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Sagot :
Solution :
The formula for the linearization of a function [tex]$f(x)$[/tex] at a point [tex]$x$[/tex] = a is given as
[tex]$L(x)=f(a)+(x-a)f'(a)$[/tex]
Assuming the time is t and the distance travelled is [tex]$f(t)$[/tex], that makes the speed as [tex]$f'(t)$[/tex].
So substituting them in the linearization formula,
A. At t = 7 minutes
f(7) = 2.5 km
f'(7) = 0.5 kpm
∴ [tex]$L_7(t)=f(7)+(t-7)f'(7)$[/tex]
[tex]$=2.5+(t-7)0.5$[/tex]
[tex]$=2.5+0.5t-3.5$[/tex]
[tex]$=0.5t-1$[/tex]
B. At t = 18 minutes
f(18) = 14.8 km
f'(18) = 0.8 kpm
∴ [tex]$L_{18}(t)=f(18)+(t-18)f'(18)$[/tex]
[tex]$=14.8+(t-18)0.8$[/tex]
[tex]$=14.8+0.8t-14.4$[/tex]
[tex]$=0.8t-0.4$[/tex]
C. Substituting the value of t as 14 in both the linearization to determine the position at 14 minutes, we get
[tex]$L_7(14)=0.5(14)-1.0$[/tex]
= 7 - 1
= 6 km
[tex]$L_{18}(14)=0.8(14)+0.4$[/tex]
= 11.2 + 0.4
= 11.6 km
D. According to the linearization at 7, the distance travelled between the 7 minutes and 14 minutes is = 6 km - 2.5 km
= 3.5 km
And between the 14 minutes and 18 minutes is = 14.8 km - 6 km
= 8.8 km
This is an average speed of 0.5 kpm in the first interval and an average speed of 2.2 kpm.
Now, according to the linearization of 18, the distance travelled between the 7 minutes and the 14 minutes is = 11.6 km - 2.5 km
= 9.1 km
And between 14 minutes and 18 minutes is = 14.8 km - 11.6 km
= 3.2 km
This gives an average speed of 1.3 kpm in the first interval and 0.8 kpm in the second interval.
Therefore, the second approximation is the better one since the average speed are closer to the actual readings in the second linearization.
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