Evaluate the line integral 𝐅 ·d 𝐫, where 𝐅(x, y)=x 𝐢+y 𝐣 and C is the curve given by 𝐫(t)=t 𝐢+t

Evaluate the line integral 𝐅 ·d 𝐫, where 𝐅(x, y)=x 𝐢+y 𝐣 and...

Key Concepts

Definite Integration

Definite integration is the process of integrating a function over a specific interval, resulting in a numerical value that represents an accumulated quantity such as area or work. In line integrals, once the vector field and the curve are expressed in terms of a common parameter, the dot product is integrated over the parameter's range to obtain the total effect of the field along the curve.

Dot Product

The dot product is an algebraic operation that takes two equal-length sequences of numbers (typically vectors) and returns a single number. In the context of line integrals, the dot product of the vector field with the differential displacement vector yields the contribution of the field in the direction of the path, which is then accumulated over the interval of integration.

Line Integral of a Vector Field

A line integral of a vector field is a way to accumulate a quantity along a curve by integrating the dot product of a vector field with an infinitesimal displacement vector. This concept is fundamental in multivariable calculus and applications such as computing work done by a force field along a path, where the integral sums the contributions of the vector field along the given trajectory.

Parametric Representation of Curves

Parametric representation involves describing a curve by specifying its coordinates as functions of a parameter, usually denoted by t. This method allows one to express both the geometry and the dynamics of the curve in a convenient way, facilitating calculations such as line integrals by converting them into integrals over the parameter's interval.

Transcript

00:01 So in this question, we're asked to determine the line integral of x, y, ds, and we're given that the curve, we have a parameterized curve, where x is equal to t squared and y is equal to t to t, and we're told that t goes from zero to one.

00:25 All right, so the first thing we're going to do is we're going to determine the arc length or ds, and we know that ds is the square root of d .y by dt squared plus dx by dt squared dt.

00:41 So if we take y and we find its derivative with respect to t, so y is just 2t.

00:47 So the derivative of that with respect to t.

00:54 So 2t, the derivative of 2t with respect to t is just 2.

00:59 And then also if we take the derivative of 2t with respect to t is just 2.

01:01 X with respect to t, so x is just t squared.

01:05 The derivative of that with respect to t is simply 2t.

01:10 All right, so finally we get it's the square root of 2 squared plus 2 t squared.

01:18 So that's just 4 plus 4 t squared.

01:23 We can pull out a 4 as a common factor.

01:26 So the square root of 4 is 2, and then what's left inside the square root is just t squared plus 1.

01:33 So we determined our ds, our ds.

01:37 We know that x is t squared, we know that y is 2t.

01:41 So simply what we're going to do is we're going to plug now everything back into this and try to determine you.

01:49 Okay.

01:50 So now when we plug everything back, we get this big mess right here.

01:57 But we're going to try to simplify this.

01:59 So if we pull out the 2, so 2 times 2 is 4, so we can pull that out as a constant.

02:03 And what we're left with is this right here.

02:08 So we have a t square plus 1 under the square root, and we have a t cubed on the outside.

02:15 Well, let's try to solve this integral using substitution.

02:19 So we're going to set u equal to t square plus 1...

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