Need Math Help?

[Varuna]

Thanks Doraki, This doesn't feel right but so could you look at what I got?

also I'm stuck on


Solve for x:

10^(x+2)=5^(7-x)

I know to

(x+2)ln(10)=(7-x)ln(5)

but I don't know what to do after that.

Once again thanks much in advance! I really appreciate it.

 

[Doraki]

no no no no no,

You don't know how to differentiate compositions.
(f o g)' (x) = g'(x).f'(g(x)), not f'(g'(x)).

(in france we call f'(x) = df/dx (x), dunno about you guys but it's shorter to write here).

If you EVER use a formula that you're thinking might be true but that you don't remember seeing in class, believe me, you're going to do something awfully wrong.

And you differentiated the exponential function like a power function and they're not the same functions at all.
don't you at least know that the derivative of x -> e^x is e^x and not some horrible abomination like xe^(x-1) ?

f1'(x) = 2x ok

2^x is also e^(x.ln 2), so the derivative of x -> 2^x is (ln 2)e^(x.ln 2), which is also (ln 2).2^x

So when you know that you can apply the composition formula :
f2(x) = 2^f1(x), so
f2'(x) = f1'(x).ln 2.2^f1(x) = 2x.ln 2.2^(x^2)

and f3'(x) = f2'(x).ln 2.2^f2(x) = 2x . (ln 2)^2 . 2^x^2 . 2^2^x^2

----

can't you solve (x+2) * a = (7-x) * b for random a and b numbers ?

 

[pikachun00b7]

What is the derivative of:

y=SQRT(x+SQRT(x)) ???

 

[Lixivium]

y = (x + x^1/2)^1/2
It's just a simple polynomial + chain rule differentiation.

 

[pikachun00b7]

y = (x + x^1/2)^1/2
It's just a simple polynomial + chain rule differentiation.

yeah but what us y'?

EDUIr: oh wait I got it.

 

[Black Waltz]

Zomg Derivatives

Find the value(s) for k such that the given line is tangent to the graph of f(x).
f(x)= x^2 - kx and the line is y=6x-25

help pl0x. im not allowed to use a calculadora. is this even possible?????

 

[GoldShadow]

Find the value(s) for k such that the given line is tangent to the graph of f(x).
f(x)= x^2 - kx and the line is y=6x-25

help pl0x. im not allowed to use a calculadora. is this even possible?????

I'm not sure if there's any easier way to do this problem, but this is my method for solving it:

First, find the derivative of f(x), which is: f'(x)=2x-k
Since the derivative gives you the slope, the value of the derivative at x must be equal to the slope of the line y at x... in other words, f'(x)=6. So you solve for k using 2x-k=6, and you get k=2x-6.

You also know that the tangent line "y" has to touch "f(x)" at one point, so you can set the two equal to each other substituting 2x-6 for k as follows:

x^2-(2x-6)x=6x-25
Solve for x and you get +/- 5. Now, you put this back into the expression for k to find the two possible values of k:

k=2(5)-6=4
k=2(-5)-6=-16

If you try out both of these values, you find that y is indeed tangent to f(x) for both of them (at x=5 if you set k to be 4, and at x=-5 if you set k equal to -16).

 

[Doraki]

Actually there is an easier way. But I wouldn't expect you to spontaneously do it like this :

The line will be tangent to the parabol if and only if the equation x^2-kx = 6x-25 has a double root.

So you only need to check when the discriminant of that equation equals 0 :
x^2-kx = 6x-25 <=> x^2-(k+6)x+25 = 0
thus the discriminant D is (k+6)^2 - 100

D = 0 <=> (k+6)^2 = 100 <=> (k+6) = +10 or -10 <=> k = 4 or -16.

Goldshadow's reasoning is what you should think about doing first though.
When you're asked about tangents you will always work with the derivatives like he did, it's a straight-forward way of working with those problems that will almost always lead to the answer.

 

[Black Waltz]

thanks goldshadow for your help! i finally understand it now! and doraki too for you input even though i don't really understand it!

 

[GoldShadow]

Hm... so I came across an interesting physics problem from a problem set. It reads:

A woman and her dog are out for a morning run to the river, which is located 4.0 km away. The woman runs at 2.5 m/s in a straight line. The dog is unleashed and runs back and forth at 4.5 m/s between his owner and the river, until the woman reaches the river. What is the total distance run by the dog?

Well the dog's first trip will be all the way to the river, so it is 4000 m (aka 4.0 km) in length.
Using the given velocities, I figured out how long this would take the dog (889 s), and how far the woman would travel in that time (2222 m).

So then for the dog's trip back to the woman, I first figured out how long it would take the two to intersect, using:

Xd (position of the dog) = -4.5t + 4000 (+4000 since it was at the river when it started its trip back to the woman)
Xw (position of the woman) = 2.5t + 2222 (because that's how far she had traveled while the dog was running to the river).

