Chapter 10.1, Problem 15E

$\text{(a)}$

To determine

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

Answer to Problem 15E

The answer:

  $\frac{dy}{dx}=4$

Explanation of Solution

Given information:

The equations are:

  $\begin{array}{c}x=\ln (2t),\\ $ y=\ln {(3t)}^4$\end{array}$

Calculation:

The given equations are

  $\begin{array}{c}x=\ln (2t),\\ $ y=\ln {(3t)}^4$\end{array}$

Then,

  $\begin{array}{c}$ \frac{dy}{dt}=\frac{1}{{(3t)}^4}\cdot 4\cdot {(3t)}^3\cdot 3$=\frac{12\cdot 3^3\cdot t^3}{3^4\cdot t^4}\\ =\frac{4}{t}\\ \frac{dx}{dt}=\frac{1}{2t}\cdot 2\\ =\frac{1}{t}\end{array}$

Consider that to use the formulas for parametric differentiation:

  $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$

To use the previous result,

  $\begin{array}{c}\frac{dy}{dx}=\frac{4/t}{1/t}\\ =\frac{4t}{t}\\ =4\end{array}$

Therefore, the required value of $\frac{dy}{dx}=4$ .

$\text{(b)}$

To determine

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

Answer to Problem 15E

The answer:

  $\frac{d^{2}y}{d{x}^2}=0$

Explanation of Solution

Given information:

The equations are:

  $\begin{array}{c}x=\ln (2t),\\ $ y=\ln {(3t)}^4$\end{array}$

Calculation:

The given equations are

  $\begin{array}{c}x=\ln (2t),\\ $ y=\ln {(3t)}^4$\end{array}$

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,

  $y^{\prime }=4$

Then,

  $d{y}^{\prime }/dt=0$

Since from part $\text{(a)}$

  $dx/dt=1/t$

To use the previous result,

  $\begin{array}{c}\frac{d^{2}y}{d{x}^2}=\frac{0}{1/t}\\ =0\end{array}$

Therefore, the required value of $\frac{d^{2}y}{d{x}^2}=0$ .