Understanding the Chain Rule in Multivariable Calculus

[Istiak]

Homework Statement

Chain rule of Calculus

Relevant Equations

Chain rule of Calculus

But, If I use chain rule than, I get that.
$\vec v_i = \frac{dr_i}{dt}=\sum_k \frac{\partial r_i}{\partial q_k} \cdot \frac{\partial q_k}{\partial t}$ But, they found that?

 

[Gaussian97]

If you have a function $f(x,y,z)$ with $x,y$ functions of $z$. How would you use the chain rule to get
$\frac{df(x,y,z)}{dz}$?

 

[Istiak]

If you have a function $f(x,y,z)$ with $x,y$ functions of $z$. How would you use the chain rule to get
$\frac{df(x,y,z)}{dz}$?

$$\frac{\partial f}{\partial x} \cdot \frac{\partial x}{\partial y} \cdot \frac{\partial y}{\partial z}$$

 

[FactChecker]

Two things to correct in your version:
1) Each function, $r_i$, is also directly a function of $t$. Your version does not include that.
2) Apparantly, each $q_i$ is a function only of $t$, so there is no partial derivative of $q_i(t)$. It should be the total derivative, $\dot{q}$.
I believe that when you correct those, your answer will match the book.

 

[Steve4Physics]

Homework Statement:: Chain rule of Calculus
Relevant Equations:: Chain rule of Calculus

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But, If I use chain rule than, I get that.
$\vec v_i = \frac{dr_i}{dt}=\sum_k \frac{\partial r_i}{\partial q_k} \cdot \frac{\partial q_k}{\partial t}$ But, they found that?

$\sum_k \frac{\partial r_i}{\partial q_k} \cdot \frac{\partial q_k}{\partial t}$ means the same as $\sum_k \frac{\partial r_i}{\partial q_k} \dot {q_k}$

It's just that you have forgotten that $r_i$ is a function of t (as well as the $q$'s):
$r_i = r_i(q_1, q_2, …. , t)$

So when using the chain rule, there is an additional term to include for the dependence on t:
Additional term = $\frac{\partial r_i}{\partial t} \cdot \frac{\partial t }{\partial t} = \frac{\partial r_i}{\partial t}\cdot 1 = \frac{\partial r_i}{\partial t}$

EDIT: I've (incorrectly) used partial derivatives (see @FactChecker's Post #4). $\frac{\partial q_k}{\partial t}$ should be $\frac{dq_k}{dt}$ and $\frac{\partial t}{\partial t}$ should be $\frac {dt}{dt}$. Comment about Post #3 removed.

 

[Istiak]

Two things to correct in your version:
1) Each function, $r_i$, is also directly a function of $t$. Your version does not include that.
2) Apparantly, each $q_i$ is a function only of $t$, so there is no partial derivative of $q_i(t)$. It should be the total derivative, $\dot{q}$.
I believe that when you correct those, your answer will match the book.

Should I take $$r_i = r_i(q_1+q_2+...+q_n+t)$$? I think I was confusing with comma. Cause, when I took this I saw my answer matched. $$v_i=\frac{dr_i}{dt}=\sum_k \frac{\partial r_i}{\partial q_k}\dot{q}_k+\frac{\partial r_i}{\partial t}$$ Here, I am adding q1,q2,...q_n by that Sigma summation. Hence, my answer matched. But, I don't know why I took comma earlier? Does comma actually represent something?

 

[FactChecker]

No. It should be a comma, meaning that $r_i$ is a function of all of those variables, but not necessarily in the simple summation way that you suggest. The change in $r_i$ is the summation of the partial changes due to all its input variables. The equation that you got for $v_i$ is correct even if all those input variables are not added together as one input.

 

[Gaussian97]

Should I take $$r_i = r_i(q_1+q_2+...+q_n+t)$$? I think I was confusing with comma. Cause, when I took this I saw my answer matched. $$v_i=\frac{dr_i}{dt}=\sum_k \frac{\partial r_i}{\partial q_k}\dot{q}_k+\frac{\partial r_i}{\partial t}$$

$$\frac{\partial f}{\partial x} \cdot \frac{\partial x}{\partial y} \cdot \frac{\partial y}{\partial z}$$

This is not correct.
I think you have some fundamental problems with the chain rule. I would suggest you to leave this particular problem and spend 1 or 2 days reviewing multivariable calculus to refresh all your knowledge about the chain rule.
If you have never studied multivariable calculus, then you should really spend time to learn it, it is a fundamental tool for almost any physics. So, you will find it impossible to learn physics in a proper way without this mathematics.

