Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 1 - Functions and Limits - 1.3 New Functions from Old Functions - 1.3 Exercises - Page 45: 59

Answer

(a) See the image (a) below. (b) $V(t)=120H(t)$ (Image (b)) (c) $V(t) = 210H(t-5)$ (Image (c))
1550171749

Work Step by Step

(a) We can easily sketch a graph for the function: $H(t) = \left\{ \begin{array}{ll} 0 & \quad if & \quad t\lt0 \\ 1 & \quad if & \quad t \geq 0 \end{array} \right. $ (Graph (a) in the image) We have two possible values for $y$, which is $0$ or $1$. If $t$ is less than $0$, $H(t)$ will be $0$, which is a line that lays on the $x$-axis from $-\infty$ to $0$ (not including $0$). But if $t$ gets a value of $0$ or higher, $H(t)$ becomes 1 and its graph is parallel to the $x$-axis, $1$ unit upwards (from $x=1$ to $\infty$). (b) The same idea is used here. If $t$ is less than $0$, this means that the switch is not turned on yet, so voltage at that point will be $0$. When we turn on the switch, we get an instant $120$ Volts for all $t$ values greater than or equal to $0$. (See the image (b)) We will have the formula: $V(t) = 120H(t)$ (c) The same idea is applied here, only the difference is that the switch is turned on at time $t=5$, so we will have an instant $240$ Volts jump at $t=5$. (See the graph (c) ) And we will have the formula: $V(t) = 240H(t-5)$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

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.