Calculating 2nd Total Derivative of u w.r.t. t

[dyn]

Homework Statement

If u = f(x,y) and x=x(t) and y=y(t) then find the 2nd total derivative of u with respect to t

Relevant Equations

chain rule

du/dt = (∂f/∂x)(dx/dt) + (∂f/∂y)(dy/dt)
So i write the operator as
d/dt = (dx/dt)(∂/∂x) + (dy/dt)(∂/∂y) and apply it to du/dt ; in the operator it is the partial derivative that acts on du/dt which involves using the product rule.
I am having a problem with the term involving (∂/∂x) (dx/dt) ; dx/dt is a function of t only so i can't figure out what the partial derivative is

 

[anuttarasammyak]

$$\frac{\partial}{\partial x}\frac{dx}{dt}=0$$
since dx/dt is not a function of x but t.
$$\frac{\partial}{\partial x}\frac{\partial u}{\partial x}=\frac{\partial^2 u}{\partial x^2}$$
$$\frac{\partial}{\partial x}\frac{\partial u}{\partial y}=\frac{\partial}{\partial y}\frac{\partial u}{\partial x}=\frac{\partial^2 u}{\partial x \partial y}$$

 

[dyn]

Thank you. You have confirmed what i thought but the answer i have written down in front of me includes 2 terms that i cannot see where they came from
(∂u/∂x) (d²x/dt²) + (∂u/∂y) (d²y/dt²)

 

[Office_Shredder]

It would probably help if you just wrote step by step what you did to get your answer

 

[dyn]

((dx/dt)(∂/∂x) + (dy/dt)(∂/∂y)) ((∂u/∂x)(dx/dt) + (∂u/∂y)(dy/dt))

= (dx/dt)²(∂²u/∂x²) + 2 (∂u/∂x∂y )(dx/dt)(dy/dt) + (dy/dt)²(∂²u/∂y²)

That is the answer i get. But the anwer i have found includes the 2 terms mentioned in #3. I cannot see where they come from

 

[PeroK]

((dx/dt)(∂/∂x) + (dy/dt)(∂/∂y)) ((∂u/∂x)(dx/dt) + (∂u/∂y)(dy/dt))

= (dx/dt)²(∂²u/∂x²) + 2 (∂u/∂x∂y )(dx/dt)(dy/dt) + (dy/dt)²(∂²u/∂y²)

That is the answer i get. But the anwer i have found includes the 2 terms mentioned in #3. I cannot see where they come from

You need to write this out properly:
$$u = f(x(t), y(t))$$
$$\frac{du}{dt}= \frac{\partial f}{\partial x}x'(t) + \dots$$
Now, to avoid any missed steps, replace the partial derivatives with funtions of $x(t), y(t)$. Because that's what they are. So you have:
$$\frac{du}{dt} = g(x(t), y(t))x'(t) + h(x(t), y(t))y'(t)$$
Now, differentiate that wrt $t$ and you'll get everything.

 

[dyn]

But the partial derivatives in the operator acting on du/dt in #5 are ∂/∂x and ∂/∂y so when they act on dx/dt and dy/dt they give zero because dx/dt and dy/dt are both only functions of t

 

[PeroK]

But the partial derivatives in the operator acting on du/dt in #5 are ∂/∂x and ∂/∂y so when they act on dx/dt and dy/dt they give zero because dx/dt and dy/dt are both only functions of t

$u$ is a function of $t$ only. $u$ is a single variable function and the "total" derivative under the covers is the plain old ordinary derivative of $u(t)$.

Whatever applies to $u$ applies to $du/dt$.

 

[vela]

You need to include a $\partial/\partial t$ in because you have an explicit dependence on $t$ now from $x'(t)$ and $y'(t)$.
$$\frac d{dt} = \frac {dx}{dt}\frac \partial{\partial x} + \frac {dy}{dt}\frac \partial{\partial y} + \frac \partial{\partial t}$$

 

[dyn]

((dx/dt)(∂/∂x) + (dy/dt)(∂/∂y)) ((∂u/∂x)(dx/dt) + (∂u/∂y)(dy/dt))

= (dx/dt)²(∂²u/∂x²) + 2 (∂u/∂x∂y )(dx/dt)(dy/dt) + (dy/dt)²(∂²u/∂y²)

That is the answer i get. But the anwer i have found includes the 2 terms mentioned in #3. I cannot see where they come from

I'm getting confused now. I will apply the 1st term in the left hand operator bracket to the right hand bracket and show what i get using the product rule of differentiation

((dx/dt)(∂/∂x)) ((∂u/∂x)(dx/dt) + (∂u/∂y)(dy/dt))

= (dx/dt)²(∂²u/∂x²) + (dx/dt)(∂u/∂x) (∂/∂x)(dx/dt) + (dx/dt)(dy/dt) (∂²u/∂x∂y) + (dx/dt)(∂u/∂y) (∂/∂x) (dy/dt)

I get the 2nd and 4th terms to be zero as (∂/∂x) ( dx/dt) and (∂/∂x) (dy/dt) are zero as in the post above #2

 

[PeroK]

I get the 2nd and 4th terms to be zero as (∂/∂x) ( dx/dt) and (∂/∂x) (dy/dt) are zero as in the post above #2

It doesn't matter how many times you assert this. It's as wrong now as it was in post #2.

