Problem 51 - Calculus and its applications

Question

In a memory experiment, Alan is able to memorize words at the rate (in words per minute) given by $$ m^{\prime}(t)=-0.009 t^{2}+0.2 t $$ In the same memory experiment, Bonnie is able to memorize words at the rate given by $$ M^{\prime}(t)=-0.003 t^{2}+0.2 t $$ a) How many more words does the person whose memorization rate is higher memorize from \(t=0\) to \(t=10\) (during the first \(10 \mathrm{~min}\) of the experiment)? b) Over the first 10 min of the experiment, on average, how many words per minute did Alan memorize? c) Over the first 10 min of the experiment, on average, how many words per minute did Bonnie memorize?

Step-by-Step Solution

Answer

Bonnie memorized 2 more words than Alan; Alan averaged 0.7 words/min, and Bonnie averaged 0.9 words/min.

Step 1: Integrate to Find Memorization Function for Alan

Alan's memorization rate is given by \( m'(t) = -0.009t^2 + 0.2t \). We need to find his memorization function \( m(t) \) by integrating \( m'(t) \) from \( t = 0 \) to \( t = 10 \).\[ m(t) = \int m'(t) \, dt = \int (-0.009t^2 + 0.2t) \, dt \]This results in:\[ m(t) = -0.003t^3 + 0.1t^2 + C \]Using \( m(0) = 0 \), we find \( C = 0 \), so \( m(t) = -0.003t^3 + 0.1t^2 \).

Step 2: Integrate to Find Memorization Function for Bonnie

Bonnie's memorization rate is given by \( M'(t) = -0.003t^2 + 0.2t \). We find her memorization function \( M(t) \) by integrating \( M'(t) \) from \( t = 0 \) to \( t = 10 \).\[ M(t) = \int M'(t) \, dt = \int (-0.003t^2 + 0.2t) \, dt \]This results in:\[ M(t) = -0.001t^3 + 0.1t^2 + C \]Using \( M(0) = 0 \), we have \( C = 0 \), so \( M(t) = -0.001t^3 + 0.1t^2 \).

Step 3: Calculate Total Words Memorized by Alan from 0 to 10

Substitute \( t=10 \) into the function \( m(t) = -0.003t^3 + 0.1t^2 \) to find the total words memorized by Alan:\[ m(10) = -0.003(10)^3 + 0.1(10)^2 = -0.003(1000) + 0.1(100) = -3 + 10 = 7 \]Alan memorized 7 words in the first 10 minutes.

Step 4: Calculate Total Words Memorized by Bonnie from 0 to 10

Substitute \( t=10 \) into the function \( M(t) = -0.001t^3 + 0.1t^2 \) to find the total words memorized by Bonnie:\[ M(10) = -0.001(10)^3 + 0.1(10)^2 = -0.001(1000) + 0.1(100) = -1 + 10 = 9 \]Bonnie memorized 9 words in the first 10 minutes.

Step 5: Calculate the Difference in Words Memorized

Bonnie memorized 9 words while Alan memorized 7 words in the first 10 minutes. The difference is:\[ 9 - 7 = 2 \]Bonnie memorized 2 more words than Alan in the first 10 minutes.

Step 6: Calculate Alan's Average Memorization Rate

To find Alan's average memorization rate over the first 10 minutes, divide the total words memorized by the time interval:\[ \text{Average rate} = \frac{7}{10} = 0.7 \text{ words per minute} \]

Step 7: Calculate Bonnie's Average Memorization Rate

To find Bonnie's average memorization rate over the first 10 minutes, divide the total words memorized by the time interval:\[ \text{Average rate} = \frac{9}{10} = 0.9 \text{ words per minute} \]

Key Concepts

Memory Experiment

In this exercise, Alan and Bonnie are participating in a memory experiment where they each memorize words over a period of time. The rates at which they memorize words are represented by polynomial functions.

The function depicting Alan's memorization rate is given by \(-0.009t^2 + 0.2t\), while Bonnie's is \(-0.003t^2 + 0.2t\).

These functions showcase how their memorization speed changes over time. Understanding these rates is crucial for comparing their memory performances over the first 10 minutes.

Integration

In calculus, integration is the process used to find the antiderivative, or the original function, from its derivative. Here, integration helps determine the total number of words memorized over a specific timeframe.

When you integrate a rate of change function, like the memorization rates for Alan and Bonnie, you find the total amount of change, or accumulation, over a given period.

To obtain Alan's memorization function, integrate \(m'(t) = -0.009t^2 + 0.2t\): \[m(t) = \int (-0.009t^2 + 0.2t) dt = -0.003t^3 + 0.1t^2 + C\] For Bonnie, integrate \(M'(t) = -0.003t^2 + 0.2t\): \[M(t) = \int (-0.003t^2 + 0.2t) dt = -0.001t^3 + 0.1t^2 + C\]

The constant \(C\) is determined by known initial conditions, such as the initial word count being zero at time zero. This yields \(C = 0\), so Alan's and Bonnie's functions sum up their total memorized words from time \(0\) to \(10\).

Average Rate of Change

The concept of the average rate of change helps to find how much, on average, a quantity changes over a particular interval. It is akin to finding an average speed when given a distance over time.

For this experiment, the average rate of change gives us the average number of words memorized per minute over the 10-minute span.

For Alan, who memorized 7 words in 10 minutes, his average rate is: \[ \text{Alan's average rate} = \frac{7}{10} = 0.7 \text{ words per minute} \]For Bonnie, who memorized 9 words: \[ \text{Bonnie's average rate} = \frac{9}{10} = 0.9 \text{ words per minute} \]

These averages provide a simplified characterization of their performance throughout the experiment's duration.

Polynomial Functions

Polynomial functions are mathematical expressions that involve powers of the variable \(t\). In this experiment's context, the memorization rates of Alan and Bonnie are both polynomial functions.

Alan's memorization rate function is: \(-0.009t^2 + 0.2t\) and Bonnie's is \(-0.003t^2 + 0.2t\).

The second-degree term, \(t^2\), suggests that the rate of memorization changes over time, not remaining constant.

These functions not only help visualize the rate at which words are memorized but also allow the prediction of performance over time through integration.

Both functions demonstrate different concavities, with each curve indicating the acceleration or deceleration of memorization rates quantitatively.