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The Math Help Thread

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So another high school girl came to me for help on her pre calc class...

ibrI7Xch93DTd2.jpg


a) this is easy. dy/dx = (6y-10x^2)/(10y-6x) subbing in x=1,y=-1 we get dy/dx=1

b) How do you do this? (6y-10x^2)/(10y-6x) = 0 And then you can equate (6y-10x^2) = 0. You get x^2=(3/5)y. Subbing that into the original equation of the ellipse....how do you solve such an equation by hand?

Another question:

iSf1Juo4kylhw.jpg


so y' = x/(4y)

say these 2 points hypothetically have coordinates (a,b) and (c,d) In the end you are left with these 2 equations using the formula y-y_1=m(x-x_1) where (x_1,y_1) is the coordinate (4,6) we end up with these 2 equations:

Equation 1: y = a/(4b)x - (a+6b)/b
Equation 2: y = c/(4d)x - (c+6d)/d

2 equations and 4 unknowns.. how do I solve this?
 
AAK said:
a) this is easy. dy/dx = (6y-10x^2)/(10y-6x) subbing in x=1,y=-1 we get dy/dx=1

b) How do you do this? (6y-10x^2)/(10y-6x) = 0 And then you can equate (6y-10x^2) = 0. You get x^2=(3/5)y. Subbing that into the original equation of the ellipse....how do you solve such an equation by hand?
Click to expand...

You didn't get the 'easy' part right. dy/dx = (6y - 10x)/(10y-6x).
 
I'm confused why I try solving this differential equation two different ways takes me to two very different answers...one that's probably right, the other maybe not.

d/dx (k * dT/dx) + q = 0

q is a constant, k = a/x

So,

d/dx (a/x * dT/dx) + q = 0

Method 1:
Code: 
d/dx(a/x * dT/dx) = -q

multiply by dx and integrate both sides

a/x * dT/dx = -qx + C1

dT/dx = -(q/a)x^2 + C1*x

multiply by dx and integrate both sides

T = -(q/(3a))*x^3 + C1*x^2 + C2

Then I can solve for the constants using the boundary conditions I have (likely correct).
Method 2:
Code: 
use product rule

dk/dx * dT/dx + k*d^2T/dx^2 = - q

(-a/x^2) dT/dx + (a/x) d^2T/dx^2 = -q

d/dx (T - x dT/dx) = (q/a)*x^2

multiply by dx and integrate both sides

T - x dT/dx = (q/(3a))*x^3 + C1
How would I solve it this way?? Can i solve it this way? If not, why not??
 
The transition between these steps are wrong.
Code: 
(-a/x^2) dT/dx + (a/x) d^2T/dx^2 = -q

d/dx (T - x dT/dx) = (q/a)*x^2
"d/dx (T - x dT/dx)".
This term is wrong, try differentiating the inside expression to see it.
 
Method 1 is the correct way (making sure to divide C1 by a as well). Method 2 fails at

d/dx (T - x dT/dx) = (q/a)*x^2. The left hand side is wrong.
 
Leezard said:
The transition between these steps are wrong.
Code: 
(-a/x^2) dT/dx + (a/x) d^2T/dx^2 = -q

d/dx (T - x dT/dx) = (q/a)*x^2
"d/dx (T - x dT/dx)".
This term is wrong, try differentiating the inside expression to see it.
Click to expand...

FortuneFaded said:
Method 1 is the correct way (making sure to divide C1 by a as well). Method 2 fails at

d/dx (T - x dT/dx) = (q/a)*x^2. The left hand side is wrong.
Click to expand...

So I guess I just cannot factor out d/dx from both terms? For some reason I thought that might be possible considering it's possible to multiply and divide and factor out differentials like dx. I guess it just doesn't work like that?
 
The Lamp said:
So I guess I just cannot factor out d/dx from both terms? For some reason I thought that might be possible considering it's possible to multiply and divide and factor out differentials like dx. I guess it just doesn't work like that?
Click to expand...

No, you cannot factor it out. It's an operator. You can "factor it out" in some special cases.
 
FortuneFaded said:
You didn't get the 'easy' part right. dy/dx = (6y - 10x)/(10y-6x).
Click to expand...