 

[vela]

I'm getting confused now. I will apply the 1st term in the left hand operator bracket to the right hand bracket and show what i get using the product rule of differentiation

((dx/dt)(∂/∂x)) ((∂u/∂x)(dx/dt) + (∂u/∂y)(dy/dt))

It doesn't make sense to do this to differentiate wrt $t$ because $x'(t)$ is a function of $t$. Instead, you can look at the problem this way:
\begin{align*}
\frac{d}{dt} \left(\frac{\partial u}{\partial x} \frac{dx}{dt}\right)
&= \left(\frac{d}{dt} \frac{\partial u}{\partial x}\right)\frac{dx}{dt} + \frac{\partial u}{\partial x} \left(\frac{d}{dt}\frac{dx}{dt}\right) \\
&= \left[\left(\frac{dx}{dt}\frac{\partial}{\partial x} + \frac{dy}{dt}\frac{\partial}{\partial y}\right) \frac{\partial u}{\partial x}\right]\frac{dx}{dt} + \frac{\partial u}{\partial x}\frac{d^2 x}{dt^2}
\end{align*}
Another way is to recognize that while $u=f(x,y)$ is a function of $x$ and $y$, its derivative $du/dt = g(t,x,y)$ is a function of $x$, $y$, and $t$. For example, if $u=x^2+y$, $x = \cos t$, and $y=\sin t$, then
$$\frac{du}{dt} = 2x(-\sin t) + \cos t.$$ Since $t$ appears explicitly in $u'$, you can't omit the $\partial/\partial t$ term in the total derivative that you could when differentiating $u$.

 

[PeroK]

Another way is to recognize that while $u=f(x,y)$ is a function of $x$ and $y$, its derivative $du/dt = g(t,x,y)$ is a function of $x$, $y$, and $t$. For example, if $u=x^2+y$, $x = \cos t$, and $y=\sin t$, then
$$\frac{du}{dt} = 2x(-\sin t) + \cos t.$$ Since $t$ appears explicitly in $u'$, you can't omit the $\partial/\partial t$ term in the total derivative that you could when differentiating $u$.

In that example, both $u$ and $\frac{du}{dt}$ are simply functions of a single variable $t$. The "total" derivative is just the ordinary derivative when it comes to it:
$$\frac{du}{dt}(t) = \lim_{h \rightarrow 0} \frac {u(t + h) - u(t)}{h}$$
In your example:
$$u(t) = \cos^2 t + \sin t, \ \ \ \frac{du}{dt}(t) = -2\cos t \sin t + \cos t$$
Neither are multivariable functions.

A lot of the mystique of the total derivative is lost when you realize it's just an ordinary derivative!

 

[dyn]

It doesn't matter how many times you assert this. It's as wrong now as it was in post #2.

Post #2 told me that for u=u( x , y ) with x=x(t) and y=y(t) ; then ∂/∂x (dx/dt) = 0 because dx/dt is not a function of x which does make sense but i am told this is wrong. Why is it wrong ?
I like the method used in #12 but i would like to know how to handle ∂/∂x (dx/dt) in this situation

 

[PeroK]

Post #2 told me that for u=u( x , y ) with x=x(t) and y=y(t) ; then ∂/∂x (dx/dt) = 0 because dx/dt is not a function of x which does make sense but i am told this is wrong. Why is it wrong ?
I like the method used in #12 but i would like to know how to handle ∂/∂x (dx/dt) in this situation

If you are differentiating with respect to $t$, then something like ∂/∂x (dx/dt) or ∂/∂y (dx/dt) don't make sense. You are trying to apply the chain rule to simple functions of $t$. Let's say you had a function $g(t)$, what you are trying to do is:
$$\frac{dg}{dt} = \frac{\partial g}{\partial x}x'(t) + \frac{\partial g}{\partial y}y'(t)$$
But, $g$ is a single-variable function of $t$. Its partial derivatives with respect to $x$ and $y$ are not defined. There are no such functions. In the same way ∂/∂x (dx/dt) is simply not defined. $dx/dt$ is not a multi-variable function of $x$ and $y$ in the first place. $dx/dt$ is a single-variable function of $t$.

