Problem 49 - Calculus Single Variable

Question

Describes the position of an object at time \(t\). Calculate the instantaneous velocity at time \(c\). $$ p(t)=\left(t^{2}+9\right) / t \quad c=2 $$

Step-by-Step Solution

Answer

The instantaneous velocity at time \( t = 2 \) is \(-1.25\).

Step 1: Understand the Definitions

The position of an object at time \( t \) is given by the function \( p(t) = \frac{t^2 + 9}{t} \). To find the instantaneous velocity at time \( c = 2 \), we need to find the derivative of the position function, which gives us the velocity function, and then evaluate it at \( t = 2 \).

Step 2: Simplify the Position Function

Simplify the expression \( p(t) = \frac{t^2 + 9}{t} \) by distributing the denominator: \( p(t) = t + \frac{9}{t} \). This makes it easier to take the derivative.

Step 3: Differentiate the Position Function

Take the derivative of \( p(t) = t + \frac{9}{t} \) with respect to \( t \). The derivative of \( t \) is \( 1 \), and the derivative of \( \frac{9}{t} \) is \( -\frac{9}{t^2} \). Thus, the velocity function is \( v(t) = 1 - \frac{9}{t^2} \).

Step 4: Evaluate the Velocity at \( t = 2 \)

Substitute \( t = 2 \) into the velocity function: \( v(2) = 1 - \frac{9}{2^2} = 1 - \frac{9}{4} = 1 - 2.25 = -1.25 \).

Key Concepts

Derivative of a Function

The concept of the derivative is central to understanding instantaneous velocity. A derivative essentially measures how a function changes as its input changes. For instance, when dealing with motion, the derivative of a position function with respect to time gives us the velocity function. In simple terms, it tells us how fast an object's position is changing at any given moment.
The process of taking a derivative involves using various rules, such as the power rule, product rule, or quotient rule, to find the derivative of functions. In the given exercise, the function \( p(t) = \frac{t^2 + 9}{t} \) is simplified to \( p(t) = t + \frac{9}{t} \). This simplification makes it easier to apply the rules of differentiation.
To find the instantaneous velocity, we needed to differentiate \( p(t) \) with respect to \( t \), resulting in \( v(t) = 1 - \frac{9}{t^2} \). Here:

• The derivative of \( t \) is \( 1 \).
• The derivative of \( \frac{9}{t} \) is \( -\frac{9}{t^2} \) using the power rule.

This process illustrates how the derivative transitions our understanding from position to velocity.

Position Function

A position function describes the exact location of an object along a particular trajectory at any given time \( t \). The position function helps us understand the object's journey through space as time progresses.
In our exercise, the position function is defined as \( p(t) = \frac{t^2 + 9}{t} \). This represents the specific path taken by the object as time varies. This function can often be simplified to make calculations easier, as was done here to \( p(t) = t + \frac{9}{t} \). This simplification shows that the position depends on two parts:

• A direct part varying as \( t \), signifying linear motion.
• A secondary part \( \frac{9}{t} \), influencing the object's movement inversely with time.

Understanding the position function is crucial because it serves as the foundation for further analysis of motion, especially when we want to calculate velocities.

Velocity Function

The velocity function is derived from the position function and indicates the rate of change of position with respect to time. It directly reveals how fast or slow an object is moving at any given moment.
Once we have simplified the position function \( p(t) = t + \frac{9}{t} \), we can derive the velocity function \( v(t) = 1 - \frac{9}{t^2} \). This expression tells us how the speed and direction (since velocity is a vector quantity) of the object varies over time. Here's the breakdown:

• The term \( 1 \) represents part of the constant speed.
• The term \( -\frac{9}{t^2} \) shows deceleration or acceleration based on the sign, affected by \( t \).

To find the instantaneous velocity at a specific time, such as \( t = 2 \) in the provided solution, we substitute this value into the velocity function \( v(2) = 1 - \frac{9}{2^2} = -1.25 \). Thus, the object is moving in the opposite direction with a rate of change of \( -1.25 \), indicating a decreasing position relative to time.