 December 09, 2019, 11:02:27 am

### AuthorTopic: Application of Calculus to the Physical World  (Read 392 times) Tweet Share

0 Members and 1 Guest are viewing this topic.

#### Jefferson ##### Application of Calculus to the Physical World
« on: July 18, 2019, 03:29:38 pm »
0
1992 Q4 (c) ii. (see attachment)

Could someone explain to me why d is a constant?

Why does it not change with length l or radius r? (i.e. a variable)

Thank you.
« Last Edit: July 18, 2019, 03:34:07 pm by Jefferson »

#### blyatman

• Trailblazer
• • Posts: 43
• Blyat
• Respect: +2 ##### Re: Application of Calculus to the Physical World
« Reply #1 on: July 18, 2019, 03:35:42 pm »
0
d is a function of l and r, both of which are constants. Hence, d is a constant.
Computational Fluid Dynamics Engineer
Research background: General Relativity (Gravitational Astrophysics Research Group, Sydney Institute for Astronomy, USYD)
-----------------------------------------------------------------------------------------------------------------------------------------------------------------------
M.S. (Aeronautics and Astronautics, Purdue University) (Major: Aerodynamics. Minor: Propulsion)
M.Phil. (Aerospace Engineering)
B.Eng. (Aerospace Engineering, Honours Class I)

#### Jefferson ##### Re: Application of Calculus to the Physical World
« Reply #2 on: July 18, 2019, 03:45:53 pm »
+1
Hi,

I think 'l' and 'r' are variables, otherwise differentiating will just return nothing (i.e. 0).

I've just read the question again and it does say

"for the given d",

So it is indeed a constant.

It was my mistake.

#### blyatman

• Trailblazer
• • Posts: 43
• Blyat
• Respect: +2 ##### Re: Application of Calculus to the Physical World
« Reply #3 on: July 18, 2019, 03:51:35 pm »
0
Sorry I didn't read the question properly. Now that I have, I don't fully understand what they're doing. They differentiated V w.r.t l, so l is a variable. But d is a function of l, so I get what you're saying now about d being a variable. Bit confusing there, but if you've figured it out then all is well.
« Last Edit: July 18, 2019, 03:53:59 pm by blyatman »
Computational Fluid Dynamics Engineer
Research background: General Relativity (Gravitational Astrophysics Research Group, Sydney Institute for Astronomy, USYD)
-----------------------------------------------------------------------------------------------------------------------------------------------------------------------
M.S. (Aeronautics and Astronautics, Purdue University) (Major: Aerodynamics. Minor: Propulsion)
M.Phil. (Aerospace Engineering)
B.Eng. (Aerospace Engineering, Honours Class I)

#### DrDusk ##### Re: Application of Calculus to the Physical World
« Reply #4 on: July 18, 2019, 05:47:54 pm »
0
Sorry I didn't read the question properly. Now that I have, I don't fully understand what they're doing. They differentiated V w.r.t l, so l is a variable. But d is a function of l, so I get what you're saying now about d being a variable. Bit confusing there, but if you've figured it out then all is well.

I guess technically then they've taken the Partial derivative
HSC/Prelim Physics tutor

#### RuiAce

• ATAR Notes Lecturer
• Honorary Moderator
• Great Wonder of ATAR Notes
•       • • Posts: 8614
• "All models are wrong, but some are useful."
• Respect: +2412 ##### Re: Application of Calculus to the Physical World
« Reply #5 on: July 19, 2019, 04:17:17 pm »
+5
I guess technically then they've taken the Partial derivative
It doesn't need to be a partial derivative here. Once you force $d$ to be a constant (as above, where it says $d$ is "given"), it essentially degenerates into a usual derivative.
1992 Q4 (c) ii. (see attachment)

Could someone explain to me why d is a constant?

Why does it not change with length l or radius r? (i.e. a variable)

Thank you.
The constancy of $d$ doesn't always have happen to be fair, and should depend on how $r$ and $l$ change. This is essentially illustrated by the result that they wanted you to prove in part i) : the result $d^2 = (2r)^2 + \left( \frac{l}{2} \right)^2$, which presumably you figured out given that your question was related to part ii).

However, that exact equation required in part i) illustrates that $d$, $r$ and $l$ are together related somehow. The reason why fixing $d$ is permissible here for part ii) is because the equation above allows the possibility that "increasing $r$ and decreasing $l$ appropriately at the same time, can be enough to create a net change of $0$," hence giving us a constancy in $d$. This also holds vice versa.

Intuitively speaking, what we're really trying to do is say that we can either pull the "pipe" longer and make the circle smaller, or collapse down the pipe and flatten the circle out to be bigger, without affecting what $d$ actually is here. We essentially wish to find the maximum volume possible, following that rule of thumb.

But of course, the question actually had to tell you that $d$ was fixed in advance, which you then spotted. This is just explaining why that is not justified.  #### Jefferson ##### Re: Application of Calculus to the Physical World
« Reply #6 on: July 25, 2019, 08:42:47 pm »
+1
It doesn't need to be a partial derivative here. Once you force $d$ to be a constant (as above, where it says $d$ is "given"), it essentially degenerates into a usual derivative. The constancy of $d$ doesn't always happen to be fair, and should depend on how $r$ and $l$ change. This is essentially illustrated by the result that they wanted you to prove in part i): the result $d^2 = (2r)^2 + \left( \frac{l}{2} \right)^2$, which presumably you figured out given that your question was related to part ii).

However, that exact equation required in part i) illustrates that $d$, $r$ and $l$ are together related somehow. The reason why fixing $d$ is permissible here for part ii) is because the equation above allows the possibility that "increasing $r$ and decreasing $l$ appropriately at the same time, can be enough to create a net change of $0$," hence giving us a constancy in $d$. This also holds vice versa.

Intuitively speaking, what we're really trying to do is say that we can either pull the "pipe" longer and make the circle smaller, or collapse down the pipe and flatten the circle out to be bigger, without affecting what $d$ actually is here. We essentially wish to find the maximum volume possible, following that rule of thumb.

But of course, the question actually had to tell you that $d$ was fixed in advance, which you then spotted. This is just explaining why that is not justified.

Thank you so much for your further insight! I'm usually much more interested in these connections than the solution to the question itself .