- #1

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[tex] a_x' = a_x cos\theta + a_y sin\theta [/tex] and

[tex] a_x' = -a_x sin\theta + a_y cos\theta [/tex]

i can't seem to find any place where they show how this is done.

- Thread starter misogynisticfeminist
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- #1

- 363

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[tex] a_x' = a_x cos\theta + a_y sin\theta [/tex] and

[tex] a_x' = -a_x sin\theta + a_y cos\theta [/tex]

i can't seem to find any place where they show how this is done.

- #2

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Draw a vector and two sets of orthogonal axes.

For each set of axes, form a right triangles with the vector as the hypotenuse, and the legs parallel to the axes.

Do a little trigononometry.

I'll post a URL with a picture, if I can find one.

Ok here's one:

http://web.umr.edu/~oci/Topic12/T12-5/cml12-5a/frame.html [Broken]

(although the vector is not drawn from the origin to the little box).

(Check the subscript on the LHS of your second equation.)

For each set of axes, form a right triangles with the vector as the hypotenuse, and the legs parallel to the axes.

Do a little trigononometry.

I'll post a URL with a picture, if I can find one.

Ok here's one:

http://web.umr.edu/~oci/Topic12/T12-5/cml12-5a/frame.html [Broken]

(although the vector is not drawn from the origin to the little box).

(Check the subscript on the LHS of your second equation.)

Last edited by a moderator:

- #3

LeonhardEuler

Gold Member

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The new x-axis is a line tilted at some angle, [itex]\theta[/tex], to the horizontal. Basically, if you want [itex]a'_y[/tex], you want the distance between a point and this line. The formula for this is

[tex]|a'_y|= \frac {|ma_x+b-a_y|} {\sqrt{m^2+1}}[/tex]

where the equation of the line is [itex]y=mx+b[/tex]. Since the line passes through the origin, b=0. [itex]tan(\theta)=m[/tex], so:

[tex]|a'_y|= \frac {|tan(\theta)a_x-a_y|} {\sqrt{tan(\theta)^2+1}}[/tex]

[tex]= \frac {|tan(\theta)a_x-a_y|} {|sec(\theta)|}[/tex]

[tex]= |tan(\theta)a_x-a_y|\times|cos(\theta)|[/tex]

[tex]=|sin(\theta)a_x-cos(\theta)a_y|[/tex]

Since, for [itex]\theta=0[/tex], we should have [itex]a'_y=a_y[/tex], this means the signs will only work out properly if:

[tex]a'_y=-sin(\theta)a_x+cos(\theta)a_y[/tex]

The other equation is found similarly. Let me know if you want to see the derivation of this distance formula.

[tex]|a'_y|= \frac {|ma_x+b-a_y|} {\sqrt{m^2+1}}[/tex]

where the equation of the line is [itex]y=mx+b[/tex]. Since the line passes through the origin, b=0. [itex]tan(\theta)=m[/tex], so:

[tex]|a'_y|= \frac {|tan(\theta)a_x-a_y|} {\sqrt{tan(\theta)^2+1}}[/tex]

[tex]= \frac {|tan(\theta)a_x-a_y|} {|sec(\theta)|}[/tex]

[tex]= |tan(\theta)a_x-a_y|\times|cos(\theta)|[/tex]

[tex]=|sin(\theta)a_x-cos(\theta)a_y|[/tex]

Since, for [itex]\theta=0[/tex], we should have [itex]a'_y=a_y[/tex], this means the signs will only work out properly if:

[tex]a'_y=-sin(\theta)a_x+cos(\theta)a_y[/tex]

The other equation is found similarly. Let me know if you want to see the derivation of this distance formula.

Last edited:

- #4

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Here's a vector-algebraic proof.

For the first set of axes, we have unit vectors [tex]\hat x[/tex] and [tex]\hat y[/tex].

For the second set, which is rotated counterclockwise by angle [tex]\theta[/tex], we have [tex]\hat x'[/tex] and [tex]\hat y'[/tex].

Note that [tex]A_x= \vec A\cdot \hat x[/tex] and so on.

