Find the lengths of the curves in Exercises $1-12 .$ If you have graphing
software, you may want to graph these curves to see what they look
like.
$$x=\left(y^{4} / 4\right)+1 /\left(8 y^{2}\right) \quad \text { from } \quad y=1 \text { to } y=2$$

Key Concepts

Integration Techniques

Integration techniques are applied to evaluate the definite integral that represents the curve's arc length. This could include algebraic manipulation, substitution, or other methods to simplify the integrand, especially when the function's form leads to complex expressions under the square root. In many cases, the integral might not have an elementary antiderivative and may require numerical methods or approximations.

Arc Length of a Curve

This concept involves finding the length of a curve by integrating an expression derived from the differential elements of the curve. For a curve defined as x = f(y), the arc length from y = a to y = b is determined by integrating the square root of (1 + (dx/dy)²) with respect to y. This formula is a fundamental result in calculus for measuring distances along curved paths.

Differentiation

Differentiation is crucial here as it is used to calculate the derivative of the function x = f(y), which is necessary for the arc length formula. The computed derivative dx/dy is squared and added to one under the square root, thereby directly influencing the integrand and the final computation of the curve's length.

Transcript

00:01 Hello, welcome to this lesson.

00:02 In this lesson, find the length of the curve x equals y to the power of 4 plus 1 on 8, 1 to the power of 2 from y equals 1 to 2.

00:12 So here we'll have the length, which is equal to the integral from a to b of square root of 1 plus dx on y.

00:28 Y that's the power 2 u y it makes sense to write a function in terms of y instead of in terms of x right so let's have x equals to y to the power 4 on 4 then plus 1 on 8 then we'll have 1 to the power negative 2 because we'll differentiate this it makes it simple to write it as a product that to write it as a numerator than a denominator right and that's we have y to the power three minus one on four because the two steps down the negative two steps down so we have negative two divided by eight which is negative one four then here we bring back the y so we have equal to the the path rate.

01:27 That's a denominator.

01:29 So let's multiply 3 by y, 4 times y to the path rate and divide 3.

01:36 We'll have dx dy, which is equal to 4y to the power 6 minus 1 all over 4y to the power 3.

01:50 Okay, so if we substitute, we'll have the length, which is equal to integral, a is 1 and b is 2.

01:58 We have 1 plus.

02:02 This becomes 4y to the power 6 minus 1 all over 4y to the power 3 squared.

02:10 And this is y.

02:17 So here we can pull down 4 times y to the power 3 from the denominator side and multiply it through.

02:24 So we would have 4 squared which is 16.

02:28 Then y to the power squared which is y to the power 6.

02:32 Then plus 4y to the power 6 minus 1 squared.

02:39 Then divided by 16y to the power 6.