For the unit vector $ˆ θ$ $hattheta$ , geometrically show that $ˆ θ = - sin θ ˆ i + cos θ ˆ j$ $hattheta = -sinthetahati + costhetahatj$ ? Essentially, converting from cartesian to polar, how would I determine the unit vector for $\to θ$ $vectheta$ in terms of $θ$ $theta$ , $ˆ i$ $hati$ , and $ˆ j$ $hatj$ ?

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For the unit vector $ˆ θ$ $hattheta$ , geometrically show that $ˆ θ = - sin θ ˆ i + cos θ ˆ j$ $hattheta = -sinthetahati + costhetahatj$ ? Essentially, converting from cartesian to polar, how would I determine the unit vector for $\to θ$ $vectheta$ in terms of $θ$ $theta$ , $ˆ i$ $hati$ , and $ˆ j$ $hatj$ ?

I've been able to show that $ˆ r = cos θ ˆ i + sin θ ˆ j$ $hatr = costhetahati + sinthetahatj$ :

$cos θ = \frac{ˆ i}{ˆ r}$ $costheta = hati/hatr$
$sin θ = \frac{ˆ j}{ˆ r}$ $sintheta = hatj/hatr$

$\Rightarrow | | ˆ r | | = \sqrt{ˆ r \cdot ˆ r ({cos}^{2} θ + {sin}^{2} θ)}$ $=> ||hatr|| = sqrt(hatrcdothatr(cos^2theta + sin^2theta))$

$= \sqrt{ˆ r \cdot ˆ r {cos}^{2} θ + ˆ r \cdot ˆ r {sin}^{2} θ}$ $= sqrt(hatrcdothatrcos^2theta + hatrcdothatrsin^2theta)$

$= \sqrt{ˆ r cos θ \cdot ˆ i + ˆ r sin θ ˆ j}$ $= sqrt(hatrcostheta * hati + hatrsintheta hatj)$

$\Rightarrow ˆ r \cdot ˆ r = {| | ˆ r | |}^{2} = ˆ r cos θ \cdot ˆ i + ˆ r sin θ \cdot ˆ j$ $=> hatrcdothatr = ||hatr||^2 = hatrcosthetacdothati + hatrsinthetacdothatj$

$= ˆ r \cdot (cos θ ˆ i + sin θ ˆ j)$ $= hatrcdot(costhetahati + sinthetahatj)$

Thus, $ˆ r = cos θ ˆ i + sin θ ˆ j$ $hatr = costhetahati + sinthetahatj$ . But how would I do it for $ˆ θ$ $hattheta$ ? I'm probably just missing something really simple, like where the $ˆ θ$ $hattheta$ vector points.

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Answer 1 · 2017-02-19T21:32:33

See below.

Explanation:

Considering $p = (r cos θ, r sin θ) = r (cos θ, sin θ) = r ˆ r$ $p = (r costheta,r sintheta)=r(costheta,sintheta)=r hat r$ where $ˆ r = (cos θ, sin θ)$ $hat r = (cos theta,sin theta)$

We have then

$. p = . r ˆ r + r . ˆ r$ $dot p = dot r hat r + r dot(hat(r))$

but $. ˆ r = (- sin θ, cos θ) . θ = . θ ˆ θ$ $dot(hat(r))=(-sintheta,costheta)dot theta = dot theta hat theta$

Here for convenience, we call $(- sin θ, cos θ) = ˆ θ$ $(-sin theta, cos theta) = hat theta$

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

so we have $. p = . r ˆ r + r . ˆ r = . r ˆ r + r . θ ˆ θ$ $dot p = dot r hat r + r dot(hat(r))=dot r hat r + r dot theta hat theta$

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 · 2017-02-20T05:07:52

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 $Q_2$ .

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