Problem 249 - Calculus Volume 1

Question

[T] The accompanying graph plots the best quadratic fit, \(a(t)=-0.70 t^{2}+1.44 t+10.44,\) to the data from the preceding table. Compute the average value of \(a(t)\) to estimate the average acceleration between \(t=0\) and \(t=5\)

Step-by-Step Solution

Answer

The average acceleration between \(t=0\) and \(t=5\) is approximately 8.206.

Step 1: Formula for Average Value

To find the average value of a quadratic function \(a(t)\) over an interval \([a, b]\), we use the formula: \(\text{Average} = \frac{1}{b-a} \int_{a}^{b} a(t) \ dt\). For this problem, \(a(t) = -0.70t^2 + 1.44t + 10.44\) and our interval is \([0, 5]\).

Step 2: Set Up the Integral

Set up the definite integral of the function \(a(t)\) over the interval \([0, 5]\): \(\int_{0}^{5} (-0.70t^2 + 1.44t + 10.44) \ dt\).

Step 3: Integrate the Function

Integrate each term of the quadratic function separately: \[\int -0.70t^2 \, dt = -0.70 \cdot \frac{t^3}{3} + C_1\],\[\int 1.44t \, dt = 0.72t^2 + C_2\],\[\int 10.44 \, dt = 10.44t + C_3\].

Step 4: Evaluate at the Endpoints

Combine the integrated terms and evaluate from 0 to 5:\[-0.70 \cdot \frac{5^3}{3} + 0.72 \cdot 5^2 + 10.44 \cdot 5\] and \[-0.70 \cdot \frac{0^3}{3} + 0.72 \cdot 0^2 + 10.44 \cdot 0\].

Step 5: Simplify the Expressions

Calculate each evaluated term:\[-\frac{0.70 \times 125}{3} = -29.17\], \[0.72 \times 25 = 18\], \[10.44 \times 5 = 52.20\].The value at \(t=5\) becomes \(-29.17 + 18 + 52.20\) and at \(t=0\), it becomes \(0\).

Step 6: Compute the Average

Plug the results into the average formula: \[\text{Average} = \frac{1}{5-0} (41.03 - 0) = \frac{41.03}{5}\].

Step 7: Final Calculation

Divide to get the average acceleration: \[\frac{41.03}{5} = 8.206\]. This represents the average value of the acceleration over the interval.

Key Concepts

Definite Integral

The definite integral is a fundamental concept in calculus, used to calculate the area under a curve within a specific interval. It is particularly useful in various applications like physics and engineering, where it helps find quantities such as distance, area, and in this case, average acceleration. In mathematical terms, the definite integral of a function \(f(t)\) over an interval \([a, b]\) is expressed as: \[ \int_{a}^{b} f(t) \, dt \] This notation represents the accumulation of the area under the curve \(f(t)\) from \(t=a\) to \(t=b\). The definite integral comprises multiple components:

• **Function \(f(t)\):** The function you want to integrate.
• **Limits of integration (\(a, b\)):** The range over which you're integrating.
• **\(dt\):** This indicates integration with respect to the variable \(t\).

By evaluating the integral, you obtain a numerical value that represents the cumulative effect of the function over the specified interval, making it incredibly valuable for precise calculations such as finding average values, like average acceleration.

Quadratic Function

A quadratic function is an algebraic expression of the format \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. In this context, the function given is \(a(t) = -0.70t^2 + 1.44t + 10.44\), which is a quadratic model used to fit a set of data points. Quadratic functions have several key features: - **Parabola Shape:** When graphed, they form a parabola, which can open upwards or downwards depending on the sign of \(a\). - **Vertex:** The highest or lowest point of the parabola, marking its peak or dip, depending on the parabola's orientation. - **Axis of Symmetry:** A vertical line that divides the parabola into two mirror-image halves. Quadratic functions are pivotal in modeling real-world scenarios where changes in acceleration or other physics aspects need to be evaluated over time, making them an integral part of calculus and its application in problems involving motion.

Calculus Problem Solving

Calculus problem solving involves using a combination of different calculus techniques to find solutions to mathematical problems. In the presented exercise, we integrate the quadratic function to determine the average acceleration. Here's a brief outline of the problem-solving approach: 1. **Understand the Problem:** Identify what information is provided and what you need to find. Here, we need the average acceleration over an interval. 2. **Set Up the Problem:** Use the appropriate mathematical formulas. For finding average acceleration, use the average value formula for a definite integral: \[ \text{Average} = \frac{1}{b-a} \int_{a}^{b} a(t) \, dt \] 3. **Solve the Integral:** Break down the function into simpler parts, integrate each part, and substitute the limits:

• Integrate terms one by one.
• Evaluate the integral at the given limits.

4. **Calculate the Average:** Substitute the results back into the average value formula to find the average acceleration. In this way, calculus provides a structured platform for addressing complex mathematical problems by leveraging integrals, derivatives, and algebraic manipulation, offering precise solutions in fields ranging from engineering to economics.