Understanding Derivative Graphs: Finding f'(x) Values and Equations

[question!]

Homework Statement

http://people.hofstra.edu/stefan_waner/Realworld/calctopic1/derivgraph.html
this is the problem solution, after I saw it, I still don't understand how did they make the chart, which have f'(x) values.
What equation should I use in this case?

Homework Equations

a=dv/dt

The Attempt at a Solution

I don't know how to do it at all.. can even find the right equation

 

[Office_Shredder]

Acceleration is the derivative of velocity. So by definition acceleration IS the slope of the velocity

 

[question!]

Thank you but could you look at my question again? I think this website would explain my question better

 

[question!!]

I am not spam!

Before I ask my question, I have to declare that I am not a spam! I am trying to get help here! Who ever banned me: I AM NOT SPAM!
I was hoping somebody answered my question already! :grumpy:

Homework Statement

I post this website is only because they have a good example! there is no viruses!
http://people.hofstra.edu/stefan_waner/Realworld/calctopic1/derivgraph.html
Or you can search "Graphing the Derivative
miscellaneous on-line topics for
Calculus Applied to the Real World" on Google, and click the first one.
My question is: in the chart, they give me values of x and f'(x), but they didn't say how to get f'(x), could somebody tell me how?

Homework Equations

i don't know

The Attempt at a Solution

don't know.

 

[sylas]

My question is: in the chart, they give me values of x and f'(x), but they didn't say how to get f'(x), could somebody tell me how?

Do you know what the derivative of a function is? What is the connection between the shape of a graph for a function, and the derivative of the function?

Cheers -- sylas

 

[question!]

Acceleration is the derivative of velocity. So by definition acceleration IS the slope of the velocity

Do you know what the derivative of a function is? What is the connection between the shape of a graph for a function, and the derivative of the function?

Cheers -- sylas

Is the slope of the tangent line.. But each point I only have one (x,y), who can I find the slope?

 

[Feldoh]

Before I ask my question, I have to declare that I am not a spam! I am trying to get help here! Who ever banned me: I AM NOT SPAM!
I was hoping somebody answered my question already! :grumpy:

Homework Statement

I post this website is only because they have a good example! there is no viruses!
http://people.hofstra.edu/stefan_waner/Realworld/calctopic1/derivgraph.html
Or you can search "Graphing the Derivative
miscellaneous on-line topics for
Calculus Applied to the Real World" on Google, and click the first one.
My question is: in the chart, they give me values of x and f'(x), but they didn't say how to get f'(x), could somebody tell me how?

Homework Equations

i don't know

The Attempt at a Solution

don't know.

They did say how they got f': "Remember that f'(x) is the slope of the tangent at the point (x, f(x)) on the graph of f."

But just looking at the graph, you notice the grid pick two points that are on the tangent line and calculate the slope.

 

[Gregg]

$$y=f(x)$$

$$\frac{dy}{dx}=f'(x) =\displaystyle\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$

If you don't know to differentiate that is going to be pretty useless and even when you can obtain derivatives it's not really used, you will use techniques to differentiate that are guided by general laws on how to do so. But that does show that it is in fact the change in y divided by the change in the x to infinite accuracy so you can find the slope at any given point on the curve.

$$\frac{\Delta y}{\Delta x} \approx \frac{dy}{dx} = \displaystyle\lim_{\Delta x\to 0} \frac{\Delta y}{\Delta x}=\tan\theta$$

 

[sylas]

I think you are making it too complicated, Gregg. Look at the link that was provided.

To complete the exercise in that link, all you need to know is what "question!" has said... the derivative is the slope.

So make a rough estimate of the slope. It only has to be very rough, and the page says that different people will get different estimates. Just eyeball the graph, and make a guess. The page has places where you can type in your guess.

Cheers -- sylas

 

[Gregg]

To get the value for an estimate of the slope you need to draw a line tangent to the slope at a given x, then make a right-angled triangle and work out, or estimate, the ratio between the y side and the x side.

http://musr.physics.ubc.ca/~jess/hr/skept/Math/img43.gif

 

[sylas]

To get the value for an estimate of the slope you need to draw a line tangent to the slope at a given x, then make a right-angled triangle and work out, or estimate, the ratio between the y side and the x side.

Yes. In the http://people.hofstra.edu/stefan_waner/Realworld/calctopic1/derivgraph.html you can mentally make this image in your head and guess the slope from that.

Here's the graph for the "interactive exercise" provided as example two.
http://people.hofstra.edu/stefan_waner/Realworld/calctopic1/pics/fig3.gif

You can enter you guesses in the page for the slope at x = -3, -2, ... 1, 2, 3 and press a button to check your answers. My guesses were a bit off, but the "check your answers" button told me "close enough". If any guess is too far off, "check your answers" will tell you which one, and whether it is too large or too small (note that if the slope is negative, "too small" means too negative).

Cheers -- sylas

 

[Gregg]

Yes; although in the http://people.hofstra.edu/stefan_waner/Realworld/calctopic1/derivgraph.html it will be enough to mentally make this image in your head and guess the slope from that.

Here's the graph for the "interactive exercise" provided as example two.
http://people.hofstra.edu/stefan_waner/Realworld/calctopic1/pics/fig3.gif

You can enter you guesses in the page for the slope at x = -3, -2, ... 1, 2, 3 and press a button to check your answers. My guesses were a bit off, but the "check your answers" button told me "close enough". If any guess is too far off, "check your answers" will tell you which one, and whether it is too large or too small (note that if the slope is negative, "too small" means too negative).

Cheers -- sylas

My post wasn't a reply to yours but an answer for the poster that I wrote after actually looking at the problem.

 

[sylas]

My post wasn't a reply to yours but an answer for the poster that I wrote after actually looking at the problem.

Sure, I realize that. And I am saying that your method is good. You can use it as a mental image in your head required to answer the problem, and that is the intent of my post. I'll edit the first line of what you have quoted to make this more clear, by removing "although" and replacing "it will be enough to" with "you can".

Cheers -- sylas