How much delta-v have I used here? What's the “official” equation for delta-v from parametric thrust?

Question

I took a break from Stack Exchange, jumped in my spacecar and flew the following squiggle:

$$a_x = \cos(10 \ t)$$ $$a_y = \sin(5 \ t)$$ $$a_z = \cos(2 \ t)$$

starting at xyz = [-0.01, 0, -0.05] and v_xyz = [0, -0.2, 0] with a total flight time of $2 \pi$.

When I got home I was told "Oh that was a lovely lissajous squiggle, but how much delta-v did you put on the car?"

I said "Oh, not much" and made a beeline to my computer to get back on Stack Exchange.

Question: How much delta-v DID I use?

1. If I have an acceleration vector (same as thrust vector; lets assume mass doesn't change) as a function of time $\mathbf{F}(t)$ what is the general integral expression for total delta-v should I use?
2. If someone looked up my trip in Horizons and got my state vectors $\mathbf{x}(t)$ and $\mathbf{v}(t)$ and had a numerical integrator and interpolator, what is the general integral expression for total delta-v should they use?

"bonus points" for including a Python script in your answer

3D plot of position (returns to origin) and plots of velocity components

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from scipy.integrate import odeint as ODEint

def deriv(X, t):
    x, v = X.reshape(2, -1)
    ax = np.cos(10*t)
    ay = np.sin(5*t)
    az = np.cos(2* t)
    return np.hstack((v, [ax, ay, az])) 

times = np.linspace(0, 2*np.pi, 1001)

X0 = np.hstack(([-0.01, 0, -0.05], [0, -0.2, 0]))

answer, info = ODEint(deriv, X0, times, full_output=True)

xyz, vxyz = answer.T.reshape(2, 3, -1)

fig = plt.figure()
ax  = fig.add_subplot(1, 1, 1, projection='3d', proj_type = 'ortho')
x, y, z = xyz
ax.plot(x, y, z)
ax.plot(x[:1], y[:1], z[:1], 'ok')
ax.plot(x[-1:], y[-1:], z[-1:], 'or')
plt.show()

for thing in vxyz:
    plt.plot(thing)
plt.show()

Answer 1

As $\Delta v$ is just change in velocity, we can just integrate the norm of the acceleration function over time:

$$\Delta v = \int|\mathbf{a}(t)| dt$$

You're out of luck getting a closed form of that integral though.

As far as analytical solutions goes, we can note that at $t = \frac{\pi}{2}$, all of $a_x$, $a_y$ and $a_z$ are maxed out, and hence $\Delta v < 2\pi\sqrt{3}$.

Similarly, the acceleration at all times is going to be greater than or equal to one of the components, and since they are trigonometric functions, their integrals are trivial.

$$4 < \Delta v < 2\pi\sqrt{3}$$

I can't see that there's much more to it from here than just putting the acceleration function into a numerical integrator. It's a smooth curve, so they are good at this.

Integral(sqrt(cos(10*x)^2 + sin(5*x)^2 + cos(2*x)^2),0,2*pi)
-> 7.5279

Or, by the definition of acceleration, if what you have is velocity data:

$$\Delta v = \int\left|\frac{d\mathbf{v}}{dt}\right| dt$$

Which if you have tabular data and don't bother with interpolation, is simply:

$$\Delta v =\sum |d\mathbf{v}|$$

Which is just summing up all the velocity differences between the discrete data points.