Why is the solution negative?

  • #1
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Summary:

I understand the formula f(x) =mx(dot) * x(double dot)

Main Question or Discussion Point

You basically just take the second derivative of the given function and multiply it by the original then multiple everything by m. I just don’t understand how the second derivative would be negative.
 

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  • #2
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Got it

(1/x)=(x^-1)=(-x^-2)=(-1/x^2)
 
  • #3
Math_QED
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Are you asking why the derivative of ##1/x## w.r.t. ##x## is ##-1/x^2##?
 
  • #4
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Summary:: I understand the formula f(x) =mx(dot) * x(double dot)
This is somewhat hard to read. This site supports LaTeX, which can be used to write equations formatted nicely.
##f(x) = m\dot x \ddot x##
The script I used for the above is ##f(x) = m\dot x \ddot x##
There's a link to our tutorial at the lower left of the text entry pane -- click the link titled LaTeX Guide.
Anonymous_ said:
You basically just take the second derivative of the given function and multiply it by the original then multiple everything by m. I just don’t understand how the second derivative would be negative.
You're given that v(x) = a/x, so v(x) is decreasing (assuming a > 0), which makes its derivative negative.
Got it
(1/x)=(x^-1)=(-x^-2)=(-1/x^2)
This makes no sense.
##\frac 1 x = x^{-1}##, but ##x^{-1} \ne -x^{-2}##
You apparently took the derivative to get from the 2nd expression to the 3rd, but you don't give any indication that that's what you did. If I were your instructor, I would mark your work down for this.
 
  • #5
10
1
This is somewhat hard to read. This site supports LaTeX, which can be used to write equations formatted nicely.
##f(x) = m\dot x \ddot x##
The script I used for the above is ##f(x) = m\dot x \ddot x##
There's a link to our tutorial at the lower left of the text entry pane -- click the link titled LaTeX Guide.
You're given that v(x) = a/x, so v(x) is decreasing (assuming a > 0), which makes its derivative negative.
This makes no sense.
##\frac 1 x = x^{-1}##, but ##x^{-1} \ne -x^{-2}##
You apparently took the derivative to get from the 2nd expression to the 3rd, but you don't give any indication that that's what you did. If I were your instructor, I would mark your work down for this.
Not saying they are equal. The equal signs are steps => basically you have to multiple by negative one.
 
  • #6
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Not saying they are equal. The equal signs are steps => basically you have to multiple by negative one.
But you said they are equal:
(1/x)=(x^-1)=(-x^-2)=(-1/x^2)
I wouldn't have complained about this if you had showed what you are doing.
##\frac d {dx}(\frac 1 x) = \frac d {dx}(x^{-1}) = -x^{-2} = -\frac 1 {x^2}##
 
  • #7
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But you said they are equal:
I wouldn't have complained about this if you had written "implies" => instead of "equals" =.
Or better yet, showed what you are doing.
##\frac d {dx}(\frac 1 x) = \frac d {dx}(x^{-1}) = -x^{-2} = -\frac 1 {x^2}##
I would just assume you would know what I mean since this is a physics site...
 
  • #8
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I would just assume you would know what I mean since this is a physics site...
Because this is a physics site, you should write what you mean.

If I were to write something like this, would that be OK? ##2x + 1 = 5 = 2x = 4 = x = 2##
I hope not, since it is saying that 5, and 4, and 2 are all the same number, among other nonsensical statements.
 
  • #9
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Because this is a physics site, you should write what you mean.

If I were to write something like this, would that be OK? ##2x + 1 = 5 = 2x = 4 = x = 2##
I hope not, since it is saying that 5, and 4, and 2 are all the same number, among other nonsensical statements.
Yes, I can understand what’s going on. I hate when people overcomplicate things like this.
 
  • #10
Math_QED
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Yes, I can understand what’s going on. I hate when people overcomplicate things like this.
Overcomplicating? As written, your post #2 is simply wrong and suggests you don't understand what an equal sign means.
 
  • #11
Math_QED
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We try to help you optimally. I asked you a question in #3 which you ignored. Could you please answer so we can move on with the question?

We are just trying to tell you that using equal signs like that is wrong (although we get what you mean). Instead use ##\implies## arrows. Most people that mark exams would deduct points because this is very sloppy notation. By saying this to you, we hope that you can avoid this to happen.
 
  • #12
jbriggs444
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This is somewhat hard to read. This site supports LaTeX, which can be used to write equations formatted nicely.
##f(x) = m\dot x \ddot x##
The script I used for the above is ##f(x) = m\dot x \ddot x##
Where did you find the fake # sign, "#"? The result is pretty slick and I'd like to be able to borrow the trick.

Edit: found it here. Just apply a textcolor to the ##, making it black like so: ##\sqrt{x}##
 
Last edited:
  • #13
jim mcnamara
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Now that you mention it, that means @Mark44 overrode the ## reserved symbol set somehow. Interesting....
 
  • #14
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Edit: found it here. Just apply a textcolor to the ##, making it black like so: ##\sqrt{x}##
Now that you mention it, that means @Mark44 overrode the ## reserved symbol set somehow. Interesting....
Yes, @jbriggs444 hit it, but I did it to the first # symbol at each end. Doing so prevents the browser from rendering the expression between the # pairs.
 
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