2. [10 x 2 = 20 points] For the first-order nonlinear...

2. [10 x 2 = 20 points] For the first-order nonlinear system: \frac{dx}{dt} = e^x - 2 a. Graph the phase flow on a line, i.e., $\frac{dx}{dt}$ as a function of x. Identify any steady states on the graph and draw arrows on the x axis to indicate when x increases or decreases. b. Sketch the detailed direction field (i.e., x versus t) using tangent vectors (arrows) specified by the model.

Transcript

00:01 Consider the following parametric equations.

00:03 We have that x is equal to cos squared theta, and y is equal to sine theta.

00:08 We want to sketch the curve using these parametric equations and indicate with an arrow the direction in which the curve is traced as theta increases.

00:18 So looking at our curves, we see that both x and y for our equations depend on variable theta, which typically we use for polar description of a curve, such as circles and ellipses.

00:32 But in addition, using a variable theta can also be used to trace conics, which will be the case in our problem.

00:45 So in our problem, if we were to trace x as a function of y, we obtain a parabola, which is a conic section.

00:54 And so here we have traced our parametric equations for theta ranging between 0 and 2 pi, where i recall x is equal to cos squared theta, and y is equal to sine theta.

01:12 Now on our graph, we want to indicate with an arrow the direction in which our graph is traced as theta increases.

01:23 So first, let's find the point on our graph which corresponds to theta equal to 0.

01:30 So when theta is equal to 0, x is equal to 1, and y is equal to 0.

01:35 So we're at the point 1, 0, which is the apex right here.

01:46 When theta increases to pi over 2, we're at the point 0, 1, which is this point right here.

01:59 So we see that for theta ranging between 0 and 2 pi, the curve is traced this way...

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