Difference quotient problem and average rate of change

In summary, a quadratic model is used to represent the height of a soccer ball kicked into the air. The average rate of change of height over a specific time interval is solved for, with the correct expression being [-4.9(2+h)^2 + 13.5(2+h) - 7.4] / h. The valid values for h in this expression are h > 0.84, with the restriction that at 2.84 seconds, h = 0 due to the ball landing on the floor. This explains why h is both used as the height of the object and a time in the problem.
  • #1
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Homework Statement



A soccer ball is kicked into the air from a platform. The height of the ball, in metres, t seconds after it is kicked is modeled by h(t) = -4.9t^2 + 13.5t + 1.2.

I solved for the expression that represents the average rate of change of height over the interval 2<= t <= (2+h) and got which is correct (according to the answer key):

[-4.9(2+h)^2 + 13.5(2+h) - 7.4] / h

For what values of h is the expression above not valid?

The answer in the back states h>0.84; at 2.84 s, h = 0.

I understood how they got h>0.84 by finding the zeroes of the quadratic equation to get 2.84 and then subbing that into the inequality to get the change (h) to be 0.84. Therefore since t !> 2.84 h !> 0.84. But how do they get at 2.84 s, h != 0 as restrictions?

How did they get 0.84 and 2.84 in the restrictions. This question is a bit confusing to me. Can anyone help me find the answer?
 
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  • #2
It is confusing that h is both used as the height of the object and a time.

h(t)=0 (the height) corresponds to the ball landing on the floor. This will certainly stop its free fall, and make the formula for h(t) invalid after that point.
 

Related to Difference quotient problem and average rate of change

1. What is the difference quotient problem?

The difference quotient problem is a mathematical concept used to find the slope of a curve at a specific point. It involves calculating the average rate of change between two points on a curve, and then taking the limit as the distance between those points approaches zero. This limit represents the instantaneous rate of change at that point, or the slope of the curve at that point.

2. How is the difference quotient problem related to the average rate of change?

The difference quotient problem is directly related to the average rate of change because it involves calculating the average rate of change between two points on a curve. The average rate of change is a measure of how much a quantity changes over a given time period, and it can be represented by the slope of a line connecting those two points on the curve.

3. What is the significance of finding the difference quotient?

Finding the difference quotient is significant because it allows us to calculate the instantaneous rate of change at a specific point on a curve. This is important in many real-world applications, such as determining the velocity of an object at a certain time or the growth rate of a population at a specific moment.

4. What are some examples of using the difference quotient and average rate of change?

The difference quotient and average rate of change can be used in various fields, such as physics, economics, and biology. For example, in physics, it can be used to determine the acceleration of an object at a certain time. In economics, it can be used to measure the rate of change of a company's profits over time. In biology, it can be used to calculate the growth rate of a population at a specific moment.

5. How can I solve a difference quotient problem?

To solve a difference quotient problem, you will need to find the average rate of change between two points on a curve, and then take the limit as the distance between those points approaches zero. This limit will give you the instantaneous rate of change at that point, which represents the slope of the curve at that point. You can then use this slope to solve for other variables in the problem or to answer specific questions about the curve.

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