Chapter 10, Problem 10RE

a)

To determine

To find: The equation for the tangent to the curve $x=1+\frac{1}{t^2}$ , $y=1-\frac{3}{t}$ at the point corresponding to $t=2$ .

Answer to Problem 10RE

The equation for the tangent to the curve $x=1+\frac{1}{t^2}$ , $y=1-\frac{3}{t}$ at the point corresponding to $t=2$ is: $y+\frac{1}{2}=3\left(x-\frac{5}{4}\right)$ .

Explanation of Solution

Given information:

The parametric equation of the curve is: $x=1+\frac{1}{t^2}$ , $y=1-\frac{3}{t}$ .

  $t=2$

Formula used:

Application of the Chain rule of Differentiation:

  $\begin{array}{l}\frac{dy}{dt}=\frac{dy}{dx}\cdot \frac{dx}{dt}\\ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\end{array}$

Tangents of Parametric curves:

When a curve is defined by parametric equations $x=f(t)$ , $y=f(t)$ , the slope $\frac{dy}{dx}$ of the tangent to the curve is computed using the Chain Rule (since, the curve is not defined as a function of $x$

):

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

Point-slope form of a line:

The equation of a line whose slope is $m$ and which passes through a point $\left(x_1,y_1\right)$ is given as: $y-y_1=m\left(x-x_1\right)$ .

Calculation:

The equation of the curve is:

  $x=1+\frac{1}{t^2}$

And,

  $y=1-\frac{3}{t}$

Substitute $t=2$ in the equations to obtain the point corresponding to $t=2$ :

  $\begin{array}{l}x=1+\frac{1}{t^2}\\ x=1+\frac{1}{2^2}\\ x=1+\frac{1}{4}\\ x=\frac{5}{4}\end{array}$

And,

  $\begin{array}{l}y=1-\frac{3}{2}\\ y=\frac{2-3}{2}\\ y=-\frac{1}{2}\end{array}$

The point corresponding to $t=2$ is $\left(\frac{5}{4},-\frac{1}{2}\right)$ .

Differentiate the equations of the curve with respect to $t$ .

  $\begin{array}{l}x=1+\frac{1}{t^2}\\ \frac{dx}{dt}=\frac{d}{dt}\left(1+\frac{1}{t^2}\right)\\ \frac{dx}{dt}=\frac{d}{dt}\left(1\right)+\frac{d}{dt}\left(\frac{1}{t^2}\right)\\ \frac{dx}{dt}=0+\frac{d}{dt}\left(t^{-2}\right)\end{array}$

And,

  $\begin{array}{l}y=1-\frac{3}{t}\\ \frac{dy}{dt}=\frac{d}{dt}\left(1-\frac{3}{t}\right)\\ \frac{dy}{dt}=\frac{d}{dt}\left(1\right)-\frac{d}{dt}\left(\frac{3}{t}\right)\\ \frac{dy}{dt}=0-\frac{d}{dt}\left(3t^{-1}\right)\end{array}$

Simplify further to obtain:

  $\begin{array}{l}\frac{dx}{dt}=\frac{d}{dt}\left(t^{-2}\right)\\ \frac{dx}{dt}=-\frac{2}{t^3}\end{array}$

And,

  $\begin{array}{l}\frac{dy}{dt}=\frac{d}{dt}\left(3t^{-1}\right)\\ \frac{dy}{dt}=-\frac{3}{t^2}\end{array}$

Substitute $\frac{dx}{dt}=-\frac{2}{t^3}$ and $\frac{dy}{dt}=-\frac{3}{t^2}$ in the formula for the slope of the tangent to the curve $x=1+\frac{1}{t^2}$ , $y=1-\frac{3}{t}$ to obtain:

  $\begin{array}{l}\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ \frac{dy}{dx}=\frac{-\frac{3}{t^2}}{-\frac{2}{t^3}}\\ \frac{dy}{dx}=\frac{-3\times t^3}{-2\times t^2}\\ \frac{dy}{dx}=\frac{3}{2}t\end{array}$

At the point corresponding to $t=2$ , the slope of the tangent is given by:

  $\begin{array}{l}{\left.m=\frac{dy}{dx}\right|}_2\\ m=\frac{3}{2}\times 2\\ m=3\end{array}$

Substitute $m=3$ in the formula of a line in point-slope form to obtain the equation of the tangent to the curve $x=1+\frac{1}{t^2}$ , $y=1-\frac{3}{t}$ at the point $\left(\frac{5}{4},-\frac{1}{2}\right)$:

  $\begin{array}{l}y-\left(-\frac{1}{2}\right)=3\left(x-\frac{5}{4}\right)\\ y+\frac{1}{2}=3\left(x-\frac{5}{4}\right)\end{array}$

Thus,

The equation for the tangent to the curve $x=1+\frac{1}{t^2}$ , $y=1-\frac{3}{t}$ at the point corresponding to $t=2$ is: $y+\frac{1}{2}=3\left(x-\frac{5}{4}\right)$ .

b)

To determine

To find:The value of $\frac{d^{2}y}{d{x}^2}$ at the point corresponding to $t=2$ .

