Problem 52 - Calculus and its applications

Question

Celia's score on a test, \(s(t),\) after \(t\) hours of studying, is given by $$ s(t)=t^{2}, \quad 0 \leq t \leq 10 $$ Dan's score on the same test is given by $$ S(t)=10 t, \quad 0 \leq t \leq 10 $$ where \(S(t)\) is his score after \(t\) hours of studying. a) For \(0

Step-by-Step Solution

Answer

a) Dan scores higher. b) 73.0, Celia's average score from 7 to 10 hours. c) 80, Dan's average score from 6 to 10 hours. d) On average, Dan scores 16.7 points more than Celia.

Step 1: Determine who scores higher for 0<t<10

For Celia: Her score function is \(s(t) = t^2\). For Dan: His score function is \(S(t) = 10t\). To determine who scores higher, we compare these two functions for \(0

Step 2: Calculate average value of s(t) over [7,10]

The average value of a function \( s(t) \) over an interval \([a, b]\) is given by \[ \text{Average} = \frac{1}{b-a} \int_{a}^{b} s(t) \, dt \]. For \(s(t) = t^2\) over \([7,10]\),\[ \text{Average} = \frac{1}{10-7} \int_{7}^{10} t^2 \, dt = \frac{1}{3} \left[ \frac{t^3}{3} \right]_{7}^{10} = \frac{1}{3} \left( \frac{10^3}{3} - \frac{7^3}{3} \right) = \frac{1}{3} \left( \frac{1000}{3} - \frac{343}{3} \right) = \frac{1}{3} \times \frac{657}{3} = \frac{657}{9} = 73.0\].This represents the average test score for Celia over the 3-hour interval from 7 to 10 hours of studying.

Step 3: Calculate average value of S(t) over [6,10]

Similarly, for \(S(t) = 10t\) over \([6,10]\),\[ \text{Average} = \frac{1}{10-6} \int_{6}^{10} 10t \, dt = \frac{1}{4} \left[ 5t^2 \right]_{6}^{10} = \frac{1}{4} \left( 5 \times 10^2 - 5 \times 6^2 \right) = \frac{1}{4} \left( 500 - 180 \right) = \frac{320}{4} = 80\].This represents the average test score for Dan over the 4-hour interval from 6 to 10 hours of studying.

Step 4: Calculate average difference in scores for t in [0,10]

To find the average score difference, calculate the average values over \([0,10]\) for both functions and subtract them.\[ \text{Average of } s(t) = \frac{1}{10-0} \int_{0}^{10} t^2 \, dt = \frac{1}{10} \left[ \frac{t^3}{3} \right]_{0}^{10} = \frac{1}{10} \times \frac{1000}{3} = 33.3 \].\[ \text{Average of } S(t) = \frac{1}{10-0} \int_{0}^{10} 10t \, dt = \frac{1}{10} \left[ 5t^2 \right]_{0}^{10} = \frac{1}{10} \times 500 = 50 \].The average score difference over \([0,10]\) is \(50 - 33.3 = 16.7\). This means that on average, Dan's score is 16.7 points higher than Celia's over the full interval.

Key Concepts

Average Value Theorem

The Average Value Theorem is an essential concept in calculus. It helps us determine the average value of a function over a specific interval. The average value of a function, like in the problems for Celia and Dan, gives an overall picture of how the function behaves on that interval. For a function \( f(t) \) over an interval \([a, b]\), the average value is calculated using:
\[ \text{Average} = \frac{1}{b-a} \int_{a}^{b} f(t) \, dt \]
This formula takes the integral of the function over the interval \([a,b]\) and divides it by the length of the interval \(b-a\). The integral captures the total area under the curve of \(f(t)\) from \(a\) to \(b\), and dividing by \(b-a\) converts this total area into an average height, resembling an average value.In practical terms, calculating the average value of \(s(t) = t^2\) and \(S(t) = 10t\) over certain intervals gives us meaningful representations of Celia and Dan's studying outcomes, like average test scores over specific study times.

