How do I find the derivative of one vector with respect to another?

[spastic]

What does it mean to operate with $\frac{d}{\vec{v}}$ where v is a vector?

Say you have another vector q, how do you do [itex]\frac{d\vec{v}} {d\vec{q}}[\itex]?

What about $\frac{d\vec{v}} {d\vec{v}}$?
(Can't remember how to do the proper font sorry)

 

[HallsofIvy]

First you will have to say what kind of vector space $\vec{v}$ is in and what kind of function of v you are talking about. If f: Rⁿ to R^m, that is if the variable, $\vec{v}$ is an n dimensional vector variable and f maps it to an m dimensional vector, the $d\vec{f}/d\vec{v}$, at $\vec{v}_0$ is "the linear transformation from Rⁿ to R^m that best approximates $\vec{f}$ in some region around $\vec{v}_0$".

More precisely, a function,$\vec{f}$, from Rⁿ to R^m, is said to be differentiable at $\vec{v}_0$ if and only if there exist a linear transformation, L, from Rⁿ to R^m, and a function $\epsilon(\vec{v})$, from Rⁿ to R^m, such that
1) $f(\vec{v})= f(\vec{v}_0)+ L(\vec{v}- \vec{v_0})+ \epsilon(\vec{v})$
2) $\lim_{\vec{v}\rightarrow \vec{0}}\epsilon(\vec{v})/||\vec{v}-\vec{v}_0||= 0$

It can be shown that the linear transformation, L, in (1), is unique and we say that L is the derivative of f at $\vec{v}_0$.

Notice that, if we reduce this to R¹ to R¹, we are saying that the derivative is NOT the slope of the tangent line y= mx+ b but, rather, the linear function y= mx.

If f is from R¹ to R³, a "vector valued function of a single real variable", then the L above is linear transformation from R¹ to R³ given by
$$Lt= \left<\frac{df_x}{dt},\frac{df_y}{dt},\frac{df_z}{dt}\right>t$$
which we can think of as being "represented" by the usual derivative vector,
$$\left<\frac{df_x}{dt},\frac{df_y}{dt}, \frac{df_z}{dt}\right>$$

If f is from R³ to R, a "real valued function of 3 real variables, x, y, z", then the derivative, in the sense above, is the linear transformation from R³ to R given by the dot product
[tex]\left<\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\left>\cdot \vec{v}[/itex]
which we can think of as represented by the gradient vector,
[tex]\left<\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\left>[/itex]

More generally, if f is from Rⁿ to R^m, it derivative, at any "point", is the linear transformation which can be represented, in some basis, as the m by n matrix having the partial derivatives of the components of f as elements.

In any case, of course, $\frac{d\vec{v}}{d\vec{v}}$, where $\vec{v}$ is a vector function from Rⁿ to itself (NOT just a single vector- derivatives are only defined for functions) is the identity transformation on Rⁿ which can be reprsented by the n by n identity matrix.

 

[spastic]

Thanks mate. I didn't expect such a long reply and one quite so theoretical. It'll take me a bit to mull over so cheers!

 

[4eburashka]

HallsofIvy

More generally, if f is from Rn to Rm, it derivative, at any "point", is the linear transformation which can be represented, in some basis, as the m by n matrix having the partial derivatives of the components of f as elements.

I want to understand exactly how do I derivate vector by vector.
My problem: I try to solve Euler equations with some method... In this method I use a linearization to the first vector $$\hat{Q}$$. I need to write the $$\partial$$F/$$\partial$$Q.
Where Q and F are vectors [1,4].
Can you help me to understand how to build the matrix of linear transformation?

Thank you, I have found the reference: http://www.met.rdg.ac.uk/~ross/Documents/VectorDeriv.html

 

[HallsofIvy]

HallsofIvy

I want to understand exactly how do I derivate vector by vector.
My problem: I try to solve Euler equations with some method... In this method I use a linearization to the first vector $$\hat{Q}$$. I need to write the $$\partial$$F/$$\partial$$Q.
Where Q and F are vectors [1,4].

I presume that you mean they are vectors in R², not just that specific vector!

Can you help me to understand how to build the matrix of linear transformation?
Thank you.

Say F= <f(x,y), g(x,y)> and Q= <p(x,y), q(x,y)>. Let r= <x,y>. Then, by the chain rule,
[tex]\frac{dF}{dQ}= \frac{dF}{dr}/\frac{dQ}{dr}[/itex].

And those two derivatives are the matrices having the partial derivatives as entries. (Strictly speaking, they are the linear transformations represented by these matrices in the <1, 0>, <0, 1> basis for R².)
$$\frac{dF}{dr}= \left[\begin{array}{cc}\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y}\end{array}\right]$$

Similarly,
$$\frac{dQ}{dr}= \left[\begin{array}{cc}\frac{\partial p}{\partial x} & \frac{\partial p}{\partial y} \\ \frac{\partial q}{\partial x} & \frac{\partial q}{\partial y}\end{array}\right]$$

So that
$$\frac{dF}{dW}= \left[\begin{array}{cc}\frac{\partial f}{\partial x} & \frac{\partial f}{\partial y} \\ \frac{\partial g}{\partial x} & \frac{\partial g}{\partial y}\end{array}\right]\left[\begin{array}{cc}\frac{\partial p}{\partial x} & \frac{\partial p}{\partial y} \\ \frac{\partial q}{\partial x} & \frac{\partial q}{\partial y}\end{array}\right]^{-1}$$

The existence of that derivative depends not only on the existence of these partial derivatives but also on dQ/dr being invertible.