Problem 33 - Thomas Calculus

Question

Exercises \(31-34\) give the position function \(s=f(t)\) of a body moving along the \(s\) -axis as a function of time \(t .\) Graph \(f\) together with the velocity function \(v(t)=d s / d t=f^{\prime}(t)\) and the acceleration function \(a(t)=d^{2} s / d t^{2}=f^{\prime \prime}(t)\) . Comment on the body's behavior in relation to the signs and values of \(v\) and \(a\) . Include in your commentary such topics as the following: a. When is the body momentarily at rest? b. When does it move to the left (down) or to the right (up)? c. When does it change direction? d. When does it speed up and slow down? e. When is it moving fastest (highest speed)? Slowest? f. When is it farthest from the axis origin? $$ s=t^{3}-6 t^{2}+7 t, \quad 0 \leq t \leq 4 $$

Step-by-Step Solution

Answer

Calculate derivatives for velocity and acceleration, find critical points for motion characteristics, analyze signs of velocity and acceleration to interpret behavior.

Step 1: Determine the Velocity Function

The velocity function is the derivative of the position function. Given the position function \(s = t^3 - 6t^2 + 7t\), the velocity function \(v(t)\) is calculated as follows:\[ v(t) = \frac{ds}{dt} = \frac{d}{dt}(t^3 - 6t^2 + 7t) = 3t^2 - 12t + 7 \]

Step 2: Determine the Acceleration Function

The acceleration function is the derivative of the velocity function. Thus, we differentiate \(v(t) = 3t^2 - 12t + 7\):\[ a(t) = \frac{d^2s}{dt^2} = \frac{d}{dt}(3t^2 - 12t + 7) = 6t - 12 \]

Step 3: Find When the Body is Momentarily at Rest

A body is momentarily at rest when \(v(t) = 0\). Solve \(3t^2 - 12t + 7 = 0\):Use the quadratic formula:\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Where \(a = 3\), \(b = -12\), \(c = 7\):\[ t = \frac{12 \pm \sqrt{(-12)^2 - 4 \times 3 \times 7}}{2 \times 3} = \frac{12 \pm \sqrt{144 - 84}}{6} = \frac{12 \pm \sqrt{60}}{6}\]Solve to find values of \(t\).

Step 4: Determine When the Body Changes Direction

A change in direction occurs when \(v(t)\) changes sign. Identify where the roots of \(v(t) = 3t^2 - 12t + 7 = 0\) correspond to sign changes by testing intervals around these roots.

Step 5: Determine When the Body Speeds Up or Slows Down

The body speeds up when \(v(t)\) and \(a(t)\) have the same sign and slows down when they have opposite signs. Check the sign of \(v(t)\) and \(a(t) = 6t - 12\) over the range of interest.

Step 6: Determine When the Body is Moving Fastest/Slowest

The body is moving fastest or slowest when the magnitude of \(v(t)\) is maximal or minimal. Evaluate \(v(t)\) at the critical points and endpoints \(t = 0\) and \(t = 4\).

Step 7: Determine When the Body is Farthest from the Origin

Determine when \(s(t)\) is maximized or minimized by finding the critical points of \(s(t)\). Evaluate \(s(t)\) at these points and the endpoints \(t = 0\) and \(t = 4\).

Key Concepts

Position Function

The position function represents the location of a body moving along an axis at any given time. In this context, the function given is \(s = t^3 - 6t^2 + 7t\). This polynomial function tells us where the body is on the \(s\)-axis as time, \(t\), progresses.
By plugging in different values of \(t\), you can find the body's position at those specific times. The position function is crucial for understanding motion because it forms the base from which we derive other functions like velocity and acceleration.
To graph this function accurately on a \(t\)-\(s\) plane, it is important to compute key positions of interest such as the beginning and end of the interval, and any critical points where the direction of motion changes.

Velocity Function

Velocity is the rate at which the body's position changes over time. The velocity function is the first derivative of the position function. Given the position function \(s = t^3 - 6t^2 + 7t\), the velocity function \(v(t)\) is derived as:

• \(v(t) = \frac{ds}{dt} = 3t^2 - 12t + 7\)

The roots of this velocity function help determine when the body is momentarily at rest, as this is when \(v(t) = 0\).
The sign of the velocity function indicates the direction of the body's movement:

• Positive \(v(t)\) means the body is moving to the right (up the \(s\)-axis).
• Negative \(v(t)\) means the body is moving to the left (down the \(s\)-axis).

Acceleration Function

Acceleration is the rate at which velocity changes over time and is the second derivative of the position function. For our problem, the acceleration function is derived from the velocity function \(v(t) = 3t^2 - 12t + 7\):

• \(a(t) = \frac{d^2s}{dt^2} = 6t - 12\)

The sign of the acceleration function indicates whether the body is speeding up or slowing down:

• If \(a(t)\) shares the same sign as \(v(t)\), the body speeds up.
• If \(a(t)\) has an opposite sign to \(v(t)\), the body slows down.

It is also insightful to know the instances when the acceleration is zero, as this can indicate a transition in speeding up or slowing down. This occurs here when \(t = 2\).

Derivatives

Derivatives in calculus allow us to understand change. They are pivotal when analyzing the position, velocity, and acceleration functions of a body. The first derivative of the position function yields the velocity function, quantifying the change in position over time:

• \(v(t) = \frac{d}{dt}(s)\)

The second derivative of the position function or first derivative of the velocity function gives us acceleration, which measures how velocity changes over time:

• \(a(t) = \frac{dv}{dt} = \frac{d^2}{dt^2}(s)\)

By analyzing these derivatives, we get a deeper understanding of motion dynamics, including the speed, direction, and the body's response over time.

Critical Points

Critical points are values of \(t\) where the derivative of a function equals zero or is undefined. These points are vital as they often indicate changes in a function's behavior. In this context, they help identify:

• When the velocity function \(v(t) = 0\), pointing to moments when the body is at rest.
• Where the position function \(s(t)\) is maximized or minimized, helping us find when the body is farthest from the origin.
• Potential points of direction change when velocity changes sign.

Using calculus, particularly derivatives, allows us to find these critical points and understand the nuances of motion by analyzing where shifts occur in speed and direction.