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 the design in the explanation and the graph.

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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Answer 2

See below.

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.
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Answer 3

To determine the unit vector (\hat{\theta}) in terms of (\theta), (\hat{i}), and (\hat{j}) in Cartesian coordinates, you can use the trigonometric relationships between Cartesian coordinates and polar coordinates.

The unit vector (\hat{\theta}) is perpendicular to the radial direction and points in the direction of increasing polar angle (\theta). In Cartesian coordinates, it can be expressed as (\hat{\theta} = -\sin(\theta)\hat{i} + \cos(\theta)\hat{j}).

This expression arises from the fact that in polar coordinates, the unit vector (\hat{\theta}) can be represented by the sine and cosine of the angle (\theta), with (\hat{i}) and (\hat{j}) representing the unit vectors along the x and y directions respectively.

So, (\hat{\theta} = -\sin(\theta)\hat{i} + \cos(\theta)\hat{j}) represents the unit vector (\hat{\theta}) in terms of Cartesian unit vectors (\hat{i}) and (\hat{j}) and the angle (\theta) in polar coordinates.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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