Problem 9 - Calculus Single Variable

Question

Describes the position of an object at time \(t .\) Calculate the instantaneous velocity at time \(c\). $$ p(t)=t^{2}-6 t+10 \quad c=2 $$

Step-by-Step Solution

Answer

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

Step 1: Understand the Problem

We need to find the instantaneous velocity of an object at a specific time. Here, the position function is given as \( p(t) = t^2 - 6t + 10 \), and we need to find the instantaneous velocity when \( t = c = 2 \). The instantaneous velocity is the derivative of the position function evaluated at \( t = c \).

Step 2: Derive the Position Function

The velocity as a function of time is the derivative of the position function \( p(t) \). Let's calculate the derivative \( p'(t) \) of \( p(t) = t^2 - 6t + 10 \).

Step 3: Apply Derivative Formula

Use the power rule for derivatives, where the derivative of \( t^n \) is \( nt^{n-1} \). For the function \( p(t) = t^2 - 6t + 10 \):- The derivative of \( t^2 \) is \( 2t \).- The derivative of \( -6t \) is \( -6 \).- The derivative of the constant 10 is \( 0 \).Therefore, the derivative \( p'(t) = 2t - 6 \).

Step 4: Evaluate the Derivative at t = c

Now, substitute \( t = 2 \) (since \( c = 2 \)) into the derivative function to find the instantaneous velocity. Calculate \( p'(2) = 2(2) - 6 = 4 - 6 = -2 \).

Step 5: Interpret the Result

The instantaneous velocity of the object at time \( t = 2 \) is \(-2\). This means the object is moving in the opposite direction at a speed of 2 units per time unit at \( t = 2 \).

Key Concepts

Position Function

Understanding the position function is crucial for calculating instantaneous velocity. A position function, denoted as \( p(t) \), gives us the position of an object at a specific time \( t \). This function helps us visualize how the object's position changes with time.
In the given exercise, the position function is \( p(t) = t^2 - 6t + 10 \). Here:

• \( t^2 \) represents how the position changes quadratically over time.
• \( -6t \) shows a linear change impacting the position.
• The constant \( 10 \) sets the initial position value, or the position offset.

These components combine to describe the precise location of the object for any time \( t \). Understanding this function is essential because it is the starting point for finding instantaneous velocity. To proceed, one needs to understand how to derive the function.

Derivative

A derivative is a fundamental concept in calculus representing the rate of change of a function. When we talk about the derivative of a position function, we are essentially finding a velocity function. Derivatives help us understand how a certain quantity changes instantaneously.
In this exercise, you are asked to find the derivative of \( p(t) = t^2 - 6t + 10 \). By taking the derivative, you translate the position function into a velocity function, which tells you how fast and in what direction the object is moving at any given time.
The derivative is denoted by \( p'(t) \) and its calculation requires applying specific rules such as the power rule. It is crucial to confidently apply these rules to move forward in solving calculus problems.

Power Rule

The power rule is a vital tool in calculus to find the derivative of functions quickly and efficiently. It states that the derivative of \( t^n \) is \( nt^{n-1} \). This simple rule makes calculating derivatives of polynomials a breeze.
In our exercise, we have the position function \( p(t) = t^2 - 6t + 10 \). Let's break it down:

• The term \( t^2 \): Applying the power rule gives \( 2t^{2-1} = 2t \).
• The term \( -6t \): Its derivative is simply \(-6\), since the power of \( t \) is 1.
• The constant 10: Its derivative is 0, as constants don't change with \( t \).

Applying the power rule systematically to each component of the position function leads us to our velocity function, \( p'(t) = 2t - 6 \). This is crucial for finding instantaneous velocity.

Derivative Evaluation

After finding the derivative, the next step is to evaluate it at a specific point. This evaluation tells us the instantaneous velocity of the object at a precise moment. Evaluating a derivative simply means substituting a specific value of \( t \) into the derivative function.
For this exercise, we evaluate \( p'(t) = 2t - 6 \) at \( t = 2 \) because \( c = 2 \). Let's break it down:

• Plug in \( t = 2 \) into the derivative: \( p'(2) = 2(2) - 6 \).
• Calculate: \( p'(2) = 4 - 6 = -2 \).

This value, \(-2\), represents the instantaneous velocity of the object at \( t = 2 \). It indicates that the object is moving at 2 units per time unit, but in the opposite direction. This demonstrates that derivative evaluation provides specific, actionable insights into an object's motion.