Hey!
just had an exam and a few questions completely baffled me, I was wondering if anyone could explain them? I know I can’t change it now, it’s just nice to understand these things 
the first one was
I feel like I shouldn’t have found that as hard as I did 
The second one was…
This is talking about the height of the cone as opposed to the slanted length, any help would be massively appreciated, these are doing my head in
First (for the way you wrote it, which is probably not how it’s on the test):
k = C
with C being any real or complex number.
Second:
If you have the formula in terms of the slanted length and radius, you can change it to use the height using the Pythagorean theorem.
No genius here.
I am going to guess that in the first question you meant t=k/(k-2) otherwise it’s pretty simple and is a constant for k not = 0. Also, by subject, I am going to guess that you are being asked to express k as the dependent variable. If so
t(k-2)=k
tk-k=2t
k(t-1)=2t
k=2t/(t-1) for t not=1
The second one, the area of the disks on the bottom are equal so they’re unimportant. The area of a sphere is 4PIx^2 so the area of a hemisphere is 2PIx^2. The area of a cone with respect to x and h is PIxSQRT(h^2+x^2).
PIxSQRT(h^2+x^2)=2PIx^2
SQRT(h^2+x^2)=2x
h^2+x^2=4x^2
h^2=3x^2
h=xSQRT(3)
What is this, an algebra class?
Thanks very much to both! I get them now 
hopefully I’ve got some marks for working out.
Cheers!
Greg the genius did more in his head than I did. In step 3 below, I combined two simple operations. Yet Greg skipped all my baby steps and went from step 2 straight to 4. Here’s what I had –
t=k/(k-2)
1: t(k-2)=k
2: tk-2t=k
3: tk-k=2t
4: k(t-1)=2t
5: k=2t/(t-1)
Also, Greg gets bonus points for remembering to specify that (not t=1). Plus, I don’t know what expressing k as a dependent variable means. Does it mean solving for k?
New question. The way the equation was originally written, it looked like it contained (k/k). I know that division by zero throws a hardware exception (hey, I’m a computer programmer). But any number divided by itself is one. So, is there an exception-to-the-rule for (0/0)?
Wait! Never mind. The answer is contained in this:
Basically, if you scroll down in the article, it says that 0/0 can be thought of as (0/0=x), meaning what does (0/0) equal? It can be rearranged as (x * 0 = 0). I’m using * to denote multiplication (hey, I’m a computer programmer). Every value satisfies that equation. The article then states, “In general, a single value can’t be assigned to a fraction where the denominator is 0 so the value remains undefined (see below for other applications).”
I’m not sure I fully understand the conclusion. But hey, I’m a computer programmer – I’m just happy that I could readily solve for k after all these years. I use simple algebra in life occasionally. I’m glad that was enough to retain some of the basic skills.
But Greg loses points for suggesting that the first equation is undefined at k=0. It’s actually undefined at k=2. But he gets points back for the disclaimer, “no genius here.”
In the equation y=f(x) y is the dependent variable determined by the value of the independent variable, x.
My answer is based on the fact that when k=0, it depends on what the application of this means. If the application is appropriate, as k approaches zero, that division by zero may be significant. Not knowing what level of math is being considered here, it leaves that question open. I’ll just assume that as k approaches zero for k/k that k/k id always equal to 1. Keep in mind that I have not kept up with this stuff and have not done a lot of this math in about 9 years.
I think you meant (t=0) above (I fixed it in your quote)? But is it undefined for BOTH k=2 and t=0 ? Aren’t the two equations identical? [edit: is ‘equivalent’ a better word?] Yet one has k in the denominator and one has t in the denominator. Why would you pick limiting the values of k instead of t? Does t automatically never become invalid (0) for all valid values of k?
Gilby – nine years? I got you beat… it’s more like 25 years for me.
t=k/(k-2) is undefined at k=2
k=2t/(t-1) is undefined at t=1
Don’t mess with my quotes, bro. I’ll come down on you like Billy with a juggling club.
But they are just two forms of the same equation.
For the first equation, may we say that k is not allowed to be 2 and t will never end up being 1?
