Problem 13 - Applied Calculus

Question

Fill in the missing values in Table \(10.2\) given that \(d y / d t=0.5 t\). Assume the rate of growth, given by \(d y / d t\), is approximately constant over each unit time interval. $$ \begin{array}{l} \text { Table } 10.2\\\ \begin{array}{l|l|l|l|l|l} \hline t & 0 & 1 & 2 & 3 & 4 \\ \hline y & 8 & & & & \\ \hline \end{array} \end{array} $$

Step-by-Step Solution

Answer

\( y(1) = 8.5 \), \( y(2) = 9.5 \), \( y(3) = 11 \), \( y(4) = 13 \).

Step 1: Understand the Relationship

First, recognize that the relationship given by \( \frac{dy}{dt} = 0.5t \) indicates that the rate of change of \( y \) with respect to \( t \) depends on the current value of \( t \). This means that as \( t \) increases by 1, \( y \) increases by \( 0.5t \).

Step 2: Initial Condition

The table shows that at \( t = 0 \), \( y = 8 \). This is the starting value of \( y \) when \( t \) begins at 0.

Step 3: Calculate Increment for \( t = 1 \)

At \( t = 1 \), the rate of change \( \frac{dy}{dt} = 0.5 \times 1 = 0.5 \). Therefore, from \( t = 0 \) to \( t = 1 \), \( y \) increases by 0.5. Thus, \( y(1) = 8 + 0.5 = 8.5 \).

Step 4: Calculate Increment for \( t = 2 \)

At \( t = 2 \), the rate of change \( \frac{dy}{dt} = 0.5 \times 2 = 1 \). Hence, from \( t = 1 \) to \( t = 2 \), \( y \) increases by 1. Thus, \( y(2) = 8.5 + 1 = 9.5 \).

Step 5: Calculate Increment for \( t = 3 \)

At \( t = 3 \), the rate of change \( \frac{dy}{dt} = 0.5 \times 3 = 1.5 \). Therefore, from \( t = 2 \) to \( t = 3 \), \( y \) increases by 1.5. Hence, \( y(3) = 9.5 + 1.5 = 11 \).

Step 6: Calculate Increment for \( t = 4 \)

At \( t = 4 \), the rate of change \( \frac{dy}{dt} = 0.5 \times 4 = 2 \). Thus, from \( t = 3 \) to \( t = 4 \), \( y \) increases by 2. Therefore, \( y(4) = 11 + 2 = 13 \).

Step 7: Complete the Table

Fill the calculated values into the table: at \( t = 1 \), \( y = 8.5 \); at \( t = 2 \), \( y = 9.5 \); at \( t = 3 \), \( y = 11 \); at \( t = 4 \), \( y = 13 \). This completes the table with all missing values filled in.

Key Concepts

Rate of Change

In differential calculus, the rate of change describes how one quantity varies with respect to another. Here, we consider the rate at which the variable \( y \) changes over time \( t \). The formulation \( \frac{dy}{dt} = 0.5t \) tells us exactly how \( y \) evolves as \( t \) progresses.

• The given derivative \( \frac{dy}{dt} = 0.5t \) indicates a direct relationship: as \( t \) increases, so does the rate at which \( y \) changes.
• If \( t \) is zero, the rate of change is zero, meaning there is no change in \( y \) at that instant.

A positive rate of change means that \( y \) will increase over time at a pace proportional to \( t \). By understanding this concept, you can predict and calculate how \( y \) increases step by step as \( t \) grows.

Initial Conditions

Initial conditions are crucial in solving differential equations as they provide a starting point. In our exercise, the initial condition is given as \( y = 8 \) when \( t = 0 \).

• The initial value serves as the baseline from which all future values of \( y \) are calculated.
• All subsequent calculations build on this value, using the rate of change defined previously.

Specifying \( y(0) = 8 \) helps us predict \( y \) for later times. It grounds the equation, offering a point of reference to properly interpret other results. This foundational power of setting initial conditions allows differential calculus to model real-world phenomena accurately.

Incremental Growth

Incremental growth is the step-by-step increase in a quantity over time. It focuses on how small changes combine to result in larger outcomes. When dealing with our exercise, we update \( y \) incrementally using the relationship \( \frac{dy}{dt} = 0.5t \).

• We calculate the growth of \( y \) one time unit at a time. For example, from \( t = 0 \) to \( t = 1 \), \( y \) increases by \( 0.5 \).
• Each new value of \( y \) depends on the previous one, plus the increment for that specific interval.

This requires repeating calculations for each subsequent time step, adapting to each new rate of change as \( t \) increases. Incremental growth thus lets us understand gradual changes and how they aggregate, offering thorough insights into dynamic systems.