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Expert-verified Found in: Page 121 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # Find an equation of the tangent line to the curve $$y = \sqrt {3 + {x^2}}$$ that is parallel to the line $$x - 2y = 1$$.

An equation of the tangent line to the curve $$y = \sqrt {3 + {x^2}}$$parallel to the line $$x - 2y = 1$$is $$y = \frac{1}{2}x + \frac{3}{2}$$

See the step by step solution

## Step 1: Important formulas

The Chain Rule for differentiation

$$\frac{{d\left( {f\left( {g\left( x \right)} \right)} \right)}}{{dx}} = \frac{{d\left( {f\left( {g\left( x \right)} \right)} \right)}}{{d\left( {g\left( x \right)} \right)}} \cdot \frac{{d\left( {g\left( x \right)} \right)}}{{dx}}$$

The equation in point-slope form $$y - {y_1} = m\left( {x - {x_1}} \right)$$

## Step 2: Given parameters

Given function $$y = \sqrt {3 + {x^2}}$$parallel to the line $$x - 2y = 1$$

## Step 3: Find the derivative of the curve

Given that $$y = \sqrt {3 + {x^2}}$$

Apply the chain rule by differentiate the square root function first then multiply by the derivative of the terms on the inside of the radical

$$\frac{{dy}}{{dx}} = \frac{{2x}}{{2\sqrt {3 + {x^2}} }}$$

Cancel $$2$$from the numerator and the denominator then take $$\frac{{dy}}{{dx}}$$equal to $$\frac{1}{2}$$then find a tangent line parallel to an equation with the slope $$\frac{1}{2}$$

$$\frac{x}{{\sqrt {3 + {x^2}} }} = \frac{1}{2}$$

Now solve for $$x$$

$$\begin{array}{c}x &=& \frac{{\sqrt {3 + {x^2}} }}{2}\\2x &=& \sqrt {3 + {x^2}} \end{array}$$

Squaring on both sides

$$\begin{array}{c}4{x^2} &=& 3 + x\\4{x^2} - x - 3 &=& 0\\4{x^2} - 4x + 3x - 3 &=& 0\\4x\left( {x - 1} \right) + 3\left( {x - 1} \right) &=& 0\\\left( {4x + 3} \right)\left( {x - 1} \right) &=& 0\\x &=& 1, - \frac{3}{4}\end{array}$$

Neglect the negative factor. Therefore, $$x = 1$$

## Step 4: Find an equation of a tangent line to the point $$x = 1$$

Now equation of a tangent line to the point $$x = 1$$

$$\begin{array}{l}y = \sqrt {3 + {{\left( 1 \right)}^2}} \\y = 4\end{array}$$

Now write the equation in point-slope form

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

Hence, an equation of the tangent line to the curve $$y = \sqrt {3 + {x^2}}$$parallel to the line $$x - 2y = 1$$is $$y = \frac{1}{2}x + \frac{3}{2}$$. ### Want to see more solutions like these? 