Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 1 - Precalculus Review - 1.2 Linear and Quadratic Functions - Exercises - Page 18: 33

Answer

No, the time period, $T$ is not linear function of length, $L$.

Work Step by Step

Using the data we get $T(20)=0.9$, $T(30)=1.1$ and $T(40)=1.27$. If the function is linear then, the change in time period with respect to change in length must be constant, i.e., $\dfrac{\Delta T}{\Delta L}$ must be constant. But we have $\dfrac{T(30)-T(20)}{30-20}=\dfrac{0.2}{10}=0.02$ while $\dfrac{T(40)-T(30)}{40-30}=\dfrac{0.17}{10}=0.017$, which clearly indicates that the function is not linear.
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