For \(t \geq 0\), the components of the velocity moving in the \(xy\)-plane are given by the parametric equations \(x'(t)=\dfrac{1}{t+1} \) and \(y'(t)=ke^{kt}\), where \(k\) is a positive constant. The line \(y=4x+3\) is parallel to the line tangent to the path of the particle where \(t=2\). What is the value of \(k\) ?

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The slope of the tangent line is \(4\).

$$ \frac{dy}{dx} = \frac{y'(t)}{x'(t)} = \frac{ke^{kt}}{\frac{1}{t+1}}$$ $$ \frac{dy}{dx}\Big|_{t=2} = 4 $$ $$ \frac{ke^{2k}}{\frac{1}{2+1}}=4 $$ $$ 3ke^{2k}=4 $$

We can use our calculator to obtain the answer:

$$ k\approx \boxed{0.495} $$