Does the slope increase if it goes from more to less negative?

  • Thread starter Ryker
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  • #1
Ryker
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Homework Statement


Linearize the equation

[tex]y = Ae^{-\frac{b}{2m}t}[/tex],

so that lny is the x variable and t is the y-variable. What happens to the slope and the intercept if you increase b?

The Attempt at a Solution


Taking ln of both sides, I get the equation

[tex]t = -\frac{2m}{b} \ln y + \frac{2m}{b} \ln A[/tex].

If you increase b, then the slope goes from more negative to less negative, whereas the intercept decreases. Is this correct or do you take the slope to be decreasing, because its absolute value decreases?
 

Answers and Replies

  • #2
SammyS
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Homework Statement


Linearize the equation

[tex]y = Ae^{-\frac{b}{2m}t}[/tex],

so that lny is the x variable and t is the y-variable. What happens to the slope and the intercept if you increase b?

The Attempt at a Solution


Taking ln of both sides, I get the equation

[tex]t = -\frac{2m}{b} \lny + \frac{2m}{b} \lnA[/tex].

If you increase b, then the slope goes from more negative to less negative, whereas the intercept decreases. Is this correct or do you take the slope to be decreasing, because its absolute value decreases?

That's t=0 .

[tex]\ln(y) = \ln(A) -\frac{b}{2m}\,t}[/tex]

Solve this for t .
 
  • #3
Ryker
1,086
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I was editing my LaTeX and you probably quoted it when it didn't show the natural logs :smile: The original post should have the proper equation.
 

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