<p>Is average value the same as average rate of change? If not, then what's the difference between the two?</p>
<p>What are the different ways you can calculate average value? o_o</p>
<p>I'm so confused. I used to know this, then my calc teacher attempted to clarify our "confusion" and now I don't get it. XD</p>
<p>Thanks,
Quesce :)</p>
<p>average value is the Mean value theorem for integrals.</p>
<p>Average rate of change is the regular mean value theorem.</p>
<p>dont get these confused, the AP -always- asks these questions.</p>
<p>im assuming u understand what i mean by regular MVT and integral MVT.</p>
<p>Yeah, about the MVT: I don't really get it. I understand that it's an existence theorem and guarantees that between a & b, there is a point c such that the tangent line to pt. c has the same slope as the secant line between a and b. But what is the significance of that, and what does that have to do with rate of change?</p>
<p>On 2002 FR Calc BC: 3d.
I think that's what you are referring to. Am I right?</p>
<p>So is average rate of change the same as instantaneous rate of change? o_o</p>
<p>//edit @austin: I don't know, i'm not referring to any specific problem at the moment i'm just seeking to clarify the terminology =s</p>
<p>Or Problem 2 on 2001 Calc BC FR.</p>
<p>*correction: i'm taking AP Calc AB, so probably not a BC question :P</p>
<p>No, it's not.
Average rate of change is the Mean Value Theorem for derivative.
somebody verify me on that.
instantaneous rate of change is the derivative/slope at that specific point.
For example, lets take 2x^2.
X=0, y = 0
x=1, y= 2
x=2, y= 8</p>
<p>Now, the average rate of change would be
f(2) - f(0) / (2-0)
so 8/2, which gives 4. This is the average rate of change.
Now the mean Value theorem for derivative says that at some point x, the instantenous slope at that point will be equal to the slope of the secant line (average rate of change) so set the derivate equal to 4.
so 4x=4,
x = 1, that's the point at which the slope is equal to that of the secant line.</p>
<p>That's the mean value theorem for derivative.</p>
<p>lol austin is a god</p>
<p>Hm..what about if a question asks you to estimate the rate of change at a specific point (where you would take [f(x2) - f(x1)] / (x2-x1) ), would this be average rate of change?</p>
<p>Therefore, if I understand you correctly, average rate of change is MVT for derivatives, </p>
<p>which is f(b)-f(a) / (b-a), </p>
<p>while average value is MVT for integrals which is 1/(b-a) * integral of f(x), </p>
<p>and instantaneous rate of change is the actual derivative?</p>
<p>XD</p>
<p>lol, thank you KrOnnik</p>
<p>Yes, instantaneous rate of change is the actual derivative.</p>
<p>yes, if the question asks you to estimate the rate of change at a specific point, you would do as you said. But it depends on what data you are given. Try to pick the points (and their values) that are close to the the point at which to you have to estimate the rate of change.</p>
<p>MVT for integrals, you are right.
that's the formula.</p>
<p>Thanks austin, you rock =D</p>
<p>Cleared my confusion right up. XDD</p>
<p>:D</p>
<p>To give an example:
t (days) W(t) (C)
0 20
3 31
6 28
9 24
12 22
15 21</p>
<p>This is from an actual AP Test. it ask: Use data from the table to find an approximation for W'(12).
this is how you would do it (and how they did it):
(21-24)/(15-9) = -3/6 = -1/2
So the point is: you have to pick points that are close to the point for which you have to estimate the slope.
Because the slope can fluctuate sharply over even a slightly wider range.</p>
<p>sure, no problem.
Good luck.
I am pulling an all-nighter tonight.</p>
<p>Wow, all the best to you. Hope you do well in all your APs :D</p>
<p>As for me, I'm turning in early for the day..been starved for sleep the past few weeks or so and will need a clear mind for calc tomorrow. <em>scared</em></p>
<p>Thank you</p>
<p>Don't be scared. Calc is easy to score high on. Especially AB.
That's a good idea. But be sure to rest properly tomorrow so that your mind is fresh during the test.</p>
<p>***?? Calculus is day after tomorrow!!!</p>
<p>haha, yes .</p>
<p>Average value is just the average function value; it's just the average "y" value for the function on a given interval. To find the average value, you do this:</p>
<ol>
<li>Take the integral of the function on the indicated domain.</li>
<li>Divide by "b-a"</li>
</ol>
<p>"b" and "a" are the limits of integration ("b" is the upper limit and "a" is the lower limit). An easy way to remember this is the concept that if you take the width of the area under the curve, let's say 8 (basically meaning your domain is from 0 to 8, or 4 to 12, etc.), then there HAS to exist some number out there that if you multiple it by the width, you get the area. You're basically forming a rectangle by multiplying width (b-a) and height (average function) value; this rectangle has the same area under the curve, but you'll notice that it'll be an over-estimate in some areas and an under-estimate in other areas (it all evens out that the area of the rectangle is equal to the integral of the function). Now, if you multiply these two (width and height), you get the area, which is the same thing as the integral. These questions are asking you for the average function value, so you just move the equation around and you just divide the area by the width to get the height, which is the same thing as dividing the integral of the function on the provided domain by "b-a."</p>
<p>If the question asks you for the average rate of change, you use this equation:</p>
<h2>f(b) - f(a)</h2>
<p>b - a</p>
<p>Which is essentially finding the average slope, so to speak.</p>
<p>EDIT: Looks like I was beat to the punch quite a while ago. Sorry!</p>