Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Additional and Advanced Exercises - Page 102: 6

Answer

$[8.8,8.9]~cm$

Work Step by Step

Step 1. Given the volume $V=36\pi h$, $V_0=1000cm^3$, and $\Delta V=0.01V_0$, we need to find the range of heights $h$ that satisfying the condition. Step 2. The range of the volume can be found as $V_1=V_0-\Delta V=0.99V_0=990cm^3$ and $V_2=V_0+\Delta V=1.01V_0=1010cm^3$. Step 3. Rewriting the equation as $h=V/(36\pi)$, we have $h_0=V_0/(36\pi)\approx8.8419$. We can calculate the corresponding heights from the above volumes as: $h_1=V_1/(36\pi)=990/(36\pi)\approx8.7535$ and $h_2=V_2/(36\pi)=1010/(36\pi)\approx8.9304$ Step 4. Thus, we need to keep the height in a range of $[8.7535,8.9304]cm$
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