(a) Sketch the curve y^3=x^2. (b) Use Formulas 3 and 4 to set up two integrals for the are lengt

(a) Sketch the curve y^3=x^2. (b) Use Formulas 3 and 4 to...

Key Concepts

Improper Integrals

An improper integral arises when the integrand becomes unbounded or the interval of integration is infinite or includes an endpoint where the function is not well-behaved. Analyzing and evaluating such integrals requires careful limit processes to ensure that the integral converges to a finite value, which is essential in properly determining quantities like arc length.

Parametrization of Curves

Parametrization involves expressing the coordinates of a curve in terms of a single parameter, which can be chosen as x, y, or another convenient variable. Depending on the form of the curve, parametrizing the curve can simplify the computation of derivatives and integrals, especially when setting up the formula for the arc length.

Implicit Curves

An implicit curve is defined by an equation relating x and y where neither variable is isolated as a function of the other. Such curves require techniques like implicit differentiation to analyze their properties, including slopes and curvature, and can exhibit unique features such as cusps or singular behavior that differ from explicitly defined functions.

Arc Length of a Curve

The arc length of a curve is the distance along the curve between two points, computed by integrating the square root of the sum of the squares of the derivatives of the coordinate functions. This method transforms the geometric problem of finding a length into an integral calculus problem, providing a way to measure the total distance along any smooth curve.

Transcript

00:01 In this question, we're asked to draw a curve, find two integrals for arc length, and then find the arc length from another point to another point.

00:13 Let's see how we do this question.

00:16 I plotted the graph for you already, and we need to find two integrals for the arc length.

00:23 Let's first remember what our arc length formulas are.

00:26 We know that the arc length is equal to 1 plus y prime of x squared in the square root.

00:41 We also know that it's equal to 1 plus x prime of y squared.

00:53 So we know both of these are true.

00:57 The issue is we need to have, we need to have y in terms of x alone or x in terms of y alone.

01:11 So let's start with the first integral.

01:13 I'll just call this one number one, and i'll call this one number two.

01:19 Okay, for integral number one, if we have y cubed equaling x squared, we're going to have to take the cube root on both sides.

01:32 So we're going to have that y equals x to the two thirds, and differentiating both sides, we get that y prime equals just two thirds times x to the one third.

01:45 Now you have to set up our integral.

01:48 Our integral will be going from 0 to 1, and we're going to have the square root of 1 plus d, y, dx squared.

01:58 So we have 2 thirds times x to the, times x to the negative 1 third, the whole thing squared.

02:09 And what that does is that will simplify to, we'll have 1 plus, we'll have 4 over 9, 9x to the 2 thirds.

02:28 And we can get a common denominator for that by having an integral from 0 to 1 of the square root.

02:35 We'll have the square root of 9, x to the 2 thirds, plus 4, all of this divided by 3x to the 1 3 .3rd, integrated with respect to x.

02:52 Now, as you can see, this one isn't proper, because i have to, if i put in 0, exactly, i'm going to be getting a division by 0 error.

03:04 And we can't have that.

03:08 So what we need to do is we need to convert this into a limit question.

03:12 This is going to become, well, we'll do that after substitutions.

03:19 Let's make a substitution first.

03:20 Our substitution will be u equals 9x to the 2 thirds plus 4, and our du will be equal to the constant goes away.

03:31 We'll have 9 times 2 thirds, which is going to be 6 times x to the minus 1 third.

03:38 Dx.

03:49 This will work out nicely.

03:54 So our limits of integration will be changing from, if i put in 0, i get 4.

04:05 If i put in 1, you will become just 9 plus 4 or 13.

04:14 Therefore, our integral becomes the integral from 4 to 13 will have the square root of u over 3x to the 1 3rd, and dx is going to be du, over 6x to the negative 1 3rd.

04:36 But the negative 1 3rd can flip up.

04:38 So we have x to the 1 3rd times 6 d u.

05:02 As you can see here.

05:09 However, since our limit is not defined, since our equation was not defined at 0, we have to use a limit at 4.

05:20 So that's our new limit.

05:22 So our question just becomes the limit, as a goes to 4, from the right -hand side, just the integral of the square root of u over 18, du.

05:42 And that's going to be from a to 13.

05:46 So this is going to be, the integral of u is just u to the three halves times two -thirds, times 18, times 1 over 18.

05:58 This is evaluated from a to 13.

06:02 And let's simplify this, this becomes 1, this becomes 9.

06:05 So we'll have the same...