Chapter 4.2, Problem 2QQ

To determine

To calculate: To find the slope of the line tangent to the curve at the point $\left(3,2\right)$ .

Answer to Problem 2QQ

The correct option is B.

Explanation of Solution

Given Information: The equation of the curve is $2x^2-3y^2=2xy-6$

Calculation:

The slope of the line tangent to the curve is equal to the derivative of the curve,

  $\begin{array}{c}2x^2-3y^2=2xy-6\left(\text{Substitute }y={\sin }^4\left(3x\right)\right)\\ 4x-6y\frac{dy}{dx}\text{=2}x\frac{dy}{dx}\text{+2}y\left(\text{Find derivative; Apply chain rule}\right)\\ 4x-2y=\left(2x+6y\right)\frac{dy}{dx}\\ \left(2x-y\right)=\left(x+3y\right)\frac{dy}{dx}\left(\text{Divide through out by 2}\right)\\ \frac{\left(2x-y\right)}{\left(x+3y\right)}=\frac{dy}{dx}\left(\text{Divide through out by }\left(x+3y\right)\right)\end{array}$

To find the slope at the point $\left(3,2\right)$ , substitute $x=3,y=2$ in $\frac{\left(2x-y\right)}{\left(x+3y\right)}=\frac{dy}{dx}$ ,

  $\begin{array}{c}\frac{\left(2x-y\right)}{\left(x+3y\right)}=\frac{dy}{dx}\\ {\left[\frac{dy}{dx}\right]}_{\left(3,2\right)}={\left[\frac{\left(2x-y\right)}{\left(x+3y\right)}\right]}_{\left(3,2\right)}\\ =\frac{2\times 3-2}{3+3\times 2}\\ =\frac{6-2}{3+6}\\ =\frac{4}{9}\end{array}$

The slope of the line tangent to the curve $2x^2-3y^2=2xy-6$ at the point $\left(3,2\right)$ is $\frac{4}{9}$