082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

10.4 Rate of change at a specific point 363
ThisisanevenbetterestimateoftherateofchangeatA.HenceatA,ifxincreases
by 1 unit then y increases by approximately 18 units. This corresponds to a steep
upward slope atA.
Example10.4illustratestheapproachofestimatingtherateofchangeatapointbyusing
the‘shrinkinginterval’method.Bytakingsmallerandsmallerintervals,betterandbetter
estimates of the rate of change of the function atx = 3 can be obtained. However, we
eventuallywanttheintervalto‘shrink’tothepointx = 3.Weintroduceasmallchange
orincrementofxdenotedby δxandconsidertheinterval[3,3+δx].Byletting δxtend
tozero,the interval [3,3 + δx] effectively shrinks tothe pointx = 3.
Example10.5 Find the rate of change ofy = 3x 2 +2 atx = 3 by considering the interval [3,3 + δx]
and letting δx tend to0.
Solution Whenx = 3,y(3) = 29. Whenx = 3 + δx then
So,
y(3 + δx) = 3(3 + δx) 2 +2
=3(9+6δx+(δx) 2 )+2
= 3(δx) 2 +18δx +29
average rate of change ofyacross [3,3 + δx] =
change iny
change inx
= (3(δx)2 +18δx +29) −29
δx
= 3(δx)2 +18δx
δx
=
δx(3δx +18)
δx
=3δx+18
We now let δx tend to0,so thatthe intervalshrinks toapoint:
rate of change ofywhenxis3 = lim
δx→0
(3δx +18) =18
We have found the rate of change ofyat a particular value ofx, rather than across an
interval.We usually write
( ) y(3 + δx) −y(3)
rate of change ofywhenxis3 = lim
δx→0 δx
( 3(δx) 2 )
+18δx
= lim
= lim (3δx +18)
δx→0 δx
δx→0
= 18