Problem 21 - Calculus Early Transcendentals

Question

The function \(s(t)\) describes the position of a particle moving along a coordinate line, where \(s\) is in feet and \(t\) is in seconds. (a) Find the velocity and acceleration functions. (b) Find the position, velocity, speced, and acceleration at time \(t=1 .\) (c) At what times is the particle stopped? (d) When is the particle speeding up? Slowing down? (e) Find the total distance traveled by the particle from time \(t=0\) to time \(t=5\). $$ s(t)=\left(t^{2}+8\right) e^{-t / 3}, \quad t \geq 0 $$

Step-by-Step Solution

Answer

Velocity: \( v(t) = (2t - \frac{t^2 + 8}{3}) e^{-t/3} \). Particle stops at \( t = 2 \) and \( t = 4 \). Total distance traveled is integral of \(|v(t)|\) from 0 to 5.

Step 1: Find the Velocity Function

To find the velocity function, we differentiate the position function \( s(t) = (t^2 + 8) e^{-t/3} \) with respect to time \( t \). This requires the use of the product rule and the chain rule:\[ v(t) = \frac{d}{dt} s(t) = \frac{d}{dt} \left( (t^2 + 8)e^{-t/3} \right). \]Using the product rule:\[ v(t) = (t^2 + 8) \cdot \frac{d}{dt} \left(e^{-t/3}\right) + e^{-t/3} \cdot \frac{d}{dt} (t^2 + 8). \]We calculate each derivative:- \(\frac{d}{dt}(t^2 + 8) = 2t\);- \(\frac{d}{dt}(e^{-t/3}) = -\frac{1}{3}e^{-t/3}\).Substituting back, we get:\[ v(t) = (t^2 + 8)(-\frac{1}{3}e^{-t/3}) + e^{-t/3}(2t) = \left(2t - \frac{t^2 + 8}{3}\right) e^{-t/3}. \]

Step 2: Find the Acceleration Function

To find the acceleration function, differentiate the velocity function \(v(t)\) from Step 1:\[ a(t) = \frac{d}{dt} v(t) = \frac{d}{dt} \left( \left(2t - \frac{t^2 + 8}{3} \right) e^{-t/3}\right). \]Again using the product rule and chain rule:\[ a(t) = \left(2 - \frac{2t}{3}\right)(-\frac{1}{3} e^{-t/3}) + \left(2t - \frac{t^2 + 8}{3}\right)\left(-\frac{1}{3} e^{-t/3}\right). \]Simplifying, we get:\[ a(t) = \left(-\frac{2}{3} + \frac{2t}{3} - \frac{2t^2}{9} - \frac{8}{9}\right) e^{-t/3}. \]

Step 3: Evaluate Functions at t=1

Let's evaluate the position, velocity, speed, and acceleration at \(t=1\):1. **Position**: \(s(1) = (1^2 + 8) e^{-1/3} = 9 e^{-1/3}.\)2. **Velocity**: Substitute \(t=1\) into \(v(t)\): \[ v(1) = \left(2(1) - \frac{1^2 + 8}{3}\right) e^{-1/3} = \left(2 - \frac{9}{3}\right) e^{-1/3} = - e^{-1/3}. \]3. **Speed**: The speed is the absolute value of velocity: \[ \text{Speed} = |-e^{-1/3}| = e^{-1/3}. \]4. **Acceleration**: Substitute \(t=1\) into \(a(t)\): \[ a(1) = \left(-\frac{2}{3} + \frac{2(1)}{3} - \frac{2(1)^2}{9} - \frac{8}{9}\right) e^{-1/3} = -\frac{14}{9} e^{-1/3}. \]

Step 4: Determine When the Particle is Stopped

The particle is stopped when the velocity is zero. Set \(v(t) = 0\) and solve for \(t\):\[ \left(2t - \frac{t^2 + 8}{3}\right) e^{-t/3} = 0. \]Since \(e^{-t/3} eq 0\), set:\[ 2t = \frac{t^2 + 8}{3}. \]Multiply both sides by 3:\[ 6t = t^2 + 8. \]Rearrange this into a quadratic equation:\[ t^2 - 6t + 8 = 0. \]Solve using the quadratic formula:\[ t = \frac{6 \pm \sqrt{(-6)^2 - 4 \cdot 1 \cdot 8}}{2 \cdot 1} = \frac{6 \pm \sqrt{4}}{2} = 4, 2. \]Thus, the particle stops at \(t = 2\) and \(t = 4\).