Setting the two equal and solving for t, I got 254 s... so I figured out that the position at which the dog and woman intersect was 2857 m, in other words the dog ran from x=4000 to x=2857, for a distance of 1143 m. Then of course, the dog turns around and runs another 1143 m back to the river.

Anyway, I continued this a few times before realizing that this would go on forever, and that the distance the dog had to run between woman/river would get smaller each time... This is what I had:

Trip 1 (dog to river): 4000 m
Trip 2 (dog back to woman): 1143 m
Trip 3 (dog back to river): 1143 m
Trip 4 (dog back to woman): 326.6 m
Trip 5 (dog back to river): 326.6 m
Trip 6: 93.3 m
Trip 7: 93.3 m

When I realized I had an infinite geometric series... the ratio between 4000 and 1143 was the same as the one between 1143 and 326 which was the same between 326 and 93... it was about .2858.

So I used the formula for sum of an infinite geometric series (a1/(1-r)) like so:

4000+2[1143(.2858)^(n-1)] --> 4000+2(1143/(1-.2858)) and got 7200 m for the total distance...


I was pretty proud of myself for using some actual calc stuff (even though it's calc-based physics, we haven't even really used derivatives, let alone series), but I couldn't shake the thought that I'd done a lot more work than I had to and there was a simpler/non-calc/mostly physics-based approached to this I overlooked.

Can anybody find an easier way to do that or was my approach alright?

 

[Doraki]

yes, the dog is moving all the time at constant speed, 4.5 m/s, while the woman is slowly going to the river.
The dog could be running anywhere in circles or in whatever shape he wants, since the dog's speed is constant, you only need to know for how long he is running, which is the same as how long it takes for the woman to reach the river :

t = 4 km / 2.5 m.s-1 = 1600 s
so the dog travels the distance d = 1600 s * 4.5 m/s = 7.2 km


Also I advise you to try and keep everything you get written with literals.
Instead of putting 4000, 2.5 and 4.5, keep them as constants D0, V0 and V1 and do all the calc stuff with them so it's easier to see what's happening in the equations.
You only need them at the end to get the numerical answer.

 

[GoldShadow]

Ah **** it, I feel stupid now, I can't believe I didn't realize that.

Many thanks for the help Doraki!

 

[Doraki]

Don't worry it's great that you could do all this complicated stuff without making a mistake.
At least you can be confident in what you write, I can't say the same about most people.

 

[Falco&Victory]

haha nice goldshadow, lol

I feel stupid though, I scored a 91% on a test, *sigh*

Oh well, it was all stupid mistakes i should have caught.

do equations that graph as a loop not classify themselves under a function, since one input can have 2 outputs?

 

[AltF4]

Jeez, I hardly get a chance to answer any questions anymore. Everybody else beats me to it. Good to see though.

F&V: Nope, a loop can't be a function. You gave the reason yourself: there are cases where one input gives you two possible outputs.

Also, stupid mistakes in math don't ever go away. In fact, stupid mistakes wind up making up the majority of points lost on your tests. Higher level math tends to get increasingly difficult conceptually, but not any more difficult on paper. That's why I maintain that Algebra 2 (aka 3-4) is the hardest math class there is. Because even in calculus and beyond, the majority of the points lost for the majority of students are Algebra mistakes, not calculus (or whatever) mistakes.

So the lessen to learn is to always slow down, take your time, write clearly, don't skip steps, and think things through. You're never going to be impervious to stupid mistakes just because you're in a new math class!

 

[Rusty Shacklefurd]

And always check your calculator to make sure it's in degrees or radians depending on what you're doing. I once took an entire Physics 12 midterm in radians and would have aced it if I didn't... shame...

 

[RazeveX]

ok...this is all WAY over my head...cept the trig stuff, i think i learnt that already...but im still in mid high school (i dont know all the american names for all the year levels)

i STILL dont get the 0=1 thing...what is it? Is it literally saying that zero is equal to one???

i am quite curious. Haha, it'd be hilarious if my parents walked in and were all like "what are you doing, stop wasting your time on the internet" and then i'd be all like "actually, i'm learning advanced calculus." and then theyd be all like "oh...k...??"

lololol

 

[AltF4]

(It's fake. Those proofs are wrong and just meant to amuse. But don't tell anybody. Try to find the mistake, it's fun.)

 

[RazeveX]

Let x = 1 – 1 + 1 – 1 + 1 …
X = (1 – 1) + (1 – 1) + (1 – 1) +…
X = 0 + 0 + 0 + …
X = 0
Let x = 1 – 1 + 1 – 1 + 1 …
X = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) …
X = 1 + 0 + 0 + 0 + 0…
X = 1
Thus,
1 = 0

Satisfied? I have an entire thread for this stuff. It's on the next page of threads.

well, isnt the whole thing wrong? I mean, the "..." thing just depends where you stop it. If you stop it at +1, then X=1. If you stop it at -1, then X=0, right?

sorry, thats the best way a simpleton like me can express it.

