# Thread: average velocity and Mean Value Theorem

1. ## average velocity and Mean Value Theorem

ok, couple problems:

they give you a graph of velocity [IMG]file:///C:/DOCUME%7E1/student/LOCALS%7E1/Temp/moz-screenshot-1.jpg[/IMG] [IMG]file:///C:/DOCUME%7E1/student/LOCALS%7E1/Temp/moz-screenshot-2.jpg[/IMG]
they ask: what is the average velocity of the plane during the two hour flight? - now, would I just use $\displaystyle \frac{1}{2-0}\int_{0}^{2}{f(x)dx}$
but that would find average distance though, wouldn't it? not velocity

next question:
$\displaystyle f(x) = 3x^3 - 5x^2 + 2x + 2$
Use graph of f(x) to estimate the value(s) of "c" guaranteed by Mean Value Theorem on interval [-1, 2]
Now, I know that $\displaystyle \int_{a}^{b}{f(x)dx = f(c)(b-a)}$
so set up is $\displaystyle \int_{-1}^{2}{f(x)dx = f(c)(3)}$ but then I'm stuck - should I integrate? how do I know what c is supposed to be?
the next question asks: Find the value(s) of "c" analytically.
Um ... help

2. Originally Posted by cassiopeia1289
ok, couple problems:

they give you a graph of velocity [IMG]file:///C:/DOCUME%7E1/student/LOCALS%7E1/Temp/moz-screenshot-1.jpg[/IMG] [IMG]file:///C:/DOCUME%7E1/student/LOCALS%7E1/Temp/moz-screenshot-2.jpg[/IMG]
they ask: what is the average velocity of the plane during the two hour flight? - now, would I just use $\displaystyle \frac{1}{2-0}\int_{0}^{2}{f(x)dx}$
but that would find average distance though, wouldn't it? not velocity

next question:
$\displaystyle f(x) = 3x^3 - 5x^2 + 2x + 2$
Use graph of f(x) to estimate the value(s) of "c" guaranteed by Mean Value Theorem on interval [-1, 2]
Now, I know that $\displaystyle \int_{a}^{b}{f(x)dx = f(c)(b-a)}$
so set up is $\displaystyle \int_{-1}^{2}{f(x)dx = f(c)(3)}$ but then I'm stuck - should I integrate? how do I know what c is supposed to be?
the next question asks: Find the value(s) of "c" analytically.
Um ... help
No it would find average velocity because you integrate losing a "time" but then remember you are dividing by time when you do $\displaystyle \frac{1}{b-a}$

So you have $\displaystyle \int_a^{b}f(t)dt$ where f(t) is velocity...we integrate and lose a time...to get s(t) but then to get average we divide by the 1/(b-a) giving us distance over time which is velocity.,..make sense?

3. Originally Posted by cassiopeia1289
ok, couple problems:

they give you a graph of velocity [IMG]file:///C:/DOCUME%7E1/student/LOCALS%7E1/Temp/moz-screenshot-1.jpg[/IMG] [IMG]file:///C:/DOCUME%7E1/student/LOCALS%7E1/Temp/moz-screenshot-2.jpg[/IMG]
they ask: what is the average velocity of the plane during the two hour flight? - now, would I just use $\displaystyle \frac{1}{2-0}\int_{0}^{2}{f(x)dx}$
but that would find average distance though, wouldn't it? not velocity

next question:
$\displaystyle f(x) = 3x^3 - 5x^2 + 2x + 2$
Use graph of f(x) to estimate the value(s) of "c" guaranteed by Mean Value Theorem on interval [-1, 2]
Now, I know that $\displaystyle \int_{a}^{b}{f(x)dx = f(c)(b-a)}$
so set up is $\displaystyle \int_{-1}^{2}{f(x)dx = f(c)(3)}$ but then I'm stuck - should I integrate? how do I know what c is supposed to be?
the next question asks: Find the value(s) of "c" analytically.
Um ... help
I think for the second part you are a little mixed up...you are thinking of the mean-value theorem for integrals....I think based on the first part of the question you are supposed to be using the mean-value theroem for slope that states there is at least one point c on an interval where the slope equals the average slope...so take your answer to a...and set it equal to your velocity equation and solve for c

4. these are actually two separate problems
the graph and velocity is number 11 and the second two questions are number 23
they're not related

5. Originally Posted by cassiopeia1289
these are actually two separate problems
the graph and velocity is number 11 and the second two questions are number 23
they're not related
Well then you will have to guess by the context of your chapter which mean-value theorem they mean....so what you do for your case is evaluate teh integral on teh left side of you equation then set it equal to f(c)(b-a) and solve for

6. ok, so it is the mean value theorem I was thinking beforehand

and then how do you find it analytically vs. setting them equal to each other (which I thought was analytically)

7. Originally Posted by cassiopeia1289
ok, so it is the mean value theorem I was thinking beforehand then?

and then how do you find it analytically vs. setting them equal to each other (which I thought was analytically)
$\displaystyle \int_{-1}^{2}f(x)dx=\frac{21}{4}$

now we ahve $\displaystyle \frac{21}{4}=f(c)(b-a)=f(c)(2+1)=3f(c)$

$\displaystyle f(c)=3c^3-5c^2+2c+2=\frac{7}{4}$
i divided by the three if you cant see how I got 7/4

now solve for c

8. fabulous - I got that part
thats not really my problem

its the difference between analytically verses "estimation"
what am I to look for for "estimation" - what, I mean, on the graph would be an indicator of the "c" value?
what you just showed was analytical