Problem 70 - Calculus and its applications

Question

In a memory experiment, the rate at which students memorize Spanish vocabulary is found to be given by $$M^{\prime}(t)=0.2 t-0.003 t^{2}$$ where \(M(t)\) is the number of words memorized in \(t\) minutes. a) Find \(M(t)\) if it is known that \(M(0)=0\). b) How many words are memorized in \(8 \mathrm{~min} ?\) (d)

Step-by-Step Solution

Answer

5.89 words are memorized in 8 minutes.

Step 1: Understanding the Problem

We are given the rate of memorizing vocabulary, which is the derivative of the number of words memorized with respect to time, described by the function \(M'(t) = 0.2t - 0.003t^2\). We need to find the original function \(M(t)\) when \(M(0) = 0\) and evaluate it at \(t = 8\) minutes.

Step 2: Integrate to Find M(t)

To find \(M(t)\), we need to integrate \(M'(t)\), the derivative of \(M(t)\). The integral of \(M'(t) = 0.2t - 0.003t^2\) is:\[M(t) = \int (0.2t - 0.003t^2) \, dt = 0.1t^2 - 0.001t^3 + C\] where \(C\) is the constant of integration.

Step 3: Apply Initial Condition

We use the initial condition \(M(0) = 0\) to solve for \(C\). Substituting \(t = 0\) into the equation:\[0 = 0.1(0)^2 - 0.001(0)^3 + C\] Thus, \(C = 0\). Therefore, \(M(t) = 0.1t^2 - 0.001t^3\).

Step 4: Calculate Number of Words at t = 8

Now that we have \(M(t) = 0.1t^2 - 0.001t^3\), substitute \(t = 8\) to find the number of words memorized in 8 minutes:\[M(8) = 0.1(8)^2 - 0.001(8)^3 = 0.1(64) - 0.001(512) = 6.4 - 0.512 = 5.888\] Therefore, approximately 5.89 words are memorized in 8 minutes.

Key Concepts

Rate of Change

In calculus, "rate of change" is a crucial concept that describes how one quantity changes in relation to another. This is often represented by the derivative. In the context of our problem, the rate of change is described by the function

• \(M'(t) = 0.2t - 0.003t^2\)

This derivative tells us how the number of Spanish words memorized changes over time \(t\). Imagine it as the speed at which the knowledge is acquired at any given moment. This rate is not constant; it changes as time progresses due to the quadratic term, \(-0.003t^2\), which introduces a slowing effect as time increases.
In practical terms, at the very start, the memorization might increase quickly, but as time goes on, it becomes harder to maintain that fast pace due to mental fatigue or complexity increasing.

Initial Value Problem

An initial value problem in calculus is a problem where you find a function based on its derivative and an initial condition, usually at \(t = 0\). In our exercise, we have a known rate of memorization

• \(M'(t) = 0.2t - 0.003t^2\)

and need to find the initial condition

• \(M(0) = 0\)

to determine the original function \(M(t)\).
The initial condition is a starting point that we use to solve for any constants that result from integrating the derivative. When we integrate to find \(M(t)\), we end up with

• \(M(t) = 0.1t^2 - 0.001t^3 + C\)

where \(C\) is a constant. Applying the initial condition \(M(0) = 0\), we find that \(C = 0\), reducing our function to

• \(M(t) = 0.1t^2 - 0.001t^3\)

This process shows how initial conditions are pivotal in determining the specific solution to a differential equation.

Definite Integral

In calculus, the definite integral of a function gives the total accumulation of a quantity over a specific interval. Unlike the indefinite integral which provides a general form without limits, the definite integral is bounded by specific limits and provides a numerical value.
Although our problem focuses on finding \(M(t)\) through the indefinite integration of its derivative, the concept of the definite integral indirectly relates in calculating the total amount of vocabulary memorized over time.

• For instance, if we wanted to calculate the total knowledge acquired from \(0\) to \(8\) minutes, we essentially substitute \(t = 8\) into \(M(t)\) to find \(M(8)\).
• This calculation renders \(M(8) = 5.89\), providing a precise number of words memorized in that interval.

This demonstrates how definite integrals can offer a concrete summary of how quantities build up over time, especially in real-world situations like educational experiments.