Q. 45 - Calculus

Question

In Exercises 45-48 use Example 6 to find $\frac{d^{2}y}{d{x}^2}$ for the parametric curve at the given value of t. Note that these are the same parametric equations as in Exercises 41-44.

$x=2 t-1, y=3 t+5, t=-1$.

Step-by-Step Solution

Answer

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

Step 1: Given information

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

Step 2: Calculation

Consider the parametric curves $x=2 t-1, y=3 t+5$ 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$\frac{d^{2}y}{d{x}^2}=\frac{\frac{dx}{dt}\times \frac{d^{2}y}{d{t}^2}-\frac{d^{2}x}{d^{2}t}\times \frac{dy}{dt}}{{\left(\frac{dx}{dt}\right)}^2}$.

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 $x=2 t-1$.

Differentiate the curve with respect to t.

Then

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

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]

Step 3: Further simplification

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

Differentiate the curve with respect to t.

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

Thus,

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

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

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

Then

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

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