SOLUTION: In a time of t hours, a particle moves a distance of s kilometers from its starting point,
where s = e^t∗ tan(t).
Find the average velocity between t = 0.25 and t = 0.25
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Question 1066029: In a time of t hours, a particle moves a distance of s kilometers from its starting point,
where s = e^t∗ tan(t).
Find the average velocity between t = 0.25 and t = 0.25 + h if
(i) h = 0.01
(ii) h = 0.001
(iii) h = 0.0001
help me please
I got to
s(.26)- s(.25)/
.01
i dont know where to start on using s = e^t
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
s=e^t+tan (t)
I'm assuming t is in radians.
e^(0.25)*tan(.25)=0.3279
e^0.26* tan(.26)=0.3450
e^0.251*tan(.251)=0.3296
e^0.2501*tan(.2501)=0.3280
so s(.26)-s(.25) all divided by 0.01 is 1.71
s(.251)-s(.25) divided by 0.001 is 1.69749
s(.2501)-s(.25) divided by 0.0001=1.69579
The actual derivative is tan(t)*e^t+e^t*sec^2(t)
this is tan(.25)*e^(0.25)*.25+e^(.25)*1/cos^2(.25)=1.69561
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