la altura en pies, h, de un modelo de cohete t segundos después del lanzamiento se da por la ecuación h(t) = 3 + 70t - 16t². la tasa media de cambio en h(t) entre t = 1 segundo y t = 3 el segundo es 6¿qué te dice la tasa de cambio promedio sobre el cohete?
el cohete está a mayor altura cuando t = 3 que cuando t = 1.
el cohete viaja a una velocidad constante de 6 pies por segundo entre t = 1 y t = 3.
el cohete es 6 pies más arriba del suelo cuando t = 3 que cuando t = 1.
el cohete viaja seis veces más rápido cuando t = 3 que cuando t = 1.

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Accepted Answer

To determine what the average rate of change tells us about the rocket, we first recall the definition of the average rate of change of a function $ h(t) $ over the interval $[a, b]$:

The average rate of change is given by:
$$
\frac{h(b) - h(a)}{b - a}
$$

In the context of a height function $ h(t) $, the average rate of change represents the average velocity of the rocket over the time interval from $ t = a $ to $ t = b $.

### Step 1: Recall the meaning of average rate of change for a height function
For a height function $ h(t) $, the average rate of change over an interval $[t_1, t_2]$ is the average velocity of the object (in this case, the rocket) over that time interval. Mathematically, it is:
$$
\text{Average rate of change} = \frac{h(t_2) - h(t_1)}{t_2 - t_1}
$$
This quantity has units of (change in height) / (change in time), which is the unit of velocity (e.g., feet per second).

### Step 2: Analyze the given average rate of change
We are told that the average rate of change of $ h(t) $ between $ t = 1 $ and $ t = 3 $ is 6. Using the formula for average rate of change:
$$
\frac{h(3) - h(1)}{3 - 1} = 6
$$
Simplifying the denominator:
$$
\frac{h(3) - h(1)}{2} = 6
$$
Multiplying both sides by 2:
$$
h(3) - h(1) = 12
$$
This tells us that the change in height from $ t = 1 $ to $ t = 3 $ is 12 feet. However, the average rate of change itself (6 feet per second) represents the average velocity of the rocket over the time interval from $ t = 1 $ to $ t = 3 $.

### Step 3: Evaluate the options
- **First option**: "El cohete está a mayor altura cuando $ t = 3 $ que cuando $ t = 1 $."
  While this is true (since $ h(3) - h(1) = 12 > 0 $), the average rate of change does not directly tell us about the "magnitude" of the height difference in terms of interpretation (it tells us about the average velocity, not just that the height is greater).

- **Second option**: "El cohete viaja a una velocidad constante de 6 pies por segundo entre $ t = 1 $ y $ t = 3 $."
  The average rate of change being 6 does not mean the velocity is constant. The average rate of change is just the average velocity over the interval; the instantaneous velocity (derivative of $ h(t) $) could be changing. However, the average rate of change does represent the average velocity over the interval. Wait, let's check the other options.

- **Third option**: "El cohete es 6 pies más arriba del suelo cuando $ t = 3 $ que cuando $ t = 1 $."
  From our calculation, $ h(3) - h(1) = 12 $, so this is false.

- **Fourth option**: "El cohete viaja seis veces más rápido cuando $ t = 3 $ que cuando $ t = 1 $."
  The average rate of change does not tell us about the ratio of the instantaneous velocities at $ t = 1 $ and $ t = 3 $; it only tells us about the average velocity over the interval.

Wait, let's re-examine the second option. The average rate of change of a position function (height, in this case) with respect to time is the average velocity. So if the average rate of change is 6 feet per second between $ t = 1 $ and $ t = 3 $, that means the rocket's average velocity over that interval is 6 feet per second. So the second option says "El cohete viaja a una velocidad constante de 6 pies por segundo entre $ t = 1 $ y $ t = 3 $." Wait, but average velocity being 6 does not mean constant velocity. However, maybe the wording is a bit loose, and among the options, this is the one that relates to the meaning of average rate of change (velocity). Let's check the first option again: "El cohete está a mayor altura cuando $ t = 3 $ que cuando $ t = 1 $." This is true (since $ h(3) - h(1) = 12 > 0 $), but does the average rate of change "tell us" that? The average rate of change being positive tells us the height increased, but the first option is a statement about the height being greater, which is implied by a positive average rate of change. However, the second option is about velocity. Let's calculate $ h(1) $ and $ h(3) $ to check:

For $ t = 1 $:
$$
h(1) = 3 + 70(1) - 16(1)^2 = 3 + 70 - 16 = 57
$$

For $ t = 3 $:
$$
h(3) = 3 + 70(3) - 16(3)^2 = 3 + 210 - 144 = 69
$$

Then $ h(3) - h(1) = 69 - 57 = 12 $, so the change in height is 12 feet over 2 seconds, so the average rate of change is $ \frac{12}{2} = 6 $ feet per second.

Now, let's re-evaluate the options:

- First option: "El cohete está a mayor altura cuando $ t = 3 $ que cuando $ t = 1 $." This is true (69 > 57), but does the average rate of change "tell us" that? The average rate of change being positive tells us the height increased, so this is a valid interpretation. But let's check the second option: "El cohete viaja a una velocidad constante de 6 pies por segundo entre $ t = 1 $ y $ t = 3 $." The average rate of change is 6, but that doesn't mean the velocity is constant. The instantaneous velocity at $ t = 1 $ is $ h'(1) $, and at $ t = 3 $ is $ h'(3) $. Let's calculate the derivative: $ h'(t) = 70 - 32t $. So $ h'(1) = 70 - 32(1) = 38 $ feet per second, and $ h'(3) = 70 - 32(3) = 70 - 96 = -26 $ feet per second. So the velocity is not constant; it's decreasing (since the derivative of $ h(t) $ is linear and decreasing). So the second option is false.

- Third option: "El cohete es 6 pies más arriba del suelo cuando $ t = 3 $ que cuando $ t = 1 $." But we saw $ h(3) - h(1) = 12 $, so this is false.

- Fourth option: "El cohete viaja seis veces más rápido cuando $ t = 3 $ que cuando $ t = 1 $." The velocity at $ t = 1 $ is 38, at $ t = 3 $ is -26 (speed is 26), so this is false.

So the first option is true: "El cohete está a mayor altura cuando $ t = 3 $ que cuando $ t = 1 $." Because the average rate of change is positive, the height at $ t = 3 $ is greater than at $ t = 1 $.

# Answer:
El cohete está a mayor altura cuando $ t = 3 $ que cuando $ t = 1 $.

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Question

Explanation:

Step 1: Recall the meaning of average rate of change for a height function

Step 2: Analyze the given average rate of change

Step 3: Evaluate the options

Answer: