Author Topic: 3U Maths Question Thread (Read 1501189 times) Tweet Share

RuiAce · « **Reply #2490 on:** July 21, 2017, 10:52:27 pm »

It's a sly trick that takes advantage of the ratio of intercepts on transversals of parallel lines. I do not advise this method unless you are absolutely certain you know what you are doing.

Of course, in this demonstration I looked at the y-coordinates instead of the x-coordinates like they did in the question. The idea is still the same, however.

junzhang · « **Reply #2491 on:** July 22, 2017, 11:21:34 am »

Hi again,
Could you help me with this circle geometry question?
Thanks so much for all your help!

RuiAce · « **Reply #2492 on:** July 22, 2017, 11:39:12 am »

Quote from: junzhang on July 22, 2017, 11:21:34 am

Hi again,
Could you help me with this circle geometry question?
Thanks so much for all your help!

Here is a diagram. I have other things to attend to now so you should give it a go (and preferably someone else as well), but if you're still stuck and receive no further help by the time I'm back I can look at it later.

That question is clearly from a source prior to 1995 because of the font used, so construction should be anticipated. Also, it is clear that $\angle AOB = 2\angle ACB$.

kiwiberry · « **Reply #2493 on:** July 22, 2017, 12:32:27 pm »

Quote from: junzhang on July 22, 2017, 11:21:34 am

Hi again,
Could you help me with this circle geometry question?
Thanks so much for all your help!

Let $\angle ACB= x $
As Rui said, $\angle AOB = 2x$
Construct AB. Triangle AOB is isosceles (OA=OB). Therefore $\angle OAB= \angle OBA = 90 -x $
Also, $\angle OAC = \angle ACB = x $ (alternate angles, OA||BC)
Therefore $\angle CAB = (90-x)-x=90-2x $
So using the angle sum of triangle KAB, $ \angle AKB = 180 - (90-2x)-(90-x) =3x $!

anotherworld2b · « **Reply #2494 on:** July 22, 2017, 01:36:12 pm »

I am having trouble with this question. I am not sure how to do it without a calculator and antidifferentiate it

I made Inx = 1 + In 2 to find where they intersect. I found x to be equal 2e. I did this using a calculator. But I'm not sure how to do this without a calculator.

So far I know that the upper limit is 2e and the lower limit is 1 but I'm not sure what to do from here.

jakesilove · « **Reply #2495 on:** July 22, 2017, 01:48:37 pm »

Quote from: anotherworld2b on July 22, 2017, 01:36:12 pm

I am having trouble with this question. I am not sure how to do it without a calculator and antidifferentiate it

I made Inx = 1 + In 2 to find where they intersect. I found x to be equal 2e. I did this using a calculator. But I'm not sure how to do this without a calculator.

So far I know that the upper limit is 2e and the lower limit is 1 but I'm not sure what to do from here.

First, let's get the intercept without a calculator

$ln(x)=1+ln(2)$
$ln(x)=ln(e)+ln(2)$
$ln(x)=ln(2e)$
$x=2e$

Now, we want to find the shaded area, which is

$\int_1^{2e}ln(x)dx$

First, we should differentiate the function in the question. Presumably that will help!

$\frac{d}{dx}(xln(x)-x)=ln(x)+\frac{x}{x}-1=ln(x)$

By the product rule. Wow, how convenient. So, we can say that

$\int_1^{2e}ln(x)dx=[xln(x)-x]_1^{2e}$

You should be able to get to an answer from there.

anotherworld2b · « **Reply #2496 on:** July 22, 2017, 02:15:24 pm »

I have some questions to ask if that's okay

1. For In (x) = In (2e)
can you drop 'In' because it is on both sides? Does this work for logs as well? ( I faintly remember it does?)
2. why do you ant differentiate In x? (isn't it the gradient of y = x In x -x )
3. I was wondering don't you have to have the integral of y = 1 + In 2 minus the integral of In x to find the shaded area?

Quote from: jakesilove on July 22, 2017, 01:48:37 pm

First, let's get the intercept without a calculator

$ln(x)=1+ln(2)$
$ln(x)=ln(e)+ln(2)$
$ln(x)=ln(2e)$
$x=2e$

Now, we want to find the shaded area, which is

$\int_1^{2e}ln(x)dx$

First, we should differentiate the function in the question. Presumably that will help!

$\frac{d}{dx}(xln(x)-x)=ln(x)+\frac{x}{x}-1=ln(x)$

By the product rule. Wow, how convenient. So, we can say that

$\int_1^{2e}ln(x)dx=[xln(x)-x]_1^{2e}$

You should be able to get to an answer from there.

Shadowxo · « **Reply #2497 on:** July 22, 2017, 02:29:03 pm »

Quote from: anotherworld2b on July 22, 2017, 02:15:24 pm

I have some questions to ask if that's okay
1. For In (x) = In (2e)
can you drop 'In' because it is on both sides? Does this work for logs as well? ( I faintly remember it does?)
2. why do you ant differentiate In x? (isn't it the gradient of y = x In x -x )
3. I was wondering don't you have to have the integral of y = 1 + In 2 minus the integral of In x to find the shaded area?

1. Yes, it's saying if you take the log of x it's the same as taking the log of 2e. Therefore, x must be 2e. Alternatively, you could solve it saying x=e^ln(2e)=2e
2. You want to find the integral of ln(x) (area under the graph). Since you know that $\frac{d}{dx}( x \ln x -x )= \ln x, \text{ it follows that } x \ln x -x = \int \ln x dx$ which is what you want to find

derivative is like the opposite of integration
3. The integral of ln x is the area under the graph (the shaded part). If you wanted to find the white part under the graph of y = 1 + In(2), you could use $\int 1 + In(2) dx - \int \ln x dx$ (with bounds) but that isn't required here.

