Position of particle in polar coordinates at t = T/8

[vaibhav garg]

1. The question
The position of a particle is given by r(t) = acos(wt) i + bsin(wt) j. Assume a and b are both positive and a > b. The plane polar coordinates of a particle at a time t equal to 1/8 of the time period T will be given by _

Homework Equations

r(t) = acos(wt) i + bsin(wt) j.

The Attempt at a Solution

at t = T/8 the value of wt=π/4.
therefore in Cartesian coordinates the vector is r= a/√2 + b/√2.
so in polar coordinates this transforms to r = √(a² + b²)/2 and tanθ = b/a. My answer does not match with the given one. Also if we were to write the polar equation of the curve how would that follow ?

 

[BvU]

Hi,
Strange. Don't see anything wrong. In what respect does your answer not match with the given one ?

 

[vaibhav garg]

Hi,
Strange. Don't see anything wrong. In what respect does your answer not match with the given one ?

The angle does not match. tanθ = b/a√2 is given
Also this is one of the part of the question, and I still can't understand how to wite the polar equation for this curve

 

[BvU]

Still strange. Where would the $\sqrt 2$ come from ?
Your $\tan\theta$ matches the expression here ($\tan t=1$).
And the answer for the polar equation is right in the next paragraph

 

[Chestermiller]

The angle does not match. tanθ = b/a√2 is given
Also this is one of the part of the question, and I still can't understand how to wite the polar equation for this curve

You have $$r\cos \theta = a \cos \omega t$$and $$r\sin \theta = b \sin \omega t$$
Just solve for r and $\theta$

 

[vaibhav garg]

Still strange. Where would the $\sqrt 2$ come from ?
Your $\tan\theta$ matches the expression here ($\tan t=1$).
And the answer for the polar equation is right in the next paragraph

I calculated the polar equation to be r = √(a²cos²(wt) + b²sin²(wt)) r^
and then I differentiated it to get the velocity as v = ((b² - a²)sin(wt)cos(wt)w)/√(a²cos²(wt) + b²sin²(wt)) r^ + √(a²cos²(wt) + b²sin²(wt))w θ^

 

[vaibhav garg]

I calculated the polar equation to be r = √(a²cos²(wt) + b²sin²(wt)) r^
and then I differentiated it to get the velocity as v = ((b² - a²)sin(wt)cos(wt)w)/√(a²cos²(wt) + b²sin²(wt)) r^ + √(a²cos²(wt) + b²sin²(wt))w θ^

Is this wrong ?

 

[Chestermiller]

Is this wrong ?

It looks correct. For aesthetic purposes, I would go to the double angle formulas:
$$2\sin \omega t\cos \omega t=\sin2\omega t$$
$$\cos^2 \omega t=\frac{1+\cos 2\omega t}{2}$$
$$\sin^2 \omega t=\frac{1-\cos 2\omega t}{2}$$

 

[BvU]

Are we sure the exercise composer wants r(t) and v(t) instead of $r(\theta)$ when asking for the polar form ?

 

[Chestermiller]

I calculated the polar equation to be r = √(a²cos²(wt) + b²sin²(wt)) r^
and then I differentiated it to get the velocity as v = ((b² - a²)sin(wt)cos(wt)w)/√(a²cos²(wt) + b²sin²(wt)) r^ + √(a²cos²(wt) + b²sin²(wt))w θ^

Oops. I don't confirm your coefficient of $\hat{\theta}$. I get $\frac{\omega ab}{r}$.

 

[vaibhav garg]

that is the problem I have been having, I can't figure out as to why I am not getting the correct angular component

 

[Chestermiller]

that is the problem I have been having, I can't figure out as to why I am not getting the correct angular component

The angular component is $rd\theta/dt$, not $r\omega$. In this problem $d\theta/dt$ is not the same thing as what they call $\omega$. You need to be able to show that $\frac{d\theta}{dt}=\frac{\omega ab}{r^2}$.

 

[Chestermiller]

$$\tan \theta = \frac{b}{a}\tan \omega t$$
$$\sec^2 \theta \frac{d\theta}{dt}=\frac{b\omega}{a}\sec^2\omega t$$$$r\cos \theta=a\cos \omega t$$

Combine 2nd and 3rd equations to eliminate $\sec \theta$.