Q. 46 - 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.

$x=t+2,y=e^t,t=0$.

Step-by-Step Solution

Answer

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

Step 1: Given information

The parametric equations $x=t+2,y=e^t,t=0$.

Step 2: Calculation

Consider the parametric curves $x=t+2,y=e^{\text{'}}$at $t=0$.

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 them in the formula.

Now take the parametric equation $x=t+2$.

Differentiate the curve with respect to t.

Then

$$\begin{gathered} \frac{dx}{dt}=\frac{d}{dt}(t+2)[sincex=t+2] \\ \frac{dx}{dt}=\frac{d}{dt}t+\frac{d}{dx}2 \\ \frac{dx}{dt}=1since\frac{d1}{dt}=0 \end{gathered}$$

Thus,

$\frac{dx}{dt}=1$

Again differentiating with respect to t,

$\frac{d^{2}x}{d{t}^2}=\frac{d}{dt}1$

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

Now take the parametric equation $y=e^t$.

Differentiate the curve with respect to t.

$$\begin{gathered} \frac{dy}{dt}=\frac{d}{dt}{e}^t \\ \frac{dy}{dt}=e^t \end{gathered}$$

Thus.

$\frac{dy}{dt}=e$

Again differentiating with respect to r.

$$\begin{gathered} \frac{d^{2}y}{d{t}^2}=\frac{d}{dt} \\ \frac{d^{2}y}{d{t}^2}=e^t \end{gathered}$$

New substitute the values in the formula $\frac{d^{2}y}{d{t}^2}=\frac{\frac{dx}{dt}\times \frac{d^{2}y}{d{t}^2}-\frac{d^{2}x}{d^{2}f}\times \frac{dy}{dt}}{{\left(\frac{dx}{dt}\right)}^2}$.

Then

$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=1\times e \\ \frac{d^{2}y}{d{x}^2}=e^t \end{gathered}$$

At t=0 the second derivative is $\frac{d^{2}y}{d{x}^2}=e^e$By substitution $\frac{d^{2}y}{d{x}^2}=1$

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