Hope this helps

Edited to put in latex code

Also never be afraid of asking questions!

jakesilove · « **Reply #2498 on:** July 22, 2017, 02:30:22 pm »

Quote from: anotherworld2b on July 22, 2017, 02:15:24 pm

I have some questions to ask if that's okay
1. For In (x) = In (2e)
can you drop 'In' because it is on both sides? Does this work for logs as well? ( I faintly remember it does?)
2. why do you ant differentiate In x? (isn't it the gradient of y = x In x -x )
3. I was wondering don't you have to have the integral of y = 1 + In 2 minus the integral of In x to find the shaded area?

Yep, two logs equal each other, then the 'inside' of those logs must equal each other, if their bases are the same. In other words

$log_ab=l_ac$
then
$b=c$

The question is just asking us to find the area under the curve of ln(x). So, we need to integrate ln(x), because that's how you find the area under the curve. That first function was just used to show you HOW to integrate ln(x), because you can't do that easily in 3U maths.

Again, unless I've misinterpreted the question, it is just asking you to find the area under the curve y=ln(x). The other function is there to tell you where the upper limit of integration is, but not to vary to actual area. It's a weird question, sure, but it comes down to a standard integration question.

anotherworld2b · « **Reply #2499 on:** July 22, 2017, 02:43:35 pm »

does it work the same for natural logs?
I see now

I believe I read the question wrong

I've been looking at some videos on youtube on U substitution and integration by parts. But I have some questions.
1. How do you determine what you should make to be U? (U substitution)
2. How do you determine what u and dv is? (integration by parts)

Quote from: jakesilove on July 22, 2017, 02:30:22 pm

Yep, two logs equal each other, then the 'inside' of those logs must equal each other, if their bases are the same. In other words

$log_ab=l_ac$
then
$b=c$

The question is just asking us to find the area under the curve of ln(x). So, we need to integrate ln(x), because that's how you find the area under the curve. That first function was just used to show you HOW to integrate ln(x), because you can't do that easily in 3U maths.

Again, unless I've misinterpreted the question, it is just asking you to find the area under the curve y=ln(x). The other function is there to tell you where the upper limit of integration is, but not to vary to actual area. It's a weird question, sure, but it comes down to a standard integration question.

jakesilove · « **Reply #2500 on:** July 22, 2017, 03:31:06 pm »

Quote from: anotherworld2b on July 22, 2017, 02:43:35 pm

does it work the same for natural logs?
I see now
I believe I read the question wrong

I've been looking at some videos on youtube on U substitution and integration by parts. But I have some questions.
1. How do you determine what you should make to be U? (U substitution)
2. How do you determine what u and dv is? (integration by parts)

Yes, because the rule applies equally when you set the base equal to e.

In 3U maths, you'll always be given the substitution, so don't worry about that. If you're not doing 4U maths, you don't need to know integration by parts.

anotherworld2b · « **Reply #2501 on:** July 22, 2017, 03:39:10 pm »

Would it be possible for you to explain to me how you would identify what to substitute for both methods please? I'm a wace student so I don't learn 3U and 4U content.

Is U substitution used for when there is a fraction and integration by parts for when two terms are multiplied?
I am not sure how to antidifferentiate tan x. I made it = to sinx/cosx but I'm not sure what to do for U substitution

Quote from: jakesilove on July 22, 2017, 03:31:06 pm

Yes, because the rule applies equally when you set the base equal to e.

In 3U maths, you'll always be given the substitution, so don't worry about that. If you're not doing 4U maths, you don't need to know integration by parts.

jakesilove · « **Reply #2502 on:** July 22, 2017, 03:56:07 pm »

Quote from: anotherworld2b on July 22, 2017, 03:39:10 pm

Would it be possible for you to explain to me how you would identify what to substitute for both methods please? I'm a wace student so I don't learn 3U and 4U content.

Is U substitution used for when there is a fraction and integration by parts for when two terms are multiplied?
I am not sure how to antidifferentiate tan x. I made it = to sinx/cosx but I'm not sure what to do for U substitution

Jamon has written a stellar guide HERE that goes through all of integration by substitution! U substitution is used where it works, and integration by parts is used where that works. Unfortunately there's no hard-and-fast rule in terms of when you should use them; Integration by substitution is probably 'easier', but integration by parts works is far more cases. Deciding which terms to define in both methods comes down to practice, and again there just aren't any rules there.

Looking to your specific question

$\int \tan(x) dx$
$\int \frac{\sin(x)}{\cos(x)}dx$
Let
$u=\cos(x)$
$\frac{du}{dx}=-\sin(x)$
$du=\cos(x) dx$
Substituting this into our integral

$\int \frac{-du}{u}=-ln(u)+C=-ln(\cos(x))+C$

However, you didn't need to do any of this; we know that

$\int \frac{f'(x)}{f(x)}dx=ln(f(x))+C$

RuiAce · « **Reply #2503 on:** July 22, 2017, 04:37:19 pm »

Quote from: jakesilove on July 22, 2017, 03:56:07 pm

However, you didn't need to do any of this; we know that

$\int \frac{f'(x)}{f(x)}dx=ln(f(x))+C$

There's a tad problem in that he does WACE and things are different... not sure if you were aware of that.

Usually when I get to his query before someone else does and it's outside the relevant HSC course he posted I just make a remark in red about it to the others.

junzhang · « **Reply #2504 on:** July 23, 2017, 05:51:46 pm »

Hi,
Could you help me with this binomial proof question?
Unfortunately, Cambridge does not give solutions
can't thank you guys enough!

Hi,
If that's ok, could you also explain this question? I have no idea where to go with the "even" and "odd" values
thanks

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