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Q52E

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Essential Calculus: Early Transcendentals

Found in: Page 121

Book edition 2nd

Author(s) James Stewart

Pages 830 pages

ISBN 9781133112280

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Short Answer

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 by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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 \(y = \frac{1}{2}x + \frac{3}{2}\).

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