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It's the end of AP Exams(finally)and my calculus teacher decides to throw in a project before the year is over(just great). Anyways, I basically have to do a project involving slope fields and most of it is easy. The one problem or question that I have is how do you find the antiderivative of ln(x) using slope fields. How would u put this into the ti-83?
My calc teacher tells me slope fields are used to find antiderivatives of functions u can't find the antiderivatives of. Like the derivative of cos(x) is sin(x) and the dervative of sin(x) is -cos(x). But u can't do that for ln(x) so that's where slope fields come into play to find the antiderviative of ln(x).
I'm not sure if this is right but feel free to add more things if u want cuz i need some more things to say about this so hopefully my presentation isn't that short either. But for the most part i don't think there are alot of things to really talk about involving slope fields.
But yea, i kinda got carried away but how would u put ln(x) to find the antiderviative of that using slope fields on the ti-83?
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lol? why would you need slope fieldssss
∫ ln(x) dx
integration by parts: uv - ∫vdu
u = ln(x)
dv = dx
xln(x) - x
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My subject that i picked is slope fields so i have to do it using that :x sucks i know
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Bill307
Canada9103 Posts
On May 19 2008 09:58 paper wrote:
lol? why would you need slope fieldssss
∫ ln(x) dx
integration by parts: uv - ∫vdu
u = ln(x)
dv = dx
xln(x) - x
Technically, it's wrong to write dv = dx. dx isn't part of the integrand: it just tells you what you're integrating with respect to. Instead, you should write dv = 1. The result is the same, though.
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Well, you could draw the slope field of lnx and be like...the anti derivative of ln x must fit the slope field.
After much trial and error (involving dividing the slopes of the slope field by x to make it look like lnx-1) you can conclude that only xlnx - x fits the slope field and that must be the anti derivative.
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On May 19 2008 10:20 thunk wrote:
Well, you could draw the slope field of lnx and be like...the anti derivative of ln x must fit the slope field.
After much trial and error (involving dividing the slopes of the slope field by x to make it look like lnx-1) you can conclude that only xlnx - x fits the slope field and that must be the anti derivative.
Is there a way u can put it into the ti-83 to do this becuase I don't know what u mean by dividing the slopes of the slope field.
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basically have a loop that calculates the slope and makes a segment
resolution = 0.5 //this is how fine you want the graph to be
scale = 10 //this is how big you want your graph to be (-scale, -scale) to (scale, scale)
x = -1 * scale
y= -1 * scale
for x <= scale {
for y<= scale {
draw a segment from point (x,y) to point (x+1, y+ln(x) )
y += resolution
}
x+= resolution
}
for directions on how to program this algorithm into the TI83, look at this guide:
http://www.ticalc.org/programming/columns/83plus-bas/cherny/
ok i busted out the old 83 and came up with this, seems to work pretty well:
Disp "Scale?"
Prompt A
Disp "Resolution?"
Prompt B
DispGraph
ClrDraw
For(C,B,A,B)
For(D,-A,A,B)
Line(C,D,C+1,D+ln(C))
End
End
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On May 19 2008 11:22 MaRiNe23 wrote:Show nested quote +On May 19 2008 10:20 thunk wrote:
Well, you could draw the slope field of lnx and be like...the anti derivative of ln x must fit the slope field.
After much trial and error (involving dividing the slopes of the slope field by x to make it look like lnx-1) you can conclude that only xlnx - x fits the slope field and that must be the anti derivative. Is there a way u can put it into the ti-83 to do this becuase I don't know what u mean by dividing the slopes of the slope field.
So you have a slope field, right?
And then you can divide the slope at point (x,y) by x and you get a new slope field (f(x) = dy/dx), showing that dy/dx = x(f(x)). It will be apparent that f(x) is equal to ln(x)-1, showing that the anti derivative of ln(x) is xln(x) - x.
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EPT![[image loading]](http://www.utpb.edu/scimath/wkfield/mod4/4gr30.gif)
