Sales The sales of a new personal computer (in thousands) are given by S(t)=100-90 e^-0.3 t wher

Sales The sales of a new personal computer (in thousands)...

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent. Such functions are widely used to model processes that involve rapid growth or decay, depending on the sign and magnitude of the exponent. In this context, the sales equation includes an exponential decay term, which captures how the influence of an initial factor decreases over time.

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the derivative of a function. The derivative represents the rate at which the function is changing at any given point. In problems like this, differentiation is used to determine the instantaneous rate of change of sales with respect to time, giving insight into how quickly sales are increasing or leveling off.

Chain Rule

The chain rule is a technique in calculus used for differentiating composite functions. It allows us to differentiate a function that is nested within another function by taking the derivative of the outer function and multiplying it by the derivative of the inner function. In this scenario, the chain rule is applied in differentiating the exponential decay term, where the exponent itself is a function of time.

Instantaneous Rate of Change

The instantaneous rate of change of a function at a specific point is given by its derivative. This concept is essential for understanding how a system evolves at any instant in time. For scenarios like sales growth or decay, calculating the derivative at certain times provides a precise measure of how quickly sales are changing at that moment.

Asymptotic Behavior

Asymptotic behavior refers to the tendency of a function as the input grows very large or approaches a particular value. In the context of exponential decay functions, this concept explains how the rate of change becomes minimal as time increases and the exponential term approaches zero, indicating that the function is approaching a steady state.

Stationary Points

Stationary points occur when the derivative of a function is zero, indicating that there is no instantaneous change at that point. Analyzing whether the rate of change of a function ever equals zero is important in understanding if and when a system reaches equilibrium or stops evolving, which is critical in many practical applications such as sales and population dynamics.

Transcript

00:01 Okay, so for number 40 in section 4 .4, we're going to a function that represents the sales of computers, and part a and b of the question are asking us to find the rate of change at certain times.

00:14 So i wrote over to the left little rules here to follow.

00:18 Anytime you're asked to find the rate of change, step one, you need to find the derivative.

00:23 And that should be your first step anytime you're asked to find the rate of change for a function.

00:30 So let me get this in blue.

00:38 So step one is to find the derivative.

00:45 And we're trying to find the derivative of the original function.

00:49 So we have s prime of t for the derivative is equal to derivative of 100.

00:55 We're not going to worry about.

00:56 And we're taking the derivative of the negative 90e to the negative .3t.

01:01 This is an exponential function that we're taking the derivative of.

01:04 So step one, we're going to write the entire exponential function as it is, and then we're going to multiply by the derivative of the exponent, which is negative 0 .3t.

01:19 This simplifies to 27e to the negative 0 .3t.

01:27 So this is satisfying step one for finding the rate of change.

01:32 The second step is going to be plug in values if needed, and that's how we're going to answer a and b for the question.

01:40 So step two is, make sure this is in blue, step two is going to be plugging in the values, and that's how we're going to answer our question.

02:03 So for a, we have after one year, so t equals 1.

02:07 We plug that in, s prime of t, which we can now.

02:13 Replace the t with 1 is equal to 27e to the negative 0 .3 times 1.

02:26 When we simplify this, you should come up with about 20.

02:30 So that means that the rate of change after a year is about 20.

02:36 For b, now we have after 5 years, so t should equal 5...