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# Find $dy/dx$ and $d^2y/dx^2$, and find the slope and concavity (if possible) at the given value of the parameter.$\underline{Parametric\ Equation}$$x=\sqrt t,\ y=3t-1$$\underline{Point}$$t=1$

Solution

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Given:

\begin{aligned} x&=\sqrt{t} \\ y&=3t-1 \end{aligned}

We need to determine the derivatives $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$, while also determining the slope and concavity at $t=1$.

The following formula can be used to determine the derivative $\frac{dy}{dx}$ of parametric equations:

\begin{aligned} \frac{dy}{dx}&=\frac{dy/dt}{dx/dt} \end{aligned}

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