Problem 8 - Calculus Single Variable

Question

Describes the position of an object at time \(t .\) Calculate the instantaneous velocity at time \(c\). $$ p(t)=2 t^{3}-3 t^{2} \quad c=-1 $$

Step-by-Step Solution

Answer

The instantaneous velocity at \( t = -1 \) is \( 12 \).

Step 1: Understand the Problem

We need to find the instantaneous velocity of an object at a specific time \( t = c \), where \( c = -1 \). The object's position is given by the function \( p(t) = 2t^3 - 3t^2 \). Instantaneous velocity is the derivative of the position function evaluated at \( t = c \).

Step 2: Differentiate the Position Function

To find the instantaneous velocity, we differentiate the position function \( p(t) = 2t^3 - 3t^2 \). Using basic differentiation rules, the derivative \( p'(t) \) is found as follows:1. The derivative of \( 2t^3 \) is \( 6t^2 \).2. The derivative of \( -3t^2 \) is \( -6t \).Therefore, the derivative of the position function is:\[ p'(t) = 6t^2 - 6t \]

Step 3: Evaluate the Derivative at \( t = c \)

Now substitute \( t = -1 \) into the derivative \( p'(t) \) to find the instantaneous velocity at \( t = c \):\[ p'(-1) = 6(-1)^2 - 6(-1) = 6(1) + 6 = 6 + 6 = 12 \]

Step 4: Conclusion

The instantaneous velocity of the object at \( t = -1 \) is \( 12 \). This means the object is moving at a velocity of \( 12 \) units per time period at that specific moment.

Key Concepts

Derivative

In calculus, a derivative represents how a function changes as its input changes. Think of it as the rate of change or the slope of the function at any given point. If you imagine a curve on a graph, the derivative at a particular point gives you the slope of the tangent line to that curve at that point.
For differentiating a simple power function like the one given in the problem, say \( f(t) = t^n \), the derivative is obtained using the formula: \( f'(t) = nt^{n-1} \). This means you multiply by the original exponent, then subtract one from the exponent. Easy, right?
To see how derivatives are used, consider our function \( p(t) = 2t^3 - 3t^2 \). We differentiate each term separately, just like peeling off layers. For example:

• The term \( 2t^3 \) becomes \( 6t^2 \).
• The term \( -3t^2 \) becomes \( -6t \).

This gives us the derivative \( p'(t) = 6t^2 - 6t \). This new function tells us exactly how fast the original function changes at every point \( t \).

Position Function

In the context of physics and motion, the position function tells us where an object is at a certain time. It's simply a mathematical formula that describes the position of an object as a function of time.
In our problem, the position function is \( p(t) = 2t^3 - 3t^2 \). Here, every value of \( t \) helps us determine where the object is along its path.
When we want to know about how fast the object is moving at a specific point in time, we need to derive this position function. By finding its derivative, we essentially determine an object's instantaneous velocity or speed at any given point.
So, the structured sequence goes like this: you start with a position function, differentiate it to get the velocity function, and then you evaluate it to find instantaneous values. Understanding how the position changes will help in predicting the motion behavior.

Differentiation Rules

Differentiation rules are guidelines that help us compute derivatives quickly and correctly. They are essential for problems involving calculus, especially those calculating instantaneous rates such as velocity.
The key rules to remember are:

• Power Rule: If \( f(t) = t^n \), then the derivative \( f'(t) = nt^{n-1} \). This is one of the most common rules and was used to differentiate the position function in our problem.
• Constants Rule: The derivative of a constant is 0. So, if a position function had a constant term (like \( +5 \)), its derivative would simply ignore that constant in calculations.
• Sum/Difference Rule: The derivative of a sum or difference of functions is the sum or difference of their derivatives. This rule is why we can differentiate \( 2t^3 - 3t^2 \) term by term.

With these rules, differentiating even complex functions becomes a simpler task. All you do is apply the relevant rule to each piece of the function, adding them up to get the full derivative. Mastering these rules aids greatly in solving real-world problems, like determining how fast something is moving at any instant in time.