Finding the second derivative of a given parametric equation

[chwala]

Homework Statement

see attached below

Relevant Equations

parametric equations differentiation

ok this is pretty straightforward to me, my question is on the order of differentiation, i know that:
$\frac {d^2y}{dx^2}=$$\frac {d}{dt}.$$\frac {dy}{dx}.$$\frac {dt}{dx}$
is it correct to have,
$\frac {d^2y}{dx^2}=$$\frac {d}{dt}$.$\frac {dt}{dx}$.$\frac {dy}{dx}$?
that is,
$\frac {d^2y}{dx^2}=$$\frac {d}{dt}$$\frac {dy}{dx}$$\frac {dt}{dx}$=[$\frac {d}{dt}$$.\frac {1}{2t}$]$\frac {1}{t}$

does the order really matter?

 

[anuttarasammyak]

You can check it for yourself by calculating the both ways.

 

[chwala]

I already checked. I realize the same result...just wanted to know if the steps is accepted mathematically...

 

[chwala]

i am wrong...i just checked with another example

the order does matter...

thanks anuttarasammyak

 

[chwala]

How did they find the critical point$(-4,6)$? in post $4$, for the other two i.e $(5,4)$ and $(-3,-4)$ its pretty straightforward ...we use $t=1$ and $t=-1$.
on my attempt of the other critical value, i used;
$t=2$, this should also give us a critical value because it is undefined at this point and therefore we ought to have critical point $(-4,4)$ and not $(-4,6)$ as shown on text. someone clarify this...

 

[anuttarasammyak]

It is the point t=2 where $\frac{dy}{dx}$ diverges to plus-minus infinities.

plot https://www.wolframalpha.com/input/?i=plot+x=t^2-4t,+y=2t^3-6t+for+-2<t<3

 

[chwala]

It is the point t=2 where $\frac{dy}{dx}$ diverges to plus-minus infinities.

plot https://www.wolframalpha.com/input/?i=plot+x=t^2-4t,+y=2t^3-6t+for+-2<t<3

I can follow this, no issue here...

kindly check on my query/question...i am specifically asking on the critical points, $(x,y)$...

 

[anuttarasammyak]

It seems to depend on how your textbook is defining "critical points".
(5,4) and (-3,-4) are stationary points where maximum and minimum of y are observed.
(-4,6) is stationary point where minimum of x is observed.

 

[chwala]

It seems to depend on how your textbook is defining "critical points".
(5,4) and (-3,-4) are stationary points where maximum and minimum of y are observed.
(-4,6) is stationary point where maximum of x is observed.

why $(x,y)=(-4,6)$? and not $(-4,4)$...my interest is on the $y$ value, why $y=6$ and not $y=4$?

 

[anuttarasammyak]

Because x(2)=-4, y(2)=6. (-4,4) is not on the graph.

 

[chwala]

$y(2)=6$?, interesting hmmmm...

 

[Mark44]

Homework Statement:: see attached below
Relevant Equations:: parametric equations differentiation

View attachment 285817

ok this is pretty straightforward to me, my question is on the order of differentiation, i know that:
$\frac {d^2y}{dx^2}=$$\frac {d}{dt}.$$\frac {dy}{dx}.$$\frac {dt}{dx}$
is it correct to have,
$\frac {d^2y}{dx^2}=$$\frac {d}{dt}$.$\frac {dt}{dx}$.$\frac {dy}{dx}$?

Your notation is incorrect, and might be the source of your confusion.
The problem is that you have the product of d/dt, dy/dx, and dt/dx. What your notation should show is the operator d/dt being applied to the product of dy/dx and dt/dx. IOW, like this:
Edited:
$\frac d{dt}\left(\frac {dy}{dt} \cdot \frac {dt} {dx}\right)$.
Inside the parentheses, the order of the product doesn't matter.

that is,
$\frac {d^2y}{dx^2}=$$\frac {d}{dt}$$\frac {dy}{dx}$$\frac {dt}{dx}$=[$\frac {d}{dt}$$.\frac {1}{2t}$]$\frac {1}{t}$

does the order really matter?

In what you wrote above, the order does matter because it's $\frac d {dt}$ of $\frac{1}{2t}$, not the product $\frac d {dt}$ times $\frac{1}{2t}$,

 

[chwala]

Your notation is incorrect, and might be the source of your confusion.
The problem is that you have the product of d/dt, dy/dx, and dt/dx. What your notation should show is the operator d/dt being applied to the product of dy/dx and dt/dx. IOW, like this:
Edited:
$\frac d{dt}\left(\frac {dy}{dt} \cdot \frac {dt} {dx}\right)$.
Inside the parentheses, the order of the product doesn't matter.

In what you wrote above, the order does matter because it's $\frac d {dt}$ of $\frac{1}{2t}$, not the product $\frac d {dt}$ times $\frac{1}{2t}$,

Yeah, I saw my confusion Mark...cheers

 

[chwala]

Your notation is incorrect, and might be the source of your confusion.
The problem is that you have the product of d/dt, dy/dx, and dt/dx. What your notation should show is the operator d/dt being applied to the product of dy/dx and dt/dx. IOW, like this:
Edited:
$\frac d{dt}\left(\frac {dy}{dt} \cdot \frac {dt} {dx}\right)$.
Inside the parentheses, the order of the product doesn't matter.

In what you wrote above, the order does matter because it's $\frac d {dt}$ of $\frac{1}{2t}$, not the product $\frac d {dt}$ times $\frac{1}{2t}$,

Your remark below is not correct Mark...
"What your notation should show is the operator d/dt being applied to the product of dy/dx and dt/dx"...

my take is that,
$\frac {d}{dt}$ ($\frac {dy}{dx}$⋅$\frac {dt}{dx}$)...will not give us the desired result...rather, the form;
$(\frac {d}{dt}$ ⋅$\frac {dy}{dx})$$\frac {dt}{dx}$ should be the correct approach on this kind of problems.

