Equal Area Property of Ellipses: Proving A'(t) = (1/2)ab

In summary, Homework Statement is about finding the area bounded by the polar curve r = f(Θ) on the interval [0,Θ]. The area is found to be (1/2) ∫ (f(u))^2 du. The function f(theta) is used to find r(Θ(t)) and r(Θ) = <acosΘ, bsinΘ>. The area is then found to be A'(Θ) = (1/2)ab.
  • #1
physics_197
27
0

Homework Statement



Consider the ellipse r(t) = <acost, bsint>, for t between 0 and 2PI, where a and b are real numbers. Let Θ be the angle between the position vector and the x-axis.

a) Recall that the area bounded by the polar curve r = f(Θ) on the interval [0,Θ] is A(Θ) = (1/2) ∫ (f(u))^2 du, where in this case f(theta) = |r(Θ(t))|. Use this fact to show that A'(t) = (1/2)ab.


Homework Equations





The Attempt at a Solution

 
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  • #2
hi physics_197! :wink:

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
  • #3
tiny-tim said:
hi physics_197! :wink:

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:

I'm not sure how to approach this problem to begin with. In earlier parts we were asked to show that tanΘ = (b/a)tan(t) and to find Θ'(t). I have those done and assume we will need to use them.

I have tried to integrate the function to find A(Θ). Then, to find A'(t), I though that I could use A'(Θ) * Θ'(t). I am not sure if that will work.

Also, I am not even sure I am using the correct function f(theta) = |r(Θ(t))|. I thought I could just plug Θ in for t for r(t) = <acost, bsint> to get r(Θ) = <acosΘ, bsinΘ> and then find the magnitude of that.
 
  • #4
physics_197 said:
a) Recall that the area bounded by the polar curve r = f(Θ) on the interval [0,Θ] is A(Θ) = (1/2) ∫ (f(u))^2 du, where in this case f(theta) = |r(Θ(t))|. Use this fact to show that A'(t) = (1/2)ab.
physics_197 said:
… tanΘ = (b/a)tan(t) and to find Θ'(t).

I have tried to integrate the function to find A(Θ). Then, to find A'(t), I though that I could use A'(Θ) * Θ'(t). I am not sure if that will work.

you need to use the chain rule dA/dt = dA/dΘ dΘ/dt = dA/dΘ Θ'

(is that what you meant?)
Also, I am not even sure I am using the correct function f(theta) = |r(Θ(t))|. I thought I could just plug Θ in for t for r(t) = <acost, bsint> to get r(Θ) = <acosΘ, bsinΘ> and then find the magnitude of that.

yes that's correct …

carry on :smile:
 
  • #5
When I tried that, I got a complicated answer (it may have simplified down to the correct answer). I was also wondering, Since we are told that A(theta) = the integral of another function, can we just say that A'(theta) = that function?
 
  • #6
physics_197 said:
When I tried that, I got a complicated answer (it may have simplified down to the correct answer).

if you want us to check it, you'll have to write it out :wink:
I was also wondering, Since we are told that A(theta) = the integral of another function, can we just say that A'(theta) = that function?

yes, if A = ∫0t f(u) du, then dA/dt = f(t) :smile:
 
  • #7
I got A'(theta) = (1/2)[ (acos(theta))2 + (bsin(theta))2 ] and Θ'(t) = (b/a)(sec(t))2 / (1 + (b/a tan(t) )2 )

After multiplying them and using some trig identities i got:

[ ab(1+(tant)2) + (b5/a3)(tant)2(1 + (tant)2) ]
---------------------------------------------
1 + ( (b/a) tant )2
 
  • #8
you need to keep replacing b/a by tanΘ/tant

and to use one of the standard trigonometirc identities 1 + tan2 = sec2 :wink:

(get rid of all the coss and sins and tans, and replace them with sec and nothing else)
 
  • #9
tiny-tim said:
you need to keep replacing b/a by tanΘ/tant

and to use one of the standard trigonometirc identities 1 + tan2 = sec2 :wink:

(get rid of all the coss and sins and tans, and replace them with sec and nothing else)

Have you worked it out? Because I am wondering if my equations are correct. Also, I understand what you are telling me, but I am having some trouble. Should I get everything in terms of t or theta? Other than what you mentioned before, are they any trig identities?
 
