Graphing with terminal velocity

Nov 2012
744
2
Hawaii
My notes didn't show how the second derivative was evalated. Would I need to plug in the same values I used in the first derivative inside the second derivative to figure out whether v'' is greater than or equal to zero?
 

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skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas


Note that the equation given for $\dfrac{dv}{dt}$ is a scalar equation, so $v$ actually represents speed rather than velocity.

Speed is bounded strictly between $0 \le v \le v_T = \dfrac{mg}{k}$, so the signs of $\dfrac{dv}{dt}$ and $\dfrac{d^2v}{dt^2}$ need only be calculated between those two bounds ...

for $v = \dfrac{mg}{2k}$ ...

$\dfrac{dv}{dt} = g - \dfrac{k}{m} \cdot \dfrac{mg}{2k} = g - \dfrac{g}{2} = \dfrac{g}{2} > 0$

$\dfrac{d^2v}{dt^2} = -\dfrac{k}{m} \cdot \dfrac{dv}{dt} = -\dfrac{k}{m} \cdot \dfrac{g}{2} < 0$

so, the graph of speed, $v$, versus time is increasing and concave down.
 
Nov 2012
744
2
Hawaii
Right, I'm actually sketching it so I want to find v'' above mg/k. Whatever we plug in dv/dt in the second derivative will always be negative since k/m is negative, which would be wrong based off of my professors work. How do you evaluate values above mg/k?
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
You can't ... why do you think they call it terminal speed?
 

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
The solution for this DE is ...

$v(t) = \dfrac{mg}{k}\left(1-e^{\frac{k}{m} t} \right)$

note $\displaystyle \lim_{t \to \infty} v(t) = \dfrac{mg}{k}$

... graph is attached.
 

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