Problem 50 - Calculus Single Variable

Question

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

Step-by-Step Solution

Answer

The instantaneous velocity at time \(t=2\) is 26 units/time.

Step 1: Understanding the Problem

To find the instantaneous velocity of the object at time \(c=2\), we need to determine the derivative of the position function \(p(t)\) and evaluate it at \(t=2\).

Step 2: Find the Derivative of p(t)

Given the position function \(p(t) = t(t+1)(t+2)\), first expand it to make differentiating easier:\[p(t) = t^3 + 3t^2 + 2t\] Now, differentiate this polynomial with respect to \(t\):\[\frac{dp}{dt} = 3t^2 + 6t + 2\]

Step 3: Evaluate the Derivative at c=2

Now that we have the derivative \(\frac{dp}{dt} = 3t^2 + 6t + 2\), substitute \(t = 2\) to find the instantaneous velocity:\[\frac{dp}{dt}\Big|_{t=2} = 3(2)^2 + 6(2) + 2\]Calculate the result:\[= 3(4) + 12 + 2 = 12 + 12 + 2 = 26\]

Step 4: Interpret the Result

The instantaneous velocity of the object at time \(t = 2\) is \(26\) units of distance per unit of time.

Key Concepts

Derivative

The concept of a derivative is one of the cornerstone principles of calculus. It describes how a function changes as its input changes. Imagine you're tracking how fast something is moving—its velocity. To find the velocity at any given instant, we use the derivative of the position with respect to time.
To get the derivative of a function, you're essentially finding the function's rate of change or its slope at any point. This is often referred to as the function's instant rate of change. So, the derivative tells us how steep the function is at any given point.
In mathematical terms, if we have a position function \( p(t) \), its derivative is denoted as \( \frac{dp}{dt} \). This function gives us the velocity or the speed of change of the position function over time.

Polynomial Differentiation

Differentiating polynomials might sound complex, but it's a straightforward process once you get the hang of it. A polynomial, like the function \( p(t) = t(t+1)(t+2) \), is composed of terms with variables raised to various power levels.
To differentiate a polynomial, you follow specific rules. One crucial rule is the power rule: if you have a term like \( t^n \), its derivative is \( nt^{n-1} \). This rule is applied individually to each term of the polynomial.
For instance, if you expand \( p(t) \) to \( t^3 + 3t^2 + 2t \), you differentiate it like so:

• \( \frac{d}{dt} (t^3) = 3t^2 \)
• \( \frac{d}{dt} (3t^2) = 6t \)
• \( \frac{d}{dt} (2t) = 2 \)

Sum these results to get \( \frac{dp}{dt} = 3t^2 + 6t + 2 \). Differentiating a polynomial lets us understand how rapidly its value changes, pinpointing the exact change rate at any moment.

Position Function

A position function describes an object's location at any given time, akin to a snapshot of where it is over time. In our example, the position function is \( p(t) = t(t+1)(t+2) \).
This function is initially provided in factored form and gives you the position based on time \( t \). When dealing with position functions, our goal is often to understand the velocity—how quickly the object's position is changing.
Breaking the function down or expanding it, as we did to \( t^3 + 3t^2 + 2t \), often simplifies differentiation, allowing us to easily find the velocity. Evaluating the derivative, or slope, at a specific point tells us the instantaneous velocity, or how fast the object is moving at that precise moment, as we did for \( t = 2 \). This approach helps us draw an accurate picture of motion, crucial for understanding dynamic systems.