Damn, thanks, didn't even occur to me to look back that far. Anyone figure out the 2nd question? I'll quote it again:

AAK said:
iSf1Juo4kylhw.jpg


so y' = x/(4y)

say these 2 points which the tangent lines touch the ellipse hypothetically have coordinates (a,b) and (c,d) In the end you are left with these 2 equations using the formula y-y_1=m(x-x_1) where (x_1,y_1) is the coordinate (4,6) we end up with these 2 equations:

Equation 1: y = a/(4b)x - (a+6b)/b
Equation 2: y = c/(4d)x - (c+6d)/d

2 equations and 4 unknowns.. how do I solve this?
Click to expand...
 
AAK said:
Damn, thanks, didn't even occur to me to look back that far. Anyone figure out the 2nd question? I'll quote it again:
Click to expand...

Q2: I didn't go through your steps, but notice (a,b) and (c,d) are all on the ellipse and the respective tangent line.
 
There is something fundamental I'm missing here, but I'll go through the steps I'm assuming you're supposed to go through:

1) get the derivative which will represent the slope for the tangent lines of the ellipse. We get y' = -x/(4y)

2) Come up with a generic line equation for the tangent lines of the ellipse: in the form of y = mx+b using the coordinate (4,6) given in the question. (the x in this equation is different from the x in the original ellipse function, so I'm going to use a and b for points of the ellipese x and y for points on the tangent.
I ended up getting: y = -(a/4b)x + (6 + a/b)

I ended up using the (4,6) coordinate to find my y intercept for the tangent line equation which I got to be (6 + a/b).

I need another coordinate to coordinate because if I just sub a and b back into the original ellipse equation to get a^2 + 4b^2 = 16 which simplifies to a = sqrt(16 - 4b^2). subbing that a back into the tangent line equation I got in 2) gives me 0=0.

The final step is having a pair of simultaneous quadratic equations that I will solve to get two answers for a and b but I just can't figure it out right now :(

I feel so stupid I don't remember how to do basic high school math.... :(
 
Instead of using (a,b) and (c,d), keep the gradient m as -x/4y and you get 4y^2 - 24y = -x^2 + 4x. This line has to intersect with the ellipse so convert the 4y^2 + x^2 into 16 so you end up getting x + 6y = 4. Now you can plug this into the equation for the ellipse to get two values for x (or y).
 
Brehs, can anyone help me with my take home test?

2) A car is traveling at 85 kilometers per hour. What is its speed in miles per minute?

Note: 1 mile = 1.61 km

I know i have to convert, but what am i missing?
 
Could someone help with this?

Not sure what the topic is called in English, but I need to find X and Y points and use the derivative of the function to find the maximum and minimum points

I have this function 2x^2+x+1/x^2

I could just use a formula, but I have an example with a similar function x^2+5x+36/x

You turn the function into x/x + 5x/x + 36/x > x+5+36/x > 1/1 - 36/x^2 > x^2 - 36

Not sure how do you do that with the first function tho :(
 
It's just saying that any number can be expressed as a sum of multiples of powers of the base.

IE 9 in binary is 1001, but you can also express it as

1*2^4 + 1*2^0

Or 19 in base 4 is 103 -> 1*4^2 + 3*4^0
 
Two questions here (in pre-calculus).

1.) The graph of the rational function crosses the horizontal asymptote once. Sketch the graph.

f(x)= (x^2-6x+8)/(x^2-4x+3)

Label the point of intersection of the curve with the horizontal asymptote.

(x,y)= (___, ___)

I've already drawn the graph and got the horizontal and vertical asymptotes, but I'm not sure how I figure out where the point of intersection is. I think I can see it on the graph, but it's not on an exact point.
_________________________________________________________________________

2.) The number of bacteria in a culture at time 'x' is given by

n=10,000[(3x^2+1)/(x^2+1)]

a.) As time increases, does the size of the bacteria colony become stable?
Y/N = Yes.

b.) If the answer to part (a) is yes, what is the stabilizing level (if the answer does not exists, put DNE)?

30,000 is the stabilizing level.

c.) How long does it take for the number of bacteria to exceed 24,000? Round your answer to one decimal place.

Don't I just change the 10,000 in the original equation to 24,000 and multiply it out or something? I'm not quite sure and could use a hand.

EDIT: Solved.
 
I'm messing around with ln and e laws because I was never comfortable with them.

