91Ó°ÊÓ

The parametric curve is $x=2t-1,y=3t+5,t=-1$.

Consider the parametric curves $x=2t-1,y=3t+5$at $t=-1$.

The objective is to find the second derivative that is$\frac{d^{2}y}{d{x}^2}$.

The formula to find the second derivative is$\frac{d^{2}y}{d{x}^2}=\frac{\frac{dx}{dt}×\frac{d^{2}y}{d{t}^2}-\frac{d^{2}x}{d^{2}t}×\frac{dy}{dt}}{{\left(\frac{dx}{dt}\right)}^2}$.

First find the derivatives $\frac{dx}{dt},\frac{dy}{dt},\frac{d^{2}y}{d{t}^2},\frac{d^{2}x}{d^{2}t}$and substitute in the formula.

Now take the parametric equation $x=2t-1$.

Differentiate the curve with respect to t.

Then

$$\begin{gathered} \frac{dx}{dt}=\frac{d}{dt}(2t-1)[\text{since}x=(2t-1\left)\right] \\ \frac{dx}{dt}=2×\frac{dt}{dt}-\frac{d1}{dt} \\ \frac{dx}{dt}=2×1-0\text{since}\frac{d1}{dt}=0 \end{gathered}$$

Thus,

$\frac{dx}{dt}=2$

Again differentiating with respect to t.

$\frac{d^{2}x}{d{t}^2}=\frac{d}{dt}2$

$\frac{d^{2}x}{d{t}^2}=0$[since derivative of a constant is 0]

Now take the parametric equation $y=3t+5$.

Differentiate the curve with respect to t.

$$\begin{gathered} \frac{dy}{dt}=\frac{d}{dt}(3t+5) \\ \frac{dy}{dt}=3\frac{d}{dt}t+\frac{d}{dt}5 \\ \frac{dy}{dt}=3×1+0 \end{gathered}$$

Thus,

$$\begin{gathered} \frac{dy}{dt}=3×1 \\ \frac{dy}{dt}=3 \end{gathered}$$

Again differentiating with respect to t,

$\frac{d^{2}y}{d{t}^2}=\frac{d}{dt}3\frac{d^{2}y}{d{t}^2}=0$

[since derivative of a constant is 0]

Now substitute the values in the formula

$\frac{d^{2}y}{d{x}^2}=\frac{\frac{dx}{dt}×\frac{d^{2}y}{d{t}^2}-\frac{d^{2}x}{d^{2}t}×\frac{dy}{dt}}{{\left(\frac{dx}{dt}\right)}^2}$.

Then

$$\begin{gathered} \frac{d^{2}y}{d{t}^2}=\frac{2×0-0×3}{2^2} \\ \frac{d^{2}y}{d{x}^2}=0 \end{gathered}$$

Therefore, the second derivative of the given parametric curve at $t=-1$is 0 .