## 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?

1. anonymous

Thanks you for showing up D:

shouldnt there be an x term in the parabola?

3. anonymous

Oh yea

so 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 $$\frac{dx}{dt}$$ $\left[\frac{dx}{dt} \right]_{y=4} = \left[\frac{df(y)}{dy}\right]_{y=4} \left[\frac{dy}{dt}\right]_{y=4}$

5. anonymous

okay, so I have to find that?

yeah find the values of the expressions in the r.h.s

7. anonymous

kk one second

8. anonymous

$\frac{ ds }{ dt }=5?$

but 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

10. anonymous

are you sure that the question given to is like that?

12. anonymous

I think i may be able to figure it out, but thank you soooo much man! :D

13. anonymous

medal for you ;)

u are welcome ;)

15. anonymous

=D