Calculus Done Right

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: \[ h(t) = - 0.25 \ \left( \frac{\text{ft}}{ \text{s}} \right) t + 30 \ \text{ft},\] like in the animation below.

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What is the rate of change of the leaf's height at any moment in time? Enter your answer as a decimal.

Rates of Change

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

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

                         

Rates of Change

Rates of change like velocity appear in all sorts of down-to-earth examples. Your wage may be in “dollars per hour”; your favorite snack may have “calories per serving” printed on the bag; your bicycle tire may be imprinted with “pounds per square inch.”

Rates measure relationships. In real life, rates like the velocity of Alice's acorn don't have to be constant. A car starts at rest, accelerates, slows down on the highway, etc. So what does it truly mean when the speedometer of an accelerating car reads 60 miles per hour?

Our goal in this unit will be to find a velocity formula for Alice's acorn. We just need some basic algebra to do it. But along the way, we'll encounter limits of difference quotients, the key idea unlocking the first major part of our course: differential calculus.

Rates of Change

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

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 and the green slider gives you control over the magnification.

True or False?

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

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Rates of Change

The last problem showed us that if we use a microscope near a point \( \big(a,f(a)\big) \)of a smooth curve \( y = f(x),\) it looks like the graph of a line. We can use the slope of this line to define the instantaneous rate of change of the function at \( x = a.\) We find a second point \( b \) very, very close to \( a\) and calculate \[ \frac{f(b)-f(a)}{b-a}, \] using the “rise-over-run” formula from algebra. This gives us an approximate value for the rate of change at \( a.\) The closer \( b \) is to \( a,\) the better the approximation.

It takes some work and a good understanding of limits to see why this works, but let's accept this on faith for now and press on.

Rates of Change

The function \( y = e^{x} \) (red) is plotted in the slider below. You can control \( (a,e^{a}) \) and \( \big(b, e^{b}\big) \) (blue), which are two points on the graph. Estimate the instantaneous rate of change of \( y = e^{x} \) at \( x = 0\) by making \( a \) and \( b\) very close together and then matching the blue line to the green.

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Rates of Change

The function \( y = x^3-x \) (red) is plotted in the slider below. You can control \( \big(a,a^3-a\big) \) and \( \big(b, b^3-b\big) \) (blue), which are two points on the graph. Estimate the instantaneous rate of change of \( y = x^3-x \) at \( x = 0\) by making \( a \) and \( b\) very close together and then matching the blue line to the green.

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Rates of Change

The function \( y = \sqrt[3]{x} \) (red) is plotted in the slider below. You can control \( \big(a,\sqrt[3]{a}\big) \) and \( \big(b, \sqrt[3]{b}\big) \) (blue), which are two points on the graph. Estimate the instantaneous rate of change of \( y = \sqrt[3]{x} \) at \( x = -1\) by making \( a \) and \( b\) very close together and then matching the blue line to the green.

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Rates of Change

To wrap up this unit, let's 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.

                         

Rates of Change

To find the instantaneous rate of change of \( h(t) \) at \( t = a \), we need to take \( b \) so close to \( a \) that they are virtually the same. More precisely, we need to find an expression for \[ \lim\limits_{b\to a} \frac{h(b)-h(a)}{b-a}, \] where the difference quotient above is given by \( -16.1 \ \left( \frac{\text{ft}}{ \text{s}^2} \right) \left( b+a \right).\)

The notation \(“\, \lim\limits_{b\to a}\,” \) just means that we are looking at the pattern of the numbers \(\frac{h(b)-h(a)}{b-a} \) as \( b \) gets very, very close to \(a.\)

What is the value of this limit as a function of \( a?\)

                         

Rates of Change

To find the formula \( v(t) = -32.2 \ \left( \frac{\text{ft}}{ \text{s}^2} \right) t \) for the velocity of Alice's acorn, we set \( b = a \) at the end of the calculation. Doing this at the beginning instead would lead to division by 0: \[ \frac{h(b) - h(a)}{b-a} \overset{b = a }{=} \frac{0}{0}.\] To get around this issue, we introduced the limit to find the velocity formula, i.e. we defined \[ v(a) = \lim\limits_{b \to a } \frac{h(b)-h(a)}{b-a}.\] Taking a limit of a quotient will be necessary for all of the rates we'll encounter in this course, so expect to put what we will learn in Chapter 2: limits to good use!

Rates of Change

To summarize, a magnified smooth curve is indistinguishable from a straight line. We can use its slope to define the instantaneous rate of change at the zoom-in point. This rate of change is approximately given by \( \frac{f(b)-f(a)}{b-a} \) for \( b\) very close to \(a,\) and is exactly given by a mysterious process we call \[\lim\limits_{b \to a } \frac{f(b)-f(a)}{b-a}. \] Limits make sense of general ratios like \( \frac{f(b)-f(a)}{b-a} \) when \( b \) approaches \( a.\) An optional quiz at the end of Chapter 3 pursues this idea. There, you can learn how the function microscope works.

Let's turn our attention now to one of the greatest applications of rates of change: finding the extreme values of a function.

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