If an arrow is shot upward on Mars with a speed of 54 m/s, its height in meters t seconds later is given by
y = 54t − 1.86t2
(i) [1, 2](ii) [1, 1.5]
(iii) [1, 1.1]
(iv) [1, 1.01]
(v) [1, 1.001]
(b) Estimate the speed when t = 1.
[f(b)-f(a)]/(b-a)
[(58(2)-1.86(2^2)) - (58(1)-1.86(1^2))]/2-1 = 52.42
To estimate the speed when t = 1, we need to calculate the derivative of the height equation with respect to time, which gives us the instantaneous velocity at any given time.
The derivative of y = 54t - 1.86t^2 with respect to t can be found by taking the derivative of each individual term:
dy/dt = d(54t)/dt - d(1.86t^2)/dt
The derivative of a constant multiplied by t (54t) is simply the constant (54), and the derivative of t raised to the power of 2 (1.86t^2) is 2 multiplied by the constant coefficient (1.86) times t raised to the power of 1 (1.86t).
Therefore, the derivative becomes:
dy/dt = 54 - 2 * 1.86t
Now, we can substitute t = 1 into the derivative to find the velocity at t = 1:
v = dy/dt at t = 1
= 54 - 2 * 1.86 * 1
Calculating this expression gives us:
v = 54 - 2 * 1.86 * 1
= 54 - 3.72
= 50.28
So, the estimated speed when t = 1 is approximately 50.28 m/s.
54.14
(a) Apparently you are supposed to find the slopes over the various intervals [a,b]. As always, that is
[f(b)-f(a)]/(b-a)
Just plug in the numbers.
(b) By looking at the way the results approach a single value, that is the estimate of the speed.
It should be 50.28