Q. 47 - 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.

$x=t^2,y=(2-f{)}^2,t=\frac{1}{2}$.

Step-by-Step Solution

Answer

The second derivative of the given parametric curve $t=\frac{1}{2}$ is 8.

Step 1: Given information

The parametric equations$x=t^2,y=(2-f{)}^2,t=\frac{1}{2}$.

Step 2: Calculation

Consider the parametric curves$x=t^2,y=(2-t{)}^2$ at $t=\frac{1}{2}$.

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 $x=t^2$.

Differentiate the curve with respect to t.

Then

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

$\frac{dx}{dt}=2t\times \frac{dt}{dt}\left[since\frac{d{t}^2}{dt}=2t\right]$

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

Again differentiating with respect to t.

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

Now take the parametric equation $y=(2-t{)}^2$.

Differentiate the curve with respect to t.

$$\begin{gathered} \frac{dy}{dt}=\frac{d}{dt}(2-t{)}^2 \\ \frac{dy}{dt}=2·(2-t\left)\frac{d}{dt}\right(2-t) \\ \frac{dy}{dt}=2(2-t\left)\right(-1) \\ \frac{dy}{dt}=-2(2-t) \end{gathered}$$

Again differentiating with respect to t.

$$\begin{gathered} \frac{d^{2}y}{d{t}^2}=-2\frac{d}{dt}(2-t) \\ \frac{d^{2}y}{d{t}^2}=-2\times -1 \\ \frac{d^{2}y}{d{t}^2}=2 \end{gathered}$$

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,

$\frac{d^{2}y}{d{x}^2}=\frac{2t\times 2-2\times -2(2-t)}{(2t{)}^2}\sin ce \frac{dx}{dt}=2t,\frac{dy}{dt}=-2(2-t)$

$\frac{d^{2}y}{d{x}^2}=\frac{4t+8-4t}{4t^2}$

$\frac{d^{2}y}{d{x}^2}=\frac{8}{4t^2}$

At $t=\frac{1}{2}$ the value of the second derivative is,

$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=\frac{1}{2}=\frac{8}{4\times \frac{1}{4}} \\ \frac{d^{2}y}{d{x}^2}=8 \end{gathered}$$

Therefore, the second derivative of the given parametric curve$t=\frac{1}{2}$ is 8.