Question 3: A force is given by F⃗(x, y) = k1xyî + k2xz^2ĵ...

Question 3: A force is given by \(\vec{F}(x, y) = k_1xy\hat{i} + k_2xz^2\hat{j}\) \((k_1\) and \(k_2\) are positive constants). Find the work done by this force when a particle moves: a) from O to A, b) from A to B, c) from O to C, d) from C to B, along the four sides of the rectangular path shown in the figure. e) Find the relationship between \(k_1\) and \(k_2\) if \(\vec{F}\) is a conservative force.

Transcript

00:01 So for this question, we are given a two -dimensional force field.

00:05 And we are given a path, which i'm labeling as c, which is y equals x cubed from the origin to the 0 .28.

00:13 So we're going to ask a couple of questions about this field.

00:16 The first is that we're going to calculate the work done by a particle traveling along this path.

00:23 So the work is going to be defined as the line integral of our force field, where c is our path.

00:34 So in order for us to do this, we're going to need a parametrization for this curve.

00:40 And whenever you have a function given in two variables, i recommend the nicest parametrization is to let your independent variable x be your parameter t.

00:54 And then we get this y component really nicely, which will be our t cubed.

01:01 And when you do a line integral of a vector field, you need the derivative of your parametrization.

01:07 So in this case, we should get the 1 and 3t squared.

01:13 So we now have a single integral, an ordinary one, where now we need to have bounds in terms of t.

01:27 So what's nice about the way we have it set up is because since our x component is t, we get that our bounds are going to be shown in the initial and endpoints.

01:39 So from 0 to 2 will be where we take our integral.

01:43 We'll have our vector field.

01:45 And we need to evaluate the vector field at our parametrization.

01:52 So we have the x and y components.

01:54 These will get substituted in to our field f.

01:57 So we have x times y, which will be t to the fourth.

02:01 And then we have 2y, which would be 2t cubed.

02:09 And this will now be dotted with the derivative, which we calculated already.

02:17 And there we go.

02:21 So now let's go ahead and evaluate the dot product.

02:28 You should get t to the fourth plus 6t to the fifth.

02:39 Great.

02:39 So now we have our single integral.

02:41 And we can calculate the work.

02:43 So we get 1 fifth t to the fifth plus t to the sixth from 0 to 2.

02:54 And substitute in your 2, you will get 32 over 5 plus 64...

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