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It is nice that you now understand all of this Ricky, but how do you know the limits of integration for y1 and y2 taking into account that the blade's length is changing by 10 centimeters.
The equations describe that change of 10 centimeters.
p0 = the vertex
p1 = point of intersection between y2 and y3
p2 = (0,0) (the end of the blade passes through this point)
||p1 - p0|| is the distance between p1 and p0:
||p1 - p0|| - 10 = ||p2 - p0||
But the wiper passes through both p1 and p2. So the wiper must be changing size.
is an articulate double module that alters in length during motion
If that line is directly out of your book, then I suggest you get another book...
Seriously, what kind of text book uses that language without explaining it? I haven't got a clue what it means.
Edit:
Whoops, missed this line:
A lorry has a downwards wiping windscreen wiper on a 40cm long metal arm.
Ok, so we start at +40, not +30. Then everything comes together.
y1= 1/50x^2 + 30
y2= 1/50x^2
y3= √(2)x + 40
y4= -√(2)x + 40
Everything else is the same.
is an articulate double module that alters in length during motion
What this means is that the distance from the vertex and the start of the rubber is constantly changing. The equations above show this distance changing, it's nothing extra. Just an explanation.
is an articulate double module that alters in length during motion
If that line is directly out of your book, then I suggest you get another book...
Seriously, what kind of text book uses that language without explaining it? I haven't got a clue what it means.
The upper end of the rubber follows a path according to the equation y= 1/50x^2 + 30.
But then the lenght of the rubber is 30. In your diagram, it is only 20.
Assuming that's supposed to be 20, what you have are:
y1= 1/50x^2 + 20
y2= 1/50x^2
y3= √(2)x + 30
y4= -√(2)x + 30
a = intersection between y2 and y3
b = intersection between y1 and y3
c = intersection between y1 and y4
d = intersection between y2 and y4
Try the link again, it's working fine for me.
but I just don't believe that many of its assumptions are universally true.
Such as?
Not all schools will require linear algebra for multivariable calculus. Mine doesn't. Multivariable is normally set in 1, 2, 3, or 4 dimensions, so you can normally get away with having non-matrix derivatives and integrals and using vectors.
If we assept the rules (a^x)(a^y)=a^(x+y) and (a^x)/(a^y)=a^(x-y)
Power rules come from group theory, they can be proven true. No acceptance is needed.
I suppose that we can use the rule n!/(n-1)!=n that is valid for all n>=1.
We can only accept that rule because 0! = 1. You can't use it to show 0! = 1.
It is basically an identity created so that the rules for factorials continue to work. Like a^0 = 1
Not true. You can prove a^0=1 using group theory (an offset of set theory). But that requires quite a bit of knowledge about sets and groups. A simpler proof I did before is:
http://www.geocities.com/rshadarack/proof.pdf
The real problem is in assuming that all mathematics are precise and factual. Like any science, no matter how stringent they appear to be regarding proofs, they all rely upon arbitrary assumptions at their foundations.
The only thing math relies on are definitions. And that is simply what we call things with certain properties. Now that isn't to say there isn't ambiguity in math. But there is far more ambiguity in any science than math.
When you have experience with proving statements like this, you know what the poster means even if (s)he doesn't say it.
That's really all it is, experience, nothing else irspow.
Oh, and I meant to say integer larger or equal to 0 in my post, since this was about integers and nothing else.
Was this a question that a teacher/professor gave you? It should be worded 9^p and 25^p for p ≥ 1, aka, natural numbers.
It's not wrong the way it is, just...weird.
If p is an integer, then p + 1 would also be an integer the equation would only produce a series like;
(27n + 25n)/4 = 54n/4 = 13n, which of course would all be integers.
Let n = 1:
So you are saying there is an integer p ≥ 1 such that (3*9^(p+1)+25^(p+1))/4 = 13?
Furthermore, you are saying that 9^(p+1) = 25^(p+1) for all p, since n = n
I was skeptical at first, but it seems that induction is the way to go on this one.
Glad you responded, I wasn't sure if you knew what an inductive proof was, so that saved me a lot of typing.
Inductive assumption:
Inductive assumption, 216, and 600 are all divisible by 4.
Edit: Ah, spotted a large error in my proof, fixing it right now.
In the end, we have developed technology which allows us to find such roots simply, there really is no need to drone away reinventing the wheel.
If you could find a root to a n'th degree polynomial in O(1) time (constant, not changing based on the values of the equation), then you could greatly speed up many different algorithms, especially for calculators and math software.
It would, indeed, be very useful.
There is no easy mathimatical way to solve that equation. You could use the quartic equation, but that's really messy. The best way to solve those is to use a computer.
Sounds like your doing calculus.
Vertical motion model is just that, you want a mathimatical model of where an object will be, how fast it will be traveling, and how fast it will be accelerating, when it is falling, or possibly being propelled (although that is quite complex).
What exactly do you want to know?
Why the heck is it 8 bits on some compilers? I thought the purpose was to save memory for switch variables. Peculiar...
This is a bit of a complex question (pun intended). Computers run much faster if all data is stored in whole bytes. To understand why, you need to understand the underlying architecture, which would take me a while to describe.
Does boolean algebra have a distributive property?
For example, I often wish I could say: a == (b or c or d)
instead of: (a == b) or (a == c) or (a == d)
...but the first form doesn't work.
Not quite. There is no equality in boolean algebra. Instead, each variable is either true or false.
So for your example, if you had:
a and (b or c or d)
the equivalent would be:
(a and b) or (a and c) or (a and d)
TON = Slang for 100 mph = 100
Isn't a ton 2000 pounds? 20 hundreds?
In C++, type "bool" is a variable that only returns true or false. It does not hold a value.
Little known fact, most compilers have a bool as 8 bits, even though only one is needed.
Hey, Ricky, is multivariable calculus hard?
If you have a very good understanding of derivatives, integration (especially finding the limits of integration), and understanding how to apply these, then no.
The only other thing to consider are converging and diverging series, but for those, all you have to do is memorize a few different tests to see which is which.
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Many math classes have long periods of learning what you should already know. This is because math takes practice and you must understand everything you have learned fully to understand more advanced things.
It does get pretty annoying though. In vector calculus, we learned how to deal with vectors (ironically having nothing to do with calculus). In multivariable calculus, we spent a month reviewing what we learned in vector calculus, then never used it again. Finally, in calculus of several variables (even more ironically, the book we are using is called Vector Calculus), we actually use vectors and calculus.
There is no way to (mathematically) explicitly solve for x. This is the case for many equations. You can, however, use computer algorithms to approximate a solution. This is what graphing calculators do.
Oh, and if you didn't know, explicitly solving for x means:
x = ____________
Where that blank does not contain x. If an equation is not explicit, it is called implicit.
What do you mean, "if the equation is really long"? By definition, a quadratic can only have 3 terms. It can't be any longer than that. All quadratic equations take the form of:
ax^2 + bx + c = 0
To solve this, use the quadratic equation, or, if you have a graphing calculator, graph it and find out when it crosses the x-axis (x = 0).
∴ We don't apply math to conceptual ideas. QED.