Question

(6) Consider the vector field F(x,y)-《22, 3y). A path is closed if it ends whiere it starts Consider the 3 closed paths starting and ending at (3,0): C1 the circle of radius 3 centered at the origin, C2 the ellipse with equation 2 +3y2-9, and Cs the flat linear path going to -3 and then going straight back. (a) Use GeoGebra to plot the vector field F (b) For each, parametrize and plot the closed path on top of F (c) Estimate the line integral for each curve, fc F -T ds. Explain your reasoning. What patterns are you observing?

answer all parts please except A if you cannot

Answer 1

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ANSWER;

b.)

Now, the three paths can be parameterised as follows:

1. Since the first path is a circle of radius 3:
   $C_1 : (x(\theta), y(\theta)) = (3 \cos \theta, 3 \sin \theta) \quad (\theta \in [0, 2\pi])$
2. Since the second part is an ellipse of semi-major axis length 3, and semi-minor axis length $\sqrt{3}$:
   $C_2 : (x(\theta),y(\theta)) = (3\cos \theta,\sqrt{3} \sin \theta) \quad (\theta \in [0,2\pi])$
3. This is a straight line path, which consists of two paths, the first which takes us from (3,0) to (-3,0), the second which takes us from (-3,0) to (3,0). By a theorem of line integrals, the second integral has the same value as the first, but with a negative value.
   The parameterisation of the first path is $C_{3_a}: (x(t),y(t)) = (3 - 6t ,0) \quad (t \in [0,1])$
   The parameterisation of the second path is $C_{3_b}: (x(t),y(t)) = (6t - 3 ,0) \quad (t \in [0,1])$

The paths are plotted below, where the black path is C1, the red one is C2 and the cyan path is C3, with the directions indicated:

Vector Field < 2 x,3 y> Vx(x,y)- 2x xmin5 xmax 5 ymin -5 ymax- 5 xn-8 yn-8 -2 V 0.02 vh 0.09

Now, in the first case:

dr dy (6 cos θ, 9 sin θ)

And, in the second case :

dx dy (6 cos θ. 3V3 sin θ) . (-3 sin θ, V3 cos θαθ sin θ cos θ dθ 0 0 similar to the case above)

In the third case, by the theorem we have mentioned the two paths exactly cancel each other out, and thus we hence have $I_3 = \int_0^{2\pi} \mathbf{F}(x,y) \cdot d\vec{s} = 0$

c.)

We see that the integral is independent of the path taken, and always equals zero. This is due to the fact that

dydr
, which implies that the line integral is path-independent.

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