Answer
a) $2{\pi}rhΔr$
b) ${\pi}(Δr)^{2}h$
Work Step by Step
a)
$V$ = ${\pi}r^{2}h$
$ΔV$ $\approx$ $dV$ = $2{\pi}rhdr$ = $2{\pi}rhΔr$
b)
The error is
$ΔV-dV$ = $[{\pi}(r+Δr)^{2}h-{\pi}r^{2}h]-2{\pi}rhΔr$ = ${\pi}(Δr)^{2}h$
Calculus 8th Edition
by Stewart, James
Published by
Cengage
ISBN 10:
1285740629
ISBN 13:
978-1-28574-062-1
Chapter 2 - Derivatives - 2.9 Linear Approximations and Differentials - 2.9 Exercises - Page 193: 35
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