Problem 169 - Calculus Volume 1

Question

Consider an astronaut on a large planet in another galaxy. To learn more about the composition of this planet, the astronaut drops an electronic sensor into a deep trench. The sensor transmits its vertical position every second in relation to the astronaut's position. The summary of the falling sensor data is displayed in the following table. $$\begin{array}{|l|l|} \hline \text { Time after dropping (s) } & \text { Position (m) } \\ \hline 0 & 0 \\ \hline 1 & -1 \\ \hline 2 & -2 \\ \hline 3 & -5 \\ \hline 4 & -7 \\ \hline 5 & -14 \\ \hline \end{array}$$ a. Using a calculator or computer program, find the best-fit quadratic curve to the data. b. Find the derivative of the position function and explain its physical meaning. c. Find the second derivative of the position function and explain its physical meaning.

Step-by-Step Solution

Answer

a. \(p(t) = -1.4t^2 - 0.6t - 0.0\). b. \(p'(t) = -2.8t - 0.6\), velocity. c. \(p''(t) = -2.8\), constant acceleration.

Step 1: Organize the Data

We are given the position of the sensor at different times. These data points are \((0, 0), (1, -1), (2, -2), (3, -5), (4, -7), (5, -14)\). Our task is to find a quadratic function \(p(t) = at^2 + bt + c\) that best fits these points.

Step 2: Use Quadratic Regression

Using a calculator or a program with quadratic regression capabilities (like Excel or a graphing calculator), input the time as \(x\) values and the positions as \(y\) values. Calculate the coefficients \(a\), \(b\), and \(c\). The best-fit quadratic curve is approximately \(p(t) = -1.4t^2 - 0.6t - 0.0\).

Step 3: Differentiate the Quadratic Function

To find the velocity as a function of time, we take the derivative of the position function \(p(t) = at^2 + bt + c\). The derivative \(p'(t)\) provides the velocity: \(p'(t) = 2at + b\). For our specific coefficients: \(p'(t) = -2.8t - 0.6\).

Step 4: Understand the First Derivative

The derivative \(p'(t) = -2.8t - 0.6\) describes the velocity of the sensor. It tells us how fast the sensor is falling at any particular time \(t\). A negative value indicates the sensor is falling downwards.

Step 5: Differentiate Again for Acceleration

Now, find the second derivative of the position function, which gives the acceleration. This is the derivative of the velocity function \(p'(t)\): \(p''(t) = -2.8\).

Step 6: Interpret the Second Derivative

The second derivative \(p''(t) = -2.8\) represents the constant acceleration due to gravity on the planet. Since the value is negative, it implies a constant downward acceleration.

Key Concepts

Derivative

In the world of mathematics and physics, the derivative is an immensely important concept. It tells us the rate at which one quantity changes with respect to another. When applied to the sensor's position function, the derivative provides the velocity, or how fast the sensor's position changes over time.

• Understanding Derivative: In our example, the position function is quadratic, given by \( p(t) = -1.4t^2 - 0.6t - 0.0 \). By differentiating this function, we find the velocity function \( p'(t) = -2.8t - 0.6 \).
• Physical Meaning: This derivative tells us the sensor's speed and direction. The negative sign in \( p'(t) \) signifies that it is moving downward.

Differentiation is a key tool for analyzing motion, as it provides instantaneous rates of change, crucial for understanding processes in physics and engineering.

Second Derivative

The second derivative of a function provides a deeper layer of understanding. For a position function, the second derivative indicates the change in velocity, known as acceleration.

• Finding the Second Derivative: When we differentiate the velocity function \( p'(t) = -2.8t - 0.6 \), we obtain the second derivative \( p''(t) = -2.8 \).
• Significance in Physics: This constant second derivative represents a constant acceleration, a straightforward scenario often encountered in classical mechanics.

Understanding the second derivative is pivotal, especially in gravity-related problems, where it directly correlates with acceleration due to gravity.

Acceleration

Acceleration measures how quickly velocity changes over time. It is fundamentally connected to the second derivative of a position function. In this exercise, the constant acceleration tells us about the gravitational pull on the sensor.

• Constant Acceleration: The calculated second derivative is \( p''(t) = -2.8 \), indicating a steady acceleration throughout the sensor's fall.
• Direction & Magnitude: The value's negativity not only informs us of the downward direction but also of the gravitational strength of the planet.

By identifying acceleration, we can understand how external forces influence an object's movement, an essential component in engineering and space exploration.