Solving Parametric Equation to Find Area - Help Needed!

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In summary, the area of the region enclosed by the parametric equation x=t^3-2t and y=9t^2 can be found by setting the two equations equal to each other and solving for y, resulting in two possible values for y: 0 and 18. Using y=9t^2, the corresponding values of t are found to be 0 and sqrt{2}. Integrating dxdy from 0 to sqrt{2} will give the area of the region enclosed by the parametric equation.
  • #1
ILoveBaseball
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Find the area of the region enclosed by the parametric equation
[tex]x=t^3-2t[/tex]
[tex]y=9t^2[/tex]

[tex]dx/dt = 3t^2-2[/tex]
9t^2 - 1 = 0
[tex]t=\pm \sqrt {1/9}[/tex]

[tex]\int_{-1/3}^{1/3} (9t^2)*(3t^2-2) dt[/tex]

= -2/5

anyone know where i went wrong?
 
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  • #2
I don't understand what you did in line 3, 4, 5...

What I recommand doing is to find a explicit relation between x and y. This can be done in the following manner.

[tex]y = 9t^2 \Rightarrow t = \pm \sqrt{y/9}[/tex]

In (1):

[tex]x(y) = \pm (y/9)^{3/2} \mp \frac{2}{3}\sqrt{y}[/tex]

Now you got two curves. See where they intersect. See at which values of t does this corespond. Integrate acordingly (you did that right, except with the wrong bounds).
 
  • #3
[tex]x(y) = \pm (y/9)^{3/2} \mp \frac{2}{3}\sqrt{y}[/tex]

what am i suppose to do after this? solve for y? can you help me out a bit more? I'm just so confuse on how to find the bounds.
 
  • #4
See where they intersect. Set them equal:

[tex](y/9)^{3/2} - \frac{2}{3}\sqrt{y} = - (y/9)^{3/2} + \frac{2}{3}\sqrt{y} [/tex]

and solve for y.

You'll find 2 y. Using [itex]y=9t^2[/itex], find to which values of t these two y corespond. These are your bounds.
 
  • #5
setting those equal to each other and solving for y, i get [tex]\pm\sqrt{973}[/tex]

those are the two y's right?

"Using y=9t^2, find to which values of t these two y corespond. These are your bounds"

are you saying to plug in [tex]\pm\sqrt{973}[/tex] into the y equation?
 
  • #6
I don't get the same thing as you do. I get

[tex](y/9)^{3/2} - \frac{2}{3}\sqrt{y} = - (y/9)^{3/2} + \frac{2}{3}\sqrt{y} [/tex]

Right there, a solution is obviously y = 0. Let's find another.

[tex]2(y/9)^{3/2} = \frac{4}{3}\sqrt{y} [/tex]

[tex]\frac{4}{729}y^3 = \frac{16}{9}y [/tex]

[tex]\frac{4}{729}y^2 = \frac{16}{9} [/tex]

[tex]y = \pm\sqrt{ \frac{16}{9}\frac{729}{4}} = \pm \sqrt{324} = \pm 18 [/tex]

This is a bit weird because [itex]y=-18[/itex] is not a selution because [itex]\sqrt{-18}[/itex] and [itex]\sqrt{(-18)^3}[/itex] are undefined. So y = +18 is our other solution. There are no other.

By "Using y=9t^2, find to which values of t these two y corespond. These are your bounds" I mean...

for which t will we have y=0 and for which t will be have y=18?

Well,

[tex]y= 0 = 9t^2 \Leftrightarrow t=0[/tex]

and

[tex]y=18 = 9t^2 \Leftrightarrow t = \pm \sqrt{2}[/tex]

We only need one. So let's discard the minus one.

Now integrate dxdy from 0 to sqrt{2} like you did before.
 

Related to Solving Parametric Equation to Find Area - Help Needed!

What is a parametric equation?

A parametric equation is a set of equations that express the coordinates of a point in terms of one or more parameters. These parameters are usually represented by letters such as t or s.

Why do we use parametric equations to find area?

Parametric equations are useful for finding the area of irregular shapes or curves, as they allow us to break down the shape into smaller segments and find the area of each segment separately. This can be especially helpful when dealing with complex shapes that cannot be easily broken down using traditional geometry methods.

How do I solve a parametric equation to find area?

To solve a parametric equation to find area, you will need to first determine the limits of integration, which are the starting and ending values for the parameter. Then, you can use the parametric equations to find the x and y coordinates of each point on the curve. Finally, you can use these coordinates to set up and solve an integral to find the area.

What are some common mistakes when solving parametric equations for area?

One common mistake is not properly setting up the integral, which can lead to incorrect answers. Another mistake is using the wrong limits of integration, which can also affect the final result. It is important to carefully check the setup and calculations to avoid these errors.

How can I check my answer when solving a parametric equation for area?

You can check your answer by using a graphing calculator or graphing software to plot the parametric equations and visually see the shape that is being calculated. You can also use the formula for the area of a specific shape, such as a circle or ellipse, to compare your answer and ensure it is reasonable.

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