The registrar of Kellogg University estimates that the student enrollment in the Continuing Educ

The registrar of Kellogg University estimates that the...

Key Concepts

Substitution Method (u-substitution)

The substitution method, or u-substitution, is a technique used in integration to simplify an integral by changing variables. By substituting part of the integrand with a new variable, the integral often becomes easier to evaluate. This technique is particularly useful when dealing with composite functions or when the integrand's structure suggests a convenient change of variables.

Integration

Integration is the mathematical process used to find the original function given its derivative. In the context of a differential equation, integration is necessary to recover the function that models the quantity of interest. Evaluating the integral while applying the correct limits or constants yields the function that describes the system over time.

Growth Rate

The growth rate describes how a quantity changes over time and is represented mathematically by a derivative. In many real-world models, including enrollment or population growth, understanding the rate at which the quantity increases or decreases is critical for predicting future behavior and making informed decisions.

Differential Equations

Differential equations involve equations that relate a function to its derivatives. They are used to describe continuous change and are fundamental in modeling various phenomena, including growth rates, decay processes, and dynamics in physics, biology, economics, and engineering. The problem involves a differential equation that expresses the rate of change of enrollment over time.

Initial Value Problems

An initial value problem specifies not only a differential equation but also an initial condition, which uniquely determines a solution. The initial value is used to integrate the differential equation correctly, ensuring that the model accurately reflects the starting state of the system, such as a given enrollment level at the current time.

Transcript

00:01 So here in this question, we're going to evaluate by using integration.

00:07 So we're supposed to do this question.

00:09 We were supposed to find an expression on giving the enrollment to t years from now and also the enrollment five years from now.

00:16 So this is the expression given here.

00:18 So i can write this as this is a variable n which is dependent on t or time where n is the number of students enrolled.

00:30 So i can write it like this 1 plus 0 .2t raised to minus 3 by 2.

00:37 I just changed 10 dash t to d n by d.

00:39 Nothing else.

00:40 So here what i'm going to tell you is that we'll be doing integration by substitution.

00:44 So we're going to substitute this particular term as you.

00:48 So then what would happen is 1 plus 0 .2 can be written as u.

00:52 So by differentiating this term, we'll be getting 0 .2 d t.

00:57 This would be 0 because that's a constant is equal to d u.

01:00 So instead of t, i'd be substituting this whole term with u.

01:06 So d n by d would be coming.

01:08 So before that i'll just show you that i'm going to rearrange this particular expression by taking dn on one side and d t on the other side.

01:14 So then this would become d n is equal to 2001 plus 0 .2t raised to minus 3 by 2 d t so this would be the expression.

01:26 Now we would be replacing this particular t we have d t we have to d u so what would happen is d t would be nothing but d u by 0 .2 so this would be the change so instead of d t i could use d u by 0 .2 so then the changed thing would happen here i just write it down so then i can write d n is equal to 2 000 instead of 1 plus 0 .2t i can write it as u u raised to minus 3 by 2 instead of d t i can write d d u by 0 .2, du by 0 .2.

02:04 So i can cancel this thing and this thing that would give us 10 ,000.

02:08 So then i can write d n is equal to 10 ,000 u raised to minus 3 by 2, du.

02:18 Yeah, du...

Please give Ace some feedback