This is how it should go. You have your first derivative:
$$\frac{du}{dt} = \frac{\partial f}{\partial x}(x(t), y(t))x'(t) + \frac{\partial f}{\partial y}(x(t), y(t))y'(t)$$
Note that I've emphasised that the partial derivatives are multi-variable functions. Then you just take the second derivative noting that the two partial derivatives must be treated like multi-variable functions when you differentiate them:
$$\frac{d^2u}{dt^2} = \frac{d}{dt}\big (\frac{\partial f}{\partial x}(x(t), y(t))x'(t)\big ) + \frac{d}{dt}\big ( \frac{\partial f}{\partial y}(x(t), y(t))y'(t)\big )$$
Now, I'll just do the first term (using the product rule):
$$\frac{d^2u}{dt^2} = \frac{d}{dt}\big (\frac{\partial f}{\partial x}(x(t), y(t))\big ) x'(t) + \big (\frac{\partial f}{\partial x}(x(t), y(t))\big )x''(t) + \dots$$
Then, you apply the chain rule to:
$$\frac{d}{dt}\big (\frac{\partial f}{\partial x}(x(t), y(t))\big )$$
Because $\frac{\partial f}{\partial x}$ is a multi-variable function of $x$ and $y$.

 

[dyn]

Thank you

 

[anuttarasammyak]

Re: my post #2

I admit this investigation is not necessary for this problem. Leaving the problem just for curiosity,

$$\frac{\partial}{\partial x}\frac{d}{dt}x=\frac{d}{dt}\frac{\partial}{\partial x}x=\frac{d}{dt} 1 = 0$$

Does this formula hold mathematically ?

 

[PeroK]

Re: my post #2

I admit this investigation is not necessary for this problem. Leaving the problem just for curiosity,

$$\frac{\partial}{\partial x}\frac{d}{dt}x=\frac{d}{dt}\frac{\partial}{\partial x}x=\frac{d}{dt} 1 = 0$$

Does this formula hold mathematically ?

If you take the example above: $x = \cos t, y = \sin t$, then $\frac{dx}{dt} = -\sin t = -y$. And, in that case if we express $\frac{dx}{dt}$ as a function of $x, y$, then $\frac{\partial}{\partial y}\frac{dx}{dt} = 1$. And it's equally easy to construct an example where $\frac{\partial}{\partial x}\frac{dx}{dt} \ne 0$.

In one sense it is valid to treat $x'(t)$ as a multi-variable function of $x, y$ - in order to differentiate it wrt $t$. But, it is far simpler to write $\frac{d}{dt}\frac{dx}{dt} = \frac{d^2x}{dt^2} = x''(t)$.

Even if we allow the roundabout method of differentiating $x'(t)$ the OP's conclusion is wrong.

 

[anuttarasammyak]

Thanks PeroK.

And it's equally easy to construct an example where ∂∂xdxdt≠0.

Say
$$x=e^t$$,
$$\frac{d}{dx}\frac{d}{dt}e^t=\frac{d}{dx}e^t=\frac{d}{dx}x=1$$
$$\frac{d}{dt}\frac{d}{dx}x=\frac{d}{dt}1=0$$
$\frac{d}{dx}$ and $\frac{d}{dt}$ are not commutable.

 

[dyn]

If you are differentiating with respect to $t$, then something like ∂/∂x (dx/dt) or ∂/∂y (dx/dt) don't make sense. You are trying to apply the chain rule to simple functions of $t$. Let's say you had a function $g(t)$, what you are trying to do is:
$$\frac{dg}{dt} = \frac{\partial g}{\partial x}x'(t) + \frac{\partial g}{\partial y}y'(t)$$
But, $g$ is a single-variable function of $t$. Its partial derivatives with respect to $x$ and $y$ are not defined. There are no such functions. In the same way ∂/∂x (dx/dt) is simply not defined. $dx/dt$ is not a multi-variable function of $x$ and $y$ in the first place. $dx/dt$ is a single-variable function of $t$.

This is how it should go. You have your first derivative:
$$\frac{du}{dt} = \frac{\partial f}{\partial x}(x(t), y(t))x'(t) + \frac{\partial f}{\partial y}(x(t), y(t))y'(t)$$

I'm getting even more confused now. You are saying that the 1st equation here is not valid but the 2nd equation is valid. They look the same to me. In my original method i used the 2nd equation and then applied the differential operator to it in order to get the 2nd derivative. But then it failed.
I am using the general method that works fine for examples such as transforming Laplace's equation in 2-D from (x,y) to (r , θ). The example in my OP should be much easier but it's causing me many more problems

 

[PeroK]

Please differentiate the function $g(t) = \sin t$:

a) Using the ordinary derivative;

b) Using the multivariable chain rule.

Even if you can make method b) work by introducing additional functions $x(t)$ and $y(t)$, why on Earth would you want to do it that way?