Answer to Problem 10RE

The value of $\frac{d^{2}y}{d{x}^2}$ at the point corresponding to $t=2$ is $-6$ .

Explanation of Solution

Given information:

The parametric equation of the curve is: $x=1+\frac{1}{t^2}$ , $y=1-\frac{3}{t}$ and $t=2$ .

Formula used:

Application of the Chain rule of Derivatives:

  $\begin{array}{l}\frac{dy}{dt}=\frac{dy}{dx}\cdot \frac{dx}{dt}\\ \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\end{array}$

Chain Rule for Higher Derivatives:

For the second derivative, $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$

Quotient rule of Derivatives:

  $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{v\frac{du}{dx}-u\frac{dv}{dx}}{v^2}$

Calculation:

The equation of the curve is:

  $x=1+\frac{1}{t^2}$ ,

  $y=1-\frac{3}{t}$

Differentiate the equations of the curve with respect to $t$ :

  $\begin{array}{l}x=1+\frac{1}{t^2}\\ \frac{dx}{dt}=\frac{d}{dt}\left(1+\frac{1}{t^2}\right)\\ \frac{dx}{dt}=\frac{d}{dt}\left(1\right)+\frac{d}{dt}\left(\frac{1}{t^2}\right)\\ \frac{dx}{dt}=0+\frac{d}{dt}\left(t^{-2}\right)\end{array}$$\begin{array}{l}y=1-\frac{3}{t}\\ \frac{dy}{dt}=\frac{d}{dt}\left(1-\frac{3}{t}\right)\\ \frac{dy}{dt}=\frac{d}{dt}\left(1\right)-\frac{d}{dt}\left(\frac{3}{t}\right)\\ \frac{dy}{dt}=0-\frac{d}{dt}\left(3t^{-1}\right)\end{array}$

Simplify further to obtain:

  $\begin{array}{l}\frac{dx}{dt}=\frac{d}{dt}\left(t^{-2}\right)\\ \frac{dx}{dt}=-\frac{2}{t^3}\end{array}$

And,

  $\begin{array}{l}\frac{dy}{dt}=\frac{d}{dt}\left(3t^{-1}\right)\\ \frac{dy}{dt}=-\frac{3}{t^2}\end{array}$

Substitute $\frac{dx}{dt}=-\frac{2}{t^3}$ and $\frac{dy}{dt}=-\frac{3}{t^2}$ in the Chain rule of Differentiation to obtain:

  $\begin{array}{l}\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ \frac{dy}{dx}=\frac{-\frac{3}{t^2}}{-\frac{2}{t^3}}\\ \frac{dy}{dx}=\frac{-3\times t^3}{-2\times t^2}\\ \frac{dy}{dx}=\frac{3}{2}t\end{array}$

Substitute $\frac{dy}{dx}=\frac{3}{2}t$ and $\frac{dx}{dt}=-\frac{2}{t^3}$, and apply Chain rule for Higher Derivatives to obtain:

  $\begin{array}{l}\frac{d^{2}y}{d{x}^2}=\frac{\frac{d}{dt}\left(\frac{3}{2}t\right)}{-\frac{2}{t^3}}\\ \frac{d^{2}y}{d{x}^2}=\frac{\frac{3}{2}\frac{d}{dt}\left(t\right)}{-\frac{2}{t^3}}\\ \frac{d^{2}y}{d{x}^2}=\frac{\frac{3}{2}\cdot 1}{-\frac{2}{t^3}}\\ \frac{d^{2}y}{d{x}^2}=-\frac{3t^3}{4}\end{array}$

The value of $\frac{d^{2}y}{d{x}^2}$ at the point corresponding to $t=2$ is:

  ${\left.\frac{d^{2}y}{d{x}^2}\right|}_2=-\frac{3{\left(2\right)}^3}{4}$

  $\begin{array}{l}{\left.\frac{d^{2}y}{d{x}^2}\right|}_2=-\frac{3\times 8}{4}\\ {\left.\frac{d^{2}y}{d{x}^2}\right|}_2=-6\end{array}$

Thus, the value of $\frac{d^{2}y}{d{x}^2}$ at the point corresponding to $t=2$ is $-6$ .