in a time of t seconds, a particle moves a distance pf s meters from it's starting point, where s=3t^2 Find the ave velocity between t =1 and t=1+h if....
a) h=.1 b)h+.01 c)h=.001
Which approximation is closest to the EXACT derivative at ...t=1?
Estimate the exact derivative at t=1?
I will do b), you do the others the same way ...
b) when h = .01
avg velocity = (3(1+.001)^2 - 3(1^2))/.01
= 6.03 m/s
thanks
To find the average velocity between t = 1 and t = 1 + h, where s = 3t^2, we need to calculate the displacement of the particle during this time interval and divide it by the length of the time interval. The displacement can be found by subtracting the initial position (at t = 1) from the final position (at t = 1 + h).
Let's calculate the average velocity for each value of h.
a) When h = 0.1:
Displacement = s(1 + 0.1) - s(1) = 3(1.1)^2 - 3(1)^2 = 3(1.21) - 3 = 0.63 meters
Average velocity = Displacement / h = 0.63 / 0.1 = 6.3 m/s
b) When h = 0.01:
Displacement = s(1 + 0.01) - s(1) = 3(1.01)^2 - 3(1)^2 = 3(1.0201) - 3 = 0.0603 meters
Average velocity = Displacement / h = 0.0603 / 0.01 = 6.03 m/s
c) When h = 0.001:
Displacement = s(1 + 0.001) - s(1) = 3(1.001)^2 - 3(1)^2 = 3(1.002001) - 3 = 0.006003 meters
Average velocity = Displacement / h = 0.006003 / 0.001 = 6.003 m/s
Now, to estimate the exact derivative at t = 1, we need to take the limit as h approaches 0 of the average velocity over the time interval (t = 1 to t = 1 + h).
Exact derivative at t = 1 = lim(h -> 0) [Average velocity]
The closest approximation to the exact derivative would be the average velocity when h is the smallest.
In this case, h = 0.001 gives an average velocity of 6.003 m/s, which would be the closest approximation to the exact derivative.