Evaluate ∫C (x^2)/(y^4 / 3) d s, where C is the curve x=t^2, y=t^3, for 1 ≤t ≤2

Name: Problem 23, Chapter 16: Calculus
Uploaded: 2020-04-09T02:53:56-08:00
Duration: 7 min 20 s
Channel: Kyle Flanagan
Description: Evaluate ∫C (x^2)/(y^4 / 3) d s, where C is the curve x=t^2, y=t^3, for 1 ≤t ≤2

Evaluate ∫C (x^2)/(y^4 / 3) d s, where C is the curve x=t^2,...

Key Concepts

Arc Length Differential (ds)

The differential element of arc length, denoted by ds, represents an infinitesimally small segment of the curve. When a curve is parameterized as x(t) and y(t), the arc length differential is computed using the formula ds = ?((dx/dt)² + (dy/dt)²) dt. This concept is crucial in converting a line integral into an integral in terms of the parameter, thus enabling integration over the curve.

Substitution in Integration

Substitution in integration refers to replacing the original variable of integration with a new variable that simplifies the integral. In the context of line integrals, once both the integrand and the differential (ds) are expressed in terms of the curve's parameter, the integration becomes a standard one-variable integral that can often be simplified using conventional substitution methods. This process is essential for evaluating integrals where the integrand involves composite functions of the parameter.

Curve Parameterization

This concept involves describing a curve in the plane by expressing the x and y coordinates as functions of a common parameter, usually denoted by t. Parameterizing a curve simplifies the evaluation of integrals along the curve, as it provides a clear method for substituting the variable of integration with a parameter and expressing both the integrand and the differential ds in terms of this parameter.

Transcript

0:00 Hello everyone.

00:01 Today we'll be doing problem 23 from chapter 16 section 1, which states to compute the line integral over the curve c of the function x squared divided by y to the four thirds with respect to arc length, where c is the curve parameterized by x equals t squared and y equals t cubed, or t ranges between one and two.

00:23 And so whenever we are computing a line integral with respect to arc length, a good first step is to change our arc length, representation, namely the differential ds to be in terms of t or change it into dt.

00:37 So to do that, recall that the derivative of the arc length, dsdt, is equal to the magnitude of the velocity vector v of t.

00:51 And this velocity vector is the derivative of the position vector r of the parameterization of our curve c.

00:59 And so to utilize this to get the correct differential, we can multiply the differential dt over with the magnitude of v to get a representation for ds.

01:16 And that'll be what we substitute into our integral above.

01:20 Now to compute exactly what this magnitude is, notice that in the statement of the problem, we are given what the position vector is.

01:31 We were told that the x component of our position vector is t squared, and the y component of our position vector is t cubed.

01:47 And so to find the velocity vector, we just take the derivative of each component.

01:53 So we get 2t for the x component and 3t squared for the y component.

02:07 And now to compute the magnitude, which is what we need, recall that the magnitude can be computed by computing the distance formula.

02:20 And the distance formula is the square root of the sums of each of the components squared.

02:26 And so in this case, we square the x component first, which will give us 4t squared, and then we will square the other term, 3t squared, which gives us 9t to the 4 .4.

02:41 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.

02:52 And so factoring out the t squared and then remembering that we need to take the square root of it, to pull it out, we get t on the outside and we're left with 4 plus 9t squared on the inside.

03:08 And so this gives us the complete substitution we'll make for the differential ds.

03:18 So now making the substitution for our integral, since everything will now turn in terms of d t, or where t is our variable, our bounds also need to be in terms of t.

03:31 And this will be 1 to 2 because we are told that our parameterization ranges between those t values...

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