# Have I finished this?

• May 21st 2008, 04:24 AM
Taylor Fylde
Have I finished this?
I'm asked to find the the times (0<t<2pi) that two vectors are perpendicular.

The two vectors are perpendicular when their Dot product is zero. Taking the Dot product, and equating to zero gives:

2.Cos(2t).Sin(2t) - 4.Cos(t).Sin(t)=0

Or:

2.Cos(2t).Sin(2t)= 4Cos(t)Sin(t)

So the vectors are perpendicular when the above holds true.

I've sketched the plot of Sin(t), Cos(t), Sin(2t) and Cos(2t).

The R.H.S. of the equation reduces to zero when either Cos(t) or Sin(t) are zero, this is at ;

t=0, pi/2, pi, 3pi/2, 2pi, and the above equation holds true.

I feel like I've done this more by intuition than by mathematical method, and I can't show that these are the only values for which the vectors are perpendicular (though I suspect they are).

There must be a better way to go about this, and I have a feeling it may be with some trig identity.

In short, have I got all the values? Is there a better way to do it?

Many thanks.
• May 21st 2008, 04:38 AM
mr fantastic
Quote:

Originally Posted by Taylor Fylde
I'm asked to find the the times (0<t<2pi) that two vectors are perpendicular.

The two vectors are perpendicular when their Dot product is zero. Taking the Dot product, and equating to zero gives:

2.Cos(2t).Sin(2t) - 4.Cos(t).Sin(t)=0

Or:

2.Cos(2t).Sin(2t)= 4Cos(t)Sin(t)

So the vectors are perpendicular when the above holds true.

I've sketched the plot of Sin(t), Cos(t), Sin(2t) and Cos(2t).

The R.H.S. of the equation reduces to zero when either Cos(t) or Sin(t) are zero, this is at ;

t=0, pi/2, pi, 3pi/2, 2pi, and the above equation holds true.

I feel like I've done this more by intuition than by mathematical method, and I can't show that these are the only values for which the vectors are perpendicular (though I suspect they are).

There must be a better way to go about this, and I have a feeling it may be with some trig identity.

In short, have I got all the values? Is there a better way to do it?

Many thanks.

The equation you solving is 2.Cos(2t).Sin(2t) - 4.Cos(t).Sin(t)=0.

Dividing both sides by 2 and applying the double angle formula in reverse:

$\displaystyle \cos (2t) \sin (2t) - \sin(2t) = 0 \Rightarrow \sin (2t) (\cos (2t) - 1) = 0$.

It follows that either $\displaystyle \sin (2t) = 0$ or $\displaystyle \cos (2t) - 1 = 0$ .....
• May 25th 2008, 02:47 AM
Taylor Fylde
Quote:

Originally Posted by mr fantastic
The equation you solving is 2.Cos(2t).Sin(2t) - 4.Cos(t).Sin(t)=0.

Dividing both sides by 2 and applying the double angle formula in reverse:

$\displaystyle \cos (2t) \sin (2t) - \sin(2t) = 0 \Rightarrow \sin (2t) (\cos (2t) - 1) = 0$.

It follows that either $\displaystyle \sin (2t) = 0$ or $\displaystyle \cos (2t) - 1 = 0$ .....

Thanks for that, but I now have a similar related problem!

I have to show a particle will pass through the origin twice in the time period (0<t<2pi), it's position described by the vector:

Sin(2t)i + 2cos(t)j = 0 (0<t<2pi), i and j are the unit vectors.

If I plot Sin(2t) and Cos(t), I see that this is at t=pi/2, t=3pi/2. This makes sense, the particle can only be at the origin if the i and j components are zero.

But if I try to solve as you showed above, I use the double angle formula and my equation becomes:

2Sin(t)Cos(t)i + 2Cos(t)j =0

I'm going to drop the i and j, I can't really justify it, but I don't think it matters, and I write:

2Cos(t){Sin(t)+1}=0

So, Cos(t) =0 at pi/2 and 3pi/2 (what I see from my trace)

Sin(t)+1 =0

Sin(t)=-1 t=3pi/4, t=7pi/4 !

How have I managed to come up with this false answer?

TF
• May 25th 2008, 02:55 AM
mr fantastic
Quote:

Originally Posted by Taylor Fylde
Thanks for that, but I now have a similar related problem!

I have to show a particle will pass through the origin twice in the time period (0<t<2pi), it's position described by the vector:

Sin(2t)i + 2cos(t)j = 0 (0<t<2pi), i and j are the unit vectors.

If I plot Sin(2t) and Cos(t), I see that this is at t=pi/2, t=3pi/2. This makes sense, the particle can only be at the origin if the i and j components are zero.

But if I try to solve as you showed above, I use the double angle formula and my equation becomes:

2Sin(t)Cos(t)i + 2Cos(t)j =0

I'm going to drop the i and j, I can't really justify it, but I don't think it matters, and I write:

2Cos(t){Sin(t)+1}=0

So, Cos(t) =0 at pi/2 and 3pi/2 (what I see from my trace)

Sin(t)+1 =0

Sin(t)=-1 t=3pi/4, t=7pi/4 !

How have I managed to come up with this false answer?

TF

Sin(2t)i + 2cos(t)j = 0i + 0j (0<t<2pi)

You require that both i and j components are zero simultaneously:

i component: sin(2t) = 0 => 2 sin(t) cos(t) = 0 .... (1)

j-component: 2 cos(t) = 0 .... (2)

Clearly the simultaneous solution to (1) and (2) is cos(t) = 0 => t = pi/2, 3pi/2.
• May 25th 2008, 03:09 AM
Taylor Fylde
Quote:

Originally Posted by mr fantastic
Sin(2t)i + 2cos(t)j = 0i + 0j (0<t<2pi)

You require that both i and j components are zero simultaneously:

i component: sin(2t) = 0 => 2 sin(t) cos(t) = 0 .... (1)

j-component: 2 cos(t) = 0 .... (2)

Clearly the simultaneous solution to (1) and (2) is cos(t) = 0 => t = pi/2, 3pi/2.

Now this, is quality service :-)

Many thanks, I could see my answer from the trace was correct, but couldn't see why my second method didn't agree.

Many thanks, I've been puzzling over this for a few days now.

TF