Chapter 4, Problem 48RE

(a)

To determine

To calculate: To find an equation for the tangent to the curve at the given point.

Answer to Problem 48RE

The equation of the line of tangent is $y-1=-\frac{5}{4}\left(x-4\right)$ .

Explanation of Solution

Given Information: The given equation is $x+\sqrt{xy}=6$ and the point is $\left(4,1\right)$ .

Calculation:

The slope of the tangent line is calculated by finding the derivative of the given equation at the given point.

The derivative of the function is,

  $\begin{array}{c}x+\sqrt{xy}=6\\ 1+\frac{1}{2\sqrt{xy}}\frac{d}{dx}\left(xy\right)=\frac{d}{dx}\left(6\right)\\ 1+\frac{1}{2\sqrt{xy}}\left(y+x\frac{dy}{dx}\right)=0\left(\because \frac{d}{dx}\left(\text{constant}\right)=0\right)\\ 1+\frac{\sqrt{y}}{2\sqrt{x}}+\frac{\sqrt{x}}{2\sqrt{y}}\frac{dy}{dx}=0\\ \frac{\sqrt{x}}{2\sqrt{y}}\frac{dy}{dx}=-\left(1+\frac{\sqrt{y}}{2\sqrt{x}}\right)\\ \frac{dy}{dx}=-\left(1+\frac{\sqrt{y}}{2\sqrt{x}}\right)\times \frac{2\sqrt{y}}{\sqrt{x}}\\ \frac{dy}{dx}=-\left(\frac{2\sqrt{x}+\sqrt{y}}{2\sqrt{x}}\right)\times \frac{2\sqrt{y}}{\sqrt{x}}\\ \frac{dy}{dx}=-\frac{2\sqrt{xy}+y}{x}\end{array}$

Derivative of the function at $\left(4,1\right)$ is,

  $\begin{array}{c}\frac{dy}{dx}=-\frac{2\sqrt{xy}+y}{x}\\ \frac{dy}{dx}=-\frac{2\sqrt{4}+1}{4}\left(\because x=4,y=1\right)\\ =-\frac{4+1}{4}\\ =-\frac{5}{4}\end{array}$

That is,

  $m_{\text{tangent}}=-\frac{5}{4}$

• Here, $m_{\text{tangent}}$ is the slope of the tangent.

Thus, equation of the line of tangent using the point-slope form is,

  $\begin{array}{c}y-y_1=m_{\text{tangent}}\left(x-x_1\right)\\ y-1=-\frac{5}{4}\left(x-4\right)\end{array}$

The equation of the line of tangent is $y-1=-\frac{5}{4}\left(x-4\right)$ .

(b)

To determine

To calculate: To find an equation for the normal to the curve at the given point.

Answer to Problem 48RE

The equation of the line of normal is $y-1=\frac{4}{5}\left(x-4\right)$ .

Explanation of Solution

Given Information: The given equation is $x+\sqrt{xy}=6$ and the point is $\left(4,1\right)$ .

Calculation:

The slope of the tangent line is calculated by finding the derivative of the given equation at the given point.

The derivative of the function is,

  $\begin{array}{c}x+\sqrt{xy}=6\\ 1+\frac{1}{2\sqrt{xy}}\frac{d}{dx}\left(xy\right)=\frac{d}{dx}\left(6\right)\\ 1+\frac{1}{2\sqrt{xy}}\left(y+x\frac{dy}{dx}\right)=0\left(\because \frac{d}{dx}\left(\text{constant}\right)=0\right)\\ 1+\frac{\sqrt{y}}{2\sqrt{x}}+\frac{\sqrt{x}}{2\sqrt{y}}\frac{dy}{dx}=0\\ \frac{\sqrt{x}}{2\sqrt{y}}\frac{dy}{dx}=-\left(1+\frac{\sqrt{y}}{2\sqrt{x}}\right)\\ \frac{dy}{dx}=-\left(1+\frac{\sqrt{y}}{2\sqrt{x}}\right)\times \frac{2\sqrt{y}}{\sqrt{x}}\\ \frac{dy}{dx}=-\left(\frac{2\sqrt{x}+\sqrt{y}}{2\sqrt{x}}\right)\times \frac{2\sqrt{y}}{\sqrt{x}}\\ \frac{dy}{dx}=-\frac{2\sqrt{xy}+y}{x}\end{array}$

Derivative of the function at $\left(4,1\right)$ is,

  $\begin{array}{c}\frac{dy}{dx}=-\frac{2\sqrt{xy}+y}{x}\\ \frac{dy}{dx}=-\frac{2\sqrt{4}+1}{4}\left(\because x=4,y=1\right)\\ =-\frac{4+1}{4}\\ =-\frac{5}{4}\end{array}$

That is,

  $m_{\text{tangent}}=-\frac{5}{4}$

• Here, $m_{\text{tangent}}$ is the slope of the tangent.

The slope of the normal line is perpendicular to the slope of the tangent line.

That is,

  $\begin{array}{c}{m}_{\text{normal}}\times m_{\text{tangent}}=-1\\ m_{\text{normal}}=\frac{-1}{m_{\text{tangent}}}\\ m_{\text{normal}}=\frac{-1}{\frac{-5}{4}}\\ =\frac{4}{5}\end{array}$

• Here, $m_{\text{normal}}$ is the slope of the normal.

Thus, equation of the line of normal using the point-slope form is,

  $\begin{array}{c}y-y_1=m_{\text{normal}}\left(x-x_1\right)\\ y-1=\frac{4}{5}\left(x-4\right)\end{array}$

The equation of the line of normal is $y-1=\frac{4}{5}\left(x-4\right)$ .