daviddoji/CalculusWithJuliaNotes.jl
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jverzani
2024-05-22 07:55:20 -04:00
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quarto/differentiable_vector_calculus/vector_valued_functions.qmd
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@@ -586,7 +586,7 @@ With this, the equation is simply $\vec{tl}(t) = \vec{f}(t_0) + \vec{f}'(t_0) \c
##### Example: parabolic motion
In physics, we learn that the equation $F=ma$ can be used to derive a formula for postion, when acceleration, $a$, is a constant. The resulting equation of motion is $x = x_0 + v_0t + (1/2) at^2$. Similarly, if $x(t)$ is a vector-valued postion vector, and the *second* derivative, $x''(t) =\vec{a}$, a constant, then we have: $x(t) = \vec{x_0} + \vec{v_0}t + (1/2) \vec{a} t^2$.
In physics, we learn that the equation $F=ma$ can be used to derive a formula for position, when acceleration, $a$, is a constant. The resulting equation of motion is $x = x_0 + v_0t + (1/2) at^2$. Similarly, if $x(t)$ is a vector-valued position vector, and the *second* derivative, $x''(t) =\vec{a}$, a constant, then we have: $x(t) = \vec{x_0} + \vec{v_0}t + (1/2) \vec{a} t^2$.
For two dimensions, we have the force due to gravity acts downward, only in the $y$ direction. The acceleration is then $\vec{a} = \langle 0, -g \rangle$. If we start at the origin, with initial velocity $\vec{v_0} = \langle 2, 3\rangle$, then we can plot the trajectory until the object returns to ground ($y=0$) as follows:
@@ -1109,7 +1109,7 @@ $$
0 = \frac{d(r^2 \dot{\theta})}{dt} = 2r\dot{r}\dot{\theta} + r^2 \ddot{\theta},
$$
which is the tranversal component of the acceleration times $r$, as decomposed above. This means, that the acceleration of the planet is completely towards the Sun at the origin.
which is the transversal component of the acceleration times $r$, as decomposed above. This means, that the acceleration of the planet is completely towards the Sun at the origin.
Kepler's first law, relates $r$ and $\theta$ through the polar equation of an ellipse:
@@ -2020,7 +2020,7 @@ Let $r$ be the radius of a circle and for concreteness we position it at $(-r, 0
* Between angles $0$ and $\pi/2$ the horse has unconstrained access, so they can graze a wedge of radius $R$.
* Between angles $\pi/2$ and until the horse's $y$ position is $0$ when the tether is taut the boundary of what can be eaten is described by the involute.
* The horse can't eat from withing the circle or radius $r$.
* The horse can't eat from within the circle or radius $r$.
```{julia}