# Graphing with terminal velocity

#### asilvester635

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

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.

#### asilvester635

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
You can't ... why do you think they call it terminal speed?

#### skeeter

MHF Helper
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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