3U Maths Question Thread

[owidjaja]

Hey!
So, when we do the substitution, we need to translate everything into the u world. Notice that dx becomes udu. So when we're integrating with respect to u, we need to replace dx with udu, and not du.

Also, notice that when we're translating everything into the u world, we need to change our original sqrt(2x+7) into the respective u component.

In this way, we can finally complete our new integral.

Hmm I don't think I'm getting it because I'm trying to follow the method for this question and it's not working out either.

[Opengangs]

Hmm I don't think I'm getting it because I'm trying to follow the method for this question and it's not working out either.

No problem! Let's take it step by step.

The first thing to note is what you will be substituting. The question gives you the substitution, so it's in our best interest to use it.

When we're differentiating both sides it's important to note that u is a function of x, so to differentiate it, we use the chain rule. That is, we differentiate the outside function (u^2 becomes 2u), and then the inside function which is just du/dx. So, the differential of the left hand side is just (du/dx) * 2u. Differentiating the right hand side should be simpler. The goal here is to make dx the subject because we're translating it from x to the u world.

The next part now is to consider the numerator. We need to translate x in terms of u. To do this, we consider u^2 = x + 2. We just simple make x the subject, and from there x = u^2 - 2. Thus, the numerator becomes u^2 - 4.

Notice that we've established dx as 2udu, sqrt(x + 2) as u, and x-2 as u^2 - 4. Do you think you could establish the new integral from here?

[owidjaja]

No problem! Let's take it step by step.

The first thing to note is what you will be substituting. The question gives you the substitution, so it's in our best interest to use it.

When we're differentiating both sides it's important to note that u is a function of x, so to differentiate it, we use the chain rule. That is, we differentiate the outside function (u^2 becomes 2u), and then the inside function which is just du/dx. So, the differential of the left hand side is just (du/dx) * 2u. Differentiating the right hand side should be simpler. The goal here is to make dx the subject because we're translating it from x to the u world.

The next part now is to consider the numerator. We need to translate x in terms of u. To do this, we consider u^2 = x + 2. We just simple make x the subject, and from there x = u^2 - 2. Thus, the numerator becomes u^2 - 4.

Notice that we've established dx as 2udu, sqrt(x + 2) as u, and x-2 as u^2 - 4. Do you think you could establish the new integral from here?

This is my working out at the moment (actually numbered the lines so it would be easier xD). I'm still not entirely sure where I'm going with this.

[Opengangs]

This is my working out at the moment (actually numbered the lines so it would be easier xD). I'm still not entirely sure where I'm going with this.

Hey there!
Be careful with your differentiation in line 2. Remember that a derivative is of the form: \( \frac{dy}{dx} \). We've introduced a new parameter, u, so we need to be careful with how we're dealing with the derivative.



From here, we have all of the pieces to transform from the x world to the u world. Do you think you could do it? Let me know if you're still confused

[owidjaja]

Hey there!
Be careful with your differentiation in line 2. Remember that a derivative is of the form: \( \frac{dy}{dx} \). We've introduced a new parameter, u, so we need to be careful with how we're dealing with the derivative.



From here, we have all of the pieces to transform from the x world to the u world. Do you think you could do it? Let me know if you're still confused

So how do I fix up line 2? Sorry, still trying to go through the steps slowly.

[Opengangs]

So how do I fix up line 2? Sorry, still trying to go through the steps slowly.

Sorry, my LaTeX messed up. Let me know if you're still struggling

[Sine]

So how do I fix up line 2? Sorry, still trying to go through the steps slowly.

Re: line 2

remember with any mathematical equation we can only do something if we do the same to the other side. You have correctly in put d/dx for both sides but if one side of the equation is in terms of u you must use the chain rule. i.e. you can't do something like d/dx(u^2) = 2u because we are differentiating with respect to x and there is no x present.

[owidjaja]

Sorry, my LaTeX messed up. Let me know if you're still struggling

So I finally fixed up line 2. Why did the numerator change from u^2-2 to u^2-4?

[Opengangs]

So I finally fixed up line 2. Why did the numerator change from u^2-2 to u^2-4?

Hey there!
Remember that our primary goal is to eliminate all of the x's so we can integrate with respect to u. Recalling that u^2 was originally x + 2, we can find x in terms of u. That is, x = u^2 - 2. This just then means the x - 2 = (u^2 - 2) - 2 = u^2 - 4

[owidjaja]

So here's where I'm up to. Did I sub it in right? (So sorry for everyone here- really wanna make sure I can do this question).

[Opengangs]

So here's where I'm up to. Did I sub it in right? (So sorry for everyone here- really wanna make sure I can do this question).

Hey there!
Don't apologise for trying to seek help
You're very close - pay attention to what you substituted for u in the question: \( u^{2} = x + 2 \). This implies that \( u = \sqrt{x + 2} \).
Hopefully, this helps with your substitution.

If you have any more questions, don't be afraid to post them here!

[owidjaja]

Hey there!
Don't apologise for trying to seek help
You're very close - pay attention to what you substituted for u in the question: \( u^{2} = x + 2 \). This implies that \( u = \sqrt{x + 2} \).
Hopefully, this helps with your substitution.

If you have any more questions, don't be afraid to post them here!

So like this?

[radnan11]

how do u do this question

"A postmaster has an unlimited supply of 3 cent and 5 cent stamps. Prove by
mathematical induction that he can make up the value of 8 cents or greater for any
postage."

[RuiAce]

how do u do this question

"A postmaster has an unlimited supply of 3 cent and 5 cent stamps. Prove by
mathematical induction that he can make up the value of 8 cents or greater for any
postage."

Hence completing the induction.

Edit: of course, this only works under the assumption we have a 5 cent coin there to begin with. If we don’t, then we can take out three 3c coins and add in two 5c coins.

[radnan11]

did u mean two 5c coins?