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Chapter 10: Problem 13

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. $$ \begin{array}{l} \text { Table } 10.2\\\ \begin{array}{l|l|l|l|l|l} \hline t & 0 & 1 & 2 & 3 & 4 \\ \hline y & 8 & & & & \\ \hline \end{array} \end{array} $$

Short Answer

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

Step by step solution

01

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 \( 0.5t \).
02

Initial Condition

The table shows that at \( t = 0 \), \( y = 8 \). This is the starting value of \( y \) when \( t \) begins at 0.
03

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, \( y(1) = 8 + 0.5 = 8.5 \).
04

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, \( y(2) = 8.5 + 1 = 9.5 \).
05

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, \( y(3) = 9.5 + 1.5 = 11 \).
06

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, \( y(4) = 11 + 2 = 13 \).
07

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Is there a value of \(n\) which makes \(y=x^{n}\) a solution to the equation \(13 x(d y / d x)=y ?\) If so, what value?

Alcohol is metabolized and excreted from the body at a rate of about one ounce of alcohol every hour. If some alcohol is consumed, write a differential equation for the amount of alcohol, \(A\) (in ounces), remaining in the body as a function of \(t\), the number of hours since the alcohol was consumed.

The amount of ozone, \(Q\), in the atmosphere is decreasing at a rate proportional to the amount of ozone present. If time \(t\) is measured in years, the constant of proportionality is \(-0.0025 .\) Write a differential equation for \(Q\) as a function of \(t\), and give the general solution for the differential equation. If this rate continues, approximately what percent of the ozone in the atmosphere now will decay in the next 20 years?

The amount of land in use for growing crops increases as the world's population increases. Suppose \(A(t)\) represents the total number of hectares of land in use in year t. (A hectare is about \(2 \frac{1}{2}\) acres.) (a) Explain why it is plausible that \(A(t)\) satisfies the equation \(A^{\prime}(t)=k A(t) .\) What assumptions are you making about the world's population and its relation to the amount of land used? (b) In 1950 about \(1 \cdot 10^{9}\) hectares of land were in use; in 1980 the figure was \(2 \cdot 10^{9} .\) If the total amount of land available for growing crops is thought to be \(3.2 \cdot 10^{9}\) hectares, when does this model predict it is exhausted? (Let \(t=0\) in \(1950 .\) )

Find particular solutions \(\frac{d B}{d t}+2 B=50, \quad B(1)=100\)
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