Chapter 4.2, Problem 64E

To determine

To identify:The correct answer for the slope of tangent for the given function at the given point.

Answer to Problem 64E

The correct answer is C: $\frac{1}{\sqrt{2}}$ .

Explanation of Solution

Given information:The function is $y^2-x^2=1$ and the point is $\left(x_1,y_1\right)=\left(1,\sqrt{2}\right)$ .

Calculation:

The function is $y^2-x^2=1$ ............. (1)

By differentiating equation (1),

  $y^2-x^2=1$

  $\begin{array}{l}\frac{d{y}^2}{dx}-\frac{d{x}^2}{dx}=0\\ 2y\frac{dy}{dx}-2x=0\\ \frac{dy}{dx}=m=\frac{x}{y}\mathrm{.........}\left(2\right)\end{array}$

Now put the value $\left(x_1,y_1\right)=\left(1,\sqrt{2}\right)$ in equation (2),

  $\begin{array}{l}m=\frac{x}{y}\\ m=\frac{1}{\sqrt{2}}\end{array}$

Here $m=\frac{1}{\sqrt{2}}$ is the slope of the tangent at $\left(x_1,y_1\right)=\left(1,\sqrt{2}\right)$ .

Hence the correct answer is C: $\frac{1}{\sqrt{2}}$ .