Parametric equations

**mlkauffman** · February 27th 2010, 08:47 AM

I am hoping someone can help me to solve this problem.

In the xy-plane, a particle moves along the parabola y = x^2 - x with a constant speed of 2 times the square root of 10, units per second. If dx/dt is greater than 0, what is the value of dy/dt when the particle is at the point (2,2)? The answer is 6, but I do not know how to get there. Thanks in advance for your help.

**nehme007** · February 27th 2010, 10:25 AM

Speed equals $\sqrt{\frac{dx}{dt}^2 + \frac{dy}{dt}^2} = 2\sqrt{10}$ .

$y = x^2 - x$ implies that $\frac{dy}{dt} = (2x-1) \frac{dx}{dt}$ .

So $\sqrt{\frac{dx}{dt}^2 +\frac{dy}{dt}^2} = \sqrt{\frac{dx}{dt}^2 + (2x-1)^2 \frac{dx}{dt}^2} = \frac{dx}{dt} \sqrt{4x^2 - 4x + 2} = 2\sqrt{10}$ .

Letting x = 2, you should get $\frac{dx}{dt} = 2$ , which means $\frac{dy}{dt} = (2*2-1)*2 = 6$ .

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