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 \frac{d\vec{v}} {d\vec{q}}[\itex]?<br /> <br /> What about \frac{d\vec{v}} {d\vec{v}}?<br /> (Can&#039;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&lt;\frac{df_x}{dt},\frac{df_y}{dt},\frac{df_z}{dt}\right&gt;t
which we can think of as being "represented" by the usual derivative vector,
\left&lt;\frac{df_x}{dt},\frac{df_y}{dt}, \frac{df_z}{dt}\right&gt;

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
\left&lt;\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\left&gt;\cdot \vec{v}[/itex]<br /> which we can think of as represented by the gradient vector,<br /> \left&amp;lt;\frac{\partial f}{\partial x},\frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}\left&amp;gt;[/itex]&lt;br /&gt; &lt;br /&gt; More generally, if f is from R&lt;sup&gt;n&lt;/sup&gt; to R&lt;sup&gt;m&lt;/sup&gt;, it derivative, at any &amp;quot;point&amp;quot;, 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.&lt;br /&gt; &lt;br /&gt; In any case, of course, \frac{d\vec{v}}{d\vec{v}}, where \vec{v} is a vector function from R&lt;sup&gt;n&lt;/sup&gt; to itself (NOT just a single vector- derivatives are only defined for functions) is the identity transformation on R&lt;sup&gt;n&lt;/sup&gt; 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 \partialF/\partialQ.
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 \partialF/\partialQ.
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,
\frac{dF}{dQ}= \frac{dF}{dr}/\frac{dQ}{dr}[/itex].<br /> <br /> 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 &lt;1, 0&gt;, &lt;0, 1&gt; basis for R².)<br /> \frac{dF}{dr}= \left[\begin{array}{cc}\frac{\partial f}{\partial x} &amp;amp; \frac{\partial f}{\partial y} \\ \frac{\partial g}{\partial x} &amp;amp; \frac{\partial g}{\partial y}\end{array}\right]<br /> <br /> Similarly,<br /> \frac{dQ}{dr}= \left[\begin{array}{cc}\frac{\partial p}{\partial x} &amp;amp; \frac{\partial p}{\partial y} \\ \frac{\partial q}{\partial x} &amp;amp; \frac{\partial q}{\partial y}\end{array}\right]<br /> <br /> So that<br /> \frac{dF}{dW}= \left[\begin{array}{cc}\frac{\partial f}{\partial x} &amp;amp; \frac{\partial f}{\partial y} \\ \frac{\partial g}{\partial x} &amp;amp; \frac{\partial g}{\partial y}\end{array}\right]\left[\begin{array}{cc}\frac{\partial p}{\partial x} &amp;amp; \frac{\partial p}{\partial y} \\ \frac{\partial q}{\partial x} &amp;amp; \frac{\partial q}{\partial y}\end{array}\right]^{-1}<br /> <br /> The existence of that derivative depends not only on the existence of these partial derivatives but also on dQ/dr being invertible.