Rate of change in calculus is a series I have been wanting to do for a while. If you are a student and just starting then your in the right place. This is the very beginning of calculus and where you want to start your learning. Since there are so many topics in school that require some calculus, I felt it was important to get some basics up here for you.

**Beginning Calculus : Rate of Change**

In the examples that are to follow I try to provide any explanations that seem necessary. I want to be clear and comment what I do and why I do it. I am hoping this method will help others understand. If it seems a little long or self-explanatory then I apologize. My goal is to be consistent and thorough. We are dealing with functions in this first chapter and the rate at which they change.

To get the most out of these examples I would encourage you to try the problem first by yourself. If you have issues then refer to how it is worked. Hopefully that, along with my comments, will fill in any gaps that you might have in your knowledge.

**Problem 1.**

Find the average rate of change of the function over the given intervals. \[f(x)= 3x+3\] Using the intervals: [2,4] [-2,2]

Use the given function and the first interval.

You are evaluating the function at both points in order to find the difference.

Once you do that you are dividing by the difference of the interval length.

For the interval [2,4]

\[ \Longrightarrow \frac{f(4)-f(2)}{4-2} \]

Substitute in all of your values to find the rate of change.

\[ \Longrightarrow \frac{((3)(4)^3 +3)-((3)(2)^3 +3)}{4-2} = 84 \]

For the interval [-2,2]

\[ \Longrightarrow \frac{f(2)-f(-2)}{2+2}\]

Evaluate the equation with those values.

\[ \Longrightarrow \frac{((3)(2)^3+3)-((3)(-2)^3+3)}{2+2} = 12\]

**Problem 2.**

Find the average rate of change of the $R\theta$ function over the given interval: [0,8]

\[ \Longrightarrow r \theta = \sqrt{3\theta+1} \]

Using this formula, we want to use the intervals given to us and evaluate them into the function. How this is done is simple. In the numerator we are using the intervals within the function itself. In the denominator, we just have the interval difference. Then we are just dividing the two.

\[ \Longrightarrow \frac{r \theta_2 – r \theta_1}{\theta_2 – \theta_1} \]

I will now go ahead and set up the equation.

\[ \Longrightarrow \frac{r (8) – r (0)}{8-0} \]

Now just substitute the intervals into the formula and start evaluating.

\[ \Longrightarrow \frac{\sqrt{3*8+1}-\sqrt{3*0+1}}{8-0} \]

This gives us a simple equation to solve.

\[ \Longrightarrow \frac {5-1}{8-0} \]

We are left with :

\[ \Longrightarrow \frac {1}{2} \]

This is your rate of change.

**Problem 3.**

Find the slope of the curve at the given point and an equation of the tangent line at that point. $$(-2,-9)$$

\[ y=7-4x^2 \]

Approach this problem as change in y divided by change in x.

\[ \Longrightarrow \frac{(7-4(-2+h)^2) -( 7-4(-2)^2)}{h} \]

Distribute everything out.

\[ \Longrightarrow \frac{16h-4h^2}{h} \]

Now simplify and set h=0.

The slope at (-2,-9) is: 16

Now to find the equation of the tangent line. You use this formula for that:

\[ \Longrightarrow y-y_1 = m(x-x_1) \]

Substitute in your points and you get:

\[ \Longrightarrow y-(-9) = 16(x-(-2) \]

Simplify a bit.

\[ \Longrightarrow y+9 = 16(x+2) \]

Keep simplifying. This is the equation of the tangent line.

\[ \Longrightarrow y = 16x + 23 \]

**Problem 4.**

Find the slope of the curve at the given point by finding the limiting value of the slope of the secants through the point. Also find an equation of the tangent line to the curve at the same point.

\[ y = x^3 – 6x \]

Point = (1,-5)

This is just another form of what we have done before. You have an equation and a point to go by. Put the point into the function and evaluate it.

\[ \Longrightarrow \frac{(1+h)^3 – 6(1+h) – (1)^3 6(1)}{h} \]

Now distribute. Pay careful attention to where everything goes as it is a lot to deal with.

\[ \Longrightarrow \frac{(1+3h+3h^2 +h^3-6-6h) – (-5)}{h} \]

Reduce it down some.

\[ \Longrightarrow \frac{h^3 + 3h^2 -3h}{h} \]

Simplify it now.

\[ \Longrightarrow h^2 + 3h – 3 \]

Then just assign h=0 and see what you have left. That is your slope of course as h approaches 0.

\[ \Longrightarrow = -3 \]

Now to find the equation of the tangent line.

We do the same process as before. We have the point-slope equation.

\[ \Longrightarrow y-y_1 = m(x-x_1) \]

Use your slope that you got above. Then put in your point when it asks for an x or a y.

\[ \Longrightarrow y-(-5) = -3(x-1) \]

Simply the equation.

\[ \Longrightarrow y + 5 = -3x + 3 \]

Move everything to one side of the equal sign. This is your equation of the tangent line.

\[ \Longrightarrow y = -3x – 2 \]

**Conclusion**

This could be your first exposure to calculus and limits. If it is then it can certainly look strange. What you do and when you do it are not always clear. The best way it to just examine problems and how to do them. Work as many as you can after that. Problems can be easy and they can be hard. Seeing several of each kind will help. The main thing to take away from this lesson is that you are finding slope and equations. Hopefully you can see how easy it is. It is just one step at a time. Don’t get intimidated if you are new. Take it step by step and enjoy the easy parts!

Perry G Ramsey says

I think you were doing fine until you got to here:

“Now simplify and set h=0.”

At this point you leapfrog over the concept of a limit and rush to an answer without explaining how you got there. It’s not that you are setting h=0, it’s that you’re letting h get as close to zero as you like, and at some point it ceases to matter that it’s not exactly zero.

Building on the foundation of Problems 1 and 2, make problem 3a “find the slope of the function y=7−4x^2 between -2 and 0. Following the same process as before, one would get 8. Then between -2 and -1 it’s 12. Put a slight twist on that: calculate the slope between -2 and -2+1. That’s exactly the same problem. Now make the leap to the abstract: calculate the slope between -2 and -2+h, where h=1. Again, exactly the same problem, but you’ve now derived that the slope is 16-4h for any arbitrary h. Now what happens if the interval is 1/2. or 1/4. What if h=1/1 000 000? Or 10^-100? This process says that you could have picked as tiny an interval as you wanted, and the tinier the interval the closer the result comes to 16.

The notion of a limit is important in itself, and one of the most useful and understandable uses is the derivation of the derivative. Don’t just leap over that intellectual canyon with your eyes closed. Take a few moments to explore it. The benefits are tremendous.

Jason Moore says

Yes you are entirely right. That was definitely a brain dead moment. Thanks for the comment, i enjoyed reading it.