Monthly sales of a particular personal computer are expected...

Monthly sales of a particular personal computer are expected to increase at the rate of $$S^{\prime}(t)=-4 t e^{0.1 t}$$ computers per month, where $t$ is time in months and $S(t)$ is the number of computers sold each month. The company plans to stop manufacturing this computer when monthly sales reach 800 computers. If monthly sales now $(t=0)$ are 2,000 computers, find $S(t) .$ How long, to the nearest month, will the company continue to manufacture the computer?

Key Concepts

Differential Equations and Initial Value Problems

This concept involves formulating and solving equations where a function’s rate of change is given in terms of the function and possibly independent variables. With an initial value provided, integration is used to find the unique solution that describes how the quantity evolves over time.

Integration Techniques

Integration techniques, especially integration by parts, are crucial for finding antiderivatives of functions that are products of different types, such as polynomials and exponentials. This method simplifies the evaluation of integrals that are not straightforward, helping in the derivation of the original function from its rate of change.

Exponential Functions in Modeling

Exponential functions play a significant role in modeling growth and decay processes. Their properties allow them to represent continuously increasing or decreasing trends, making them a natural component in the solutions of differential equations that describe processes over time.

Solving Threshold Conditions in Mathematical Models

This concept refers to determining the point in time when a modeled quantity reaches a specific value. Setting the function equal to a threshold and solving for the variable is a common technique in applied mathematics, providing insights into when certain events or transitions occur within a modeled system.

Transcript

00:01 We are given the function s prime of t.

00:05 So s prime of t is equal to negative 4 t to the 0 .1 t.

00:13 And this is the rate of change in number of computers sold per month.

00:19 We're also given that monthly sales right now is 2000.

00:25 So s of t is the number of computers sold per month.

00:31 And now being time zero, so that would be s of zero, is equal to 2000.

00:37 So the question is, we want to find to the nearest month.

00:41 So we want to find t.

00:43 So let's write that in blue.

00:45 Want to find t when the number, when the monthly sales reach 800.

00:56 So we want to find t when s of t is equal to 800.

01:03 So what we need to do is we need to first find a function for s of t.

01:08 So s of t is going to be the integral of s prime of t.

01:19 Okay, so let's go ahead and do that.

01:21 So the integral of s prime of t, that would be negative 4 t, e to the 0 .1 t vt.

01:28 So now to find this, we can see that we're going to need to use integration by parts.

01:32 So we'll let u be equal to t, and we can let dv be equal to e to the 0 .1 t d t.

01:42 So now d, d .u is going to be dt, and b is going to be 10 times e to the 0 .1t.

01:52 Okay, now putting this into our integration by parts formula, so we're going to pull the negative four out to the front, and then open bracket.

02:01 So that will be u times v so that's 10 t b to the 0 .1 t and then we have minus the integral of vd u so that would be 10 e to the 0 .1 t vt okay then we know how to do that integral so let's copy the front part down so that would be minus and that will be 100 e to the make sorry sorry e to 0 .1t.

02:36 And then don't forget our plus c at the very end because this is an indefinite integral.

02:42 So now this ends up being, so let's multiply the negative 4th route.

02:48 So that would be negative 40t, e to the 0 .1t, plus 400e to the 0 .1t, and then plus a constant at the very end.

03:04 So this is our function for s of t.

03:08 So how do we find c? well, this is what this initial condition is, right? because we're told that s of zero is equal to 2000.

03:15 So this tells us if we put in t is equal to zero.

03:20 So if we put in s of zero, this should give us 2000...