Note
You can download this example as a Python script: :jupyter-download:script:`differentiation` or Jupyter Notebook: :jupyter-download📓`differentiation`.
.. jupyter-execute:: import sympy as sm import sympy.physics.mechanics as me sm.init_printing(use_latex='mathjax')
If a vector \bar{v} is a function of n scalar variables q_1,q_2,\ldots,q_n in reference frame A then the first partial derivatives of \bar{v} in A with respect to q_r where r=1\ldots n can be formed by applying the product rule of differentation and taking into account that the mutually perpendicular unit vectors fixed in A do not change when observed from A. The partial derivatives are then:
\frac{{}^A\partial \bar{v}}{\partial q_r} = \sum_{i=1}^3 \frac{\partial
v_i}{\partial q_r} \hat{a}_i \textrm{ for } r=1\ldots n
where v_i are the measure numbers of \bar{v} expressed in A associated with the mutually perpendicular unit vectors \hat{a}_1,\hat{a}_2,\hat{a}_3.
Given \bar{v}=v_x\hat{a}_x+v_y\hat{a}_y+v_z\hat{a}_z the above definition expands to:
\frac{{}^A\partial \bar{v}}{\partial q_r} =
\frac{\partial v_x}{\partial q_r} \hat{a}_x +
\frac{\partial v_y}{\partial q_r} \hat{a}_y +
\frac{\partial v_z}{\partial q_r} \hat{a}_z
Many of the vectors we will work with in multibody dynamics will be a function of a single variable, most often time t. If that is the case, the partial derivative reduces to a single variate derivative:
\frac{{}^A d \bar{v}}{dt} := \sum_{i=1}^3 \frac{d v_i}{dt} \hat{a}_i
Warning
A derivative written as \frac{\partial \bar{v}}{\partial q_r} is meaningless because no reference frame is indicated. The derivative is dependent on which reference frame the change is observed from, so without a reference frame, the derivative cannot be calculated. This is not the case for partial derivatives of scalar expressions, as no reference frame is involved.
The above definition implies that a vector must be expressed in the reference frame one is observing the change from before calculating the partial derivatives of the scalar measure numbers. For example, here is a vector that is expressed with unit vectors from three different reference frames:
.. jupyter-execute::
alpha, beta = sm.symbols('alpha, beta')
a, b, c, d, e, f = sm.symbols('a, b, c, d, e, f')
A = me.ReferenceFrame('A')
B = me.ReferenceFrame('B')
C = me.ReferenceFrame('C')
B.orient_axis(A, alpha, A.x)
C.orient_axis(B, beta, B.y)
v = a*A.x + b*A.y + c*B.x + d*B.y + e*C.x + f*C.y
v
To calculate \frac{{}^A\partial\bar{v}}{\partial \alpha} we first need to project the vector \bar{v} onto the unit vectors of A and take the partial derivative of those measure numbers with respect to \alpha. The dot product provides the projection and the resulting scalar is differentiated:
.. jupyter-execute:: dvdax = v.dot(A.x).diff(alpha) dvdax
.. jupyter-execute:: dvday = v.dot(A.y).diff(alpha) dvday
.. jupyter-execute:: dvdaz = v.dot(A.z).diff(alpha) dvdaz
We can then construct the vector \frac{{}^A\partial \bar{v}}{\partial \alpha} from the new measure numbers know that the A unit vectors are fixed:
.. jupyter-execute:: dvda = dvdax*A.x + dvday*A.y + dvdaz*A.z dvda
SymPy Mechanics vectors have a special
:external:py:meth:`~sympy.physics.vector.vector.Vector.diff` method that
manages taking partial derivatives from different reference frames. For the
vector .diff() method you provide first the variable \alpha
followed by the reference frame you are observing from:
.. jupyter-execute:: dvdalpha = v.diff(alpha, A) dvdalpha
The result is not so simplified because SymPy attempts to express the derivative in the same components as the vector was, so you can use the vector :external:py:meth:`~sympy.physics.vector.vector.Vector.simplify` method, which applies :external:py:func:`~sympy.simplify.trigsimp.trigsimp` to each measure number:
.. jupyter-execute:: v.diff(alpha, A).simplify()
This multi reference frame form can be shown to be the same as we calculated above by expressing it fully in A and simplifying:
.. jupyter-execute:: v.diff(alpha, A).express(A).simplify()
Consider again vector \bar{v}=v_x\hat{a}_x+v_y\hat{a}_y+v_z\hat{a}_z. Previously, only the measure numbers of this vector were scalar functions of q_r. Now consider a reference frame N that is oriented relative to A such that the relative orientation also depends on q_r. This means, that when observed from N, the unit vectors \hat{a}_x,\hat{a}_y,\hat{a}_z may be a function of q_r. With both the measure numbers and unit vectors dependent on q_r the derivative of \bar{v} in N requires the use of the product rule when taking the partial derivative. For example:
\frac{{}^N\partial \bar{v}}{\partial q_r} =
\frac{{}^N\partial v_x}{\partial q_r}\hat{a}_x + v_x \frac{{}^N\partial \hat{a}_x}{\partial q_r} +
\frac{{}^N\partial v_y}{\partial q_r}\hat{a}_y + v_y \frac{{}^N\partial \hat{a}_y}{\partial q_r} +
\frac{{}^N\partial v_z}{\partial q_r}\hat{a}_z + v_z \frac{{}^N\partial \hat{a}_z}{\partial q_r}
The three similar terms with scalar derivatives have the same interpretation of the ones in the prior section.
