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.

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- Mar 5th 2007, 02:49 AMIdeasmanParallel 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. - Mar 5th 2007, 03:31 AMtopsquark
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 - Mar 5th 2007, 05:42 AMIdeasman
- Mar 5th 2007, 06:07 AMtopsquark