Temperature cooling function

So I know that if a function is continuous then f(a) must be defined and that as lim(x->a) then f(x) must exist, but I’m not sure how to apply it to this question. Or am I overthinking it?

I’m also not sure how to start with question ii. Is it just the lim(t->infinity) or do I need to calculate any of the constants?

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Reply 1

mqb2766

21

I doubt there is any great analysis required for i) as an exponential is obviously continuous. Maybe verify the initial condition so
T(0) = T_0
which is easy enough. Cant say for certain though.

for ii) its the steady state temperature so as t->inf, so what happens to the exponential ... so the temperature tends to ...

(edited 6 months ago)

Reply 2

username6352379

OP

5

Original post by mqb2766

I doubt there is any great analysis required for i) as an exponential is obviously continuous. Maybe verify the initial condition so
T(0) = T_0
which is easy enough. Cant say for certain though.

for ii) its the steady state temperature so as t->inf, so what happens to the exponential ... so the temperature tends to ...

For the first one I literally just wrote T(t) is continuous as T(t) is defined for all values of t, I just thought it seemed a bit too simple for the answer to just be that.

For the second one is it just Ta? As e^-(k)infinity then (T0-Ta)(e^-k(infinity)) will tend to 0 so it will just become Ta+0 = Ta. Is that right?

(edited 6 months ago)

Reply 3

mqb2766

21

Original post by kwikmaffs

For the first one I literally just wrote T(t) is continuous as T(t) is defined for all values of t, I just thought it seemed a bit too simple for the answer to just be that.

For the second one is it just Ta? As e^-(k)infinity then (T0-Ta)(e^-k(infinity)) will tend to 0 so it will just become Ta+0 = Ta. Is that right?

For ii) I agree, the temperature will cool to the ambient temperature. Its worth noting the form of the temperature function as its
ambient + initial difference * decaying exponential

For i) The function is a simple linear transformation (scale by iniital difference and translate by ambient temperature) of an exponential so as an exponential is continuous, so is a linear transformation of it. Saying "its defined" means little in terms of being continuous, and discontinuous functions are defined as well.

(edited 6 months ago)

Reply 4

DFranklin

18

Original post by kwikmaffs

For the first one I literally just wrote T(t) is continuous as T(t) is defined for all values of t, I just thought it seemed a bit too simple for the answer to just be that.

That's not a valid argument. If I define f(x) = 0 except where x = 0, f(0) = 1, then f is defined for all values of t, but it's not continuous at x = 0.

Realistically, I don't think there's a sensible answer here other than "Yes, T is continuous"; if you really wanted to, you could say "T is continuous because it is formed by the addition, subtraction, multiplication and composition of continuous functions", but TBH I don't feel that adds a lot.

For the second one is it just Ta? As e^-(k)infinity then (T0-Ta)(e^-k(infinity)) will tend to 0 so it will just become Ta+0 = Ta. Is that right?

The limit is Ta, but your justification is "not even wrong" TBH.

"As e^-(k)infinity"? What is this even supposed to mean? You haven't said what you think it equals, or what it tends to.
Also, if you are attempting to be precise about limits, continuity, etc then infinity is not a value that can be put into a formula (values can tend to infinity, they can't equal infinity).
"(T0-Ta)(e^-k(infinity)) will tend to 0" See the comment above about infinity. But If you are being informal enough to treat infinity as a value, then "(T0-Ta)(e^-k(infinity))" is a constant (that equals 0).

Reply 5

username6352379

OP

5

Original post by mqb2766

For ii) I agree, the temperature will cool to the ambient temperature. Its worth noting the form of the temperature function as its
ambient + initial difference * decaying exponential

For i) The function is a simple linear transformation (scale by iniital difference and translate by ambient temperature) of an exponential so as an exponential is continuous, so is a linear transformation of it. Saying "its defined" means little in terms of being continuous, and discontinuous functions are defined as well.

Okay thank you, that makes a lot of sense. I haven’t done any maths for 3.5 years and just recently started an engineering degree so it’s a bit of a shock to the system. Thank you again :smile:

Reply 6

DFranklin

18

Original post by kwikmaffs

Okay thank you, that makes a lot of sense. I haven’t done any maths for 3.5 years and just recently started an engineering degree so it’s a bit of a shock to the system. Thank you again :smile:

For future info: It's useful to give the context that it's an engineering degree.

There are typically quite different expectations for how "correct" you need to be when talking about limits, continuity etc. in engineering v.s. mathematics.

Reply 7

username6352379

OP

5

Original post by DFranklin

That's not a valid argument. If I define f(x) = 0 except where x = 0, f(0) = 1, then f is defined for all values of t, but it's not continuous at x = 0.

Realistically, I don't think there's a sensible answer here other than "Yes, T is continuous"; if you really wanted to, you could say "T is continuous because it is formed by the addition, subtraction, multiplication and composition of continuous functions", but TBH I don't feel that adds a lot.

The limit is Ta, but your justification is "not even wrong" TBH.

"As e^-(k)infinity"? What is this even supposed to mean? You haven't said what you think it equals, or what it tends to.
Also, if you are attempting to be precise about limits, continuity, etc then infinity is not a value that can be put into a formula (values can tend to infinity, they can't equal infinity).
"(T0-Ta)(e^-k(infinity)) will tend to 0" See the comment above about infinity. But If you are being informal enough to treat infinity as a value, then "(T0-Ta)(e^-k(infinity))" is a constant (that equals 0).

I mostly just wrote that because I didn’t know if the simple answer of “T(t) is continuous” would be sufficient. I see now that it would be.

And yeah I skipped over some words when typing my explanation of the second part. But on my worksheet irl I’ve written: as t-> infinity, (T0-Ta)(e^-kt) tends to 0, therefore final temperature = Ta.

Thanks for the precise explanation, I need to get better at articulating myself with limits and things.

Reply 8

username6352379

OP

5

Original post by DFranklin

For future info: It's useful to give the context that it's an engineering degree.

There are typically quite different expectations for how "correct" you need to be when talking about limits, continuity etc. in engineering v.s. mathematics.

Okay sorry, I’ll mention it next time. And thank you again, I appreciate the help.

Reply 9

DFranklin

18

Original post by kwikmaffs

Okay sorry, I’ll mention it next time. And thank you again, I appreciate the help.

No need to apologise; it's just I probably wouldn't have been quite so dogmatic about "Ur doin' it 'rong" if I'd known.

What you wrote the second time (i.e. what you wrote in the notebook) is absolutely fine, by the way.

[It is probably worth rereading stuff you transcribe before you post to check it still makes sense. People on here are actually fairly accepting of "I wrote this funny because I'm using a keyboard", but if there's ambiguity, you'll typically get asked to clarify rather than people wanting to guess at what you meant.

The forum also supports LaTeX (maths typesetting system). You might find it worth getting familiar with (it's also used for most published mathematics).

For example: "as t-> infinity, (T0-Ta)(e^-kt) tends to 0" could be entered as
[noparse]