Problem 25 - Calculus

Question

If \(f(t)=t^{2}\) then \(v(t)=2 t .\) Does the speeded-up function \(f(4 t)\) have velocity \(v(4 t)\) or \(4 v(t)\) or \(4 v(4 t) ?\)

Step-by-Step Solution

Answer

Speeded-up function has velocity 32t.

Step 1: Understanding Function Composition

The problem requires determining the velocity of the speeded-up function \(f(4t)\). First, understand that speeded-up means substituting \(4t\) into the original function \(f(t) = t^2\), so the new function becomes \(f(4t) = (4t)^2 = 16t^2\).

Step 2: Calculating the Derivative for Velocity

The velocity of a function \(g(t)\) is the derivative of \(g(t)\) with respect to \(t\). We know \(v(t) = 2t\) for \(f(t) = t^2\). For the speeded-up function \(f(4t) = 16t^2\), find the derivative: \(\frac{d}{dt}[16t^2] = 32t\).

Step 3: Comparing Velocity Expressions

Now, compare the calculated velocity \(32t\) with the provided options. We have \(v(4t) = 2(4t) = 8t\) and \(4v(t) = 8t\), and \(4v(4t) = 16t\). Note that \(32t\) matches none directly, but if multiplied by 4, it can resemble one of the options, suggesting both scaling and dependency on \(t\).

Key Concepts

Understanding Derivatives

The concept of a derivative is a cornerstone in calculus. It measures how a function changes as its input changes. In simpler terms, a derivative tells us the rate of change of a function. For example:

• If we have a function that represents the position of a car over time, the derivative of that function will represent the speed of the car at any given moment.
• Mathematically, the derivative of a function at a point is the slope of the tangent line at that point.

In the problem, the derivative of the original function, \(f(t) = t^2\), is \(v(t) = 2t\). This derivative provides us with the velocity function, which we will discuss in the next section.

Relating Velocity to Derivatives

Velocity is a term that describes the speed of an object in a specific direction. It is essentially the derivative of a position function in physics. When you have a function that represents position, like \(f(t) = t^2\), its derivative \(v(t) = 2t\) gives the velocity:

• Velocity tells us how quickly and in what direction an object's position is changing over time.
• In the context of the exercise, the velocity for \(f(4t)\) was found through derivative calculations.

So, when we derived \(f(4t) = 16t^2\), the resulting velocity function was \(rac{d}{dt}[16t^2] = 32t\). This signifies that, as we scale time with a factor in the input of our position function, we need to reconsider its output changes in terms of velocity.

Using Substitution in Calculus

Substitution is a technique used to simplify calculus problems by replacing one variable with another. In function composition, substitution involves inserting a function into another function:

• This is particularly useful in problems where a function needs to be evaluated at a new expression of the input variable.
• For the given problem, we replaced \(t\) with \(4t\) in the function \(f(t) = t^2\) to get \(f(4t) = 16t^2\).

This step is crucial because it changes the behavior of the function with respect to time, which consequently affects its derivative or velocity. This substitution method is common in many calculus problems, helping to navigate complex functions and equations.

Problem Solving in Calculus

Calculus problem-solving often requires systematic steps and a good understanding of concepts like derivatives and substitution. Here’s how to approach these problems effectively:

• Identify known elements: What is given in the problem? In this case, we know \(f(t) = t^2\) and its derivative velocity \(v(t) = 2t\).
• Look for what is unknown: Determine what you need to find, like the new velocity for \(f(4t)\).
• Apply relevant calculus concepts: Use derivatives and substitution appropriately to find the solution.
• Compare results: After solving, match your answer against provided options to verify accuracy.

In our exercise, understanding function composition and the effect of scaling inputs allowed us to calculate and compare different velocity expressions accurately. With practice, these steps become intuitive in solving calculus problems.