Can someone either confirm or deny this one

ln(ln(e^e^10)) =

ln(e^10) =

10
 
Hello math gaf. I'm having a hard time understanding the substitution rule dealing with definite integrals in Cal II. One problem in particular I've been working on got me scratching my head.
∫0,1 (4r)/√(4+2r^2)

The answer our professor gave us is 2√(6-4) I can't seem to get that by working it out.
I let u=4+2r^2, 4r dx=du; rdx=du/4

r=0 u=4+2(0)^2 = 4
r=1 u=4+2(1)^2 = 6

When I plug it back in I deff don't get the correct answer. If anybody could point me into the right direction it'll be a big help. Thanks!
 
Hello again! I've been asked to show how to express [0,1] as the intersection of infinitely many open intervals. I was thinking like (0,1), (0,2), (0,3) and so on but how do I make sure 0 and one are included? Can anyone help?
 
_illmatic_ said:
Hello math gaf. I'm having a hard time understanding the substitution rule dealing with definite integrals in Cal II. One problem in particular I've been working on got me scratching my head.
∫0,1 (4r)/√(4+2r^2)

The answer our professor gave us is 2√(6-4) I can't seem to get that by working it out.
I let u=4+2r^2, 4r dx=du; rdx=du/4

r=0 u=4+2(0)^2 = 4
r=1 u=4+2(1)^2 = 6

When I plug it back in I deff don't get the correct answer. If anybody could point me into the right direction it'll be a big help. Thanks!
Click to expand...

Chain Rule. The (4r) is a differential of what is inside the square root bracket. No substitution required.
 
VegiHam said:
Hello again! I've been asked to show how to express [0,1] as the intersection of infinitely many open intervals. I was thinking like (0,1), (0,2), (0,3) and so on but how do I make sure 0 and one are included? Can anyone help?
Click to expand...

intersection of (-1/n,1+1/n)
 
luoapp said:
intersection of (-1/n,1+1/n)
Click to expand...

Oh wow, that's totally it! Now I feel dumb. n is an element of the natural numbers here, right? Cus the negatives make things confusing. Thanks again Iuoapp!

I will forever be making the face in my Avatar, yes.
 
RELAYER said:
I'm messing around with ln and e laws because I was never comfortable with them.

Can someone either confirm or deny this one

ln(ln(e^e^10)) =

ln(e^10) =

10
Click to expand...
A more developed solution.

ln(ln(e^e^10)) =

ln ( e^10 * ln (e) ) =

ln ( e^10 * ln (e) ) = ln ( e^10 )

= 10 * ln (e) = 10
 
I don't have a specific question, but I was just looking for some resources. I'm taking a course on Metric Topology this semester, and I'm quickly realizing that some of the material is out of what I'm comfortable with. I'm sure I could grasp it, but I think I was just missing out on some prerequisite information. Does anyone know of any good resources for studying things like topology on my own?
 
Guys, I need major help. I'm trying to find the computational complexity of a question.

Once I've reached the final step, I got that I needed this many basic operations to complete the algorithm: 23*a^4*b^4 + 3*b^5. I need to express the computational complexity in "big-O" notation.

Unfortunately, I have no idea which of the two terms grows faster unless I start arbitrarily setting constrains for a & b. Can anyone please help me out?

Edit: Could I do this? I specify conditions for a & b in which 23*a^4*b^4 is less than or equal to 23b^5 or vice-versa? Then, the algorithm can actually take on different complexities depending on the value of the two parameters.
 
EskimoJoe said:
I don't have a specific question, but I was just looking for some resources. I'm taking a course on Metric Topology this semester, and I'm quickly realizing that some of the material is out of what I'm comfortable with. I'm sure I could grasp it, but I think I was just missing out on some prerequisite information. Does anyone know of any good resources for studying things like topology on my own?
Click to expand...
Here's a book on Real Analysis with Basic Metric Topology.
Lonely1 said:
O(a^4*b^4 + b^5) would be a correct answer imo.
Click to expand...
I'm not a computer scientist but if a and b are both arithmetic steps, wouldn't they take the same amount of time? Isn't the expression the number of steps? O(8) (chopping second term, combining exponents in first) or something in the form of f(a,b) in O(g(a,b)) but I honestly don't know. My speculative side would suggest a piecewise approach and list the conditions when true; something like: O(5) if a^4 < b for all a. Again, I'm not sure.
 