Can someone post a math problem solving thing thats fun?? That dog and woman one wasnt bad.

 

[GoldShadow]

well, isnt the whole thing wrong? I mean, the "..." thing just depends where you stop it. If you stop it at +1, then X=1. If you stop it at -1, then X=0, right?

The "..." means it goes on forever, to infinity.

 

[AltF4]

How about this:

There is a hotel called "The Infinite Hotel", which has an infinite number of rooms available. It's the holiday season, however, and wouldn't you know it? The entire Infinite Hotel is booked up, completely full.

That evening, another infinite number of people show up in the infinite lobby and each want a room to themselves. How can the hotel fit an infinite number of people into the already full Infinite Hotel?

 

[Rici]

Spoiler

Because infinity+infinity=infinity, right?

Or in other words, an hotel with infinite room can't never be full

Sigh, infinity is such a stupid concept .

 

[RazeveX]

How about this:

There is a hotel called "The Infinite Hotel", which has an infinite number of rooms available. It's the holiday season, however, and wouldn't you know it? The entire Infinite Hotel is booked up, completely full.

That evening, another infinite number of people show up in the infinite lobby and each want a room to themselves. How can the hotel fit an infinite number of people into the already full Infinite Hotel?

well if its full, then its capacity obviously isn't infinite. Instead of saying it was full, you should have said there was already an infinite number of people staying there. ooooh....cryptic....

woah! check me out! smash JOURNEYMAN. Whats the next one, and at what post count??

 

[Rusty Shacklefurd]

Let x = 1 – 1 + 1 – 1 + 1 …
X = (1 – 1) + (1 – 1) + (1 – 1) +…
X = 0 + 0 + 0 + …
X = 0
Let x = 1 – 1 + 1 – 1 + 1 …
X = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) …
X = 1 + 0 + 0 + 0 + 0…
X = 1
Thus,
1 = 0

Satisfied? I have an entire thread for this stuff. It's on the next page of threads.

The problem is it only works if you assume an ending to the infinite series being a positive 1. Then, and only then, can you restrict the series to an infinite pattern of (-1+1) But since original pattern starts with 1 and ends with -1, you can't assume the series ends halfway through a repitition.

I'm just guessing here.

 

[Doraki]

this is why you need absolute convergence when you think about the sum of an infinite set of numbers.

 

[Xephalon]

uhh my teacher said to answer this question a couple years ago..concept it pretty easy, this was a question with a 20 dollar reward:

a/2 + b/3 + c/4=1

solve for a,b,and c

 

[Doraki]

a,b,c have to be natural integers ?

 

[Bli33ard]

I didn't go any further than grade 12 maths, heh.

One of the questions in my exam had a triforce in it :S

 

[Lixivium]

I didn't go any further than grade 12 maths, heh.

One of the questions in my exam had a triforce in it :S

You mean this thing?

 

[Xephalon]

a,b,c have to be natural integers ?

i dont think so

 

[Rici]

uhh my teacher said to answer this question a couple years ago..concept it pretty easy, this was a question with a 20 dollar reward:

a/2 + b/3 + c/4=1

solve for a,b,and c

A=1
B=12
C=1

Something like this?

 

[Haruka's DNA]

(AltF4 posting on Alicia's account to save my 1000th post):

Well, obviously there's no such thing as an infinite hotel, so you have to use your imagination. The answer isn't something like "there's no such thing as infinity, lol!".

Xephalon: There must be some sort of restriction on A, B, and C in that equation. Or something else to make it harder. Because otherwise, there's an infinite number of answers. That would be a very easy problem to solve, definitely not a $20 problem.

Oh, and for those looking at that 0=1 proof. That series is a well known series called "Grandi's Series". It is also divergent, which means it has no sum. Not that it sums to infinity, or it sums to zero, but it has no sum. So once you even try to even treat it like it has a set "value", you've made your mistake.

 

[plasmawisp6633]

Galileo's Paradox: Although most numbers are not perfect squares, there are just as many perfect squares as there are numbers (because every single number can be squared).

I had fun thinking about that for an hour.

P.S. I just started digging into the meat and potatoes of Pre-calc, so keep this thread open, because I might need help!

 

[Jammer]

Um, I'm pretty much a genius in math, and I've been told I have excellent teaching skills, so if anyone isn't getting the answer they want here, they can just PM me, and I'll answer within 20 minutes of seeing the PM. Actually, only ask if it's a high school math question. Or an easier college math question. You know, since I'm so knowledgeable, ask me absolutely any question you have in a high school curriculum. Try it; I won't disappoint.

