0
$\begingroup$

Let $a,b : I \to \mathbb{R}$ be differentiable functions, and $m: I \to \mathbb{R}^2$ be a function such that $m(t)=(a(t),b(t))$.

Further, let there be a differentiable function $f:\mathbb{R}^2\to\mathbb{R}$, and let $z:I\to\mathbb{R}$, $z(t)=f(a(t),b(t))$.

We wish to find $\frac{dz}{dt}$.

We know that $\frac{dz}{dt}=D(z)(t)$. (Total derivative)

So $$D(z)(t)=D(f\circ m)(t)=D(f)(m(t)).D(m)(t)$$ (Chain rule) $$D(f)(m(t))=[\frac{\partial f}{\partial x}(m(t)) \space\space\space\frac{\partial f}{\partial y}(m(t))]$$ By this I mean, partial derivative of $f$ w.r.t $x$ at $m(t)$. $$ D(m)(t)= \left[ {\begin{array}{c} \frac{da}{dt} \\ \frac{db}{dt} \\ \end{array} } \right] $$. Multiplying both gives, $$\frac{dz}{dt}=\frac{\partial f}{\partial x}.\frac{da}{dt}+\frac{\partial f}{\partial y}.\frac{db}{dt}$$ Now how to express this in terms of $z$ only not $f$?

According to the book I have, the answer is $$\frac{dz}{dt}=\frac{\partial z}{\partial a}.\frac{da}{dt}+\frac{\partial z}{\partial b}.\frac{db}{dt}$$ Also, what is the meaning of $\frac{\partial z}{\partial a}$ according to the definition?

$\endgroup$
  • 1
    $\begingroup$ I don't know but your notation here and the book's notation seem to be different. If I try to reconcile the 2, it seems like the book, from your perspective, is using notation such as $z = z(a,b)$ where $a = a(t)$ and $b = b(t)$. That is, they give the 3 functions a name $z, a , b$ then call the output of the function, the same as the function name. This is commonly done in science classes, but causes confusion sometimes. It's better to do as in this question, to call the output of $a(t) = x$, the output $b(t)$ as $y$. $\endgroup$ – DWade64 Oct 4 '18 at 13:57
  • $\begingroup$ @DWade64, can you please elaborate a little bit more? Maybe an answer? I took $x$ and $y$ as the variables which go inside $f$. That is, $f(x,y)$. $\endgroup$ – Aditya Agarwal Oct 4 '18 at 14:10
  • $\begingroup$ Thank you for the reply but I'm not sure what more to say since I'm getting confused by what seems to be 2 different notations between you and the book. Your work looks correct. You seem to call the composition, you give it the name, $f \circ m = : z$. Which I think is good. The outside function is $f$, the inside function is $m$, giving altogether a new function $z$. However the book seems to call the outside function $z$, with direct input variables labeled as $a$ and $b$. I wish I could help more, but I'm getting confused as well $\endgroup$ – DWade64 Oct 4 '18 at 14:36
  • 1
    $\begingroup$ @DWade64 Okay. Well I could wait for an answer which tells what is correct then. :) $\endgroup$ – Aditya Agarwal Oct 4 '18 at 14:37
2
$\begingroup$

The formula

$$\frac{dz}{dt}=\frac{\partial z}{\partial a} \cdot \frac{da}{dt}+\frac{\partial z}{\partial b} \cdot \frac{db}{dt}$$

given in your book does not make sense. In fact, $z$ is a function of one variable $t$ and there are no partial derivatives $\frac{\partial z}{\partial a} ,\frac{\partial z}{\partial b}$ with respect to two variables $a, b$. The correct formula is nothing else than

$$\frac{dz}{dt}=\frac{\partial f}{\partial x} \cdot \frac{da}{dt}+\frac{\partial f}{\partial y} \cdot \frac{db}{dt}$$

or, written pointwise,

$$\frac{dz}{dt}(t_0) =\frac{\partial f}{\partial x}(a(t_0),b(t_0)) \cdot \frac{da}{dt}(t_0)+\frac{\partial f}{\partial y}(a(t_0),b(t_0)) \cdot \frac{db}{dt}(t_0) .$$

It is impossible to eleminate $f$ from this formula. The only "explanation" is that your book abuses notation and writes

$$z(t) = z(a(t),b(t))$$

which is not correct.

$\endgroup$
  • $\begingroup$ Some programming languages support the concept of "polymorphic functions". See for example en.wikipedia.org/wiki/Ad_hoc_polymorphism. Mathematically it is an abuse of notation to denote different things by the same symbol, but perhaps it is what your book intends. $\endgroup$ – Paul Frost Oct 4 '18 at 15:03
  • $\begingroup$ This answer was useful to me, reminding me that we need unambiguous notation in mathematics as also seen here in this question. Compositions should always be given a different function name as you point out. Also the wikipedia page on the chain rule is horrendous in this regard. If I say $y = f(x)$, technically, I should not write down $dy/dx$ as y is a coordinate. Because on my $xy$ coordinate plane I could have $y = f(x)$ and $y = g(x)$ making $\endgroup$ – DWade64 Oct 4 '18 at 15:14
  • $\begingroup$ $dy/dx$ ambiguous $\endgroup$ – DWade64 Oct 4 '18 at 15:14
  • $\begingroup$ Thanks so much for the answer! $\endgroup$ – Aditya Agarwal Oct 4 '18 at 17:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.