Partial Gamification said:
I'm not a computer scientist but if a and b are both arithmetic steps, wouldn't they take the same amount of time? Isn't the expression the number of steps? O(8) (chopping second term, combining exponents in first) or something in the form of f(a,b) in O(g(a,b)) but I honestly don't know. My speculative side would suggest a piecewise approach and list the conditions when true; something like: O(5) if a^4 < b for all a. Again, I'm not sure.
Click to expand...

I'm not sure if I get what you are saying. Assuming that Kieli reached to the correct expression for numbers of steps required to complete a computation, then it doesn't matter if they are arithmetic steps, we only see the expression as a function of the variables a and b. Given the definition of big-O notation, which only deals with an upper bound of the rate of growth up to a linear constant, O(5) and O(8) are the same as O(1).
 
Lonely1 said:
I'm not sure if I get what you are saying. Assuming that Kieli reached to the correct expression for numbers of steps required to complete a computation, then it doesn't matter if they are arithmetic steps, we only see the expression as a function of the variables a and b. Given the definition of big-O notation, which only deals with an upper bound of the rate of growth up to a linear constant, O(5) and O(8) are the same as O(1).
Click to expand...
Sorry, I meant O(n^5) and O(n^8), polynomial growth and my bad notation...

I was thinking that if both steps were arithmetic, then they could be considered the same "step" for the algorithm, and the 23*a^4*b^4 + 3*b^5
would be considered 23*(n^4)*n^4+3*n^5 for the O(n^8) (but this just screams violation).

Like I said, I don't know but it just doesn't seems right that the number of steps expression (a finite series) just gets slapped into the big-O notation but it is a polynomial growth so I'm comfortable just taking your word for it (its not my homework).
 
thequickandthedead said:
I'm really stumped on these two questions :/
http://i.imgur.com/IaXG8ly.jpg
Click to expand...

Hours later, I'm sorry. not sure what exactly you are struggling with, so I will give you a strategy, but please have a go yourself.

For question 3 at least:
(i) you have y(x)= k(x^2)exp(-2x)
find y'(x) and y''(x) - the derivative and 2nd derivative of y w.r.t x
place them in the ODE: y'' + 4y' + 4y = 2exp(-2x)
terms will cancel out and you should find k.
Answer:
k=1
from my working (please check)

(ii) we want to find a solution to the 2nd order (non-homogeneous) ODE and we know y(0)=1 and y'(0)=0

Strategy: Find the complementary function y_c(x) - the general solution to the corresponding homogeneous equation - that is:
y'' + 4y' +4y = 0
we do this by: writing the auxiliary equation: &#955;^2 + 4 &#955; + 4 = 0
and finding its roots
Answer:
we should have a repeated root &#955;_1= &#955;_2= &#955; = -2 which implies y_c(x) = Aexp(&#955;x) + Bxexp(&#955;x) (A and B are constants to be found)
Again please check

then find the particular solution y_p(x) of the given non-homogeneous equation (there are some easy to remember rules here based on what the additional function is)

but essentially we try y_p(x) = C(x^2)exp(-2x) (where C is a constant)

Note: y_p(x) is the same as part (i) thus we already know the value of C

The general solution is therefore y(x) = y_c(x) + y_p(x)

then using out boundary conditions y(0)=1 and y'(0) = 0 we can find constants A and B
and thus the complete solution.

Answer: Simplified
y(x) = (1 + 2x + x^2 )*exp(-2x)
(check)

(iii) again pretty simple, you should have the answer from part (ii), you can easily see the dominant term so as x becomes very large y tends to
0

edit: took a break - second part.

4) we have the homogeneous ODE: x'' +2kx' + 4x =0 (x = x(t), x is a function of t, k is a real constant)

again like above, solving the ODE we write the Auxiliary equation: &#955;^2 + 2k &#955; + 4 = 0
then we look at our 3 possibilities:
a) |k| > 2
b) |k| < 2
c) k = 2

a) =>
we have real distinct roots
thus the general solution is
x(t) = Aexp(&#955;_1*x) + Bexp(&#955;_2*x) (A and B are constants)
b) =>
we have complex conjugate roots &#955; = &#945; +or- &#946;i
thus the general solution is
y(x) = Aexp(&#945;x)cos(&#946;x) + Bexp(&#945;x)sin(&#946;x)
c) =>
repeated roots
thus the general solution is
y(x) = Aexp(&#955;x) + (Bx)exp(&#955;x)

Hopefully all this is familiar to you, I couldn't imagine trying this without it being taught previously in lectures.

