Math Problem Statement

answer both questions

A projectile’s path is represented by the function h(t) = -4.9t^2 + 25t, where h is its height in metres, after t seconds. Find the average rate of change of the projectile’s height between 0 and 2 seconds. How does changing the value of h in the difference quotient bring the slope of the secant closer to the slope of the tangent? Do you think there is a limit to how small h can be? Explain.

Solution

Let's solve both of the problems step by step:

Question 7:

Given:
The height of a projectile is represented by the function: h(t)=4.9t2+25th(t) = -4.9t^2 + 25t where hh is the height in meters, and tt is the time in seconds.

We need to find the average rate of change of height between t=0t = 0 and t=2t = 2.

Formula for average rate of change:

The average rate of change of a function between t1t_1 and t2t_2 is given by: Average rate of change=h(t2)h(t1)t2t1\text{Average rate of change} = \frac{h(t_2) - h(t_1)}{t_2 - t_1} Here, t1=0t_1 = 0 and t2=2t_2 = 2.

Step-by-step calculation:

  1. Find h(0)h(0): h(0)=4.9(0)2+25(0)=0h(0) = -4.9(0)^2 + 25(0) = 0

  2. Find h(2)h(2): h(2)=4.9(2)2+25(2)=4.9(4)+50=19.6+50=30.4h(2) = -4.9(2)^2 + 25(2) = -4.9(4) + 50 = -19.6 + 50 = 30.4

  3. Use the average rate of change formula: Average rate of change=h(2)h(0)20=30.4020=30.42=15.2m/s\text{Average rate of change} = \frac{h(2) - h(0)}{2 - 0} = \frac{30.4 - 0}{2 - 0} = \frac{30.4}{2} = 15.2 \, \text{m/s}

Thus, the average rate of change of the projectile's height between 0 and 2 seconds is 15.2 m/s.


Question 8:

Given:
We are asked to discuss how changing the value of hh in the difference quotient brings the slope of the secant closer to the slope of the tangent line. We are also asked if there is a limit to how small hh can be.

Explanation:

The difference quotient is the formula used to estimate the slope of the secant line between two points on a curve: f(x+h)f(x)h\frac{f(x+h) - f(x)}{h} As hh becomes smaller, the second point on the secant line moves closer to the first point, making the secant line better approximate the tangent line at xx.

  1. How changing hh affects the slope:

    • As hh decreases, the two points (at xx and x+hx+h) come closer together.
    • This causes the secant line (which passes through both points) to better approximate the tangent line at xx.
    • The smaller hh becomes, the closer the slope of the secant line becomes to the slope of the tangent line.
  2. Is there a limit to how small hh can be?

    • As hh approaches 0, the secant line becomes indistinguishable from the tangent line, and the slope of the secant line approaches the derivative at xx, which is the slope of the tangent line.
    • However, while conceptually hh can become arbitrarily small, in practice, there is a limit. If hh becomes too small (especially in numerical computations), round-off errors or precision issues can arise.
    • Mathematically, hh approaching zero gives us the derivative, but h=0h = 0 itself is undefined in the difference quotient because it would involve division by zero.

In summary, the smaller hh gets, the closer the secant slope approximates the tangent slope, but there are practical limits due to numerical accuracy.


Would you like further clarification on any part of this? Here are 5 related questions you might want to explore:

  1. How is the difference quotient related to the derivative?
  2. Can you find the instantaneous rate of change of the projectile at t=2t = 2 seconds?
  3. What happens to the average rate of change if the interval is increased to t=3t = 3 seconds?
  4. Why do secant lines provide only an approximation of the tangent line’s slope?
  5. How does the second derivative relate to the curvature of the projectile's path?

Tip: When finding average rates of change, always remember that you're calculating the slope of the secant line between two points on the curve. For the instantaneous rate of change, use derivatives!

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Math Problem Analysis

Mathematical Concepts

Calculus
Rates of Change
Secant Line
Tangent Line

Formulas

Average rate of change = (h(t2) - h(t1)) / (t2 - t1)
Difference quotient = (f(x+h) - f(x)) / h

Theorems

Definition of Average Rate of Change
Limit Definition of the Derivative

Suitable Grade Level

Grades 11-12