Calculus (3rd Edition)

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

Chapter 3 - Differentiation - 3.4 Rates of Change - Exercises - Page 129: 14


The object is cooling or decreasing at a rate of $7.5^{\circ}/minute$.

Work Step by Step

$T(t) = \frac{3}{8}t^2 - 15t + 180$ To find the rate of change, first we need to find the derivative which we can do using the power rule: $T'(t) = \frac{3}{4}t - 15$ Now, we substitute t=10 into the equation at that specific point: $T'(10) = \frac{3}{4}*10 - 15 = 7.5 - 15 = -7.5$ To interpret this data, we note that the rate is negative which implies the temperature goes down or cools down. So the object is cooling or decreasing at a rate of $7.5^{\circ}/minute$.
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