Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 10 - Parametric Equations and Polar Coordinates - 10.2 Exercises - Page 675: 20

Answer

There are horizontal tangents of that curve at $(1,e)$ and at $(1, \frac 1 e)$, and vertical tangents at $(e, 1)$ and at $(\frac 1 e, 1)$

Work Step by Step

1. Calculate $\frac{dy}{dt}$ and $\frac{dx}{dt}$: $\frac{d(e^{cos(\theta)})}{dt} = -sin(\theta)e^{cos(\theta)}$ $\frac{d(e^{sin(\theta)})}{dt} = cos(\theta)e^{sin(\theta)}$ 2. There will be a horizontal tangent when $-sin(\theta)e^{cos(\theta)}= 0$ and $ cos(\theta)e^{sin(\theta)} \neq 0$: $-sin(\theta)e^{cos(\theta)} = 0$ $(-sin(\theta))(e^{cos(\theta)}) = 0$ $e^{cos(\theta)}$ will never equal $0$. $-sin(\theta) = 0$ $sin(\theta) = 0$ So, $\theta$ must be equal to $0, \pi$or $2\pi$. (Values between $0$ and $2\pi$) Check if $ cos(\theta)e^{sin(\theta)} \neq 0$ - The exponential part can never equal $0$, so we can just test $cos(\theta)$. $cos(0) = 1$ $cos(\pi) = -1$ $cos(2\pi) = 1$ - Therefore, all these values for $\theta$ are valid. 3. There will be a vertical tangent when $cos(\theta)e^{sin(\theta)}= 0$ and $-sin(\theta)e^{cos(\theta)} \neq 0$: $cos(\theta)e^{sin(\theta)}= 0$ $e^{sin(\theta)}$ will never equal $0$. $cos(\theta) = 0$ So, $\theta$ must be equal to $\frac {\pi} 2$ or $\frac{3\pi}{2}$ - Check if $-sin(\theta)e^{cos(\theta)} \neq 0$: The exponential part will can never equal 0, so we can just test $-sin(\theta)$: $-sin(\frac{\pi} 2) = -1$ $-sin(\frac{3\pi}{2}) = 1$ - Therefore, all these values for $\theta$ are valids. 4. Find the points coordinates: Horizontal tangents: $\theta = 0$ $x = e^{sin(0)} = e^0 = 1$ $y = e^{cos(0)} =e^1 =e$ $\theta = \pi$ $x = e^{sin(\pi)} = e^0 = 1$ $y = e^{cos(\pi)} =e^{-1} = \frac{1}{e}$ $\theta = 2\pi$ $x = e^{sin(2\pi)} = e^0 = 1$ $y = e^{cos(2\pi)} =e^1 =e$ Vertical tangents: $\theta = \frac{\pi} 2$ $x = e^{sin(\frac{\pi} 2)} = e^1 = e$ $y = e^{cos(\frac{\pi} 2)} =e^0 =1$ $\theta = \frac{3\pi} 2$ $x = e^{sin(\frac{3\pi} 2)} = e^{-1} = \frac 1 e$ $y = e^{cos(\frac{3\pi} 2)} =e^0 =1$
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