The altitude (in feet) attained by a model rocket t seconds into flight is
given by the function
h(t) = -1/3t^3 + 4t^2 + 20t + 2,for t _> 2
a)Find the maximum altitude attained by the rocket.
b)Why does it not make sense to use this function after t = 16
seconds? Use the graph of h(t) that is given
dh/dt = -t^2 + 8t + 20
= -(t-10)(t+2)
max h when t=10, h(10) = 806/3
h(16)=0, so the rocket has landed by then. No change in h thereafter.
a) The maximum altitude attained by the rocket can be found by finding the vertex of the function. The vertex of a cubic function can be found using the formula x = -b / (3a), where a, b, and c are the coefficients of the function.
In this case, the function is h(t) = -1/3t^3 + 4t^2 + 20t + 2. Comparing this to the general form of a cubic function h(t) = at^3 + bt^2 + ct + d, we can see that a = -1/3, b = 4, c = 20, and d = 2.
Using the formula, the t-coordinate of the vertex is t = -b / (3a) = -4 / (3 * -1/3) = -4 / -1 = 4.
To find the maximum altitude, we substitute t = 4 into the function h(t):
h(4) = -1/3 * 4^3 + 4 * 4^2 + 20 * 4 + 2
= -1/3 * 64 + 4 * 16 + 80 + 2
= -64/3 + 64 + 80 + 2
= -64/3 + 192 + 2
= -64/3 + 194
= 191 1/3 feet
Therefore, the maximum altitude attained by the rocket is approximately 191 1/3 feet.
b) It does not make sense to use this function after t = 16 seconds because the function is only defined for t ≥ 2. The given function h(t) is only valid for time after 2 seconds, so it cannot be used to make predictions or calculations beyond that point.
According to the graph of h(t), it shows that the rocket's altitude begins to decrease after approximately t = 16 seconds. It is possible that the rocket has reached its maximum altitude at that point and is on its way back down to the ground. Therefore, using the given function beyond t = 16 seconds would not accurately represent the rocket's altitude.
To find the maximum altitude attained by the rocket, we need to find the vertex of the quadratic function h(t). The vertex of a quadratic function in the form of f(t) = at^2 + bt + c is given by the formula t = -b/2a.
In this case, the function h(t) = -1/3t^3 + 4t^2 + 20t + 2 is a cubic function, but we only need to consider the quadratic term, which is 4t^2.
Comparing it to the standard form of a quadratic function f(t) = at^2 + bt + c, we have a = 4, b = 0, and c = 0.
Using the vertex formula, we calculate:
t = -b/2a = -0/(2*4) = 0
So, the maximum altitude is attained at t = 0.
To determine the maximum altitude value, we substitute t = 0 into the function h(t):
h(0) = -1/3(0)^3 + 4(0)^2 + 20(0) + 2 = 2
Therefore, the maximum altitude attained by the rocket is 2 feet.
b) Looking at the graph of h(t), it is clear that after t = 16 seconds, the altitude starts to decrease rapidly. This suggests that the rocket is descending or has already landed, and it would not make sense to use the function h(t) to calculate the altitude beyond t = 16 seconds. The graph provides visual evidence of this trend, showing a significant downturn in altitude after t = 16 seconds.
To find the maximum altitude attained by the rocket, we need to determine the highest point on the graph of the function h(t). The maximum or minimum points of a function occur where the derivative of the function is zero or does not exist.
a) Maximum Altitude:
1. Differentiate the function h(t) with respect to t: h'(t) = -t^2 + 8t + 20.
2. Set h'(t) equal to zero and solve for t:
-t^2 + 8t + 20 = 0.
Rearranging the equation: t^2 - 8t - 20 = 0.
3. Solve the quadratic equation. You can use the quadratic formula: t = (-b ± √(b^2 - 4ac)) / (2a).
Plugging in the values a = 1, b = -8, and c = -20, we get:
t = (-(-8) ± √((-8)^2 - 4 * 1 * -20)) / (2 * 1).
Simplifying further, we have:
t = (8 ± √(64 + 80)) / 2.
t = (8 ± √144) / 2.
t = (8 ± 12) / 2.
This gives us two values for t: t = 10 or t = -2.
In this case, t = -2 is extraneous since the function is defined for t ≥ 2.
Therefore, the maximum altitude is reached at t = 10 seconds.
4. To find the maximum altitude, substitute t = 10 into the function h(t):
h(10) = -(1/3)(10^3) + 4(10^2) + 20(10) + 2.
h(10) = -1000/3 + 400 + 200 + 2.
Simplifying, we get: h(10) = 600 + 202/3.
The maximum altitude attained by the rocket is 2000/3 feet.
b) After t = 16 seconds, it does not make sense to use this function because the graph of h(t) might not accurately represent the rocket's altitude. The given function h(t) is only defined for t ≥ 2, which means it is only valid for time intervals starting at or after t = 2 seconds. Beyond that, the rocket's behavior might not follow the same pattern as described by this function. Therefore, using h(t) after t = 16 seconds would not provide reliable information about the rocket's altitude.