# Average and instantaneous rates of change

For $y=x^2+1$
Find the average rate of change of $y$ with respect to $x$ over the interval [3,5].

I did it $5=3^2+1$ =5
I think there must be formula for average rate of change but doesn't give the formula in the book.

Find the instantaneous rate of change of y with respect to $x$ at the point $x=-4$

I have no idea to start this question, the answer is =-8

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For the first question: when $x=3$ we have $y=10$, when $x=5$ we have $y=26$. The variable $x$ changed by $\Delta x=5-3=2$. The variable $y$ changed by $\Delta y=26-10=16$. So the average rate of change over this interval must be $$\frac{\Delta y}{\Delta x}=\frac{16}2=8.$$ –  Jyrki Lahtonen Apr 22 '12 at 11:40
These are very basic examples of application of integrals and derivatives. I strongly suggest that you study a bit the theory and the examples provided in class/by the textbook before you attempt to solve problems on your own. –  Andrea Mori Apr 22 '12 at 11:40
Now, Sb, I know you don't believe that $5=3^2+1$, but I don't know what you meant to write when you wrote that. Can you go back to edit your question? –  Gerry Myerson Apr 22 '12 at 13:05
Just a general piece of advice, Sb Sangpi --- unless you have decided you do not want to learn this material (which means that you have decided that you are not going to pursue math or physics beyond this class), you should really, really not just look up a formula in the book when you run across an exercise. Instead, you should do your best to try to understand why that formula is relevant to this exercise. Where does the formula come from? What is it communicating? Just looking for a formula will dull your intellect as well as make the work boring instead of interesting. –  Neal Apr 22 '12 at 13:17

You do not need to differentiate to find the average rate of change. In fact, that would rather over-complicate things.

More simply: The average rate of change = total change / length on x-axis (this comes from the definition of mean average)

When $x = 3, y = (3)^2 + 1 = 10$ When $x = 5, y = (5)^2 + 1 = 26$

Thus, the total change is $26-10 = 16$

The length on the x-axis is 2. And the average rate of change is $16/2 = 8$

To find the rate of change at a single point (instantaneous rate of change), you do need to differentiate first, then substitute $x=-4$ to find the gradient at that particular point.

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You know, you could compute the average rate of change by $\frac{1}{b-a}\int_a^b y'(\tau)d\tau$ ... :) –  Neal Apr 22 '12 at 13:15
yes :) but by the Fundamental Theorem of Calculus, that simplifies to my method... ;) –  Ronald Apr 22 '12 at 19:02