Calculus and Parametric Equations: An object is moving along a curve in the xy-plane has position (x(t), y(t)) at time t with dx/dt = 2sin(t^3) and dy/dt = cos(t^2) for 0

Question

Calculus and Parametric Equations: An object is moving along a curve in the xy-plane has position (x(t), y(t)) at time t with dx/dt = 2sin(t^3) and dy/dt = cos(t^2) for 0 <= t <= 4. At time t = 1, the object is at the position (3,4). Find the position of the object at time t = 2.

anonymous · Answer

not sure
are you supposed to integrate $\frac{dy}{dx}$?

anonymous · Answer

that would be awesome, but I'm not sure how to do that while dy/dx is still in terms of t..

anonymous · Answer

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

anonymous · Answer

oh i see the problem it is still in terms of $t$

anonymous · Answer

yup :)

anonymous · Answer

I wish both dy/dt and dx/dt had simple solutions.. but in this case they're not, so I think there's another way to do it using parametrics..

anonymous · Answer

yeah, you are not going to find an anti derivative of that anyway
hmm i wonder what you can do?
i would bump this question up, maybe some smart person can answer it

anonymous · Answer

bumped.. thanks. I know you're one of the smartest/most helpful people around here since I've been using openstudy for some years now and you're one of the people who could always answer my questions. :)

anonymous · Answer

You going to close this?  You still stuck?

anonymous · Answer

still stuck, yes

anonymous · Answer

I think you should leave the answer in terms of definite integrals.

anonymous · Answer

$$
\mathbf r(2) = \left\langle \int_{1}^{2}\cos(t^2)\;dt + 3, \int_{1}^{2}2\sin(t^3)\;dt + 4  \right\rangle
$$

anonymous · Answer

There isn't any particular theorem here that I know of that will make these integrals closed.

anonymous · Answer

Thanks, well I can carry out numeric integration using my TI-84.. and doing that seems to yield the correct answer (well only that you switched the places of the cos and sin terms [x is sin and y is cos in the prob]). The answer turns out to be (3.437, 3.557)

anonymous · Answer

Could you please explain to me the thought process of how you set up those integrals?

anonymous · Answer

$$
\mathbf r(b) - \mathbf r(a) = \int_a^b\mathbf r'(t)\;dt
$$

anonymous · Answer

$a=1$, $b=2$

anonymous · Answer

$$
\mathbf r(2) = \int_1^2\mathbf r'(t)\;dt+\mathbf r(1)
$$

anonymous · Answer

Thanks, that confirms what I've thought. Thanks so much!