Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.1 - Derivatives of Polynomials and Exponential Functions - 3.1 Exercises: 38

Answer

The equation of the tangent line: $$(l):y=\frac{3}{2}x-\frac{1}{2}$$ The equation of the normal line: $$(m):y=-\frac{2}{3}x+\frac{5}{3}$$

Work Step by Step

$$y^2=x^3$$$$y=\sqrt{x^3}$$ or $$y=-\sqrt{x^3}$$ Now we see that there would be 2 situations for $y$, one that is $y\gt0$ (since $\sqrt{x^3}\gt0$ for all $x$ in its domain) and one that is $y\lt0$ (since $-\sqrt{x^3}\lt0$ for all $x$ in its domain). However, here we consider the tangent line and normal line at point $(1,1)$, which means we consider a case where $y=1\gt0$. Therefore, we only consider the situation when $y\gt0$. So the function considered here is $$y=f(x)=\sqrt{x^3}=x^{3/2}$$ 1) Find the derivative of $y$$$y'=\frac{d}{dx}(x^{3/2})$$ $$y'=\frac{3}{2}x^{1/2}$$ $$y'=\frac{3}{2}\sqrt x$$ 2) Find $y'(1)$ $$y'(1)=\frac{3}{2}\sqrt1$$$$y'(1)=\frac{3}{2}$$ 3) $y'(1)$ is the slope of the tangent line $(l)$ to the curve $f(x)$ at point $(1,1)$ Knowing the slope of the tangent line $(l)$, which is $y'(1)$ and 1 point of (l), which is $(1,1)$, the equation of the tangent line $(l)$ would be: $$(l):(y-1)=y'(1)(x-1)$$ $$(l):y-1=\frac{3}{2}(x-1)$$ $$(l):y-1=\frac{3}{2}x-\frac{3}{2}$$ $$(l):y=\frac{3}{2}x-\frac{1}{2}$$ 4) The normal line $(m)$ at $(1,1)$ is the line perpendicular to the tangent line $(l)$ at $(1,1)$. So, the slope $a$ of $(m)$ is: $$a=\frac{-1}{y'(1)}=\frac{-1}{3/2}=-\frac{2}{3}$$ (the product of the slopes of 2 perpendicular lines equals $-1$) Knowing the slope of $(m)$, which is $a$ and 1 point of $(m)$, which is $(1,1)$, the equation of the normal line $(m)$ would be $$(m):(y-1)=a(x-1)$$ $$(m):y-1=-\frac{2}{3}(x-1)$$ $$(m):y-1=-\frac{2}{3}x+\frac{2}{3}$$ $$(m):y=-\frac{2}{3}x+\frac{5}{3}$$
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