Calculus: Early Transcendentals, 8th Edition
Authors: James Stewart
ISBN-13: 978-1285741550
See our solution for Question 7E from Chapter 2.7 from Stewart's Calculus, 8th Edition.
Problem 7E
Chapter:
Problem:
Find an equation of the tangent line to the curve at the given point.
Step-by-Step Solution
Given information
We are given with the curve\[y = \sqrt x \]We have to find tangent to this curve at the point (1,1).
Step 1: Slope of the tangent at given point
\[\begin{array}{l}m = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {a + h} \right) - f\left( a \right)}}{h}\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {1 + h} \right) - f\left( 1 \right)}}{h}\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sqrt {\left( {1 + h} \right)} - \sqrt 1 }}{h}\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sqrt {\left( {1 + h} \right)} - 1}}{h} \times \dfrac{{\sqrt {\left( {1 + h} \right)} + 1}}{{\sqrt {\left( {1 + h} \right)} + 1}}\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{{1 + h - 1}}{{h\sqrt {\left( {1 + h} \right)} + 1}}\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{h}{{h\sqrt {\left( {1 + h} \right)} + 1}}\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{1}{{\sqrt {\left( {1 + h} \right)} + 1}}\\ = \dfrac{1}{{\sqrt {\left( {1 + 0} \right)} + 1}}\\ = \dfrac{1}{{1 + 1}}\\ = \dfrac{1}{2}\end{array}\]Therefore, the slope at (1,1) is \[{m_{\left( {1,1} \right)}} = \dfrac{1}{2}\]
Step 2: Equation of tangent
The general form of straight line passing through a point($x_1, y_1$) and having a slope, $m$ is given by:\[y - {y_1} = m\left( {x - {x_1}} \right)\]Apply the point (1,1) and slope (m 1/2)\[\begin{array}{l}y - {y_1} = m\left( {x - {x_1}} \right)\\y - 1 = \dfrac{1}{2}\left( {x - 1} \right)\\y - 1 = \dfrac{1}{2}x - \dfrac{1}{2}\\y = \dfrac{x}{2} - \dfrac{1}{2} + 1\\y = \dfrac{x}{2} + \dfrac{1}{2}\end{array}\]Therefore, \[y = \dfrac{x}{2} + \dfrac{1}{2}\]
We are given with the curve\[y = \sqrt x \]We have to find tangent to this curve at the point (1,1).
Step 1: Slope of the tangent at given point
\[\begin{array}{l}m = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {a + h} \right) - f\left( a \right)}}{h}\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{{f\left( {1 + h} \right) - f\left( 1 \right)}}{h}\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sqrt {\left( {1 + h} \right)} - \sqrt 1 }}{h}\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{{\sqrt {\left( {1 + h} \right)} - 1}}{h} \times \dfrac{{\sqrt {\left( {1 + h} \right)} + 1}}{{\sqrt {\left( {1 + h} \right)} + 1}}\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{{1 + h - 1}}{{h\sqrt {\left( {1 + h} \right)} + 1}}\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{h}{{h\sqrt {\left( {1 + h} \right)} + 1}}\\ = \mathop {\lim }\limits_{h \to 0} \dfrac{1}{{\sqrt {\left( {1 + h} \right)} + 1}}\\ = \dfrac{1}{{\sqrt {\left( {1 + 0} \right)} + 1}}\\ = \dfrac{1}{{1 + 1}}\\ = \dfrac{1}{2}\end{array}\]Therefore, the slope at (1,1) is \[{m_{\left( {1,1} \right)}} = \dfrac{1}{2}\]
Step 2: Equation of tangent
The general form of straight line passing through a point($x_1, y_1$) and having a slope, $m$ is given by:\[y - {y_1} = m\left( {x - {x_1}} \right)\]Apply the point (1,1) and slope (m 1/2)\[\begin{array}{l}y - {y_1} = m\left( {x - {x_1}} \right)\\y - 1 = \dfrac{1}{2}\left( {x - 1} \right)\\y - 1 = \dfrac{1}{2}x - \dfrac{1}{2}\\y = \dfrac{x}{2} - \dfrac{1}{2} + 1\\y = \dfrac{x}{2} + \dfrac{1}{2}\end{array}\]Therefore, \[y = \dfrac{x}{2} + \dfrac{1}{2}\]