Math

[LOLKILLERTOTHEDEATH 1,921]

Waiting for more people to attempt to solve it before I show my working! 

I have no idea on how to solve that, but I do have lots of time. As such, would it be too much to ask for you to type the equation on Desmos? I'm not sure what the equation is exactly.

[Chronicle 164]

I'm not familiar with the equation on desmos, I'm only an expert with exponential functions and logarithms, I'll start with an explanation to my question so I don't keep you waiting, bear with me.

[Chronicle 164]

How did you solve it?

 

^ is like saying, this is to the power of... Eg. 5 ^ 2, 5 to the power of 2. There can also be fractions, so 5 ^ 2/5. And there can also be negatives, they are called the laws of indices. 

 

In this question we are asked to find what y is, given that y = 35 at x = 4. I've learnt that to do a question like this you have to work backwards (we have to integrate dy/dx). 

 

We start by saying that y = the integral of the other conditions of the equation. 

 

We have to write the conditions in brackets because we have multiple terms. (5x ^ -1/2 + x^3/2) 

 

Also, we can't write "x√x". Instead we write it using the law of Indices. We write the "√x" as x ^ 1/2". We have another term which is x and we simply add it to the power, so we get a top-heavy fraction in the form "x ^ 3/2".

We then close the bracket with the terms, and complete it by putting the dx at the end as we are integrating with respect to x, its just to show them we are integrating with respect. Here is what we get once we complete all the steps:  

 

y = ∫ (5x ^ -1/2 + x^3/2) dx 

 

Don't let the "∫" scare you, its only an integral symbol to denote the integral. 

 

We then add 1 to the power and divide the term by the new power, we do this to both terms inside the brackets. Here's what we get: 

 

y = ∫ (5x ^ -1/2 + x^3/2) dx  (So here we add 1 to -1/2, and that gives us a half, hence the " 1/2), we do the same to the other, so 1 + 3/2 is 5/3)

 

y = 5 (x^1/2) / 1/2 + (x^5/2) / 5/2  +  c. (You can't forget the C, its the constant of integration)

 

Because everything is scattered around in the equation (y = 5 (x^1/2) / 1/2 + (x^5/2) / 5/2  +  c), we clean things up. We can do this by multiplying the top and bottom of each term by the denominators of the fractions. (So we multiply 5x^1/2 by 2 which is from the 1/2, we do the same with x^5/2, it depends on the denominator of course, in this case both are two. We get: 

 

y = 10x ^1/2 + (2x ^ 5/2) / 5 + c.

 

Now we need to work out this constant of integration ©. This is where " y = 35 at x = 4" comes in handy. 

 

We know when x = 4, y = 35, we can substitute these values into our latest equation. Therefore:

35 = 10 (4) ^1/2 + (2 (4) ^ 5/2) / 5 + c. 

 

we simplify everything down to make it easier, and we get.

 

35 = 20 + 64/5 + c. Now we have to make "c" the subject, once this is done, it gives us: 

 

C = 11/5. We substitute this value into our original equation. 

 

y = 10x^1/2 + (2x^5/2)/5 + 11/5. <- this is your answer. 

[LOLKILLERTOTHEDEATH 1,921]

^ is like saying, this is to the power of... Eg. 5 ^ 2, 5 to the power of 2. There can also be fractions, so 5 ^ 2/5. And there can also be negatives, they are called the laws of indices. 

 

In this question we are asked to find what y is, given that y = 35 at x = 4. I've learnt that to do a question like this you have to work backwards (we have to integrate dy/dx). 

 

We start by saying that y = the integral of the other conditions of the equation. 

 

We have to write the conditions in brackets because we have multiple terms. (5x ^ -1/2 + x^3/2) 

 

Also, we can't write "x√x". Instead we write it using the law of Indices. We write the "√x" as x ^ 1/2". We have another term which is x and we simply add it to the power, so we get a top-heavy fraction in the form "x ^ 3/2".

We then close the bracket with the terms, and complete it by putting the dx at the end as we are integrating with respect to x, its just to show them we are integrating with respect. Here is what we get once we complete all the steps:  

 

y = ∫ (5x ^ -1/2 + x^3/2) dx 

 

Don't let the "∫" scare you, its only an integral symbol to denote the integral. 

 

We then add 1 to the power and divide the term by the new power, we do this to both terms inside the brackets. Here's what we get: 

 

y = ∫ (5x ^ -1/2 + x^3/2) dx  (So here we add 1 to -1/2, and that gives us a half, hence the " 1/2), we do the same to the other, so 1 + 3/2 is 5/3)

 

y = 5 (x^1/2) / 1/2 + (x^5/2) / 5/2  +  c. (You can't forget the C, its the constant of integration)

 

Because everything is scattered around in the equation (y = 5 (x^1/2) / 1/2 + (x^5/2) / 5/2  +  c), we clean things up. We can do this by multiplying the top and bottom of each term by the denominators of the fractions. (So we multiply 5x^1/2 by 2 which is from the 1/2, we do the same with x^5/2, it depends on the denominator of course, in this case both are two. We get: 

 

y = 10x ^1/2 + (2x ^ 5/2) / 5 + c.

 

Now we need to work out this constant of integration ©. This is where " y = 35 at x = 4" comes in handy. 

