Question

Suppose that there are 110 words to be learned and that k = 0.3 per hour for this student. Recall that the solution to y' = k(M − y) with y(0) = 0 is given by y(t) = M(1 − e−kt).

Answer 1

The answer describes how to use the provided differential equation to model the student's learning process of more than 100 words over time, using a specific learning rate. Numerical substitution into the provided formulation can yield the amount of words learned after a specific amount of time.

Step-by-step explanation:

This question revolves around the concept of exponential growth or decay, typically observed in the field of Mathematics, particularly in the subject of Calculus. We are provided with a differential equation y' = k(M - y) which defines the rate of memory retention or learning of some information, in this case, the learning of more than 100 words, precisely 110 words. The variable 'k' represents the learning rate, 'M' is the total content to be learned, and 'y' is the amount of content learned at a given time 't'.

Now, suppose we want to know how many words have been learned after, say, 2 hours, we need to substitute the values into given function as follows;

y(t) = M(1 - e-kt)
y(2) = 110(1 - e-0.3*2)

So, the precise number of words that our student has learned after 2 hours can be calculated using the above expression. This method allows us to model situations where the quantity of something is constantly changing in a way that is directly related to its current amount.

Learn more about Exponential Growth