Problem 7 - Single Variable Calculus

Question

In Exercises, \(s(t)\) is the position function of a body moving along a coordinate line; \(s(t)\) is measured in feet and \(t\) in seconds, where \(t \geq 0 .\) Find the position, velocity, and speed of the body at the indicated time. $$ s(t)=\left(t^{2}-1\right)^{2} ; \quad t=1 $$

Step-by-Step Solution

Answer

At time \(t=1\), the position of the object is 0 feet, the velocity is 0 feet per second, and the speed is also 0 feet per second.

Step 1: Find the Position at \(t = 1\)

Plug in the value \(t = 1\) into the position function to find the position at this time: $$ s(1) = (1^2 -1)^2 = (0)^2 = 0 $$ So, the position of the object at time \(t = 1\) is 0 feet.

Step 2: Find the Velocity Function

To find the velocity function, take the derivative of the position function with respect to time: $$ v(t) = \frac{ds(t)}{dt} = \frac{d}{dt}[(t^2 -1)^2] $$ Here, we can use the chain rule for differentiation. Let \((t^2 - 1) = u\), then we have \(s(t) = u^2\). So, we need to find \(\frac{du}{dt}\) and \(\frac{d(u^2)}{du}\): $$ \frac{du}{dt} = \frac{d}{dt}(t^2 - 1) = 2t $$ and $$ \frac{d(u^2)}{du} = 2u $$ Now, by using the chain rule, we have $$ \frac{ds(t)}{dt} = \frac{d(u^2)}{du} \cdot \frac{du}{dt} = 2u \cdot 2t = 2(t^2 - 1) \cdot 2t = 4t(t^2 - 1) $$ Thus, the velocity function is $$ v(t) = 4t(t^2 - 1) $$

Step 3: Find the Velocity at \(t = 1\)

Plug in the value \(t = 1\) into the velocity function: $$ v(1) = 4(1)(1^2 - 1) = 4(0) = 0 $$ So, the velocity of the object at time \(t = 1\) is 0 feet per second.

Step 4: Find the Speed at \(t = 1\)

Find the speed at time \(t = 1\) by taking the absolute value of the velocity at \(t = 1\): $$ \text{Speed} = |v(1)| = |0| = 0 $$ So, the speed of the object at time \(t = 1\) is 0 feet per second. #Result# At time \(t=1\), the position of the object is 0 feet, the velocity is 0 feet per second, and the speed is also 0 feet per second.

Key Concepts

Position Function

The position function, denoted as \(s(t)\), gives the location of a moving object along a coordinate line at any time \(t\). In our exercise, the position function \(s(t) = (t^2 - 1)^2\) measures the location in feet while \(t\) is measured in seconds. Understanding how to evaluate this function is the first step in analyzing movement.

To find the position at a specific time, you simply plug the value of \(t\) into the position function. For this problem, when we evaluated the position at \(t = 1\), we got:

• \(s(1) = (1^2 - 1)^2 = 0^2 = 0\)


• The object is at 0 feet at \(t = 1\).

Knowing the position helps us understand where the object is on the coordinate line at any given time.

Velocity Function

The velocity function is an essential concept when studying motion. It represents how quickly the position changes with respect to time. In mathematical terms, it's the derivative of the position function \(s(t)\).

Taking the derivative of \(s(t) = (t^2 - 1)^2\) gives us the velocity function:

• \(v(t) = \frac{d}{dt}[(t^2 - 1)^2]\)


• Using the chain rule, we find \(v(t) = 4t(t^2 - 1)\).

The velocity can be positive, negative, or zero, indicating the direction of motion. For our specific case at \(t = 1\), we find:

• \(v(1) = 4(1)(1^2 - 1) = 0\)
• The object is momentarily at rest.

Knowing the velocity function allows us to determine how the object's speed and direction change over time.

Chain Rule

The chain rule is a powerful tool in calculus used to differentiate composite functions. It allows us to find the derivative of a function composed of two or more functions. In the exercise, the position function \((t^2 - 1)^2\) is an example of a composite function.

To apply the chain rule, follow these steps:

• Identify the inner function, in this case, \(u = t^2 - 1\).
• The outer function is \(u^2\).
  
• Differentiate the outer function: \(\frac{d(u^2)}{du} = 2u\).
• Differentiate the inner function: \(\frac{du}{dt} = 2t\).


• Combine them: \(\frac{d}{dt}[(t^2 - 1)^2] = 2u \cdot 2t = 4t(t^2 - 1)\).

This approach breaks down complex differentiation into manageable parts, simplifying the process.