Now, compute [tex]A_{x'}= \vec A\cdot \hat x'[/tex] where one writes

[tex] \vec A=(\vec A \cdot \hat x) \hat x + (\vec A \cdot \hat y) \hat y=A_x\hat x + A_y\hat y[/tex].

So, let's plug it in:

[tex]\begin{align*}

A_{x'}

&= \vec A\cdot \hat x' \\

&= \left( A_x\hat x + A_y\hat y \right) \cdot \hat x' \\

&=(A_x) \hat x\cdot \hat x' + (A_y) \hat y\cdot \hat x' \\

&=(A_x) \cos\theta + (A_y) \cos(90^\circ - \theta) \\

&=(A_x) \cos\theta + (A_y) \sin\theta \\

\end{align}

[/tex]

Do the same for [tex]A_{y'}[/tex], and note that [tex] \hat x\cdot \hat y' =\cos(90^\circ+\theta)= -\sin\theta [/tex] and [tex] \hat y\cdot \hat y' =\cos\theta[/tex].

For the first set of axes, we have unit vectors [tex]\hat x[/tex] and [tex]\hat y[/tex].

For the second set, which is rotated counterclockwise by angle [tex]\theta[/tex], we have [tex]\hat x'[/tex] and [tex]\hat y'[/tex].

Note that [tex]A_x= \vec A\cdot \hat x[/tex] and so on.

Now, compute [tex]A_{x'}= \vec A\cdot \hat x'[/tex] where one writes

[tex] \vec A=(\vec A \cdot \hat x) \hat x + (\vec A \cdot \hat y) \hat y=A_x\hat x + A_y\hat y[/tex].

So, let's plug it in:

[tex]\begin{align*}

A_{x'}

&= \vec A\cdot \hat x' \\

&= \left( A_x\hat x + A_y\hat y \right) \cdot \hat x' \\

&=(A_x) \hat x\cdot \hat x' + (A_y) \hat y\cdot \hat x' \\

&=(A_x) \cos\theta + (A_y) \cos(90^\circ - \theta) \\

&=(A_x) \cos\theta + (A_y) \sin\theta \\

\end{align}

[/tex]

Do the same for [tex]A_{y'}[/tex], and note that [tex] \hat x\cdot \hat y' =\cos(90^\circ+\theta)= -\sin\theta [/tex] and [tex] \hat y\cdot \hat y' =\cos\theta[/tex].

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- #5

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When the original vector gets rotated the original x-component, let’s call that X, will not be only in the direction of the x-axis anymore, the component of the rotated “original x-component“ that will still be in the direction of the x-axis is:

[tex]X\cos(\theta)[/tex]

but it will now also have a component in the direction of the y-axis, which will be:

[tex]-X\sin(\theta)[/tex]

The same happens with the original y component. The component of the rotated “original y component“ that will still be in the direction of the y-axis is:

[tex]Y\cos(\theta)[/tex]

and the component in the direction of the x-axis will be:

[tex]Y\sin(\theta)[/tex]

To get the x-component of the new vector add the parts in the direction of the x-axis:

[tex]X\cos(\theta) + Y\sin(\theta)[/tex]

To get the y-component of the new vector add the parts in the direction of the y-axis:

[tex]-X\sin(\theta) + Y\cos(\theta)[/tex]

- #6

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hey thanks alot for the help. Yup the subscript on the LHS should be y. Here's what i have done, a little algebra,

[tex] a_x' cos \theta = a_x[/tex]

[tex] a_x' sin \theta= a_y [/tex]

what i did was to multiply sin and cosine to both equations respectively and add them up, to get the expression for [tex] a_x'[/tex]

for [tex] a_y' [/tex] i did the same thing, except that this time, comlementary angles were involved and the vector a_x points in the opposite direction.

Is that right?

edit: i'll look at the rest of the methods too. : )

[tex] a_x' cos \theta = a_x[/tex]

[tex] a_x' sin \theta= a_y [/tex]

what i did was to multiply sin and cosine to both equations respectively and add them up, to get the expression for [tex] a_x'[/tex]

for [tex] a_y' [/tex] i did the same thing, except that this time, comlementary angles were involved and the vector a_x points in the opposite direction.

Is that right?

edit: i'll look at the rest of the methods too. : )

Last edited:

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