Chapter 10, Problem 9RE

a)

To determine

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

Answer to Problem 9RE

The equation for the tangent to the curve $x=\frac{1}{2}\tan t$ , $y=\frac{1}{2}\sec t$ at the point corresponding to $t=\frac{\pi }{3}$ is:

  $y-1=2\sqrt{3}\left(x-\frac{\sqrt{3}}{2}\right)$ .

Explanation of Solution

Given information:

The parametric equation of the curve is: $x=\frac{1}{2}\tan t$ , $y=\frac{1}{2}\sec t$ and $t=\frac{\pi }{3}$ .

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=\frac{1}{2}\tan t$$y=\frac{1}{2}\sec t$

Substitute $t=\frac{\pi }{3}$ in the equations to obtain the point corresponding to $t=\frac{\pi }{3}$ :

  $\begin{array}{l}x=\frac{1}{2}\tan \frac{\pi }{3}\\ x=\frac{1}{2}\cdot \sqrt{3}\\ x=\frac{\sqrt{3}}{2}\end{array}$$\begin{array}{l}y=\frac{1}{2}\sec \frac{\pi }{3}\\ y=\frac{1}{2}\cdot 2\\ y=1\end{array}$

The point corresponding to $t=\frac{\pi }{3}$ is $\left(\frac{\sqrt{3}}{2},1\right)$ .

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

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

And,

  $\begin{array}{l}y=\frac{1}{2}\sec t\\ \frac{dy}{dt}=\frac{d}{dt}\left(\frac{1}{2}\sec t\right)\\ \frac{dy}{dt}=\frac{1}{2}\frac{d}{dt}\left(\sec t\right)\\ \frac{dy}{dt}=\frac{1}{2}\sec t\tan t\end{array}$

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

  $\begin{array}{l}\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ \frac{dy}{dx}=\frac{\frac{1}{2}\sec t\tan t}{\frac{1}{2}{\sec }^{2}t}\\ \frac{dy}{dx}=\frac{\tan t}{\sec t}\\ \frac{dy}{dx}=\frac{\frac{\sin t}{\cos t}}{\frac{1}{\cos t}}\end{array}$

Simplify further to obtain:

  $\frac{dy}{dx}=\frac{\sin t}{{\cos }^{2}t}$

At the point corresponding to $t=\frac{\pi }{3}$ , the slope of the tangent is given by:

  $\begin{array}{l}{\left.m=\frac{dy}{dx}\right|}_{\frac{\pi }{3}}\\ m==\frac{\sin \frac{\pi }{3}}{{\cos }^2\frac{\pi }{3}}\\ m=\frac{\frac{\sqrt{3}}{2}}{{\left(\frac{1}{2}\right)}^2}\\ m=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{4}}\\ m=2\sqrt{3}\end{array}$

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

  $y-1=2\sqrt{3}\left(x-\frac{\sqrt{3}}{2}\right)$

Thus,

The equation for the tangent to the curve $x=\frac{1}{2}\tan t$ , $y=\frac{1}{2}\sec t$ at the point corresponding to $t=\frac{\pi }{3}$ is: $y-1=2\sqrt{3}\left(x-\frac{\sqrt{3}}{2}\right)$ .

b)

To determine

To find:The value of $\frac{d^{2}y}{d{x}^2}$ at the point corresponding to $t=\frac{\pi }{3}$ .

Answer to Problem 9RE

The value of $\frac{d^{2}y}{d{x}^2}$ at the point corresponding to $t=\frac{\pi }{3}$ is 7.

Explanation of Solution

Given information:

The parametric equation of the curve is: $x=\frac{1}{2}\tan t$ , $y=\frac{1}{2}\sec t$ and $t=\frac{\pi }{3}$ .

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=\frac{1}{2}\tan t$$y=\frac{1}{2}\sec t$

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

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

And,

  $\begin{array}{l}y=\frac{1}{2}\sec t\\ \frac{dy}{dt}=\frac{d}{dt}\left(\frac{1}{2}\sec t\right)\\ \frac{dy}{dt}=\frac{1}{2}\frac{d}{dt}\left(\sec t\right)\\ \frac{dy}{dt}=\frac{1}{2}\sec t\tan t\end{array}$

Apply the Chain rule of Differentiation to obtain:

  $\begin{array}{l}\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}\\ \frac{dy}{dx}=\frac{\frac{1}{2}\sec t\tan t}{\frac{1}{2}{\sec }^{2}t}\\ \frac{dy}{dx}=\frac{\tan t}{\sec t}\\ \frac{dy}{dx}=\frac{\frac{\sin t}{\cos t}}{\frac{1}{\cos t}}\end{array}$

Simplify further to obtain:

  $\frac{dy}{dx}=\frac{\sin t}{{\cos }^{2}t}$

Substitute $\frac{dy}{dx}=\frac{\sin t}{{\cos }^{2}t}$ and $\frac{dx}{dt}=\frac{1}{2}{\sec }^{2}t$, and apply Chain rule for Higher Derivatives to obtain:

  $\frac{d^{2}y}{d{x}^2}=\frac{\frac{d}{dt}\left(\frac{\sin t}{{\cos }^{2}t}\right)}{\frac{1}{2}{\sec }^{2}t}$

Apply the Quotient rule of derivatives to compute the value of $\frac{d^{2}y}{d{x}^2}$ :

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

Simplify further:

  $\begin{array}{l}\frac{d^{2}y}{d{x}^2}=\frac{2\left({\cos }^{2}t+2{\sin }^{2}t\right)}{\cos t}\\ \frac{d^{2}y}{d{x}^2}=\frac{2\left({\cos }^{2}t+{\sin }^{2}t+{\sin }^{2}t\right)}{\cos t}\end{array}$

Apply the Trigonometric Identity ${\cos }^2\theta +{\sin }^2\theta =1$ :

  $\frac{d^{2}y}{d{x}^2}=\frac{2\left(1+{\sin }^{2}t\right)}{\cos t}$

The value of $\frac{d^{2}y}{d{x}^2}$ at the point corresponding to $t=\frac{\pi }{3}$ is:

  $\begin{array}{l}{\left.\frac{d^{2}y}{d{x}^2}\right|}_{\frac{\pi }{3}}=\frac{2\left(1+{\sin }^2\frac{\pi }{3}\right)}{\cos \frac{\pi }{3}}\\ {\left.\frac{d^{2}y}{d{x}^2}\right|}_{\frac{\pi }{3}}=\frac{2\left(1+{\left(\frac{\sqrt{3}}{2}\right)}^2\right)}{\frac{1}{2}}\\ {\left.\frac{d^{2}y}{d{x}^2}\right|}_{\frac{\pi }{3}}=\frac{2\left(1+\frac{3}{4}\right)}{\frac{1}{2}}\\ {\left.\frac{d^{2}y}{d{x}^2}\right|}_{\frac{\pi }{3}}=4\times \frac{7}{4}\\ {\left.\frac{d^{2}y}{d{x}^2}\right|}_{\frac{\pi }{3}}=7\end{array}$

Thus, the value of $\frac{d^{2}y}{d{x}^2}$ at the point corresponding to $t=\frac{\pi }{3}$ is $7$ .