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. 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.

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. The green slider gives you control over the magnification: zoom \( =1\) let's 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.

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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.

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. What is the approximate instantaneous rate of change of \( y = e^{x} \) at \( x = 0?\) You can estimate by making \( a \) and \( b\) be 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

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

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

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, and so we'll spend a fair amount of time on them in Chapter 2.

In the next unit we'll put rates of change to use finding extreme values of a function, one of the great applications of calculus.

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