# Homework Help: Cartesian to polar unit vectors + Linear Combination

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1. Mar 20, 2015

### Christina909

I've been trying to solve this question all day. If somebody could point me in the right direction I would really appreciate it!

(ii) A particle’s motion is described by the following position vector r(t) = 4txˆ + (10t − t)ˆy Determine the polar coordinate unit vectors ˆr and ˆθ for r. [4]

v(t) is given as: v(t) = 4xˆ + (10t − 1)ˆy
(iv) Using the ˆr and ˆθ you found in (ii) above, write v(t) as a linear combination of rˆ and ˆθ. [4]

(v) Differentiate the expression for r(t) you got in part (ii) (in terms of ˆr and ˆθ, and using the expressions ˙rˆ = ˙θ ˆθ , ˙ˆθ = − ˙θrˆ derived in the lectures, show that you obtain the same answer as in part (iv)

I understand that for θ^ = -sinθ + cosθ and r^=sinθ + cos θ I'm just unsure how to do part (ii) without it being really messy e.g with cos(tan^-1(10t-1/4)) where tan^-1(10t-1/4)=θ especially knowing I have to find a linear combination afterwords.

The attempt at a solution
My current answer for part (ii):
r^= cos(tan^-1(10t-1/4))x^ + sin(tan^-1(10t-1/4))y^
θ^= -sin(tan^-1(10t-1/4))x^ + cos(tan^-1(10t-1/4))y^

And for part (iv) am I am right in saying that I need to find, a and b scalars for the following:
4 = a*cos(tan^-1(10t-1/4)) +b*-sin(tan^-1(10t-1/4))
10t-1 = a*sin(tan^-1(10t-1/4))+b*cos(tan^-1(10t-1/4))

I'm really not sure about this one. It appears like it's going to difficult to eliminate the t value.
Anyway, thanks in advance, I'm going to keep working on it now.

2. Mar 20, 2015

### Svein

I am pretty sure there's at least one typo here. Why would anybody specify (10t - t)y?

3. Mar 20, 2015

### Christina909

Yes that was a typo, sorry.
But when I say y^ -'y' hat (Or ^y - the typo) I mean it as the component, and it's the same with the others, it was just a formatting error

4. Mar 20, 2015

### Goddar

Hi. This all seems a bit confusing, partly because of notation and possible typos... (What's the correct r(t) you are given, by the way?)
Now regardless of that, you seem to have taken a wrong track in the beginning. So to avoid the hats i'm going to define: r-hat = er ; θ-hat = eθ , and: x-hat = ex ; y-hat = ey

You know that in polar coordinates (because you mention it at some point):
er = cosθ ex + sinθ ey , eθ = –sinθ ex + cosθ ey , and this is always true.
You also know that: cosθ = x/r = x⋅(x2+ y2)–1/2, and: sinθ = y/r = y⋅(x2+ y2)–1/2 ,
And this is also always true.
Now you are given an expression: r(t) = x(t) ex + y(t) ey,
So what does all that make of er and eθ?

5. Mar 21, 2015

### vela

Staff Emeritus
I think the typo referred to was writing the y-component as $10t-t$. Why not simply write it as $9t$? Plus it's inconsistent with what you later said was the y-component of the velocity, $10t-1$. That's not the derivative of $10t-t$.

6. Mar 22, 2015

### BvU

Chris, can you clear this up ?

Either

$\vec r(t) = 4t\hat x + (10 t^2 - t )\hat y\ \$, but then $\vec v(t) = 4\hat x + (20 t - 1)\hat y$,

or

$\vec v(t) = 4\hat x + (10 t - 1)\hat y\ \$ and then $\vec r(t) = 4t\hat x + (5 t^2 - t )\hat y$.

Unless, of course, there is more than one typo harassing us...