## 44 Reputation

17 years, 36 days

## Thanks again for your time....

Thanks again for your time. I'm starting to see the relation, but the reasoning behind it is still a bit ahead of me. :) Cheers :D edit - Oh my. I watched the animations a few times more, and it clicked. It totally makes sense now. Funny how you tend to get most of your 'eureka' moments at 5:45 AM. Great work and thank you, Alec - I hope you don't mind me noting you for credit for this assignment.

## Aha, I think the first one...

Aha, I think the first one was enough. I know you're not a math teacher, but I was wondering if you could help me understand what you're doing. The solve({func1, func2}, {range1=a..b, range2=c..d}) is simple enough, and if what I understand about the diff function is correct, you're finding the first and second derivatives of h3d(x,k) in relation to x, and the same with the second and third. So to sum it up, you're finding the first and second derivatives of h3d(x,k) in relation to x, then maple solves for x, substitutes, finds a value for k, and plugs that in to one of the original functions to find x? I understand why you're solving in relation to x, but why compare the first and second and then second and third? Why not third and fourth, or further, fourth and fifth? If I'm totally off on what's going on here, don't waste the effort on trying to explain a few weeks worth of course work in one reply :) I'll just submit the trial and error, and learn the relations of 2-variable functions more gradually :)

## Her exact words were 'a...

Her exact words were 'a value as accurately as possible.' Well, it's nice to know some math junkies were as perplexed as I was (I usually catch on to concepts fairly quickly). I suppose trial and error is the best way. I'll have to ask her if there is an exact way afterwards. Curiosity will kill me on this one for sure. Thank you both for your promt help though. I have my values to 4 decimal places, which I think is enough (units are metres). edit - Wow, didn't even notice alec's addition. Thanks for that, I'll have to read up on how you actually got those numbers :D

## Yes, Thomas, that's exactly...

Yes, Thomas, that's exactly what I'm looking to find. Values for k (there should be two) in which the 2d plot of `h3d := x -> -5 * ( 1 - x^2 / 1600 ) * sin ( ( Pi * x ) / 1600 * ( x / 2 - k ) * ( x / 2 - 20 ) )^2;` changes amount of minimas where x=0..40. As stated before, we are given when k = 3 there are 4 minias, when k = 6 there are 3 minimas, and when k=9 there are 2. The animation was pretty good at helping me trial-and-error them (it's a pretty cool function as well), but I was curious if there was a way to grab exact answers. I may have to be tip-toed through this, as the 2nd variable tends to overwhelm me...

## So the best way is trial and...

So the best way is trial and error pretty much? Thanks for the help thus far :)
 Page 1 of 1
﻿