Similarly, if you have a function $x(t)$, then $\frac{dx}{dt} = x'(t)$ and $\frac{d^2x}{dt^2} = x''(t)$. There is absolutely no reason to introduce a multivariable chain rule.

Moreover, when you have unnecessarily involved a multi-variable chain rule instead of a simple ordinary derivative, you've done it wrong.

Why do you refuse to use the ordinary derivative when it's appropriate?

 

[vela]

If you have a function $f(x,y,t)$, then its derivative wrt $t$ is given by
$$\frac{df}{dt} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial t} + \frac{\partial f}{\partial t}.$$ If $f$ has no explicit dependence on $t$, the last term vanishes, and the derivative reduces to
$$\frac{df}{dt} = \frac{\partial f}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial f}{\partial y}\frac{\partial y}{\partial t}.$$ In your problem, $u$ depends on $t$ only indirectly though $x$ and $y$, so you can use the latter expression to calculate the first derivative. Its derivative $du/dt$ does, however, depend on $t$ directly because of the factors $x'(t)$ and $y'(t)$, so you have to use the more general expression.

I'll note this is perhaps the third time I've made this point, but for some reason, you keep ignoring it and insist on using the wrong operator to calculate the second derivative and wonder why you get the wrong answer.

 

[PeroK]

So i write the operator as
d/dt = (dx/dt)(∂/∂x) + (dy/dt)(∂/∂y)

This is the root of the problem. That is only valid when applied to multivariable functions of $x$ and $y$. That doesn't apply to the ordinary derivative of single-variable function of $t$. For example:
$$\frac{d}{dt}\sin t = [(dx/dt)(∂/∂x) + (dy/dt)(∂/∂y)]\sin t$$
is nonsensical.

 

[vela]

Moreover, when you have unnecessarily involved a multi-variable chain rule instead of a simple ordinary derivative, you've done it wrong.

Why do you refuse to use the ordinary derivative when it's appropriate?

Because it's not appropriate here. The problem the OP was given is to find the derivative by applying the multivariable chain rule, and the OP does need to understand how to solve problems using this method.

 

[PeroK]

Because it's not appropriate here. The problem the OP was given is to find the derivative by applying the multivariable chain rule, and the OP does need to understand how to solve problems using this method.

But not to differentiate $x(t)$. That's where it's all going wrong. Trying to treat $x(t)$ as a multivariable function of $x$ and $y$.

 

[dyn]

Thank you for your time everyone. I understand that it is a single variable problem but i was trying to use a different approach

 

[PeroK]

Thank you for your time everyone. I understand that it is a single variable problem but i was trying to use a different approach

It's not a single variable problem. $f(x,y)$ is a multivariable function. But $x(t)$ and $y(t)$ are single variable functions of $t$. This is where you go wrong by trying to apply tge chain rule to those.

 

[PeroK]

Homework Statement:: If u = f(x,y) and x=x(t) and y=y(t) then find the 2nd total derivative of u with respect to t

Let's go back to the beginning. $u(t)$ is a single-variable function of $t$. But, it is constructed using the composition of a multi-variable function $f(x, y)$ and two single-variable functions $x(t)$ and $y(t)$. Therefore, we have:
$$u(t) = f(x(t), y(t))$$
In order to differentiate $u$ with respect to $t$ we must use the multi-variable chain rule (which you know):
$$\frac{du}{dt} \equiv u'(t) \equiv \dot u(t) = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}$$
Now, let's analyse the function $\frac{du}{dt}$. Note that $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are multi-variable functions of $x,y$; and, $\frac{dx}{dt}$ and $\frac{dy}{dt}$ are single-variable functions of $t$.

We must take this into account when we take the derivative $\frac{d^2u}{dt^2} \equiv \frac{d}{dt}\frac{du}{dt}$.

Now, many people make the mistake of not treating the partial derivatives as multi-variable functions. You are making the opposite mistake of trying to treat $\frac{dx}{dt}$ and $\frac{dy}{dt}$ as some sort of multi-variable functions. With the conclusion that for all functions $x(t)$ we have $\frac{d^2x}{dt^2}= 0$.

In a nutshell that is what you have done. Taken a function $x(t)$ and by a misuse of multi-variable calculus on a simple single-variable function concluded that $\frac{d^2x}{dt^2}= 0$ in all cases.

Instead, you have simply $\frac{d}{dt}\frac{du}{dt} = \frac{d^2u}{dt^2} \equiv u''(t) \equiv \ddot u(t)$. You cannot express that derivative in any other terms.

You do, of course, have to apply the multi-variable chain rule when differentiating $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$.

 

[dyn]

I thank everyone for their replies but I've had enough of this problem. I wish i'd never seen it !