Posted by **John** on Sunday, October 18, 2009 at 6:16pm.

An object is moving along the parabola y = 3x^2. (a) When it passes through the point (2,12), its horizontal velocity is dx/dt = 3. what is its vertical velocity at that instant? (b) If it travels in such a way that dx/dt = 3 for all t, then what happens to dy/dt as t --> +infinity? (c) If, however, it travels in such a way that dy/dt remains constant, then what happens to dy/dt as t --> +infinity?

- calculus -
**bobpursley**, Sunday, October 18, 2009 at 6:17pm
What was wrong with my previous response? I still make no sense of the last part.

- calculus -
**Andre**, Sunday, October 18, 2009 at 6:39pm
For question a, y = 3x^2, so

dy/dt = 6x dx/dt

dy/dt = 6(2) * 3

dy/dt = 36

For question b, as t approacing infinity, x is also approaching infinity (the object is moving along the track of the function, remember(, then dy/dt is also approaching infinity in vertical sense...

For question c, I think you ask for dx/dt, instead of dy/dt since dy/dt always constant... If you ask for dx/dt, dx/dt will be approaching 0 as dt approaching infinity. This is the reason:

dy/dt = 6x dx/dt

as t approaching infinity, x will also be approaching infinity.. Since we hold dy/dt to be constant, dx/dt will have to be a small number to counter the growth of 6x as it approaches infinity.. Therefore, dx/dt will have to approach 0

Hope it helps

- calculus -
**John**, Sunday, October 18, 2009 at 8:55pm
No, nothing was wrong with your last response. That was my mistake for posting it up again. Thank you for all your responses.

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