Analyze the curve in parametric form

[emergency9177]

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I've solved the two tangents at the self-intersection point:





I've solved the intersections with the:
x axis at points: P1 (0,0), P2 (3,0)
y axes at point: P3 (0,0)

And also the:
point where the tanget is vertical P6 (0,0)
points where the tangent is horizontal P4 (1,-2) and P5 (1,2)

Now I have to plot the curve and find the area of the region inside the loop formed by the curve.

Can someone help me please?

Thank you!

 

[blamocur]

Draw a graph.
Find the min and max values for x coordinates of the loop.
Express the upper and the lower curves of the loop as [imath]y=f_1(x)[/imath] and [imath]y=f_2(x)[/imath].
Now the problem should yield to simple integration.

 

[Deleted member 4993]

View attachment 30198
_______________________________________________________________________________________________________________________________________________________

I've solved the two tangents at the self-intersection point:

View attachment 30195
View attachment 30200


I've solved the intersections with the:
x axis at points: P1 (0,0), P2 (3,0)
y axes at point: P3 (0,0)

And also the:
point where the tanget is vertical P6 (0,0)
points where the tangent is horizontal P4 (1,-2) and P5 (1,2)

Now I have to plot the curve and find the area of the region inside the loop formed by the curve.

Can someone help me please?

Thank you!

First, you have to show that the curve is self-intersecting? If you did - what are the co-ordinates of the point of intersection?

 

[emergency9177]

Draw a graph.
Find the min and max values for x coordinates of the loop.
Express the upper and the lower curves of the loop as [imath]y=f_1(x)[/imath] and [imath]y=f_2(x)[/imath].
Now the problem should yield to simple integration.

Not sure if is the correct graph...no loop here.

Accoring to the graph the, limits for integration are:




It is correct?

Thanks

 

[blamocur]

View attachment 30202

Not sure if is the correct graph...no loop here.

Accoring to the graph the, limits for integration are:

View attachment 30203
View attachment 30204

It is correct?

Thanks

Correct, but somewhat redundant: you only need the limits for [imath]x[/imath]. Can you find [imath]f_1[/imath] and [imath]f_2[/imath]?

Side note: my computer did a slightly better job at drawing the loop -- see the attached.

 

[emergency9177]

Correct, but somewhat redundant: you only need the limits for [imath]x[/imath]. Can you find [imath]f_1[/imath] and [imath]f_2[/imath]?

Side note: my computer did a slightly better job at drawing the loop -- see the attached.

Thank you for the loop.

Actually I don't know if I have tu use computation of line integrals along a parametrized curve in the form:

or Green's theorem in vectors form:



I already try simple integral bounded by tle limits and simple double integrals bounded by limits...I'm confused.

Thank you again for your explanation.

 

[blamocur]

Thank you for the loop.

Actually I don't know if I have tu use computation of line integrals along a parametrized curve in the form:
View attachment 30216
or Green's theorem in vectors form:

View attachment 30217

I already try simple integral bounded by tle limits and simple double integrals bounded by limits...I'm confused.

Thank you again for your explanation.

Have you tried finding [imath]f_1,f_2[/imath] as I suggested in posts #2 and #5?
How did you try simple integral?

 

[emergency9177]

Have you tried finding [imath]f_1,f_2[/imath] as I suggested in posts #2 and #5?
How did you try simple integral?

From the given parametric function curve:

>

>than substitute into y>

Is the correct way?

Now I have to integrate the last y from x=0 to x=3

Correct?

Thanks!

 

[blamocur]

From the given parametric function curve:

View attachment 30228 > View attachment 30229

View attachment 30225 >than substitute into y> View attachment 30226

Is the correct way?

Now I have to integrate the last y from x=0 to x=3

Correct?

Thanks!

"Half-correct" -- this way you only get the area between the top half of the loop and the horizontal axis [imath]y=0[/imath].

 

[emergency9177]

"Half-correct" -- this way you only get the area between the top half of the loop and the horizontal axis [imath]y=0[/imath].

Sice the loop is symmetric I have to multip. x 2 I guess...



= −13.29630792

But the plotted curve is not a loop...

 

[blamocur]

Sice the loop is symmetric I have to multip. x 2 I guess...

View attachment 30234

= −13.29630792

But the plotted curve is not a loop...

This why I mentioned [imath]f_1[/imath] and [imath]f_2[/imath], one for the upper curve and one for the lower one. Hint : [imath]t = \;\pm\;\; \sqrt{x}[/imath]

 

[emergency9177]

This why I mentioned [imath]f_1[/imath] and [imath]f_2[/imath], one for the upper curve and one for the lower one. Hint : [imath]t = \;\pm\;\; \sqrt{x}[/imath]

I've try to solve this expression:



The area of the region inside the loop formed by the curve is: 6.92820323…

It is correct now?

Thank you!

 

[blamocur]

I've try to solve this expression:

View attachment 30276

The area of the region inside the loop formed by the curve is: 6.92820323…

It is correct now?

Thank you!

No, not correct yet. We are not computing area for [imath]\sqrt{x}[/imath].
What I meant is that you computed the area between the curve [imath]y= f_1(x) = x^{\frac{3}{2}} - 3x^{\frac{1}{2}}[/imath] and the horizontal axis [imath]y=0[/imath]. Here [imath]f_1(x)[/imath] corresponds to the solution [imath]t=+\sqrt{x}[/imath] and represents the lower curve of the loop. But there is the upper curve [imath]y=f_2(x)[/imath], which corresponds to the solution [imath]t=-\sqrt{x}[/imath]. You need to compute the area between [imath]y=f_1(x)[/imath] and [imath]y=f_2(x)[/imath].

 

[emergency9177]

No, not correct yet. We are not computing area for [imath]\sqrt{x}[/imath].
What I meant is that you computed the area between the curve [imath]y= f_1(x) = x^{\frac{3}{2}} - 3x^{\frac{1}{2}}[/imath] and the horizontal axis [imath]y=0[/imath]. Here [imath]f_1(x)[/imath] corresponds to the solution [imath]t=+\sqrt{x}[/imath] and represents the lower curve of the loop. But there is the upper curve [imath]y=f_2(x)[/imath], which corresponds to the solution [imath]t=-\sqrt{x}[/imath]. You need to compute the area between [imath]y=f_1(x)[/imath] and [imath]y=f_2(x)[/imath].

Ok, now is much more clear:

The area is: 8.31384

Thanks a lot and best wishes!

 

[blamocur]

Perfect!

 

[BigBeachBanana]

Ok, now is much more clear:

The area is: 8.31384

Thanks a lot and best wishes!

Just a small typo. In your first line of the solution, you define f(x) twice, should be f(x) and g(x)