# Math Help - Parallel Vectors functions

1. ## Parallel Vectors functions

Hello,

I have to find every value t such that the vector r'(t) is parallel to the yz plane, given that r(t) = <cos(t^2), e^(sqrt(t)), t>

Well, I know how to find the derivative of that, but then don't know what to do from there.

2. Originally Posted by Ideasman
Hello,

I have to find every value t such that the vector r'(t) is parallel to the yz plane, given that r(t) = <cos(t^2), e^(sqrt(t)), t>

Well, I know how to find the derivative of that, but then don't know what to do from there.
Any unit vector parallel to the yz plane will take the form:
<0, cos(phi), sin(phi)>
So r(t) must be proportional to one of these in general.

Comparing the x components we see that
cos(t^2) = 0
Thus
t^2 = (2n + 1)pi/2 for n = 0, 1, 2, ...

t = (+/-)sqrt{(2n + 1)pi/2}

Additionally, dividing the z component by the y component gives us:
e^(sqrt(t))/t = tan(phi)

Since tan(phi) can be any angle t can be any value, so this doesn't give any restrictions on t. So the solution is:

t = (+/-)sqrt{(2n + 1)pi/2}

-Dan

3. Originally Posted by topsquark
Any unit vector parallel to the yz plane will take the form:
<0, cos(phi), sin(phi)>
So r(t) must be proportional to one of these in general.

Comparing the x components we see that
cos(t^2) = 0
Thus
t^2 = (2n + 1)pi/2 for n = 0, 1, 2, ...

t = (+/-)sqrt{(2n + 1)pi/2}

Additionally, dividing the z component by the y component gives us:
e^(sqrt(t))/t = tan(phi)

Since tan(phi) can be any angle t can be any value, so this doesn't give any restrictions on t. So the solution is:

t = (+/-)sqrt{(2n + 1)pi/2}

-Dan
Where in the world is the 2n+1 from?

4. Originally Posted by Ideasman
Where in the world is the 2n+1 from?
cos(t^2) = 0

So
t^2 = pi/2, 3pi/2, 5pi/2, etc.

Now, 2n + 1 = 1, 3, 5, ... for n = 0, 1, 2, ...

So
t^2 = (2n+1)pi/2 for n = 0, 1, 2, ...

-Dan