4)(ii) same as 3)(ii)
 
Could someone help me with this? (calc 2)

Find the centroid of the region bounded by the two curves

y=x^3
y=0
x+y=2
 
Tater Tot said:
Could someone help me with this? (calc 2)

Find the centroid of the region bounded by the two curves

y=x^3
y=0
x+y=2
Click to expand...
Is there a given density for this region?
graphxnu37.png

edit: note the region is cut-off, y=0 is the x-axis.
Click to expand...

&#961;: density, M: Mass

M =&#961; &#8747; f(x) - g(x) dx

I'm feeling mistake prone, lazy, and tired. If you have any more questions, don't hesitate to ask.
momentdouk1.png

source
Click to expand...
 
MLH said:
Hours later, I'm sorry. not sure what exactly you are struggling with, so I will give you a strategy, but please have a go yourself.

For question 3 at least:
(i) you have y(x)= k(x^2)exp(-2x)
find y'(x) and y''(x) - the derivative and 2nd derivative of y w.r.t x
place them in the ODE: y'' + 4y' + 4y = 2exp(-2x)
terms will cancel out and you should find k.
Answer:
k=1
from my working (please check)

(ii) we want to find a solution to the 2nd order (non-homogeneous) ODE and we know y(0)=1 and y'(0)=0

Strategy: Find the complementary function y_c(x) - the general solution to the corresponding homogeneous equation - that is:
y'' + 4y' +4y = 0
we do this by: writing the auxiliary equation: &#955;^2 + 4 &#955; + 4 = 0
and finding its roots
Answer:
we should have a repeated root &#955;_1= &#955;_2= &#955; = -2 which implies y_c(x) = Aexp(&#955;x) + Bxexp(&#955;x) (A and B are constants to be found)
Again please check

then find the particular solution y_p(x) of the given non-homogeneous equation (there are some easy to remember rules here based on what the additional function is)

but essentially we try y_p(x) = C(x^2)exp(-2x) (where C is a constant)

Note: y_p(x) is the same as part (i) thus we already know the value of C

The general solution is therefore y(x) = y_c(x) + y_p(x)

then using out boundary conditions y(0)=1 and y'(0) = 0 we can find constants A and B
and thus the complete solution.

Answer: Simplified
y(x) = (1 + 2x + x^2 )*exp(-2x)
(check)

(iii) again pretty simple, you should have the answer from part (ii), you can easily see the dominant term so as x becomes very large y tends to
0

edit: took a break - second part.

4) we have the homogeneous ODE: x'' +2kx' + 4x =0 (x = x(t), x is a function of t, k is a real constant)

again like above, solving the ODE we write the Auxiliary equation: &#955;^2 + 2k &#955; + 4 = 0
then we look at our 3 possibilities:
a) |k| > 2
b) |k| < 2
c) k = 2

a) =>
we have real distinct roots
thus the general solution is
x(t) = Aexp(&#955;_1*x) + Bexp(&#955;_2*x) (A and B are constants)
b) =>
we have complex conjugate roots &#955; = &#945; +or- &#946;i
thus the general solution is
y(x) = Aexp(&#945;x)cos(&#946;x) + Bexp(&#945;x)sin(&#946;x)
c) =>
repeated roots
thus the general solution is
y(x) = Aexp(&#955;x) + (Bx)exp(&#955;x)

Hopefully all this is familiar to you, I couldn't imagine trying this without it being taught previously in lectures.

4)(ii) same as 3)(ii)
Click to expand...
Thanks! Really appreciate it. Apart from a few things here and there i did generally ok :)
 
RELAYER said:
my god...
i'm an idiot
Click to expand...
Nah, you were looking for something else and just overlooked the algebra. Its like the optical illusions in this thread.
campfireweekend said:
Could you explain that further? I can see how it might help with change in temp part, but not where the bee is. I may be missing something incredibly obvious.
Click to expand...

r(t) is a function of the mutant bee's position, wouldn't r(1/4) give the position (x,y,z)?
When t=(1/4), ( 5sin(pi/2), 0, (3/5)cos(pi/2) ) = ( 5, 0, 0 )
 
Wow, I feel stupid. So I would just plug that back into my gradient vector and get the answer the second part right?

Edit: wait, I still need the direction for rate of change, how would I extract that from the r(t) function?
 
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