Application of Calculus to the Physical World

[Jefferson]

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.

[blyatman]

d is a function of l and r, both of which are constants. Hence, d is a constant.

[Jefferson]

Hi,
Thanks for replying.

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]

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.

[DrDusk]

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

[RuiAce]

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]

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 .