Calculus with Parametrics problem

[ChickenChakuro]

Hey guys, I'm having difficulty understanding exactly what this problem is asking:

The position vector of an object for t\geq0 is r(t) = <cos(3t),cos(2t)>.
Find all possible values of \frac{dy}{dx} at the point (0,1/2).

Seriously, what does this mean?

 

[Data]

Well, you have y = \cos{2t} and x = \cos{3t}. You can use the chain rule

\frac{dy}{dx} = \frac{dy}{dt}\frac{dt}{dx}

to find the derivative of y, then you'll need to solve \cos{2t}=\frac{1}{2}, \ \cos{3t}=0 to find the values of t corresponding to the point (0, 1/2), substitute those into the derivative that you found (which should be in terms of t) and find all its possible values.

 

[ChickenChakuro]

Thanks, here's what I did:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2sin(2t)}{3sin(3t)}

Then,

cos(2t) = \frac{1}{2} when t = \frac{\pi}{4}, \frac{3\pi}{4}, etc and

cos(3t) = 0 when t = \frac{\pi}{6}, \frac{3\pi}{6}, etc, and

\frac{2sin(2t)}{3sin(3t)} has possible values: \frac{2}{3},\frac{-2}{3}.

Is that right?

 

[xanthym]

Thanks, here's what I did:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2sin(2t)}{3sin(3t)}

Then,

cos(2t) = \frac{1}{2} when t = \frac{\pi}{4}, \frac{3\pi}{4}, etc and

cos(3t) = 0 when t = \frac{\pi}{6}, \frac{3\pi}{6}, etc, and

\frac{2sin(2t)}{3sin(3t)} has possible values: \frac{2}{3},\frac{-2}{3}.

Is that right?

Valid "t" values are {t = (π/6), (5π/6), (7π/6), & (11π/6)}, for which:
{dy/dx = (+√3)/3=(+0.577) :OR: (-√3)/3=(-0.577)}


~~

 

[HallsofIvy]

The point is that you can't say "cos(2t)= 1/2)" for t= ...
and "cos(3t)= 0 for t= ...". Each value of t must give both cos(2t)= 1/2 and sin(3t)= 0.

 

[ChickenChakuro]

Thanks guys. Got it :)