Asked by jessica
2. 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 dt
is its “vertical” velocity at that instant?
b)If it travels in such away that dx/dt=3 for all t,then what happens to dy/dt as t→+∞?

c) If, however, it travels in such a way that dy/dt remains constant, then what happens
to dx/dt as t→+∞?
a) When it passes through the point (2, 12), its “horizontal” velocity is dx/dt= 3. What dt
is its “vertical” velocity at that instant?
b)If it travels in such away that dx/dt=3 for all t,then what happens to dy/dt as t→+∞?

c) If, however, it travels in such a way that dy/dt remains constant, then what happens
to dx/dt as t→+∞?
Answers
Answered by
Steve
well, since y=3x^2
dy/dt = 6x dx/dt
= 6*2*3 = 36
since dx/dt remains constant, dy/dt gets ever faster as x gets larger.
If, however,
k = 6x dx/dt, then dx/dt gets ever slower as x gets larger
dy/dt = 6x dx/dt
= 6*2*3 = 36
since dx/dt remains constant, dy/dt gets ever faster as x gets larger.
If, however,
k = 6x dx/dt, then dx/dt gets ever slower as x gets larger
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