The height of a ball t seconds after it is thrown upward from a height of 6 feet and with an initial velocity of 80 feet per second is f (t) = -16t^2 + 80t + 6.
According to Rolle's Theorem, what must be the velocity at some time in the interval (2, 3)?
ft/sec?
& Find that time t?
I believe i need to take the 1st derivative which is -32t+80 equal to zero so t=-80/32? and for ft/sec plug in 2 for t? Anyhow i get wrong answers so need help!
Rolle's Theorem states that on the interval (2,3) there will be a point c where f'(c) = the slope of the chord from f(2) to f(3)
So, f(2) = 102
f(3) = 102
so, the chord has slope 0.
so, at some point c in (2,3) f'(c) = 0
f'(t) = 80 - 32t
f'(t) = where t = 80/32 = 2.5
That's what you'd expect, since parabolas are symmetric about their axis.
THNKS :)I really appreciate your help!
so what would be the velocity at some time in the interval (2, 3)?
To find the velocity at some time in the interval (2, 3) according to Rolle's Theorem, you are correct that you need to find the first derivative of the function f(t). However, the first derivative of f(t) represents the rate of change of height with respect to time, known as the velocity, rather than the actual velocity.
Let's start by finding the first derivative of the function f(t) = -16t^2 + 80t + 6. The derivative of -16t^2 is -32t, the derivative of 80t is 80, and the derivative of 6 (a constant) is 0. Therefore, the first derivative, which represents the velocity, is:
f'(t) = -32t + 80
To find the time at which the velocity is zero, you are correct in setting f'(t) equal to zero and solving for t:
-32t + 80 = 0
Solving this equation, we get:
-32t = -80
t = -80 / -32
t ≈ 2.5
So, at some point in the interval (2, 3), the velocity is zero.
To find the exact time t, you can substitute the value of t ≈ 2.5 back into the original function f(t):
f(2.5) = -16(2.5)^2 + 80(2.5) + 6
f(2.5) = -16(6.25) + 200 + 6
f(2.5) = -100 + 200 + 6
f(2.5) = 106
Therefore, at t = 2.5 seconds, the height of the ball is 106 feet.
Note: It's important to remember the context of the problem when interpreting the results. In this case, the negative value of t indicates that the ball passed the point of upward motion and started falling again.