Problem 37 - Calculus Early Transcendentals: Pearson New International Edition

Question

An object moves along a horizontal coordinate line in such a way that its position at time \(t\) is specified by \(s=t^{3}-3 t^{2}-24 t-6\). Here \(s\) is measured in centimeters and \(t\) in seconds. When is the object slowing down; that is, when is its speed decreasing?

Step-by-Step Solution

Answer

The object is slowing down on the intervals \((-2, 1)\) and \((4, \infty)\).

Step 1: Determine the Velocity Function

To find when the object is slowing down, we first need its velocity function. Velocity is the derivative of the position function. Given \(s(t) = t^3 - 3t^2 - 24t - 6\), differentiate to get the velocity: \(v(t) = \frac{ds}{dt} = 3t^2 - 6t - 24\).

Step 2: Find the Acceleration Function

The next step is to find the acceleration, which is the derivative of the velocity. Differentiate the velocity function to get the acceleration: \(a(t) = \frac{dv}{dt} = 6t - 6\).

Step 3: Determine When Velocity is Positive or Negative

Solve the velocity function \(3t^2 - 6t - 24 = 0\) to find its roots, which will help identify intervals of increasing or decreasing velocity. Using the quadratic formula, \(t = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 3 \cdot (-24)}}{2 \cdot 3}\), we find the roots are \(t = -2\) and \(t = 4\). The velocity changes sign at these points.

Step 4: Determine When Acceleration is Positive or Negative

Solve the acceleration function \(6t - 6 = 0\) to determine where acceleration changes sign. Solving gives \(t = 1\).

Step 5: Analyze Speed Changes

An object slows down when the velocity and acceleration have opposite signs. Determine the sign of both velocity \(v(t)\) and acceleration \(a(t)\) in each interval determined by the critical points: \((-\infty, -2)\), \((-2, 1)\), \((1, 4)\), \((4, \infty)\).

Step 6: Evaluate Intervals for Speed Decrease

- On \((-\infty, -2)\): both \(v(t) < 0\) and \(a(t) < 0\), speed increases.- On \((-2, 1)\): \(v(t) > 0\), \(a(t) < 0\), speed decreases.- On \((1, 4)\): \(v(t) > 0\), \(a(t) > 0\), speed increases.- On \((4, \infty)\): \(v(t) < 0\), \(a(t) > 0\), speed decreases.

Key Concepts

Velocity

In calculus, velocity is a key concept used to describe how fast an object's position is changing over time. For the problem at hand, velocity is derived from the position function, which represents the object's movement along a line. The position function in this case is \(s(t) = t^3 - 3t^2 - 24t - 6\), where \(s\) is the position in centimeters and \(t\) is the time in seconds. To find the velocity, we differentiate the position function with respect to time, resulting in the velocity function \(v(t) = 3t^2 - 6t - 24\).

Here are some key points about velocity:

• It tells us whether the object is moving forward or backward, depending on whether the velocity is positive or negative.
• Changes in sign of the velocity function indicate where the object changes direction.
• To find when the object's velocity is decreasing, you'll need to analyze where the velocity function changes from positive to negative, or vice versa.

This helps determine intervals where the object is speeding up or slowing down.

Acceleration

Acceleration refers to how quickly the velocity of an object changes as time goes by. In terms of calculus, acceleration is the derivative of the velocity function, representing the rate of change of velocity. In the context of the exercise, the acceleration function is derived by differentiating the velocity \(v(t) = 3t^2 - 6t - 24\), leading us to \(a(t) = 6t - 6\).

Here are some fundamental concepts about acceleration:

• Positive acceleration means that an object's speed is increasing, while negative acceleration suggests a slowdown.
• Acceleration shifting from positive to negative, or vice versa, points to changes in movement speed.
• In physics, acceleration could also mean the change in direction of motion even if the speed magnitude doesn't change.

By using acceleration, one can identify specific intervals where the object changes speed in the given timeframe.

Derivative

Understanding derivatives is essential in calculus as they provide the tool used for finding both velocity and acceleration. A derivative represents a function's rate of change, effectively describing how one quantity changes in response to another. For instance, the derivative of the position function \(s(t)\) results in the velocity function \(v(t)\). If you differentiate again, the result becomes the acceleration function \(a(t)\).

Important points about derivatives:

• They help to determine slopes of tangent lines on a curve.
• In physics, derivatives help translate physical phenomena into mathematical models.
• Each successive derivative relates to a higher order of change, such as position, velocity, acceleration, and so forth.

Through derivatives, we can analyze and predict the motion and speed changes of objects in a given problem.

Quadratic Formula

The quadratic formula is a crucial tool in mathematics used to solve quadratic equations—equations that take the standard form \(ax^2 + bx + c = 0\). In this exercise, the quadratic formula is used to find the roots of the velocity function \(3t^2 - 6t - 24 = 0\). These roots are the critical points where the velocity changes sign.

Here's how the quadratic formula is applied:

• It is given by \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a\), \(b\), and \(c\) are coefficients from the equation.
• This formula helps find precise points at which the function equals zero, marking changes in direction or speed.
• In the velocity context, solving the quadratic equation provides times \(t\) at which the velocity changes direction.

By finding these roots, the quadratic formula permits analysis of intervals of speed changes.