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 March 2nd, 2018, 07:40 AM #1 Newbie   Joined: Sep 2017 From: xxxx Posts: 8 Thanks: 0 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 07:43 AM.
March 2nd, 2018, 07:47 AM   #2
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 Originally Posted by New123 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 $-\dfrac{1}{s^2} \cdot s = - \dfrac{1}{s}$

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

March 2nd, 2018, 10:08 AM   #3
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 Originally Posted by skeeter 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 $-v$

March 2nd, 2018, 01:11 PM   #4
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 Originally Posted by New123 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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