Calculate the arc length of y=x^2 over [0, a] . Hint: Use trigonometric substitution. Evaluate f

Calculate the arc length of y=x^2 over [0, a] . Hint: Use...

Key Concepts

Arc Length Formula

The arc length formula is a fundamental concept in calculus used to compute the length of a curve defined by a function over a given interval. It involves integrating the square root of one plus the square of the derivative of the function with respect to the independent variable. This formula provides a method to measure the distance along a continuously curved path.

Differentiation

Differentiation is a key tool in calculus that involves computing the derivative of a function. In the context of finding an arc length, obtaining the derivative of the function is imperative because the formula requires squaring the derivative and adding one to it under the square root. This step translates the geometric notion of slope into an analytical expression.

Trigonometric Substitution

Trigonometric substitution is an integration technique used to simplify integrals that involve square roots of quadratic expressions. It involves substituting a trigonometric function for a variable to transform a complicated algebraic integrand into a form that is easier to integrate, often by exploiting Pythagorean identities.

Definite Integration and Evaluation

Definite integration is the process of computing the integral of a function over a specific interval to obtain a numerical value. After applying techniques such as differentiation and trigonometric substitution, definite integration is used to evaluate the length of a curve between two points. The evaluation step may also include substituting specific parameter values to obtain a precise numerical result.

Transcript

00:01 In this problem, we're going to compute the arc length of y equals x squared from 0 to a.

00:07 And so that's just the length of the writ portion of the curve.

00:12 And to find that arc length, we're going to use the arc length formula, which is the integral from 0 to a of the square root of 1 plus the y, dx squared dx.

00:36 And since y is equal to x squared in this problem, that's our curve, we know that dydx is 2 times x.

00:49 So this is the integral from 0 to a of the square root of 1 plus the quantity 2x squared dx.

01:04 And now from here we can use a trig substitution to handle this integral.

01:09 So in this case, we let 2x be equal to tangent of theta.

01:21 So then we have that dx is one -half times secant squared theta, d -theta.

01:38 And so this integral becomes the integral from 0 to a of the square root of 1 plus, let's see, 2x is tangent of theta.

01:53 So that's 1 plus tangent square theta.

01:59 Multiplied by one half times sequence square theta d theta.

02:14 And then we can actually simplify this part a bit more because we know that, i'll write it over here, we know that cosine squared theta plus sine square theta is equal to one.

02:44 So we can divide by cosine square theta and we get that this is one plus tangent squared theta is equal to 1 over cosine square theta, which is just secant squared theta.

03:05 And that's what we have on the inside of this square root.

03:09 So if we replace that by secant squared theta, they cancel out, and we get that this is the integral from 0 to a of 1⁄2 times secant cubed theta.

03:23 But we will leave this as secant theta times secant squared theta.

03:36 D -theta.

03:39 And our reason for doing that is so that we can set up, so that we can set up integration by parts.

03:49 So here we can let you be equal to one -half times secan -theta, and then d -v can be siquant -theta d -theta.

04:23 So then we have that d -u is equal to one -half times secant -theta -tangent -theta.

04:31 And v is equal to tangent theta.

04:52 And remember with integration by parts, we get that this integral here, that that is equal to u times v minus the integral of vdu.

05:18 So that works out to be, this is equal to one -half times secant theta tangent theta.

05:38 And we also have to remember our limits of integration.

05:41 Which is from 0 to a minus 1 half.

05:49 I forgot the integral sign right.

05:53 This is uv minus the integral of vdu.

05:57 So this is 1 half secant theta, tangent theta from 0 to a, minus the integral from 0 to a of 1 half secant theta times tangent.

06:18 So 1 half secant theta, tangent theta times tantal.

06:21 Tangent theta again, which i will write as just tangent squared theta, d theta.

06:38 Alright, so this is equal to one half times secant theta, tangent theta, and then including our limits of integration, minus the integral from 0 to a...

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