Question

In the xy-plane, a particle moves along the parabola y=x^2-x with a constant speed of 2(10)^1/2 units per second. If dx/dt>0, what is the value of dy/dt when the particla is at the point (2,2)?

Answer 1

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Answer 2

use the chain rule\[\frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt}\]\[\frac{dx}{dt} = \frac{\frac{dy}{dt}}{\frac{dy}{dx}}\]the speed,\[v = \sqrt{\left(\frac{dx}{dt}\right)^{2} + \left(\frac{dy}{dt}\right)^2}\]solve for dy/dt at (2,2)