Silly calculus problem I can't seem to do

[MaxMath]

Besides asking the author of that solution, how about telling us your thoughts: What limits of integration would you consider using, and why?

Or, look it up. Here's a nice explanation I found for the general concept:

https://math.mit.edu/~jorloff/suppnotes/suppnotes03/d.pdf

By intuition, these are the right ones to be used. However, I was after mathematic rigour I guess!

The course note is very useful and interesting -- one aspect of the two forms of the answer to an IVP very well distilled. Thank you.

By the way, after reading that, I realised at least I had problems in my formula, in terms of the use of symbols

[imath]\displaystyle s(t) = 1 + \int_{0}^{t} v(t) \ dt[/imath]

Perhaps it should be written as

[imath]\displaystyle s(t) = 1 + \int_{0}^{t} v(T) \ dT[/imath]

This however brings about another question: what does it mean by saying [imath]v[/imath] is a function of [imath]t[/imath], when it seems we can also say [imath]v[/imath] is a function of [imath]s[/imath] (apparently), which seems to lead to [imath]v(s, t)[/imath]? What are the implications of we looking at it in these different ways?

 

[Dr.Peterson]

By the way, after reading that, I realised at least I had problems in my formula, in terms of the use of symbols

[imath]\displaystyle s(t) = 1 + \int_{0}^{t} v(t) \ dt[/imath]

Perhaps it should be written as

[imath]\displaystyle s(t) = 1 + \int_{0}^{t} v(T) \ dT[/imath]

This however brings about another question: what does it mean by saying [imath]v[/imath] is a function of [imath]t[/imath], when it seems we can also say [imath]v[/imath] is a function of [imath]s[/imath] (apparently), which seems to lead to [imath]v(s, t)[/imath]? What are the implications of we looking at it in these different ways?

Some people (including me) object to the use of the same variable in the integrand and in the limits; but I've chosen not to point that out. I do prefer it with this change.

As for v being a function of both s and t, no, it isn't really. It's the derivative of a function of t, and therefore itself a function of t. But there are a multitude of relationships, and you do have to be careful. I don't have time at the moment to try to dig into that.

 

[MaxMath]

Some people (including me) object to the use of the same variable in the integrand and in the limits;

I'm with you.

As for v being a function of both s and t, no, it isn't really. It's the derivative of a function of t, and therefore itself a function of t. But there are a multitude of relationships, and you do have to be careful.

That's exactly (part of) why this problem is tricky.

 

[Agent Smith]

I'm getting ...

[imath]\frac{\sqrt s}{6} - \frac{1}{6} = t[/imath]

For t = 1, [imath]s = \sqrt 7[/imath]

 

[Agent Smith]

I mean s = 49