 

[Mark44]

Your remark below is not correct Mark...
"What your notation should show is the operator d/dt being applied to the product of dy/dx and dt/dx"...

my take is that,
$\frac {d}{dt}$ ($\frac {dy}{dx}$⋅$\frac {dt}{dx}$)...will not give us the desired result...

Typo on my part. I inadvertently wrote dy/dx, but meant dy/dt inside the parentheses. I'll go back and correct what I wrote.

rather, the form;
$(\frac {d}{dt}$ ⋅$\frac {dy}{dx})$$\frac {dt}{dx}$ should be the correct approach on this kind of problems.

No, that's not it, either. Too be honest, I didn't check all your work -- I just spotted something that was in error and commented on it.
Since the problem is to find $\frac{d^2y}{dx^2}$, you need to take this derivative: $\frac d{dx}\left( \frac{dy}{dx}\right) = \frac d{dx}\left(\frac{dy}{dt} \cdot \frac{dt}{dx}\right)$.

To do this, you're going to need to use the chain rule and the product rule.

Your expression $\left(\frac d{dt} \cdot \frac{dy}{dx} \right)\frac{dt}{dx}$ is incorrect because it doesn't use the product rule.

Also, it's not a good idea to write $\frac d{dt} \cdot \frac{dy}{dx}$. To the reader it looks like you're multiplying the differentiation operator times dy/dx. That's not at all what's happening. What you're doing is applying the differentiation operator to dy/dx.

 

[chwala]

Typo on my part. I inadvertently wrote dy/dx, but meant dy/dt inside the parentheses. I'll go back and correct what I wrote.

No, that's not it, either. Too be honest, I didn't check all your work -- I just spotted something that was in error and commented on it.
Since the problem is to find $\frac{d^2y}{dx^2}$, you need to take this derivative: $\frac d{dx}\left( \frac{dy}{dx}\right) = \frac d{dx}\left(\frac{dy}{dt} \cdot \frac{dt}{dx}\right)$.

To do this, you're going to need to use the chain rule and the product rule.

Your expression $\left(\frac d{dt} \cdot \frac{dy}{dx} \right)\frac{dt}{dx}$ is incorrect because it doesn't use the product rule.

Also, it's not a good idea to write $\frac d{dt} \cdot \frac{dy}{dx}$. To the reader it looks like you're multiplying the differentiation operator times dy/dx. That's not at all what's happening. What you're doing is applying the differentiation operator to dy/dx.

Thanks Mark, i will not use the symbol {⋅}... in future working.

 

[chwala]

Thanks Mark, i will not use the symbol {⋅}... in future working.

we are seeking to find the second derivative and have already found the values of,
$\frac{dy}{dx}$ and $\frac{dx}{dt}$, now to find the desired second derivative, i think its just a straightforward method, like I've indicated...no need for further chain rule or product rule...
This is straight to the point,
$(\frac {d}{dt}$ of $\frac {dy}{dx})$times $\frac {dt}{dx}$... same as the textbook suggests as in my post $1$ above.

 

[chwala]

Typo on my part. I inadvertently wrote dy/dx, but meant dy/dt inside the parentheses. I'll go back and correct what I wrote.

No, that's not it, either. Too be honest, I didn't check all your work -- I just spotted something that was in error and commented on it.
Since the problem is to find $\frac{d^2y}{dx^2}$, you need to take this derivative: $\frac d{dx}\left( \frac{dy}{dx}\right) = \frac d{dx}\left(\frac{dy}{dt} \cdot \frac{dt}{dx}\right)$.

To do this, you're going to need to use the chain rule and the product rule.

Your expression $\left(\frac d{dt} \cdot \frac{dy}{dx} \right)\frac{dt}{dx}$ is incorrect because it doesn't use the product rule.

Also, it's not a good idea to write $\frac d{dt} \cdot \frac{dy}{dx}$. To the reader it looks like you're multiplying the differentiation operator times dy/dx. That's not at all what's happening. What you're doing is applying the differentiation operator to dy/dx.

The expression is correct...

 

[Mark44]

The expression is correct...

Yes

i think its just a straightforward method, like I've indicated...no need for further chain rule or product rule...

Maybe no further chain rule or product rule, but in going from this:
$$\frac d{dx}\left(\frac{dy}{dt} \cdot \frac{dt}{dx}\right) \dots$$
to this:
$$\frac {dy}{dt} \cdot \frac d{dx}\left( \frac{dt}{dx}\right)$$
you have to use 1) the product rule (one of the terms in the product turns out to be zero), and 2) the chain rule.

You don't show that work, so it's not clear to me that you realize this.

 

[chwala]

YesMaybe no further chain rule or product rule, but in going from this:
$$\frac d{dx}\left(\frac{dy}{dt} \cdot \frac{dt}{dx}\right) \dots$$
to this:
$$\frac {dy}{dt} \cdot \frac d{dx}\left( \frac{dt}{dx}\right)$$
you have to use 1) the product rule (one of the terms in the product turns out to be zero), and 2) the chain rule.

You don't show that work, so it's not clear to me that you realize this.

I fully understand what you are saying, its pretty obvious that in finding the first derivative, one has to use chain rule... I know that.
My interest was on dealing with the part of finding the second derivative. Thanks Mark, great day...

 

[Mark44]

My interest was on dealing with the part of finding the second derivative.

Which is what I was explaining...