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

[physics_197]

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

 

[tiny-tim]

hi physics_197!

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

 

[physics_197]

hi physics_197!

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

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.

 

[tiny-tim]

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.

… 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 ​

 

[physics_197]

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?

 

[tiny-tim]

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

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 = ∫₀^t f(u) du, then dA/dt = f(t)

 

[physics_197]

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

 

[tiny-tim]

you need to keep replacing b/a by tanΘ/tant

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

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

 

[physics_197]

you need to keep replacing b/a by tanΘ/tant

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

(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?

 

[tiny-tim]

Have you worked it out?

yes

start with a²cos²t + b²sin²t …

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

 

[physics_197]

yes

start with a²cos²t + b²sin²t …

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

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

Then I get it to:

[ ( a² + b²sec²Θ - b² ) / sec²Θ ] * [ (b/a)sec²t / sec²Θ ]

 

[tiny-tim]

ah, now i see where you're going wrong …

where did you get that a²cos²Θ + b²sin²Θ from?

shouldn't it be a²cos²t + b²sin²t ?

 

[physics_197]

I thought it was theta because it is A'(Θ).

Using t, I get:

[ ( a² + b²sec²t - b² ) ] * [ (b / asec²Θ ]

 

[tiny-tim]

and b²sec²t - b² = … ?

 

[physics_197]

and b²sec²t - b² = … ?

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?

 

[tiny-tim]

hi physics_197!

(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