I won't be looking at this thread, by the way.

I'm going to change my signature to reflect my new helpfulness.

@plasmawisp^: I never heard that before. That's a good one, but it totally leaves out negative numbers. For each perfect square, there is both a positive and negative number that is its root. So there are twice as many numbers as perfect squares (not counting the special case of zero, or any non-integer number).

 

[shadydentist]

Every number has a square. Therefore, at least half of all numbers are squares. But of those numbers, some are squares themselves. Therefore, there has to be MORE square numbers than not. Disregarding negative numbers.

 

[Doraki]

Aw come on

Jammer > "So there are twice as many numbers as perfect squares"

the set of numbers and the set of squares are both infinite.
saying "twice as infinitely many" doesn't really make sense.
Just because you have a function x -> x^2 from the set of numbers into the set of squares and such that every square can be obtained from 2 numbers, you can't say there is twice as many numbers as perfect squares.
If I do this with the function x -> (x/2) (say.. rounded down), from the set of numbers into the set of numbers, it has the same property, so you're going to say that there's twice as many numbers as numbers ?

You can't really compare infinite sets by saying "half of" or "twice as much".
You could do it with finite sets, but it makes no sense with infinite sets.
A set being twice as big as itself is pretty weird right ?

do you know there is a function f from N to N such that every number y in N can be obtained as an f(x) from not only 2 values of x, but an infinite subset of N ? would you say that N is infinitely times bigger than itself ?

woops I almost told the answer to the infinite hotel riddle.

 

[Jammer]

Aw come on

Jammer > "So there are twice as many numbers as perfect squares"

the set of numbers and the set of squares are both infinite.
saying "twice as infinitely many" doesn't really make sense.

While you are completely right that it doesn't make sense to say "twice as infinitely many" (although you can compare different infinities--for example, the infinity of real numbers from zero to one and the infinity of numbers from zero to infinity*), that's not what's really going on here. Plasmawisp was pointing out that there was a one-to-one mapping between numbers and their squares, so there are the same number of each. I showed that there is a two-to-one mapping (a negative and a positive number for each square); i.e. there are twice as many numbers as perfect squares.

Infinity never even entered the picture.

I assume you were thinking about infinity because you thought the concept of infinity was necessary to understand Galileo's paradox. It isn't, but I'm not sure how to explain that beyond what has already been said.

If you'd like to continue this discussion, I'm having fun. No feelings will be hurt.

*The second infinity is "bigger" in set theory.

 

[Doraki]

Well generally, when people do this they stick to positive numbers to avoid this little concern.

In the case of infinite sets, having a countable-to-one mapping doesn't change the cardinality anyway so having a 2-to-1 mapping instead isn't bothersome.

Saying there's twice as many numbers as perfect squares is confusing because then people may think it's not the same as saying there's as many numbers as perfect squares, whereas there's no difference.

 

[Jammer]

Doraki, why are you talking about infinite sets? Sure, there are an infinite number of numbers and an infinite number of squares. What I'm saying is that for each individual square there are two roots. It seems to me that that means there are twice as many roots as squares. How does the fact that both roots and squares are in infinite sets make that invalid? I'm comparing the sets element by element, not as a whole.

Also, keep in mind that I have absolutely no real education in this area. I only read a book called "Introduction to Discrete Mathematics" last summer. I'm just going on intuition here. So this could be a good learning experience for me.

Well generally, when people do this they stick to positive numbers to avoid this little concern.

I figured that would be the case. I guess I was just trying to stir up trouble in an unfamiliar area...

 

[Doraki]

It all depends on how you link the roots with the squares.
But if you forget that squares are obtained from roots by squaring them, there's a lot more ways to link the set of square numbers with the set of numbers.

When you're saying that there's twice as many roots as squares, you're implicitly thinking about this 2-to-1 mapping. Maybe because you're calling them roots and squares you think it's ok to forget about the mapping when you say that there's twice as many roots as squares.

When you deal with finite sets, having a 2-to-1 mapping between A and B really means that there's twice as many objects in A than in B. And because of this there can't be a 1-to-1 or 3-to-1 mapping between A and B if there is a 2-to-1 mapping already.

This doesn't hold when the sets are infinite, and this is why I'm talking about infinite sets.
There's a canonic 2-to-1 mapping between the set of numbers and the set of square numbers ; but there're also 1-to-1 or 3-to-1 mappings between them and so on.
If you feel you can say there's twice as many roots as squares because you have a 2-to-1 mapping, then I can also say there's ten times as much roots as squares, or whatever number I want instead of 10.
So it becomes kinda stupid because we're saying things that seem different and that would be impossible with finite sets ; but that only mean the same thing, that the set of roots and the set of squares are infinite.


also I'm done for today because it's good to be lazy.