Support NeoGAF

[TUSR]

Find the slope of the tangent line to the curve xy^3 + 2y + 0.384 = 0 at the point (-2,-0.2).


I took the derivative of the equation and got
-y^3/(3xy^2 + 2)

To calculate the slope at that specific point, do I just plug those coordinates into my slope?
I tried doing that and my answer was wrong.

i got -20

Spoiler

did it in my head so dont be amazed if its wrong

 

[youngwerther]

Just tried it, that's wrong too.

But more than just an answer, I'm looking for where I'm going wrong.
The only things that could be happening is I either didn't take the derivative correctly, or I fudged the algebra when calculating the slope.

 

[TUSR]

Just tried it, that's wrong too.

But more than just an answer, I'm looking for where I'm going wrong.
The only things that could be happening is I either didn't take the derivative correctly, or I fudged the algebra when calculating the slope.

You derivative is right

y' = -(y^3)/(3xy^2+2)

 

[youngwerther]

Well then what the hell.

The answer I got was .004545.

I tried inputing both .004545 and -.004545 just in case some wacky exponent stuff was going on with the negatives, and both were wrong.

 

[TUSR]

Well then what the hell.

The answer I got was .004545.

I tried inputing both .004545 and -.004545 just in case some wacky exponent stuff was going on with the negatives, and both were wrong.

Ill fix this

I assume you're doing this with your calculator, which your order of operations and bracket control is goofy.

 

[youngwerther]

This is probably a really, really bad question at this point, but why are you substituting y's for x's and x's for y's?

 

[TUSR]

This is probably a really, really bad question at this point, but why are you substituting y's for x's and x's for y's?

What?

Find the slope of the tangent line to the curve xy^3 + 2y + 0.384 = 0 at the point (-2,-0.2).

Hahahaha ohhhhhhh yeah

Sorry I was more focused on getting the derivative down, which you had. Should just be a plug and play question at this point.

 

[Leezard]

Well then what the hell.

The answer I got was .004545.

I tried inputing both .004545 and -.004545 just in case some wacky exponent stuff was going on with the negatives, and both were wrong.

I got this answer as well when I tried it. However, the exact answer is 1/220, 0.004545 is rounded. Perhaps you should try that.

 

[youngwerther]

By the way, when you are taking the derivative of that 2y, what is the proper way to denote that?

 

[MikeDip]

By the way, when you are taking the derivative of that 2y, what is the proper way to denote that?

2(dy/dx) or 2y`

If you mean like before doing the derivative it's
d(2y)/dx or (2y)`

 

[thequickandthedead]

Yeah I do, do you need help with something?

Nothing as of yet, although i'm curious to know what exam board you did for Decision maths.

 

[youngwerther]

How do you show that implicit and explicit differentiation gives the same result?

For example I am given cosx+sqrt =5

implicitly the derivative is y' = 2sqrt *(sinx)

explicitly - y' = 2sinx(5-cosx)

Now my next step is to prove that they are the same. How do I do that?

 

[MikeDip]

How do you show that implicit and explicit differentiation gives the same result?

For example I am given cosx+sqrt =5

implicitly the derivative is y' = 2sqrt *(sinx)

explicitly - y' = 2sinx(5-cosx)

Now my next step is to prove that they are the same. How do I do that?

You just transform one of your answers into the other. For this question, take y' = 2sqrt *(sinx). Now plug in what y is, to get it all in terms of x. So from the question since cosx+sqrt =5, then y = (5-cosx)^2. Plug this in and we get

2sqrt((5-cosx)^2))*sinx
Which equals 2(5-cosx)sinx.

This is equal to what we want it to be, so done.

 

[Lion Heart]

I'm getting back into math after a looong hiatus. Currently doing simple factoring, and while its going fine, two questions have me stumped.

1) a^2-1+4 b^2-4 a^2 b^2

And the answer is -(a-1) (a+1) (2b-1) (2b+1).

I get the first two brackets but (2b-1)(2b+1)? No idea. Group factoring sucks.

