4. Calculate average speed of particle which position at time t is given by the function g(t) = t3, on the intervals
(a) [2,3]; (b) [2,2.1];
(c) [2,2.0001];
(d) [2, 2 + h] , where h > 0 is arbitrary positive number.
Find g′(2).
To calculate the average speed of the particle on the given intervals, we need to find the displacement of the particle and divide it by the duration of the interval. Since the function g(t) = t^3 gives the position of the particle at time t, we can find the displacement by subtracting the initial position from the final position.
(a) Interval [2,3]:
The initial position at t = 2 is g(2) = 2^3 = 8.
The final position at t = 3 is g(3) = 3^3 = 27.
Therefore, the displacement is 27 - 8 = 19.
The duration of the interval is 3 - 2 = 1.
So, the average speed is 19/1 = 19 units per time.
(b) Interval [2,2.1]:
The initial position at t = 2 is g(2) = 2^3 = 8.
The final position at t = 2.1 is g(2.1) = (2.1)^3 = 9.261.
The displacement is 9.261 - 8 = 1.261.
The duration of the interval is 2.1 - 2 = 0.1.
Therefore, the average speed is 1.261/0.1 = 12.61 units per time.
(c) Interval [2,2.0001]:
The initial position at t = 2 is g(2) = 2^3 = 8.
The final position at t = 2.0001 is g(2.0001) = (2.0001)^3 = 8.000240004.
The displacement is 8.000240004 - 8 = 0.000240004.
The duration of the interval is 2.0001 - 2 = 0.0001.
Therefore, the average speed is 0.000240004/0.0001 = 0.00240004 units per time.
(d) Interval [2, 2 + h]:
The initial position at t = 2 is g(2) = 2^3 = 8.
The final position at t = 2 + h is g(2 + h) = (2 + h)^3.
The displacement is g(2 + h) - 8.
The duration of the interval is 2 + h - 2 = h.
Therefore, the average speed is (g(2 + h) - 8)/h units per time.
To find g'(2), we need to take the derivative of the function g(t). The derivative of g(t) = t^3 is given by g'(t) = 3t^2. Therefore, g'(2) = 3(2)^2 = 12.