Chapter 10.1, Problem 11E

$\text{(a)}$

To determine

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

Answer to Problem 11E

The answer:

  $\frac{dy}{dx}=\frac{3t^2}{2t-3}$

Explanation of Solution

Given information:

The equations are:

  $\begin{array}{c}x=t^2-3t,\\ y=t^3\end{array}$

Calculation:

The given equations are

  $\begin{array}{c}x=t^2-3t,\\ y=t^3\end{array}$

Consider that to use the formulas for parametric differentiation:

  $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$

To use the previous result,

  $\frac{dy}{dx}=\frac{3t^2}{2t-3}$

Therefore, the required value of $\frac{dy}{dx}=\frac{3t^2}{2t-3}$ .

$\text{(b)}$

To determine

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

Answer to Problem 11E

The answer:

  $\frac{d^{2}y}{d{x}^2}=\frac{6t^2-18t}{{(2t-3)}^3}$

Explanation of Solution

Given information:

The equations are:

  $\begin{array}{c}x=t^2-3t,\\ y=t^3\end{array}$

Calculation:

The given equations are

  $\begin{array}{c}x=t^2-3t,\\ y=t^3\end{array}$

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,

  $y^{\prime }=3t^2/(2t-3)$

Then,

  $\begin{array}{c}\frac{d{y}^{\prime }}{dt}=\frac{6t\cdot (2t-3)-3t^2\cdot 2}{{(2t-3)}^2}\\ =\frac{12t^2-18t-6t^2}{{(2t-3)}^2}\\ =\frac{6t^2-18t}{{(2t-3)}^2}\end{array}$

To use the previous result,

  $\begin{array}{c}\frac{d^{2}y}{d{x}^2}=\frac{\left(6t^2-18t\right)/{(2t-3)}^2}{2t-3}\\ =\frac{6t^2-18t}{{(2t-3)}^3}\end{array}$

Therefore, the required value of $\frac{d^{2}y}{d{x}^2}=\frac{6t^2-18t}{{(2t-3)}^3}$ .