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:

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