 

We know when x = 4, y = 35, we can substitute these values into our latest equation. Therefore:

35 = 10 (4) ^1/2 + (2 (4) ^ 5/2) / 5 + c. 

 

we simplify everything down to make it easier, and we get.

 

35 = 20 + 64/5 + c. Now we have to make "c" the subject, once this is done, it gives us: 

 

C = 11/5. We substitute this value into our original equation. 

 

y = 10x^1/2 + (2x^5/2)/5 + 11/5. <- this is your answer. 

Well, that doesn't make too much sense to me yet. Thanks for explaining it though, once I hopefully get a hang of that concept, I'll try to see if I can get that solution myself.

[Chronicle 164]

Well, that doesn't make too much sense to me yet. Thanks for explaining it though, once I hopefully get a hang of that concept, I'll try to see if I can get that solution myself.

Yes, It takes time, you need to get your head around the basics first. Try easier integration questions, I'll gladly send you some here if you'd like

[Chronicle 164]

Wait, my mistake. That was an integration question... under the title integration (equations of curves). 

[Chronicle 164]

I am yet to learn about graphing parametric equations in desmos, once I understand the concept I'll try explaining it to you. 

[LOLKILLERTOTHEDEATH 1,921]

I see. I spent the past few days learning about limits, so I can do some very basic limits questions now. I hope to move on to derrivatives and such.

[thethiefofvictory 339]

I see. I spent the past few days learning about limits, so I can do some very basic limits questions now. I hope to move on to derrivatives and such.

Do teach me about derivatives! I've taken an internet lesson on limits so I too can do some basic limits questions.

[QuicKK 975]

Guys please,do not copy explanations about math's subjects from the websites, just do your own explanation, this might help others understand easily, beacuse our type of language is familiar, so that's why it's not hard to understand.

That's my opinion.

[Chronicle 164]

Guys please,do not copy explanations about math's subjects from the websites, just do your own explanation, this might help others understand easily, beacuse our type of language is familiar, so that's why it's not hard to understand.

That's my opinion.

Well that's a provocative claim, how do you know that they've copied from a maths website? Who have you seen copy so far? 

[QuicKK 975]

Well that's a provocative claim, how do you know that they've copied from a maths website? Who have you seen copy so far? 

Well, in the future I meant. And it's not provocative.

[Chronicle 164]

Well, to me it is! But in a way I do concur, our explanations are easier to comprehend so explaining maths in our own words should help readers understand better. 

[QuicKK 975]

Well, to me it is! But in a way I do concur, our explanations are easier to comprehend so explaining maths in our own words should help readers understand better. 

Great, I was thinking you'd say something negative but now it's cool, thanks for your opinion :D

[Chronicle 164]

Great, I was thinking you'd say something negative but now it's cool, thanks for your opinion :D

Yea, no problem G, I gotcha! 

[girl_from_kurdistanX 383]

 

Communicate the name of the Google site with math

Do you know what google means? Google is taken from Googol. Googol is also the nickname of a number called by Milton Sirota. equivalent to ten raised to the power of a hundred (10 100 ).(look at the size of this number).

Google's choice is a slogan, in the sense that Google intends to expand its services and services to the whole world.

googolplex---->equivalent to ten raised to the power of a googol. a googolplex is therefore a 1 with zeroes, whereby already a googol is clearly larger than number of protons in the visible universe.

Edited June 26, 2018 by girl_from_kurdistanX

[ThirdOnion 4,898]

Guys please,do not copy explanations about math's subjects from the websites, just do your own explanation, this might help others understand easily, beacuse our type of language is familiar, so that's why it's not hard to understand.

"Your own explanation" can and most likely will result in loss of rigor and can raise misconceptions. I am wary of personal explanations for that reason.

[LOLKILLERTOTHEDEATH 1,921]

"Your own explanation" can and most likely will result in loss of rigor and can raise misconceptions. I am wary of personal explanations for that reason.

Thanks for the explanation of the derivative!

I think instead of making this just a Math topic, perhaps we should turn this into a math club. All in favor of joining please post accordingly.

[girl_from_kurdistanX 383]

I know this site is too childish.

But nevertheless I will defeat you all.

Today at 05:00 pm  UTC   I will be online in this site .

https://www.arcademics.com/games

 

 

find this game --------->  

 

i will play with this name today in there--->zhina

Edited June 27, 2018 by girl_from_kurdistanX

[LOLKILLERTOTHEDEATH 1,921]

 

I know this site is too childish.

But nevertheless I will defeat you all.

Today at 05:00 pm I will be online in this site .

https://www.arcademics.com/games

 

 

find this game --------->  

 

i will play with this name today in there--->zhina

 

5pm in what time-zone though?

[girl_from_kurdistanX 383]

5pm in what time-zone though?

utc  

:)

[railguniz4noobz 2,394]

My tractor will destroy you n00bs.

[LOLKILLERTOTHEDEATH 1,921]

utc  

:)

Ah I see. Much as I'd like to, I won't be able to make it as I'll be busy then.

[QuicKK 975]

Why is everyone so stubborn :(

[LOLKILLERTOTHEDEATH 1,921]

Does anybody want to be part of the math club? No commitments involved.