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A particle moves at constant speed 1630 units along the curve of intersection of the two surfaces y=x^2 and z= (2/3)x^3 in the direction of increasing x. Find it's velocity when it is at the point (-9, 81, -486).

I know how to find velocity vectors at a point, however I'm very confused by it being the intersection of two surfaces. How would I approach this? Thank you.

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    I don't think moving 1630 units is enough to tell us anything about speed. – 2017-01-24
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    yes I was also confused by that, but that's all the question states.. – 2017-01-24
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    I agree with Ahmed. You need to know something about the time to say anything about the velocity. – 2017-01-24

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You have been given the speed of the particle: $$ \frac{ds}{dt}=1630.$$ Supposing this was presented as a math problem rather than a physics problem, it is reasonable to assume that if specific units are not given, we're intended to use "math units" in which one (unnamed) unit of speed equals one unit of distance per unit of time. So I would not worry about the units.

What you do want to be concerned with is how to relate the speed $\frac{ds}{dt}$ to the three components of velocity, namely $\frac{dx}{dt}$, $\frac{dy}{dt}$, and $\frac{dz}{dt}$. To do this, you can use the equation $$\left(\frac{ds}{dt}\right)^2 = \left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2.$$ To simplify this, you can use what you know about the two surfaces the particle is riding on to determine $\frac{dy}{dx}$ and $\frac{dz}{dx}$, which will enable you to rewrite $\frac{dy}{dt}$ and $\frac{dz}{dt}$ in terms of $\frac{dx}{dt}$.