smc-1159-70-Parametric

The position of a particle moving in the Cartesian plane, at time \(t\), is given by the parametric equations

\(x(t)=\dfrac{6 t}{t+1}\) and \(y(t)=\dfrac{-8}{t^2+4}\), where \(t \geq 0\).

What is the slope of the tangent to the path of the particle when \(t=2\) ?

\(-\dfrac{1}{3}\)
\(-\dfrac{1}{4}\)
\(\dfrac{1}{3}\)
\(\dfrac{3}{4}\)
\(\dfrac{4}{3}\)

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\(D\)

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\(\text{At}\ \ t=2\ \text{(by calc):}\)

\(\dfrac{dx}{dt}=\dfrac{2}{3}, \ \dfrac{dy}{dt}=\dfrac{1}{2} \)

\(\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{dx} = \dfrac{1}{2} \times \dfrac{3}{2} = \dfrac{3}{4} \)

\(\Rightarrow D\)