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/dt0, what is the value of dy/dt when the particle is at the point (2,2)?

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 particle is at the point (2,2)?

anonymous · Answer

The choices I was given are 
a) 2/3
b) 2(10^.5)/3
c) 6

dumbcow · Answer

$$v^{2} = (\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}$$
$$\frac{dx}{dt} = \frac{dx}{dy}*\frac{dy}{dt}$$
$$\frac{dx}{dy} = \frac{1}{2x-1}$$
...what is the speed? 2*sqrt10   , i am not getting any of your answers

anonymous · Answer

Yeah, that's it...It's from one of the AP Calculus BC Released Exams.

dumbcow · Answer

anyway the equation then becomes
$$v = \frac{dy}{dt}\sqrt{1+\frac{1}{(2x-1)^{2}}}$$
v = 2*sqrt10
solve for dy/dt

anonymous · Answer

Thank you!!

dumbcow · Answer

your welcome..ahh i see what i did wrong...answer is 6

anonymous · Answer

cool. just plugged it in & i got that too.