Problem 8 - 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. $$ s(t)=\frac{t^{3}}{t^{3}+1} ; \quad t=1 $$

Step-by-Step Solution

Answer

At \(t=1\), the body is at position \(\frac{1}{2}\) feet along the coordinate line, its velocity is \(\frac{3}{2}\) feet per second, and its speed is also \(\frac{3}{2}\) feet per second.

Step 1: Find the position at time \(t=1\)

We need to find the position of the body at \(t=1\) by evaluating the function \(s(t)\) at \(t=1\). So, we will replace \(t\) with \(1\) in the function: $$ s(1) = \frac{1^3}{1^3+1} = \frac{1}{2} $$ At \(t=1\), the body is at position \(\frac{1}{2}\) feet along the coordinate line.

Step 2: Calculate the velocity function

To calculate the velocity function, we need to find the time derivative of the position function, \(s(t)\). We will use the quotient rule to differentiate: $$ \frac{d}{dt}\left(\frac{t^3}{t^3+1}\right)=\frac{(t^3+1)\cdot(3t^2)-(t^3)\cdot(3t^2)}{(t^3+1)^2} $$ Simplifying the derivative, we get: $$ v(t)=\frac{3t^2(t^3+1)-3t^5}{(t^3+1)^2} $$ Now, we have the velocity function, \(v(t)\).

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

Now, we will find the velocity of the body at \(t=1\) by evaluating the velocity function, \(v(t)\), at \(t=1\): $$ v(1)=\frac{3(1)^2((1)^3+1)-3(1)^5}{((1)^3+1)^2}=\frac{6}{4}=\frac{3}{2} $$ At \(t=1\), the body has a velocity of \(\frac{3}{2}\) feet per second.

Step 4: Calculate the speed at time \(t=1\)

To calculate the speed of the body at time \(t=1\), we will take the absolute value of the velocity at that time: $$ \text{Speed at } t=1 = |v(1)|=|\frac{3}{2}|=\frac{3}{2} $$ So, the speed of the body at \(t=1\) is \(\frac{3}{2}\) feet per second. In conclusion, at \(t=1\), the body is at position \(\frac{1}{2}\) feet along the coordinate line, its velocity is \(\frac{3}{2}\) feet per second, and its speed is also \(\frac{3}{2}\) feet per second.

Key Concepts

Position Function

A position function, often denoted as \(s(t)\), represents the location of a moving object along a path, at any given time \(t\). The position function is a fundamental concept in calculus used to determine where an object is located in relation to a reference point. In our example, the function is given by:\[s(t) = \frac{t^3}{t^3 + 1}\]This function tells us how the position of the body changes with time, where \(t\) is measured in seconds and \(s(t)\) in feet. To find the exact position of the body at a specific time, you can substitute the time value into the position function. For instance, when \(t = 1\), the position calculated is \(\frac{1}{2}\) feet, indicating the body's location on the coordinate line.

Velocity Function

The velocity function, denoted as \(v(t)\), provides insight into how fast and in which direction an object's position is changing with respect to time. It is derived by taking the derivative of the position function with respect to time \(t\). In our case, the derivative involves using the chain rule and quotient rule. Calculating the derivative of the position function \(s(t) = \frac{t^3}{t^3 + 1}\) involves:

• The quotient rule, which is \(\frac{u}{v}\)'s derivative: \((v\cdot u' - u \cdot v')/(v^2)\).
• Applying this to \(s(t)\), we get:\[v(t) = \frac{3t^2(t^3 + 1) - 3t^5}{(t^3 + 1)^2}\]

Evaluating this velocity function at \(t = 1\) results in \(v(1) = \frac{3}{2}\) feet per second, indicating not only the speed but also the direction of the body's movement at that moment.

Quotient Rule

The quotient rule is an essential tool in calculus for differentiating functions that are expressed as the division of two other functions. It's a formula used when you encounter a ratio or a fraction of two differentiable functions. For a function given by \(\frac{u(t)}{v(t)}\), where both \(u(t)\) and \(v(t)\) are differentiable functions, the quotient rule states:\[\left(\frac{u}{v}\right)' = \frac{v(t)\cdot u'(t) - u(t)\cdot v'(t)}{v(t)^2}\]In our specific example, \(u(t)\) is \(t^3\) and \(v(t)\) is \(t^3 + 1\). Applying the quotient rule helps differentiate the position function to find the velocity function. Remember that correctly identifying \(u\) and \(v\), and calculating their derivatives accurately, is crucial for applying the quotient rule effectively.

Speed Calculation

The speed of an object is the magnitude of its velocity and gives us an idea of how fast the object is moving regardless of its direction. Speed is always a non-negative value, representing the absolute value of the velocity. It is calculated using the formula: \[\text{Speed} = |v(t)|\]In our problem, once the velocity at \(t = 1\) was determined to be \(\frac{3}{2}\) feet per second, the speed is simply the absolute value of this velocity: \[\text{Speed at } t=1 = \left| \frac{3}{2} \right| = \frac{3}{2}\]This means that at \(t = 1\) second, the body moves at a speed of \(\frac{3}{2}\) feet per second. Speed gives us practical and real-world insight, such as understanding how quickly an object travels across a certain distance.