2) x-y-xy+1

Usually these questions are easy for me but the 1 coefficient F's with my mind. I know the answer, and I can expand to see it but not sure how to properly figure it out.

 

[MikeDip]

I'm getting back into math after a looong hiatus. Currently doing simple factoring, and while its going fine, two questions have me stumped.

1) a^2-1+4 b^2-4 a^2 b^2

And the answer is -(a-1) (a+1) (2b-1) (2b+1).

I get the first two brackets but (2b-1)(2b+1)? No idea. Group factoring sucks.

2) x-y-xy+1

Usually these questions are easy for me but the 1 coefficient F's with my mind. I know the answer, and I can expand to see it but not sure how to properly figure it out.

Hi, here's what I did;
a^2-1+4 b^2-4 a^2 b^2 I factored out the -4b^2 from the last two terms. This gives
a^2 - 1 + 4b^2(a^2 - 1) Now we can factor out (a^2 - 1) to give us
(a^2 - 1)(1- 4b^2)

Now with both of these, we can use difference of squares.
(a^2 - 1) = (a+1)(a-1)
(1-4b^2) = (1-2b)(1+2b) (since sqrt(1) is 1, and sqrt (4b^2) is 2b. so we have (1^2 - (2b)^2) and can use difference of squares.

So the final answer is (a+1)(a-1)(1-2b)(1+2b). This is the same as your answer, the front minus sign is just multiplied with the (1-2b) term.

For number 2,

x-y - xy + 1
rearrange to get x-xy-y+1. Now I will factor out the x from the first two terms
x(1-y) - y+1 Note how similar those last two terms look compared with what is multiplying with x. Let's try to get them in that form by simply rearranging.

x(1-y) +1 - y
Now we can factor out (1-y)
(1-y)(x+1)

Do you have any questions?

 

[thequickandthedead]

Does anyone know a quick way to do (α+β+&#947^3 with having to expand?

 

[GavinGT]

Does anyone know a quick way to do (α+β+&#947^3 with having to expand?

Just plug your values into the final formula?

(x+y+z)^3 = x^3+3 x^2 y+3 x^2 z+3 x y^2+6 x y z+3 x z^2+y^3+3 y^2 z+3 y z^2+z^3

 

[thequickandthedead]

Just plug your values into the final formula?

(x+y+z)^3 = x^3+3 x^2 y+3 x^2 z+3 x y^2+6 x y z+3 x z^2+y^3+3 y^2 z+3 y z^2+z^3

My mind just went blank. Which formula did you use?

 

[GavinGT]

My mind just went blank. Which formula did you use?

http://www.wolframalpha.com/input/?i=%28x%2By%2Bz%29%5E3

I just asked Wolfram Alpha.

Binomial theorem ?
http://en.wikipedia.org/wiki/Binomial_theorem

Technically it's a trinomial. I guess he could use this if he understood it:

 

[GavinGT]

Well to be fair you can always rewrite (x+y+z)^(3) as (x+(y+z))^3.

Good point! Here's a more user friendly version:

 

[thequickandthedead]

You know how (α+β+&#947^2 can be easily solved by substituting values into:
((α+β+&#947^2)-(2(αβ+αγ+β&#945)
(i already know the values of (α+β+&#947 and (αβ+αγ+β&#945 beforehand)

I was wondering if there was an equivalent for (α+β+&#947^3.

 

[GavinGT]

You know how (α+β+&#947^2 can be easily solved by substituting values into:
((α+β+&#947^2)-(2(αβ+αγ+β&#945)

That can't be the correct formula. That says that a quantity equals itself minus another quantity.

 

[Lion Heart]

Hi, here's what I did;
a^2-1+4 b^2-4 a^2 b^2 I factored out the -4b^2 from the last two terms. This gives
a^2 - 1 + 4b^2(a^2 - 1) Now we can factor out (a^2 - 1) to give us
(a^2 - 1)(1- 4b^2)

Now with both of these, we can use difference of squares.
(a^2 - 1) = (a+1)(a-1)
(1-4b^2) = (1-2b)(1+2b) (since sqrt(1) is 1, and sqrt (4b^2) is 2b. so we have (1^2 - (2b)^2) and can use difference of squares.

