## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 3 - Section 3.1 - Derivatives of Polynomials and Exponential Functions - 3.1 Exercises - Page 181: 61

#### Answer

The equation of the normal line that satisfies the requirements is $$(m_0):y=-2x+3$$

#### Work Step by Step

$$(s): y=\sqrt x$$ $$(d): 2x+y=1$$ 1) Find the derivative of $y$ of curve $(s)$: $$y'=\frac{d}{dx}(\sqrt x)$$ $$y'=\frac{d}{dx}(x^{1/2})$$ $$y'=\frac{1}{2}x^{-1/2}$$ $$y'=\frac{1}{2\sqrt x}$$ 2) The slope of the tangent line $(l)$ of curve $(s)$ at point $A(a,b)$ equals $y'(a)$. The slope of the normal line $(m)$ at point $A(a,b)$ we would call $z(a)$. The normal line is perpendicular with the tangent line. So, the product of their slope equals -1. In other words, $$z(a)\times y'(a)=-1$$ $$z(a)\times\frac{1}{2\sqrt x}=-1$$ $$z(a)=-2\sqrt x$$ Consider line $(d)$: $$(d): 2x+y=1$$ $$(d):y=-2x+1$$ So the slope of line $(d)$ equals -2. For 2 lines to be parallel with each other, their slopes must be equal. That means, for the normal line $(m)$ and line $(d)$ to be parallel with each other, $$z(a)=-2$$ $$-2\sqrt a=-2$$ $$\sqrt a=1$$ $$a=1$$ So, $b=\sqrt 1=1$ Therefore, the normal line at point $A(1,1)$ is parallel with line $(d)$. The equation of the normal line $(m_0)$ at $A$ would be $$(m_0):(y-1)=-2(x-1)$$ $$(m_0):y-1=-2x+2$$ $$(m_0):y=-2x+3$$

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.