Chapter 11, Problem 10RE

(a)

To determine

To determine: The equation for the tangent to the curve at the pint corresponding to the given value of t .

Answer to Problem 10RE

The required equation is $y=-3x+\frac{13}{4}$ .

Explanation of Solution

Given information:

The equations: $x=1+\frac{1}{t^2}$

  $y=1-\frac{3}{t}$

The value of t = 2

Formula used:

Point-slope form of a line:

  $y-y_1=m\left(x-x_1\right)$

Formulas of derivatives:

  $\frac{d}{dx}\tan x={\sec }^{2}x$

  $\frac{d}{dx}\sec x=\sec x\tan x$

Calculation:

The given equations are $x=1+\frac{1}{t^2}$  ..... (1)

And, $y=1-\frac{3}{t}$  ..... (2)

Substitute 2 for t in equation (1) and simplify.

  $\begin{array}{c}x=1+\frac{1}{{\left(2\right)}^2}\\ =1+\frac{1}{4}\\ =\frac{5}{4}\end{array}$

Substitute 2 for t in equation (2) and simplify.

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

Thus, the point through which the tangent passes is $\left(\frac{5}{4},-\frac{1}{2}\right)$ .

Find derivative of equation (1) and equation (2) with respect to t .

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

And, $\begin{array}{c}\frac{dy}{dt}=\frac{d}{dt}\left(1-\frac{3}{t}\right)\\ =\frac{d}{dt}\left(1\right)-\frac{d}{dt}\left(3t^{-1}\right)\\ =0-\left(-1\right)3t^{-2}\\ =\frac{3}{t^2}\end{array}$

Now,

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

Substitute 2 for t in the above derivative.

  $\begin{array}{c}\frac{dy}{dx}=-\frac{3}{\cancel{2}}\left(\cancel{2}\right)\\ =-3\end{array}$

So, the slope of the tangent is $-3$ .

To find the equation of the tangent line, use the formula for the point-slope form of the equation of a line $y-y_1=m\left(x-x_1\right)$ .

Substitute $\frac{5}{4}$ for $x_1$ , $-\frac{1}{2}$ for $y_1$ and $-3$ for m in the above formula.

  $\begin{array}{c}y-\left(-\frac{1}{2}\right)=-3\left(x-\left(\frac{5}{4}\right)\right)\\ y+\frac{1}{2}=-3x+\frac{15}{4}\\ y=-3x+\frac{15}{4}-\frac{1}{2}\\ y=-3x+\frac{13}{4}\end{array}$

Hence, the required equation is $y=-3x+\frac{13}{4}$ .