Chapter 10.1, Problem 10E

$\text{(a)}$

To determine

To find: The value of $dy/dx$ .

Answer to Problem 10E

The answer:

  $\frac{dy}{dx}=-t$

Explanation of Solution

Given information:

The equations are:

  $\begin{array}{c}x=1/t,\\ y=-2+\ln t\end{array}$

Calculation:

The given equations are

  $\begin{array}{c}x=1/t,\\ y=-2+\ln t\end{array}$

Consider that to use the formulas for parametric differentiation:

  $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$

To use the previous result,

  $\begin{array}{c}\frac{dy}{dx}=\frac{1/t}{-1/t^2}\\ =-\frac{t^2}{t}\\ =-t\end{array}$

Therefore, the required value of $\frac{dy}{dx}=-t$ .

$\text{(b)}$

To determine

To find: The value of $d^{2}y/d{x}^2$ in terms of $t$ .

Answer to Problem 10E

The answer:

  $\frac{d^{2}y}{d{x}^2}=t^2$

Explanation of Solution

Given information:

The equations are:

  $\begin{array}{c}x=1/t,\\ y=-2+\ln t\end{array}$

Calculation:

The given equations are

  $\begin{array}{c}x=1/t,\\ y=-2+\ln t\end{array}$

To use the formulas for parametric differentiation, see that

  $\frac{d^{2}y}{d{x}^2}=\frac{d{y}^{\prime }/dt}{dx/dt}$

Where,

  $y^{\prime }=dy/dx$

Hence from part $\text{(a)}$ know that,

  $y^{\prime }=-t$

Then,

  $d{y}^{\prime }/dt=-1$

To use the previous result,

  $\begin{array}{c}\frac{d^{2}y}{d{x}^2}=\frac{-1}{-1/t^2}\\ =t^2\end{array}$

Therefore, the required value of $\frac{d^{2}y}{d{x}^2}=t^2$ .