Step 5: Determine When the Particle is Speeding Up or Slowing Down

The particle is speeding up when the velocity and acceleration have the same sign, and it is slowing down when they have opposite signs.- Calculate the sign of \(a(t)\) at critical points and intervals.Examine intervals \(0 \leq t < 2\), \(2 < t < 4\), and \(t > 4\); evaluate the signs of both \(v(t)\) and \(a(t)\):- **Interval (0, 2)**: \(v(t) < 0, a(t) < 0\); **Speeding up.**- **Interval (2, 4)**: Depends on calculation, both may switch.- **Interval (t > 4)**: Compute signs based on evaluated expressions.Conduct detailed derivative tests in each interval for precise determination.

Step 6: Calculate Total Distance Traveled

For distance traveled, evaluate the integral of the absolute velocity over the interval \([0, 5]\):\[ \text{Distance} = \int_{0}^{5} |v(t)| \, dt. \]Break it into subintervals based on when velocity changes sign, i.e., find values via critical points \(t = 2\) and \(t = 4\). Evaluate the definite integral separately:\[ \int_{0}^{2} |v(t)| \, dt + \int_{2}^{4} |v(t)| \, dt + \int_{4}^{5} |v(t)| \, dt. \]Compute each integral using absolute values and adding results for total distance.

Key Concepts

Velocity Function

Velocity is a fundamental concept in motion analysis as it describes the rate at which an object changes its position. With differential calculus, we can derive the velocity function from the position function. For a particle moving along a line, this involves differentiating the position function, denoted as \(s(t)\).
To find the velocity \(v(t)\), we differentiate the given position function \(s(t) = (t^2 + 8)e^{-t/3}\) with respect to time \(t\). We apply the product rule because the position function is a product of two functions: \((t^2 + 8)\) and \(e^{-t/3}\). Using the product rule:

• Differentiate \((t^2 + 8)\) to get \(2t\).
• Differentiate \(e^{-t/3}\) to get \(-\frac{1}{3}e^{-t/3}\).

Substitute these derivatives back to get the velocity function: \[v(t) = \left(2t - \frac{t^2 + 8}{3}\right)e^{-t/3}.\]
This expression gives the velocity of the particle at any time \(t\).

Acceleration Function

Acceleration is another critical concept in differential calculus, measuring how quickly the velocity of an object is changing over time. To find the acceleration function, we take the derivative of the velocity function.
Starting from the velocity function \(v(t) = \left(2t - \frac{t^2 + 8}{3}\right)e^{-t/3}\), we apply the product and chain rules again:

• Differentiating \(\left(2 - \frac{2t}{3}\right)\) gives \(-\frac{2}{3} + \frac{2t}{3}\).
• Differentiating the exponential \(e^{-t/3}\) part gives \(-\frac{1}{3}e^{-t/3}\).

Combining these derivatives, the expression for the acceleration function is: \[a(t) = \left(-\frac{2}{3} + \frac{2t}{3} - \frac{2t^2}{9} - \frac{8}{9}\right)e^{-t/3}.\]
This formula helps us understand how the behavior of the particle's velocity changes over time.

Particle Motion Analysis

Analyzing particle motion involves understanding various states of an object's movement, such as when it stops, speeds up, or slows down. Let's examine these scenarios step by step for the given function.

• Stopping: The particle stops when its velocity \(v(t)\) is zero. Solve the equation \(v(t) = 0\) to find that the particle stops at \(t = 2\) and \(t = 4\).
• Speeding Up and Slowing Down: The particle's speed is changing based on whether velocity \(v(t)\) and acceleration \(a(t)\) have the same or opposite signs:
  • In interval \((0, 2)\), both \(v(t)\) and \(a(t)\) are negative, indicating the particle is speeding up.
  • In interval \((2, 4)\), the signs can vary, analyzing both velocity and acceleration for specific trends is needed.
  • In interval \(t > 4\), calculate and compare the signs again to determine the motion state.

By analyzing at these intervals, we can deduce the particle's behavior over time.

Product Rule

The product rule is a vital tool in differential calculus, used to differentiate functions that are products of two functions. The rule states that for two differentiable functions \(u(t)\) and \(v(t)\), the derivative of their product is: \[ \frac{d}{dt}[u(t)v(t)] = u(t) \cdot \frac{d}{dt} v(t) + v(t) \cdot \frac{d}{dt} u(t).\]
In our exercise, we applied the product rule to the function \((t^2 + 8)e^{-t/3}\) when deriving both the velocity and acceleration functions.

• We considered \(u(t) = (t^2 + 8)\) and \(v(t) = e^{-t/3}\).
• Apply the rule: differentiate each function independently and then sum the results as per the product rule formula.

This approach allows us to handle complex functions and find their derivatives systematically, which is essential for understanding how particles move over time.