Question

For the unit vector `hattheta`, geometrically show that `hattheta = -sinthetahati + costhetahatj`? Essentially, converting from cartesian to polar, how would I determine the unit vector for `vectheta` in terms of `theta`, `hati`, and `hatj`?

I've been able to show that `hatr = costhetahati + sinthetahatj`:

`costheta = hati/hatr`
`sintheta = hatj/hatr`

`=> ||hatr|| = sqrt(hatrcdothatr(cos^2theta + sin^2theta))`

`= sqrt(hatrcdothatrcos^2theta + hatrcdothatrsin^2theta)`

`= sqrt(hatrcostheta * hati + hatrsintheta hatj)`

`=> hatrcdothatr = ||hatr||^2 = hatrcosthetacdothati + hatrsinthetacdothatj`

`= hatrcdot(costhetahati + sinthetahatj)`

Thus, `hatr = costhetahati + sinthetahatj`. But how would I do it for `hattheta`? I'm probably just missing something really simple, like where the `hattheta` vector points.

Answer 1

See below.

Explanation:

Considering `p = (r costheta,r sintheta)=r(costheta,sintheta)=r hat r` where `hat r = (cos theta,sin theta)`

We have then

`dot p = dot r hat r + r dot(hat(r))`

but `dot(hat(r))=(-sintheta,costheta)dot theta = dot theta hat theta`

Here for convenience, we call `(-sin theta, cos theta) = hat theta`

`hat r, hat theta` form a basis of orthogonal unit vectors. They can be also called `hat n, hat (tau)` instead.

so we have `dot p = dot r hat r + r dot(hat(r))=dot r hat r + r dot theta hat theta`

deriving again

`ddot p = ddot r hat r + dot r dot theta hat theta + dot r dot theta hat theta + r ddot theta hat theta+r dot theta dot(hat(theta))`

Here `hat theta = (-sin theta, costheta)` so

`dot(hat(theta))=-(costheta,sin theta) dot theta = -dot theta hat r`

and finally

`ddot p = (ddot r-r(dot theta)^2)hat r + (2dot r dot theta+r ddot theta)hat theta`

Concluding `hat r, hat theta` are used for convenience. They have a strong geometric appeal. Also they obey all rules of vector and differential calculus.

Answer 2

See the design in the explanation and the graph.

Explanation:

As a matter of convenience, I use `alpha`, instead of `theta`.

The unit vector in the direction `theta = alpha` is

`cosalpha veci+sinalpha vecj`

`= < x, y > = < cosalpha, sinalpha >`, in Cartesian form.

The parallel position vector through the origin O ( r = 0 ) is

`vec (OP)`, where `P (cosalpha, sinalpha)`, is the radius vector of

the unit circle r = 1 , in the direction `theta= alpha`.

P' is at `(cos(pi/2+alpha), sin(pi/2+alpha))=(-sinalpha, cosalpha)`.

`vec(OP') = <-sinalpha, cosalpha>` is constructed as the radius

vector of the unit circle, in the direction `theta = pi/2+alpha.`.

graph{(x^2+y^2-1)(y-x/sqrt3)(y+sqrt3x)=0 [-1, 1, -.05, 1]}

Here `theta = pi/6`. Any vector of length 1, in the direction

`alpha = theta +pi/2=2/3pi` (shown as a radius ),

would represent `-sin(pi/6)vec i + cos (pi/6) vecj`.

In brief, `vec(OP')` is `vec(OP)` turned through `pi/2`, in the (

`theta uarr`) anti-clockwise sense. OP is in `Q_1`and OP' is in `Q_2`.