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--- gibson:teaching:spring-2016:math445:lab10 [2016/04/12 06:18]
gibson created
+++ gibson:teaching:spring-2016:math445:lab10 [2016/04/12 06:52] (current)
gibson
@@ Line 20: / Line 20: @@
-**Problem 3.** Superimpose on your previous plot the path of a raindrop that falls at
+**Problem 3.** Superimpose on your previous plot the path $\vec{x}(t)$ of a raindrop that falls at
-$\vec{x} = (x,y) = (0.9,0.9)$. We're assuming the velocity of the raindrop as it drips down is proportional to the negative of the gradient of the height. That is, $\vec{v} = -c \vec{\nabla} h $,
+$\vec{x}(0) = (x_0,y_0) = (0.9,0.9)$. We're assuming the velocity of the raindrop as it drips down is proportional to the negative of the gradient of the height. That is, $\vec{v} = -c \vec{\nabla} h $,
 where $\vec{v} = d\vec{x}/dt$ is the $x,y$ velocity of the raindrop. For convenience set $c=1$ (this won't change the path). Then compute the path of the raindrop numerically using //forward Euler time-stepping//,
 \begin{eqnarray*}
-\vec{x}(t+\Delta t) = \vec{x}(t) + \Delta t \, \vec{v}(\vec{x})
+\vec{x}(t+\Delta t) = \vec{x}(t) + \Delta t \, \vec{v}(\vec{x}(t))
 \end{eqnarray*}
+The above formula is accurate only if the time step $\Delta t$ is small. The value $\Delta = 0.1$ is small enough for the given system. (You can experiment with larger and smaller time steps if you like.)
 {{:gibson:teaching:spring-2015:math445:montesol_path.png?direct&400|}}

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