And for the second equation, may we say that t is not allowed to be 1 and k will never end up being 2?
In other words, how can two forms of the same equation have different restrictions? There are a set of values of t and a corresponding set of values for k that will satisfy BOTH equations – because they are just mixed up variations of the same equation.
I’m plugging in the restricted value from the second equation into the first equation.
t=1
t=k/(k-2)
Baby steps, because I don’t want to screw this up:
1=k/(k-2)
1(k-2)=k
k-2=k
Ouch.
So what gives? Is it interesting that t=1 will never happen? Is there a word for it? I think that t=k/(k-2) is undefined at t=1 as well as k=2, but why are we not required to say that? Why do we only care about division by zero? Seems a little short-sighted.
Is it because we only care about the independent variables? The inputs? The outputs, the dependent variables, will do whatever they will do. And in the case of t, will never equal 1. If I plug t=1 into the first equation, then k becomes the dependent variable. In order to compute its value, we must solve for k, yielding the second equation with its restriction of t.
Am I on the right track? I didn’t touch your quote this time.
My dear Dave-
Think about what the plots of these functions look like.
t=k/(k-2), where t is dependent and on the vertical axis.
Think of k being a large negative number. t approaches 1 from below but never reaches 1. Let k become smaller negative and go to 0 at which point t=0. Now, as k becomes positive, t becomes larger negative until k approaches 2 at which point t approaches some huge negative number. Let k skip over 2 and become just slightly greater than 2. Now t starts at some huge positive number and decreases as k increases positive. As k becomes large and positive, t approaches 1 this time from above.
Now, rotate this plot in your mind (or draw it and rotate it) 90 degrees and flip it and you will see the plot of
k=2t/(t-1), where k is dependent and on the vertical axis.
and that it exhibits very similar behavior. This time k approaches 2 asymptotically from above or below depending on whether or not t is large positive or negative. As t approaches 1 from above or below then k approaches a huge positive or negative number.
At this point, maybe look at the definitions for function, continuous function, and discontinuous function which you may remember when you see them again.
OMG, when you see those equations, do you see – in your mind’s eye – something like the graph below? And can you actually rotate and flip it in your mind?
Because I see, as a computer programmer, something more like the following…
float t( float k )
{
if ( k != 2.0 )
return k / (k - 2.0);
else
return NaN; /* not a number */
}
float k( float t )
{
if ( t != 1.0 )
return (2.0 * t) / (t - 1.0);
else
return NaN;
}
In my algorithmic “visualization”, I don’t see what you see. It’s like I’m blind. I just see numbers go in, numbers come out. In fact, it never would have occurred to me to graph them!
I like to say that computer programming teaches you how to think properly and that everyone should learn. But now I see that its applicability is somewhat more narrowly defined (to include only “most” of life).
I can follow your explanation to a point, which would have been hopeless without the graphing software. I’m still working on interpreting the rotated and flipped version (courtesy of Photoshop). Even with the Graph software and Photoshop, I’m still having trouble and must stop to rest. And I’m sure I’ll never get the big picture – like why would you want such an equation and why does the graph matter? If I were developing some business application that required some money-related calculations, I wouldn’t see the value of graphing the results. As long as the inputs produced the correct outputs, why would I care what it looked like?
Obviously, and with a bit of sadness, I am professing my profound ignorance of basic concepts of mathematics. I’ve retained just enough brute-force, simple algebra to easily solve for k, as was the original assignment. But I know I’m missing the beauty that lies beyond. It’s just been too long. Algebra 1, 2, 3, trig, calculus. Almost all gone now…
Ooooh. Nice. A picture’s worth a thousand words.
Yeah, but I think you saw that picture in your mind first, which must be very convenient. And, you should note, I still needed your verbal description of the travels of t and k in order for the picture to really sink in.
And a sound is worth a thousand pictures.
But you lose them again for not reading your own original quote
The way I understood it, you were pointing out that the equation in its original form as posted by the OP (i.e. containing the “k/k”) is undefined at k=0.
Dang. Caught again. I will defer to the sure thing quote:
Was this an exam by any chance? Maybe GCSE’s and exam board Edexcel?