So the final answer is (a+1)(a-1)(1-2b)(1+2b). This is the same as your answer, the front minus sign is just multiplied with the (1-2b) term.

For number 2,

x-y - xy + 1
rearrange to get x-xy-y+1. Now I will factor out the x from the first two terms
x(1-y) - y+1 Note how similar those last two terms look compared with what is multiplying with x. Let's try to get them in that form by simply rearranging.

x(1-y) +1 - y
Now we can factor out (1-y)
(1-y)(x+1)

Do you have any questions?

Hey thanks that helped me out a lot. My problem is group factoring, I know on my test Monday, I'll choose the wrong things to group and that will lead to a wrong answer but when you point it out it makes sense. Although for the first example, I think you missed a negative so "a^2 - 1 + 4b^2(a^2 - 1)" should be a^2 - 1 -4b^2(a^2 - 1). Anyway, good stuff.

 

[GavinGT]

Although for the first example, I think you missed a negative so "a^2 - 1 + 4b^2(a^2 - 1)" should be a^2 - 1 -4b^2(a^2 - 1). Anyway, good stuff.

Actually, he just didn't pull out a '-1' from the final (1-2b) term. His answer was acceptable.

(a+1)(a-1)(1-2b)(1+2b) = -(a+1)(a-1)(2b-1)(1+2b)

EDIT: I guess you were talking about the first line.

 

[MikeDip]

Actually, he just didn't pull out a '-1' from the final (1-2b) term. His answer was acceptable.

(a+1)(a-1)(1-2b)(1+2b) = -(a+1)(a-1)(2b-1)(1+2b)

Yeah The answer is fine, but I had a typo on the second line, the plus should have been a negative.

Hey thanks that helped me out a lot. My problem is group factoring, I know on my test Monday, I'll choose the wrong things to group and that will lead to a wrong answer but when you point it out it makes sense. Although for the first example, I think you missed a negative so "a^2 - 1 + 4b^2(a^2 - 1)" should be a^2 - 1 -4b^2(a^2 - 1). Anyway, good stuff.

Yeah, sorry about that, I typed that up at like 5 am so I didn't catch the typo. In regards to choosing the "wrong things to group", just pair up things that look similar. Like in the first question, the last two terms had that 4, so that's a nice place to start looking. Also, what I usually do is put all similar variables next to each other, it makes it easier to see what is going on. Just do small factoring between those similar terms and worry about the big picture later. Good luck on your test though!

 

[Lion Heart]

Yeah The answer is fine, but I had a typo on the second line, the plus should have been a negative.




Yeah, sorry about that, I typed that up at like 5 am so I didn't catch the typo. In regards to choosing the "wrong things to group", just pair up things that look similar. Like in the first question, the last two terms had that 4, so that's a nice place to start looking. Also, what I usually do is put all similar variables next to each other, it makes it easier to see what is going on. Just do small factoring between those similar terms and worry about the big picture later. Good luck on your test though!

Yeah I'm starting to see patterns, hopefully everything clicks. And thanks!
Heres one last question, its sort of a bonus Q the teacher gave.

Not sure how to approach it.

 

[cpp_is_king]

Yeah I'm starting to see patterns, hopefully everything clicks. And thanks!
Heres one last question, its sort of a bonus Q the teacher gave.

Not sure how to approach it.

What 2 numbers multiply together to give 1/4, but add together to give -1?

Since they add together to give -1, you know that at least one of them is negative.

Since the multiply together to give 1/4, you know that either both are positive or both are negative.

Combining these two things together, you now know that both are negative. So two negative numbers that multiply to 1/4 and add to -1. If you took a random guess, your first guess would probably be correct. Then verify it works

 

[Lion Heart]

Yeah I get (h-1/2)^2, thats cool but I would like to know how to get

Edit: eh its probably not necessary, the teacher will accept my answer.

