# Calculus with Parametrics problem

1. Mar 29, 2005

### 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?

Last edited: Mar 29, 2005
2. Mar 29, 2005

### 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.

3. Mar 29, 2005

### 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?

4. Mar 29, 2005

### xanthym

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)}

~~

5. Mar 29, 2005

### HallsofIvy

Staff Emeritus
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.

6. Mar 30, 2005

### ChickenChakuro

Thanks guys. Got it :)