Problem 28 - Calculus

Question

Involve numbers \(\int_{0}, f_{1}, f_{2}, \ldots\) and their differences \(v_{j}=f_{j}-f_{j-1}\), They give practice with subscripts \(0, \ldots, j\). If \(f(t)=t^{2}+t,\) compute \(f(99)\) and \(f(101)\). Between those times, what is the increase in \(f\) divided by the increase in \(t\) ?

Step-by-Step Solution

Answer

The ratio of the increase in \(f\) over the increase in \(t\) is 201.

Step 1: Calculate f(99)

First, we need to plug 99 into the function \(f(t) = t^2 + t\). Calculate \(f(99)\) as follows.\[f(99) = 99^2 + 99\]Computing \(99^2 = 9801\),\[f(99) = 9801 + 99 = 9900\].

Step 2: Calculate f(101)

Now, plug 101 into the function \(f(t) = t^2 + t\). Calculate \(f(101)\) as follows.\[f(101) = 101^2 + 101\]Computing \(101^2 = 10201\),\[f(101) = 10201 + 101 = 10302\].

Step 3: Determine the Increase in f

The increase in \(f\) is the difference between \(f(101)\) and \(f(99)\). \[\Delta f = f(101) - f(99) = 10302 - 9900 = 402\].

Step 4: Calculate the Increase in t

The increase in \(t\) from 99 to 101 is:\[\Delta t = 101 - 99 = 2\].

Step 5: Compute the Ratio

Finally, compute the increase in \(f\) divided by the increase in \(t\):\[\frac{\Delta f}{\Delta t} = \frac{402}{2} = 201\].

Key Concepts

Function Evaluation

Evaluating a function involves finding the value of a function for a specific input. In the context of the given exercise, we start by identifying the function, which is given by \(f(t) = t^2 + t\). To find the values of \(f(99)\) and \(f(101)\), we substitute each number separately into the function.

• For \(f(99)\), substitute 99 for \(t\), giving \(f(99) = 99^2 + 99\).
• For \(f(101)\), substitute 101 for \(t\), resulting in \(f(101) = 101^2 + 101\).

This method of plugging in values allows us to compute the specific output values for the function at given points. Through this straightforward substitution and arithmetic calculation, we derive numerical results that provide insights into the function's behavior at these points.

Difference Quotient

The difference quotient is a fundamental concept in calculus often used to find the average rate of change of a function between two points. This is closely tied to the concept of a derivative. In this exercise, the difference quotient is used to measure how much the function \(f(t)\) changes as \(t\) increases from 99 to 101.

The difference quotient is calculated as follows:

• First, determine the change in function values, \(\Delta f = f(101) - f(99)\).
• Next, determine the change in the input values, \(\Delta t = 101 - 99\).

Then, the difference quotient can be represented by \(\frac{\Delta f}{\Delta t}\). This quotient essentially gives us the average rate of change of the function over the interval from \(t = 99\) to \(t = 101\). In this particular exercise, the computation results in \(201\), pointing to the overall change in \(f\) with respect to \(t\).

Algebraic Manipulation

Algebraic manipulation involves rewriting and solving equations to simplify or evaluate them. In the exercise, manipulation is crucial when applying operations to numbers and expressions in function evaluation and difference quotient calculations.

To evaluate \(f(99)\) and \(f(101)\), follow these steps:

• First, perform the squared operation, then add the linear component for each value of \(t\).
• Use basic arithmetic to simplify: for example, calculating \(99^2\) and adding \(99\) gives \(9900\).
• Similarly, work with \(101^2 + 101\) to result in \(10302\).

This process involves sequence execution of multiplication, addition, and subtraction to achieve clearer, simpler forms. Once these values are determined, the differences and quotients in further steps allow more complex problem-solving using the structured rearrangement of these numbers.

Calculus Concepts

Calculus concepts such as derivatives and integrals often start with understanding a function's rate of change. In this context, examining how much \(f(t)\) changes between two discrete points helps intuition regarding instantaneous changes.

Understanding the average rate of change stage involves calculating \(\frac{\Delta f}{\Delta t}\), similar to evaluating a derivative at simple discrete points. This foundational comparison serves as an introductory look into understanding how calculus will explore changes between smaller and more 'instantaneous' segments in functions.

These concepts shine light on how we can analyze situations and predict future values, building on the engine that powers many real-world mathematical applications. This exercise itself is elemental yet bridges to the broader understanding of how functions change over continuous and discrete domains.