\frac{{}^N\partial v_x}{\partial q_r}\hat{a}_x,
\frac{{}^N\partial v_y}{\partial q_r}\hat{a}_y,
\frac{{}^N\partial v_z}{\partial q_r}\hat{a}_z
But the part with unit vector derivatives is more interesting. The partial derivative of a unit vector depends on how it changes. But unit vectors do not change in length, only in orientation.
v_x \frac{{}^N\partial \hat{a}_x}{\partial q_r},
v_y \frac{{}^N\partial \hat{a}_y}{\partial q_r},
v_z \frac{{}^N\partial \hat{a}_z}{\partial q_r}
You will learn in the next chapter how to interpret and use these terms to simplify the calculations of common derivatives. But for now, just be aware of the nature of this partial derivative in N.
The product rule also applies to the dot and cross products:
\frac{\partial}{\partial q_r}(\bar{v} \cdot \bar{w}) = &
\frac{\partial \bar{v}}{\partial q_r} \cdot \bar{w} +
\bar{v} \cdot \frac{\partial \bar{w}}{\partial q_r}
\frac{\partial}{\partial q_r}(\bar{v} \times \bar{w}) = &
\frac{\partial \bar{v}}{\partial q_r} \times \bar{w} +
\bar{v} \times \frac{\partial \bar{w}}{\partial q_r}
and generalizes to any series of products. Let G=f_1 \cdots f_n be a series of products, then:
\frac{\partial G}{\partial q_r} =
\frac{\partial f_1}{\partial q_r}\cdot f_2 \cdots f_n +
f_1 \cdot\frac{\partial f_2}{\partial q_r}\cdot f_3 \cdots f_n +
\dots +
f_1 \cdots f_{n-1} \cdot \frac{\partial f_n}{\partial q_r}
\frac{{}^A\partial \bar{v}}{\partial q_r} is also a vector and, just like \bar{v}, may be a vector function. We can thus calculate the second partial derivative with respect to q_s where s=1\ldots n. This second partial derivative need not be taken with respect to the same reference frame as the first partial derivative. If we first differentiate with respect to A and then with respect to B, the second partial derivative is:
\frac{{}^B\partial}{\partial q_s} \left(\frac{{}^A\partial\bar{v}}{\partial
q_r}\right)
Second partials in different reference frames do not necessarily commute:
\frac{{}^B\partial}{\partial q_s} \left(\frac{{}^A\partial\bar{v}}{\partial
q_r}\right)
\neq
\frac{{}^A\partial}{\partial q_r} \left(\frac{{}^B\partial\bar{v}}{\partial
q_s}\right)
If the reference frames of each partial derivative are the same, then mixed partials do commute.
.. todo:: Make an example of second derivatives not commuting.
In multibody dynamics we are primarily concerned with how motion changes with respect to time t and our vectors and measure numbers will often be implicit functions of time, i.e. q_r(t). When that is the case the chain rule can be used to take total derivatives:
\frac{{}^A d\bar{v}}{dt} =
\sum_{i=1}^n \frac{{}^A\partial \bar{v}}{\partial q_r(t)} \frac{d q_r(t)}{dt} +
\frac{{}^A \partial \bar{v}}{\partial t}
\textrm{ where } r=1,\ldots,n
Note
We will typically use the "dot" notation for time derivatives, i.e. \frac{dq}{dt} as \dot{q} and \frac{d^2q}{dt^2} as \ddot{q} and so on.
In SymPy Mechanics, scalar functions of time can be created like so:
.. jupyter-execute::
t = sm.symbols('t')
q_of = sm.Function('q')
q = q_of(t)
q
And these scalar functions can be differentiated:
.. jupyter-execute:: q.diff(t)
SymPy Mechanics provides the convince function
:external:py:func:`~sympy.physics.vector.functions.dynamicsymbols` to create
scalar functions of time just like symbols():
.. jupyter-execute::
q1, q2, q3 = me.dynamicsymbols('q1, q2, q3')
q1, q2, q3
The time variable used in q1,q2,q3 can be accessed like so:
.. jupyter-execute:: t = me.dynamicsymbols._t
SymPy Mechanics also provide a special printing function :external:py:func:`~sympy.physics.vector.printing.init_vprinting` which enables the dot notation on functions of time:
.. jupyter-execute:: me.init_vprinting(use_latex='mathjax') q1.diff(t), q2.diff(t, 2), q3.diff(t, 3)
Now these scalar functions of time can be used to formulate vectors:
.. jupyter-execute::
A = me.ReferenceFrame('A')
B = me.ReferenceFrame('B')
B.orient_body_fixed(A, (q1, q2, q3), 'ZXZ')
v = q1*A.x + q2*A.y + t**2*A.z
v
And the time derivative can be found with:
.. jupyter-execute:: v.diff(t, A)
Lastly, vectors have a :external:py:meth:`~sympy.physics.vector.vector.Vector.dt` method that calculates time derivatives, saving a few characters of typing:
.. jupyter-execute:: v.dt(A)