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anonymous
 one year ago
a point moves on the parabola y^2=8 in such a way that the rate of change of the ordinate is always 5 units/second. how fast is the abscissa changing when the ordinate is 4?
anonymous
 one year ago
a point moves on the parabola y^2=8 in such a way that the rate of change of the ordinate is always 5 units/second. how fast is the abscissa changing when the ordinate is 4?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Thanks you for showing up D:

BAdhi
 one year ago
Best ResponseYou've already chosen the best response.1shouldnt there be an x term in the parabola?

BAdhi
 one year ago
Best ResponseYou've already chosen the best response.1so in general what you should do is differentiate the equation wrt to time. For example if the equation is \(x = f(y)\) \[\frac {dx}{dt} = \frac{df(y)}{dt}\] which further can be represented as, \[\frac{dx}{dt} = \frac{df(y)}{dy}\frac{dy}{dt}\] so when \(y=4\) find the \(\frac{dx}{dt}\) \[\left[\frac{dx}{dt} \right]_{y=4} = \left[\frac{df(y)}{dy}\right]_{y=4} \left[\frac{dy}{dt}\right]_{y=4} \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay, so I have to find that?

BAdhi
 one year ago
Best ResponseYou've already chosen the best response.1yeah find the values of the expressions in the r.h.s

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ \frac{ ds }{ dt }=5?\]

BAdhi
 one year ago
Best ResponseYou've already chosen the best response.1but according to your given problem, the question itself contradicts. since it is stated that \(y^2=8\implies y = \pm \sqrt 8\) which is a constant and later it talks about a rate of change in the value of y wrt time.. Can you state the equation correctly

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have no clue about this :/

BAdhi
 one year ago
Best ResponseYou've already chosen the best response.1are you sure that the question given to is like that?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think i may be able to figure it out, but thank you soooo much man! :D
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