Problem 70 - Calculus and its applications

Question

The rate of memorizing information initially increases. Eventually, however, a maximum rate is reached, after which it begins to decrease. Suppose another experiment finds that the rate of memorizing is given by $$M^{\prime}(t)=-0.003 t^{2}+0.2 t$$ where \(M^{\prime}(t)\) is the memory rate, in words per minute. How many words are memorized in the first \(10 \mathrm{~min}\) \((\) from \(t=0\) to \(t=10) ?\)

Step-by-Step Solution

Answer

In the first 10 minutes, 9 words are memorized.

Step 1: Define the Integration Bounds

To find out how many words are memorized in the first 10 minutes, we need to calculate the integral of the rate function from time \( t=0 \) to \( t=10 \). This means we will integrate \( M'(t) \) from \( t=0 \) to \( t=10 \).

Step 2: Set up the Integral

The formula to find the total number of memorized words is given by the integral of \( M'(t) \) with respect to \( t \, , M(t) = \int_{0}^{10} (-0.003t^2 + 0.2t) \, dt \).

Step 3: Integrate the Terms

Calculate the integral of each term separately. The integral of \(-0.003t^2\) is \(-0.003 \cdot \frac{t^3}{3}\), and the integral of \(0.2t\) is \(0.2 \cdot \frac{t^2}{2}\). So the integrated function becomes: \( M(t) = \left[-0.003 \cdot \frac{t^3}{3}\right] + \left[0.2 \cdot \frac{t^2}{2}\right] + C\).

Step 4: Simplify the Integral Function

Simplifying the integrated terms, we get \( M(t) = -0.001t^3 + 0.1t^2 + C \). Since we're calculating the number of words memorized and starting from \( t=0 \), the constant \( C \) is 0 (i.e., \( M(0) = 0 \)).

Step 5: Evaluate the Definite Integral

To find the number of words memorized from \( t=0 \) to \( t=10 \), we evaluate \( M(10) - M(0) \). Using the function we found, substitute \( t=10 \): \( M(10) = -0.001(10)^3 + 0.1(10)^2 = -1 + 10 = 9 \).

Step 6: Conclude the Calculation

Therefore, the total number of words memorized in the first 10 minutes is 9 words.

Key Concepts

Definite Integrals

Definite integrals are a fundamental concept in calculus, particularly when you want to determine the total accumulation of a quantity. In our exercise, we deal with the rate of memorizing information and want to know how many words are memorized over a time period of 10 minutes.
To calculate this, we set up a definite integral from the start time to the finish time. This integral evaluates the accumulated change in memorizing rate over this specific interval.
The bounds of the integral are crucial because they define the starting and ending points of our calculation, in this case from time \( t = 0 \) to \( t = 10 \). By solving this definite integral, we capture the total words memorized within that time span.

Rate of Change

The concept of the rate of change is central in calculus and can be particularly intuitive if you think about it in terms of everyday phenomena. In this problem, the rate at which words are memorized is given as a function of time, expressed as a polynomial: \( M'(t) = -0.003t^2 + 0.2t \).
A rate of change function like this one tells us how fast or slow a process is occurring at any given time. The negative and positive coefficients in the function inform us about the acceleration and deceleration of the memorizing rate over time.

• The negative term \(-0.003t^2\) suggests a reduction in the rate over time, indicating a slowdown.
• The positive term \(0.2t\) shows an initial increase, suggesting an initial ramp-up in memorization speed.

By analyzing this rate of change, we can predict that the memorization process speeds up initially, reaches a peak, and then starts slowing down.

Polynomial Integration

Polynomial integration involves integrating expressions made up of terms with variables raised to whole number exponents. The process itself can be broken down into integrating each term separately. In our problem, \( M'(t) \) is a polynomial function consisting of two terms: \(-0.003t^2\) and \(0.2t\).
When integrating polynomials, we use the basic rule that the integral of \( t^n \) is \( \frac{t^{n+1}}{n+1} \). Applying this rule to each term, we compute:

• The integral of \(-0.003t^2\) becomes \(-0.003 \cdot \frac{t^3}{3}\)
• The integral of \(0.2t\) turns into \(0.2 \cdot \frac{t^2}{2}\)

Bringing these together, you construct the integrated function \( M(t) = -0.001t^3 + 0.1t^2 + C \). Since we're evaluating a definite integral over a certain timeframe, the constant of integration \( C \) becomes irrelevant, simplifying our final calculation.