# Rates of Change

Let's start with a problem to ease us into the rate of change concept.

Alice is enjoying a beautiful summer day sitting on the limb of an old oak tree. When she looks over, she sees a leaf fall from the branch opposite her. She measures the fall of the leaf and finds the height as function of time: $h(t) = - 0.25 \ \left( \frac{\text{ft}}{ \text{s}} \right) t + 30 \ \text{ft},$ as in the animation below.

What is the rate of change of the leaf's height at any moment in time?

## Rates of Change

### Introduction to Calculus

# Rates of Change

Later that same afternoon, Alice sees an acorn fall from the branch across from her. She now measures the height as $h(t) = -16.1 \ \left( \frac{\text{ft}}{ \text{s}^2} \right) t^2 + 30 \ \text{ft},$ as in the animation below.

Will the acorn's velocity remain constant, or will it change as a function of time?

## Rates of Change

### Introduction to Calculus

# Rates of Change

Rates of change like velocity appear in all sorts of down-to-earth examples. They measure relationships and, like the velocity of Alice's acorn, they don't have to be constant.

We'll use Alice's acorn to start us on the path to understanding rates that change with time. We just need some basic algebra to do it. Along the way, we'll encounter limits of difference quotients, the key idea unlocking the first major part of our course.

To find a velocity formula for Alice's acorn given $$h(t)$$, we need to change our perspective. Imagine we are in outer space, far from Earth.

The planet looks quite round, but if we take a spaceship ride to the surface, it looks flatter the closer we approach.

It turns out that graphs of smooth functions like $$h(t)$$ behave in a similar way: from afar, they generally have curves, but when you “zoom in” close up, they appear flat.

## Rates of Change

### Introduction to Calculus

# Rates of Change

On the left (blue), there's a plot of $$y = x^3-x$$. Think of the right plot (red) as an adjustable microscope. The red slider positions the graph under the microscope. The green slider gives you control over the magnification: zoom $$=1$$ lets you view the graph at normal distance, and taking zoom to 0 brings you closer and closer to the curve.

True or False?

No matter where we look, the magnified images $$(\text{zoom} \to 0)$$ always look like lines.

## Rates of Change

### Introduction to Calculus

# Rates of Change

Since the graph $$y=f(x)$$ looks like a line close up, the “rise-over-run” formula $\frac{f(b)-f(a)}{b-a}$ gives us the instantaneous rate of change at $$x = a$$ if $$b$$ is very, very close by. The closer $$b$$ is to $$a,$$ the better the approximation.

This difference quotient is just the slope of the line we see close up, which is also its rate of change.

The function $$y = x^3-x$$ (blue) is plotted in the slider below. You can control $$\big(b, b^3-b\big)$$ (green), which sits on a line through the origin (red) and also on the curve. Find the instantaneous rate of change of $$y = x^3-x$$ at $$a = 0$$ by making $$b$$ very close to 0 and reading off the slope.

## Rates of Change

### Introduction to Calculus

# Rates of Change

Let's use what we uncovered to find a velocity formula for Alice's acorn. Remember that $h(t) = -16.1 \ \left( \frac{\text{ft}}{ \text{s}^2} \right) t^2 + 30 \ \text{ft}.$ To find the velocity at $$t = a$$, we need to first simplify $$\frac{h(b)-h(a)}{b-a},$$ where $$b \neq a$$ is a close-by point. Choose the correct option from those provided.

Hint: $$b^2-a^2 = (b+a)(b-a).$$

## Rates of Change

### Introduction to Calculus

# Rates of Change

The instantaneous rate of change of $$h(t)$$ at $$t = a$$ is $\lim\limits_{b\to a} \frac{h(b)-h(a)}{b-a},$ where the difference quotient is given by $-16.1 \ \left( \frac{\text{ft}}{ \text{s}^2} \right) \left( b+a \right).$ The notation $$“\, \lim\limits_{b\to a}\,”$$ tells us to find the pattern of the numbers $$\frac{h(b)-h(a)}{b-a}$$ as $$b$$ gets very, very close to $$a,$$ and is called the limit.

What is the value of this limit as a function of $$a?$$

## Rates of Change

### Introduction to Calculus

# Rates of Change

We found the velocity of Alice's acorn by calculating the limit of a difference quotient: $v(a) = \lim\limits_{b \to a } \frac{h(b)-h(a)}{b-a}.$ Limits are central to any rate problem, so we'll spend a fair amount of time on them in Chapter 2.

In the remainder of this unit, our focus will be on a particularly powerful application of rates of change: finding extreme values.

Entrepreneurs are concerned with maximizing profits and minimizing losses. Engineers want to minimize shear stress and waste energy production. Chemists wish to maximize the amount of product in an expensive reaction.

Finding extreme values of a function is one of the most powerful tools calculus has to offer people working in the real world. We'll showcase only the very basic concepts here to whet your appetite for derivatives, a special kind of limit explored in great depth starting with Chapter 3.

## Rates of Change

### Introduction to Calculus

# Rates of Change

Let's revisit the function $$f(x) = x^3 - 3 x, - \sqrt{3} \leq x \leq \sqrt{3}$$ pictured below.

From the plot on the left, we see that the graph has both a maximum (highest value) and a minimum (lowest value).

Center the graph's maximum and minimum values in turn below the function microscope and zoom in. What distinguishes those two points on the curve from the others?

## Rates of Change

### Introduction to Calculus

# Rates of Change

In the last problem we used our function microscope to explore the graph near the maximum and minimum values of $f(x) = x^3-3 x,\quad - \sqrt{3} \leq x \leq \sqrt{3}.$ These appear to occur at $$\pm 1,$$ respectively, precisely where the magnified curve looks like a perfectly horizontal line.

Let's compute the rate of change at each of these points directly without the function microscope.

Simplify $$\frac{f(b)-f(a)}{b-a}$$ as much as possible.

Hint: $$b^3 - a^3 = (b-a) \big( b^2 + ab + a^2 \big)$$

## Rates of Change

### Introduction to Calculus

# Rates of Change

Through algebra we found that $$f(x) = x^3-3 x$$ implies $$\frac{f(b)-f(a)}{b-a} = b^2 + ab + a^2 -3.$$ To find the instantaneous rate of change at $$a,$$ take $$b = a.$$

Use the result to find the instantaneous rate of change at $$a = \pm 1.$$ Enter your result when you're done.

## Rates of Change

### Introduction to Calculus

# Rates of Change

Later chapters will show how we can streamline extreme value computations by taking the derivative of a function. We still have much ground to cover, but the essential idea is in place: if we are interested in finding extreme values of a function without a graph, we should look for points where the rate of change vanishes. Rates of change, which are computed with limits, are central to one of the most important calculus applications of all.

We've reached the end of the beginning. This short chapter gave us a glimpse of the fundamental ideas of calculus. Through calculus, we can compute rates of change, which give us the chance of finding extreme function values. Calculus also opens up the way to computing areas and understanding what an infinite sum truly means.

Behind all of these marvelous notions is the limit, to which we turn our attention in the next chapter.

## Rates of Change

### Introduction to Calculus

×