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Am I finding the equation of the slope of the tangent line at c(t)?

$\frac{dy/dt}{dx/dt}$ = $\frac{2t}{3t^2-8}$

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Quaxton Hale
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  • Isn't the velocity=$\sqrt{\left(\dfrac{\mathrm{d}y}{\mathrm{d}t}\right)^2+(\dfrac{dx}{dt})^2} $, using Pythagoras Theorem?? – K. Rmth Nov 24 '13 at 20:40

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The position of the particle at any particular time will be given by the vector

r(t) = (x(t),y(t))

You want to differentiate r(t), i.e,

 v(t) = dr(t)/dt = (dx(t)/dt,dy(t)/dt)

Note that the speed will be given by

 s(t) = d/dt of sqrt(x(t)^2+y(t)^2), but that is not the velocity or a vector
Penguino
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  • To make it a vector, we need the direction. $\arctan \dfrac{dy/dt}{dx/dt}$ is the direction angle w.r.t the horizontal. – K. Rmth Nov 24 '13 at 20:45