  • #10
physics_197 said:
Have you worked it out?

yes :smile:

start with a2cos2t + b2sin2t …

write that in terms of sect and secΘ without any cos or sin or tan :wink:
 
  • #11
tiny-tim said:
yes :smile:

start with a2cos2t + b2sin2t …

write that in terms of sect and secΘ without any cos or sin or tan :wink:

( a2cos2Θ + b2sin2Θ ) * [ (b/a)sec2t / sec2Θ ]

Then I get it to:

[ ( a2 + b2sec2Θ - b2 ) / sec2Θ ] * [ (b/a)sec2t / sec2Θ ]
 
  • #12
ah, now i see where you're going wrong …

where did you get that a2cos2Θ + b2sin2Θ from?

shouldn't it be a2cos2t + b2sin2t ?
 
  • #13
I thought it was theta because it is A'(Θ).

Using t, I get:

[ ( a2 + b2sec2t - b2 ) ] * [ (b / asec2Θ ]
 
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  • #14
and b2sec2t - b2 = … ? :smile:
 
  • #15
tiny-tim said:
and b2sec2t - b2 = … ? :smile:

b2tan2t
But I am thinking your looking for something else.

EDIT: Never mind, it also equals a2tan2Θ

Thanks, got it now :)

The next part asks me to show that the area swept out by any angle theta is equal to the area swept out by the same angle (doesn't depend on where it starts).
I was thinking of integrating A(Θ) = (1/2) ∫ (f(u))^2 du over T to T+Θ where T is just some angle. And then somehow my answer will not have any T in it so I can clude that the area only depends on the angle swept out and not the starting position?
 
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  • #16
hi physics_197! :smile:

(don't use T, it looks too much like t … use something like ψ or Θ0)

in the first part, you proved that dA/dΘ = ab/2,

so A(Θ) = abΘ/2,

and the area between Θ0 and Θ1 is ab(Θ1 - Θ0)/2 :wink:
 

FAQ: Equal Area Property of Ellipses: Proving A'(t) = (1/2)ab

What is the equal area property of ellipses?

The equal area property of ellipses states that if a line is drawn from the center of an ellipse to any point on the ellipse's perimeter, the area of the ellipse between this line and the center will be equal to half of the area of the entire ellipse.

How is the equal area property of ellipses related to the derivative of the ellipse's area function?

The derivative of the ellipse's area function, denoted as A'(t), represents the rate of change of the area of the ellipse at a specific point in time. The equal area property of ellipses is proven by showing that A'(t) is equal to half of the product of the ellipse's semi-major axis (a) and semi-minor axis (b) at any point in time (t).

What is the significance of proving A'(t) = (1/2)ab in the equal area property of ellipses?

Proving that A'(t) is equal to half of the product of the ellipse's semi-major and semi-minor axes confirms the equal area property of ellipses. This property has important applications in various fields such as astronomy, physics, and engineering.

What are the steps involved in proving A'(t) = (1/2)ab for the equal area property of ellipses?

The steps involved in the proof include setting up the equation for the area of an ellipse using calculus, finding the derivative of this equation, simplifying the derivative using algebraic manipulations, and showing that the resulting equation is equal to half of the product of the ellipse's semi-major and semi-minor axes.

Are there any limitations to the equal area property of ellipses?

While the equal area property of ellipses holds true for all ellipses, it does have some limitations. For example, it does not apply to degenerate ellipses (such as a circle) or to ellipses with complex or imaginary axes. Additionally, the property only holds true for ellipses in a two-dimensional plane and does not apply to ellipses in higher dimensions.

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