 

[GavinGT]

Yeah I get (h-1/2)^2, thats cool but I would like to know how to get

Factor a 1/2 out of the expression in the parentheses.

(h-1/2)^2 = [(1/2)(2h-1)]^2 = (1/4)(2h-1)^2

 

[Leezard]

Yeah I get (h-1/2)^2, thats cool but I would like to know how to get

Edit: eh its probably not necessary, the teacher will accept my answer.

factor out 1/2^2. done.

 

[MikeDip]

Yeah I'm starting to see patterns, hopefully everything clicks. And thanks!
Heres one last question, its sort of a bonus Q the teacher gave.

Not sure how to approach it.

If fractions are getting you down, you could always just multiply by 1, (4/4) in this case. In math, a lot of the time it is very useful to multiply by 1. Then the question becomes

(4h^2 - 4h + 1) / 4 and factoring the top should be simple.
(2h-1)(2h-1) / 4

Other ways are fine too, of course.

 

[f0rk]

I have 2 random variables on different intervals and a joint probability density distribution, how do I calculate the two 1-dimensional PDDs?

 

[Psycho_Mantis]

Guys I have a problem with FP2 complex number, roots and exponents.

Considering (cos@ +jsin@)^4 express tan4@ in terms of tan@

@ is theta.

 

[cpp_is_king]

Guys I have a problem with FP2 complex number, roots and exponents.

Considering (cos@ +jsin@)^4 express tan4@ in terms of tan@

@ is theta.

Im confused. Tan^4 is already expressed in terms of tan

 

[Leezard]

Guys I have a problem with FP2 complex number, roots and exponents.

Considering (cos@ +jsin@)^4 express tan4@ in terms of tan@

@ is theta.

use de moivres formula, expand, divide by cos. You should be quite close after that.

 

[MikeDip]

I have 2 random variables on different intervals and a joint probability density distribution, how do I calculate the two 1-dimensional PDDs?

Taking partial derivatives of the joint pdf with respect to each random variable should give you their 1 dimensional pdfs if I recall? It's been a while since I looked at stats.

 

[Leezard]

I have 2 random variables on different intervals and a joint probability density distribution, how do I calculate the two 1-dimensional PDDs?

Taking partial derivatives of the joint pdf with respect to each random variable should give you their 1 dimensional pdfs if I recall? It's been a while since I looked at stats.

If the variables are independent, you should be able to calculate the 1-dimensional pdf for one varable by integrating over all space for the other variable. It was quite some time since I did statistics as well, but intuitively that should work out well since the important part is everything integrates to unity.

 

[f0rk]

Might as well just give the question as I'm not sure I understand

Two random variables a and b take values in [1,2] and [4,5]. The joint PDD is p(a,b)=(a+b)/6.
a) Find 1D PDDs for a and b
b) are a and b statistically dependant

I'm guessing they are statistically dependent as independence => p(a,b)=p(a)*p(b) which isn't possible in this case?

 

[Leezard]

Well, what are the limitations; is it discrete or continuous?

 

[f0rk]

They are continuous on [1,2] and [4,5] respectively

 

[siddhu33]

Guys I have a problem with FP2 complex number, roots and exponents.

Considering (cos@ +jsin@)^4 express tan4@ in terms of tan@

@ is theta.

Expand binomial, separate into real and complex parts, divide complex by real to form expression for tan4x in terms of sin x & cos x, and then divide by (cos x)^4 top and bottom to turn all sin and cos into tan.

 

[Leezard]

They are continuous on [1,2] and [4,5] respectively

try integrating p(a,b) over b, you get (a/6) + 9/12. integrating p(a,b) over a, you get (3/12) + b/6. Now, I don't know what form the answer is supposed to be in. If you integrate the earlier equations you get unity. If you don't want addition in the answer you could take the a or b part and scale it to unity. seeing as a/6 integrated over the relevant interval is 3/12, you could scale it to 4a/6. In the end, it depends on what form the answer is supposed to be in.

