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March 2nd, 2018, 06:40 AM   #1
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Substitution

Hi All,

Im running into the following problem

The equation is something like this:
$\displaystyle dv=(-1/s^2*s)$

and s = 1/v

How do I substitute 1/v (v^-1) into the equation? Can someone explain that to me in simple steps?

I thought: $\displaystyle dv=(-1/v^-1*v^-1)$

But I think that is wrong.

Thanks!!!

Last edited by New123; March 2nd, 2018 at 06:43 AM.
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March 2nd, 2018, 06:47 AM   #2
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Originally Posted by New123 View Post
Hi All,

Im running into the following problem

The equation is something like this:
$\displaystyle dv=(-1/s^2*s)$

and s = 1/v

How do I substitute 1/v (v^-1) into the equation? Can someone explain that to me in simple steps?

I thought: $\displaystyle dv=(-1/v^-1*v^-1)$

But I think that is wrong.
note $-\dfrac{1}{s^2} \cdot s = - \dfrac{1}{s}$

$s = \dfrac{1}{v} \implies v = \dfrac{1}{s} \implies dv = -\dfrac{1}{s} = -v$
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March 2nd, 2018, 09:08 AM   #3
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Originally Posted by skeeter View Post
note $-\dfrac{1}{s^2} \cdot s = - \dfrac{1}{s}$

$s = \dfrac{1}{v} \implies v = \dfrac{1}{s} \implies dv = -\dfrac{1}{s} = -v$
Okay, that clarifies a bit. One last question:

$dv = -\dfrac{1}{s} = -v$

Can you explain the above? I simply do not understand why $dv = -\dfrac{1}{s}$ becomes $-v$

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March 2nd, 2018, 12:11 PM   #4
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Okay, that clarifies a bit. One last question:

$dv = -\dfrac{1}{s} = -v$

Can you explain the above? I simply do not understand why $dv = -\dfrac{1}{s}$ becomes $-v$

$s$ and $v$ are reciprocals ...

$s = \dfrac{1}{v} \implies v=\dfrac{1}{s}$


you are given $s = \dfrac{1}{v}$

algebraically ...

(1) multiply both sides by $v$

$v \cdot s = v \cdot \dfrac{1}{v}$

$v \cdot s = \cancel{v} \cdot \dfrac{1}{\cancel{v}}$

$v \cdot s = 1$

(2) divide both sides by $s$ ...

$\dfrac{v \cdot s}{s} = \dfrac{1}{s}$

$\dfrac{v \cdot \cancel{s}}{\cancel{s}} = \dfrac{1}{s}$

$v = \dfrac{1}{s}$
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