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Second derivatives...

Started by December 31, 2002 08:57 PM
20 comments, last by LewieM80 22 years, 1 month ago
If a derivative is really the slope of a tangent, what does it mean to pull off a second derivative on a parametric funciton... This book I have doesn''t really explain it too well and it only gave me one problem to solve. So it''s still very vague... input? Thanx! -Lewis [m80] Play QUADz MX @ www.m80produxions.com
Lewis [m80]Interactive Designerhttp://ismstudios.com
Think of a derivative as a rate of change. The first derivative is the rate of change of the original function. The second derivative would therefore be the rate of change of the first derivative.

Think of the original function as the position of some object as a function of time (i.e f(x) is the position, x is the time). The derivative (rate of change) of position-time is velocity. Velocity decribes how fast the position changes. Now, what's the rate of change of velocity? Acceleration. It describes how fast an object's velocity changes. It can be thought of as the second derivative.

So when it says to take the second derivative, it's the derivative of the first derivative. In the case of a parametric function, you take the derivative of y(t) and the derivative of x(t) seperately, as many times as you need. Now, that only describes the rate of change in respect to each coordinate seperately. In respect to X and Y, you can solve for t in x(t), and plug that into y(t), and derive.

[edited by - Zipster on December 31, 2002 10:07:39 PM]
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This second derivative is basically used to find the local maxiums and minimums on a graph. And to tell if it is concave up or down. It can also be used to find inflection points in a graph ( ie where it changes from concave up to concave down )

In pratice it can be used for serveral applied things such as the second derivative of a postion equation will give the accelearation.


--------
Andrew


Thanx fellaz... I really that like acceleration usage part of it... definitely makes sense.

It''s not hard to understand but sometimes all the details can kill a person.

Thanx for the link acraig... good stuff on that!

I do have another question... there are some quadratic equations that are really difficult to find the inverse for. Any suggestions for an algorithm for a program that can work this out for me? I have no idea how to go about it!

-Lewis [m80]
Play QUADz MX @
www.m80produxions.com
Lewis [m80]Interactive Designerhttp://ismstudios.com
Hi,

Aren''t most (all?) quadratic functions "non-inversable"?
or?

/Mankind gave birth to God.
/Mankind gave birth to God.
I have this calculus book that got an answer from the equation from a graphing calculator... he did not explain how to get it though. I was just wondering if anybody knew how to do this without one... I seems very unconventional. Definitely not a common process.

-Lewis [m80]
Play QUADz MX @
www.m80produxions.com
Lewis [m80]Interactive Designerhttp://ismstudios.com
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The second derivative is the derivative of the derivative.

f''(x) = dx/dy[f(x)]

f''''(x) = dx/dy[f''(x)]

In regards to parametric equations, I don`t think its any different really, just a hair more complex.

It tends to be either rather simple or rather complex when compared with the original function.

What a second derivative means is dependant on the initial function and what it describes, of course.

And yes, you can keep on taking the derivative of derivatives until you are blue in the face.


~V''lion
~V'lionBugle4d
The second derivative is useful for finding relative maxima and minima. For a given linear function, the first derivative will give you the slope of the function. Solving that function for its zeros will give you the relative minima and maxima of that function (peaks and valleys). The sign of the second derivative at each zero point will tell you whether it was a relative maximum or minimum.

Think Liberally.
quote:

I do have another question... there are some quadratic equations that are really difficult to find the inverse for. Any suggestions for an algorithm for a program that can work this out for me? I have no idea how to go about it!



try this on for size...

you've get a quadratic equation, so y = Ax^2 + Bx^2 + C. Now you can solve this when y=0, thats just
    -B +- sqrt(B^2 - 4AC)x = ---------------------             2A 

so what about when y!=0? easy, change C!
y = Ax^2 + Bx^2 + C0 = Ax^2 + Bx^2 + D 

look familiar? so...
    -B +- sqrt(B^2 - 4AD)x = ---------------------             2A    -B +- sqrt(B^2 - 4A(C-y))x = -------------------------               2A 

and hey presto, the inverse of a quadratic.....almost

now the only problem is should that +- be a + or a -? This bit i'm not totally sure on, but it seems to work (it did for a couple of test I did in matlab anyway). You use both the + and - results. Lets take an example, for
x = -10 -> 10y = 10x^2 + 5x - 50    -5 +- sqrt(5^2 - 4*10*(-50-y))x = ------------------------------               2*10    -5 +- sqrt(2025 + 40y)x = ----------------------              20 

splitting up the +-
     -5 + sqrt(2025 + 40y)x1 = ---------------------              20x1 = 9.5  8.5  7.5  6.5  5.5  4.5  3.5  2.5  1.5  0.5  0  1  2  3  4  5  6  7  8  9 10     -5 - sqrt(2025 + 40y)x2 = ---------------------              20x2 = -10  -9  -8  -7  -6  -5  -4  -3  -2  -1  -0.5  -1.5  -2.5  -3.5  -4.5  -5.5  -6.5  -7.5  -8.5  -9.5  -10.5 

as you can see, if you combine the results of x1 and x2 you get your original x. This is the bit i'm not sure of, so I'll let you do some fiddling of your own and work it out for yourself.


DISCLAIMER
----------

I only made this up just then, so I could be totally wrong. If searching for an algorithmic way to combine x1 and x2 into your original x is the direct or indirect cause of the loss of your sanity, I may not be held responsible. If anyone thinks I'm a babbling idiot, please feel free to step in.




[edit]
stupid formatting....
[/edit]

[edited by - joanusdmentia on January 1, 2003 6:01:13 PM]
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The understood range of the square root operation is all non-negative numbers. Technically, the "square root" of 4 is 2 and -2, but when you do the operation, it''s assumed to be the positive root only (2). The +/- is simply there to indicate that the negative square root should also be used, since it is mathamatically true, yet simply not assumed when you do the operation.

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