Quadratic Functions

Quadratic functions are mathematical expressions that feature a term in which the variable is squared. This gives the function a distinctive parabolic shape in its graph. Celia’s test score, given by \(s(t) = t^2\), is a classic example of a quadratic function.
Quadratic functions generally have three characteristic components:

• A squared term (\(t^2\) in this case)
• A linear term that may or may not be present
• A constant term

In the equation \(s(t) = t^2\), you notice it’s a pure quadratic since there is no linear or constant term, which makes it simpler as the vertex is at the origin. Quadratic functions are unique because:

• They create parabolas that can open upwards or downwards.
• They can represent a variety of real-world situations, including projectile motion and area problems.

The rapid growth of Celia's score through repeated multiplications (squaring) shows how her score increases significantly the longer she studies. This exponential growth, however, is slower initially compared to linear functions like Dan's score.

Linear Functions

Linear functions are among the simplest forms of functions, characterized by their straight line graphs. Dan's score with \(S(t) = 10t\) represents a linear function where the score increases steadily as a proportion of the time studied. This is typical of linear functions, which can be expressed generally as \(y = mx + b\), where:

• \(m\) is the slope of the line, dictating how steeply it rises or falls.
• \(b\) is the y-intercept, indicating where the line crosses the y-axis.

For Dan, \(m = 10\) and \(b = 0\), which means his score increases by 10 points for every hour he studies, starting from zero when he doesn't. Because of this direct proportionality, linear functions are straightforward:

• They map inputs directly to outputs in a one-to-one manner.
• The rate of change is constant, unlike quadratic functions that accelerate growth.

Dan’s consistent increase exemplifies this linear relationship, making predictions straightforward and interpretations clear for understanding his progress with time.

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Other exercises in this chapter

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Evaluate. $$ \int_{0}^{b} 2 e^{-2 x} d x $$

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Problem 52

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Problem 53

Evaluate. $$ \int_{0}^{b} m e^{-m x} d x $$

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Key Concepts

Average Value Theorem

The Average Value Theorem is an essential concept in calculus. It helps us determine the average value of a function over a specific interval. The average value of a function, like in the problems for Celia and Dan, gives an overall picture of how the function behaves on that interval. For a function \( f(t) \) over an interval \([a, b]\), the average value is calculated using:
\[ \text{Average} = \frac{1}{b-a} \int_{a}^{b} f(t) \, dt \]
This formula takes the integral of the function over the interval \([a,b]\) and divides it by the length of the interval \(b-a\). The integral captures the total area under the curve of \(f(t)\) from \(a\) to \(b\), and dividing by \(b-a\) converts this total area into an average height, resembling an average value.In practical terms, calculating the average value of \(s(t) = t^2\) and \(S(t) = 10t\) over certain intervals gives us meaningful representations of Celia and Dan's studying outcomes, like average test scores over specific study times.

Quadratic Functions

Quadratic functions are mathematical expressions that feature a term in which the variable is squared. This gives the function a distinctive parabolic shape in its graph. Celia’s test score, given by \(s(t) = t^2\), is a classic example of a quadratic function.
Quadratic functions generally have three characteristic components:

• A squared term (\(t^2\) in this case)
• A linear term that may or may not be present
• A constant term

In the equation \(s(t) = t^2\), you notice it’s a pure quadratic since there is no linear or constant term, which makes it simpler as the vertex is at the origin. Quadratic functions are unique because:

• They create parabolas that can open upwards or downwards.
• They can represent a variety of real-world situations, including projectile motion and area problems.

The rapid growth of Celia's score through repeated multiplications (squaring) shows how her score increases significantly the longer she studies. This exponential growth, however, is slower initially compared to linear functions like Dan's score.

Linear Functions

Linear functions are among the simplest forms of functions, characterized by their straight line graphs. Dan's score with \(S(t) = 10t\) represents a linear function where the score increases steadily as a proportion of the time studied. This is typical of linear functions, which can be expressed generally as \(y = mx + b\), where:

• \(m\) is the slope of the line, dictating how steeply it rises or falls.
• \(b\) is the y-intercept, indicating where the line crosses the y-axis.

For Dan, \(m = 10\) and \(b = 0\), which means his score increases by 10 points for every hour he studies, starting from zero when he doesn't. Because of this direct proportionality, linear functions are straightforward:

• They map inputs directly to outputs in a one-to-one manner.
• The rate of change is constant, unlike quadratic functions that accelerate growth.

Dan’s consistent increase exemplifies this linear relationship, making predictions straightforward and interpretations clear for understanding his progress with time.