Problem 45 - Calculus Single Variable

Question

Describes the position of an object at time \(t\). Calculate the instantaneous velocity at time \(c\). $$ p(t)=t(t+1) \quad c=2 $$

Step-by-Step Solution

Answer

The instantaneous velocity at time \( c = 2 \) is 5.

Step 1: Understanding the Function

We are given the position function of an object: \[ p(t) = t(t+1) \]. This represents the position of the object at any time \( t \). We need to find the instantaneous velocity of the object at \( c = 2 \).

Step 2: Differentiate the Position Function

The instantaneous velocity is the derivative of the position function with respect to time, which gives us velocity as a function of time. To find the derivative \( p'(t) \), apply the product rule: \[ \frac{d}{dt}[t(t+1)] = \frac{d}{dt}[t] \times (t+1) + t \times \frac{d}{dt}[t+1].\]Using the derivatives \( \frac{d}{dt}[t] = 1 \) and \( \frac{d}{dt}[t+1] = 1 \), we get:\[ p'(t) = 1\times(t+1) + t\times 1 = t + 1 + t = 2t + 1. \]

Step 3: Calculate the Instantaneous Velocity at c = 2

Now that we have the velocity function as \( p'(t) = 2t + 1 \), substitute \( t = c = 2 \) into this equation to find the instantaneous velocity at that moment:\[ p'(2) = 2(2) + 1 = 4 + 1 = 5. \]

Key Concepts

Position Function

In the study of physics and calculus, understanding an object's position at any given time is often crucial. This is where the position function comes in. The position function typically denoted as \( p(t) \) tells us where an object is located along a particular axis, say a straight line, at any time \( t \).
Consider a runner dashing down a track. The runner's position changes over time, and this change can be modeled by a position function. For the exercise example, the function is \( p(t) = t(t+1) \). This function connects time and position directly, detailing precisely where the object will be at any given second.

• Time \( t \): The independent variable, typically measured in seconds.
• Position \( p(t) \): The dependent variable varies with time.

Such functions are foundational in computing instantaneous velocity or determining how fast an object's position changes, which is the next step in our analysis.

Differentiation

Differentiation is a concept from calculus that gives us a detailed view of how a function behaves. When we differentiate a function, we're essentially finding the rate at which it changes. For motion-related problems, this rate of change corresponds to velocity.
To differentiate a function, we use specific rules. Differentiation can turn a complex description of motion into a simpler expression, which can then be used for various calculations.

• Derivative: A function that describes the rate of change of the original function.
• Function Transform: Differentiation turns one function into another, often simpler one.

In the context of the exercise, differentiating the position function \( p(t) = t(t+1) \) helps us ascertain how the object's position changes over time. This leads us to the concept of the derivative, which we calculate by using differentiation rules like the product rule.

Derivative

The derivative is the workhorse of calculus when it comes to analyzing motion. It provides us a new function that describes the rate of change of the original function with respect to time. In this scenario, the derivative of the position function gives us the velocity function.
For a position function \( p(t) \), its derivative is noted as \( p'(t) \). It represents the instantaneous velocity or speed of the object at each moment in time.

• Velocity Function: The derivative of a position function captures how the position changes per unit of time.
• Instantaneous Velocity: The derivative evaluated at a specific time gives us the object's speed at that exact moment.

In our exercise, the derivative after applying the product rule was \( p'(t) = 2t + 1 \), which perfectly encapsulates the object's velocity at any given time \( t \). Calculating this at \( t = 2 \) provides the instantaneous speed at that moment.

Product Rule

The product rule is a fundamental tool from calculus used when differentiating functions that are products of two or more simpler functions. When a position function such as \( p(t) = t(t+1) \) is built from such a composition, the product rule comes into play to correctly find its derivative.
The formula for the product rule is critical:
\[ \left( f(t)g(t) \right)' = f'(t)g(t) + f(t)g'(t)\]Here, \( f(t) \) and \( g(t) \) are two functions multiplied together, and their derivatives are \( f'(t) \) and \( g'(t) \), respectively.

• Application: Use the rule whenever you encounter a product of functions that needs differentiating.
• Result: Gives a new simplified expression related to velocity or acceleration in physical terms.

For our task, applying this rule simplified the position function derivative into a manageable form, resulting in \( p'(t) = 2t + 1 \). This easing effect transforms complex relationships into understandable models, revealing how positions change across time.