 

[Istiak]

This is not correct.
I think you have some fundamental problems with the chain rule. I would suggest you to leave this particular problem and spend 1 or 2 days reviewing multivariable calculus to refresh all your knowledge about the chain rule.
If you have never studied multivariable calculus, then you should really spend time to learn it, it is a fundamental tool for almost any physics. So, you will find it impossible to learn physics in a proper way without this mathematics.

If you have a function $f(x,y,z)$ with $x,y$ functions of $z$. How would you use the chain rule to get
$\frac{df(x,y,z)}{dz}$?

Is $$\frac{df(x,y,z)}{dz}=\frac{\partial f}{\partial x}\cdot \frac{\partial x}{\partial z}+\frac{\partial f}{\partial y} \cdot \frac{\partial y}{\partial z}+\frac{\partial f}{\partial z}$$? I am confused with the last one. I am not sure if that's correct one...

 

[Gaussian97]

Ok, this looks much better.
Just keep an eye on the terms $\frac{\partial x}{\partial z}$, $\frac{\partial y}{\partial z}$. If $x$ and $y$ are functions of only $z$ there's no problem. But what would happen if $x$ has the form $x=x\left(z, f(z)\right)$, for example if $x$ depends also on $y$. Does this affect the result?

 

[FactChecker]

Is $$\frac{df(x,y,z)}{dz}=\frac{\partial f}{\partial x}\cdot \frac{\partial x}{\partial z}+\frac{\partial f}{\partial y} \cdot \frac{\partial y}{\partial z}+\frac{\partial f}{\partial z}$$? I am confused with the last one. I am not sure if that's correct one...

Yes, that is correct. The last one is actually the simplest since there is no intermediate variable (like x and y) between f and z.

 

[Istiak]

Ok, this looks much better.
Just keep an eye on the terms $\frac{\partial x}{\partial z}$, $\frac{\partial y}{\partial z}$. If $x$ and $y$ are functions of only $z$ there's no problem. But what would happen if $x$ has the form $x=x\left(z, f(z)\right)$, for example if $x$ depends also on $y$. Does this affect the result?

Just writing what I got in my last equation for partial x.

$$\frac{\partial f}{\partial x} \cdot \frac{\partial x}{\partial z}=\frac{\partial f}{\partial x} (\frac{\partial x}{\partial y} \cdot \frac{\partial y}{\partial z}+\frac{\partial x}{\partial z})$$

I think it does.

 

[Istiak]

Yes, that is correct. The last one is actually the simplest since there is no intermediate variable (like x and y) between f and z.

Thanks! I thought chain rule is same as "normal" differentiation. That's why I didn't study that. I had studied Multivariable calculus. But, I haven't completed some topics in Multivariable calculus which is Taylor, Maclaurin and Leibniz.

 

[Gaussian97]

Just writing what I got in my last equation for partial x.

$$\frac{\partial f}{\partial x} \cdot \frac{\partial x}{\partial z}=\frac{\partial f}{\partial x} (\frac{\partial x}{\partial y} \cdot \frac{\partial y}{\partial z}+\frac{\partial x}{\partial z})$$

I think it does.

Nop, that's not correct notice that you can cancel the $\frac{\partial f}{\partial x}\frac{\partial x}{\partial z}$ term and obtain the relation
$$\frac{\partial f}{\partial x} \frac{\partial x}{\partial y} \frac{\partial y}{\partial z}=0$$
which, of course, is not true in general.

To avoid this, the $\frac{\partial x}{\partial z}$ in the LHS and the RHS must have different meanings, (in such case we would no longer be able to cancel them).
Do you know what the difference should be?

 

[FactChecker]

Thanks! I thought chain rule is same as "normal" differentiation. That's why I didn't study that.

They are closely related. If y changes twice as fast as x and z changes three times as fast as y, then z changes 6 (=2*3) times as fast as x.

 

[Leo Liu]

I could be wrong but here is my ¥0.20:
As noted above, a chain rule concerning n+1 vars (extra 1 due to time var) is applied. Here I am assuing a $q_i$ is a function of time.
Whenever I take the derivative of a multivariable function I draw a diagram looks like the one below to help me figure out the relationship between each set of variables. In such a diagram the direction of an arrow connects a dependent variable (tail) and a independent variable (head).

_{(I love my macbook's trackpad. I didn't even need a stylus pen to draw it.)}
$$\mathbf v=\frac{\partial \mathbf r}{\partial t}=\frac{\partial \mathbf r}{\partial q_1}\cdot \frac{d q_1}{dt}+\frac{\partial \mathbf r}{\partial q_2}\cdot \frac{d q_2}{dt}+\dots+\frac{\partial \mathbf r}{\partial q_n}\cdot \frac{d q_n}{dt}+\frac{\partial\mathbf r}{\partial t}$$