 

[f0rk]

Sorry the notes and previous problems I've been given don't really help in context with this question. I can't find anything referencing how to deal with this one. There are a lot of integrations between - and +infinity in my notes but they don't make sense in this case, I think there is a 'discrete' missing in a lot of the descriptions.

So is the PDD p1(a)=a/6+9/12? Which is dependent on b because (a/6+9/12)*(b/6+3/12) =/= (a+b)/6?
I also need to find the mean of a, here I should multiply p1(a) by a and then integrate between 2 and 1 right?

 

[leroidys]

Cross-posting this from the programming thread...

Can anyone explain how this captures every possible binary tree for n nodes? I don't understand the recurrence relation. (It's the catalan series of numbers).

EDIT: herp derp.... as soon as i post it I think I finally figure it out. Also cross-posting...

EDIT: I think I FINALLY got it (I worked on this problem for several hours yesterday. Sometimes you just need to focus on something else for a while...)

So the two factors represent the left and right child of the parent node. The children range from 0 to n-1 nodes. So if we know how arrangements of 0..n-1 nodes are possible, then we can just multiply the results together to find the total number of combinations.

If anyone has any other input on this or catalan numbers in general though, I would love to hear it. I love learning about new sequences.

 

[Partial Gamification]

If anyone has any other input on this or catalan numbers in general though, I would love to hear it. I love learning about new sequences.

pic is link

also: if you love sequences.

edit: and to be clear for any others, your summation produces a series; and a "list" is a sequence.

 

[leroidys]

pic is link

also: if you love sequences.

edit: and to be clear for any others, your summation produces a series; and a "list" is a sequence.

OMG that site is amazing. Thanks!

 

[Leezard]

Sorry the notes and previous problems I've been given don't really help in context with this question. I can't find anything referencing how to deal with this one. There are a lot of integrations between - and +infinity in my notes but they don't make sense in this case, I think there is a 'discrete' missing in a lot of the descriptions.

So is the PDD p1(a)=a/6+9/12? Which is dependent on b because (a/6+9/12)*(b/6+3/12) =/= (a+b)/6?
I also need to find the mean of a, here I should multiply p1(a) by a and then integrate between 2 and 1 right?

Remember that integrating between -/+ infinity is the same as integrating between [1,2] (or [4,5]) since by convention the probability density is zero outside the defined interval.

I cannot answer your first question, but you should be able to calculate the expectation value for a by the procedure you described in the second question, provided that the found pdf is actually the correct one. I cannot guarantee that.

 

[MikeDip]

pic is link

also: if you love sequences.

edit: and to be clear for any others, your summation produces a series; and a "list" is a sequence.

That's an awesome link, thanks!

 

[cpp_is_king]

Remember that integrating between -/+ infinity is the same as integrating between [1,2] (or [4,5]) since by convention the probability density is zero outside the defined interval.

I cannot answer your first question, but you should be able to calculate the expectation value for a by the procedure you described in the second question, provided that the found pdf is actually the correct one. I cannot guarantee that.

If the question is to just find a 1 dimensional probability density distribution, just integrate with respect to one of the two variables (but not the other), and you're done. Right?

 

[Leezard]

If the question is to just find a 1 dimensional probability density distribution, just integrate with respect to one of the two variables (but not the other), and you're done. Right?

Yeah, it's just that I think the pdf looks a bit awkward when you have a constant term added to it. I cannot recall that I've seen a pdf like that before.

The reasoning for the integration part is that if you have a multidimensional pdf, it has to become 1 as you integrate over all space for all the variables. In the 2-dimensional case, you have to integrate over both x and y, to use standard notation. Imagine that for each constant x-value, you add the probabilities of all different y-values. In a way, you take the 2-dimensional space and press it together along the y axis to have a 1-dimensional curve that only depends on x. That's integrating with the respect of one of the two variables.