Answer
a) $-0.434906$ $kilopascals$
b)
Actual change = -0.418273$ $kilopascals$
Percentage error = $3.98$%
Work Step by Step
a)
$P'(h)$ = $128(-0.157)e^{-0.157h}$ = $-20.096e^{-0.157h}$
$P'(h)$ = $\frac{ΔP}{Δh}$
at $h = 20, Δh =0.5$
$ΔP$ = $P'(h)Δh$ = $-20.096e^{-0.157(20)}(0.5)$ = $-0.434906$ $kilopascals$
b)
Actual change
$P(20.5) = 128e^{-0.157(20.5)} = 5.121925$
$P(20) = 128e^{-0.157(20)} = 5.540198$
$P(20.5) - P(20) = 5.121925 - 5.540198 = -0.418273$ $kilopascals$
Percentage error = $\frac{-0.434906-(-0.418273)}{-0.418273}\times100$ = $3.98$%
