3U Maths Question Thread

[michaelalt]

Where does the 3 (on the LHS) in line 2 go? You've kept it on the RHS, but I don't understand why its disappeared on the LHS in line 2.

[jamonwindeyer]

Where does the 3 (on the LHS) in line 2 go? You've kept it on the RHS, but I don't understand why its disappeared on the LHS in line 2.

So you've got \(3^{n-5}\) on the LHS, and \(3^{n-4}\) on the RHS. If you think about these as just a whole bunch of 3's, there is one extra 3 on the RHS compared to the LHS, because the power is one higher. So, Rui divided by \(3^{n-5}\), which removed the 3 on the LHS but left a single 3 on the RHS, if that makes sense

[michaelalt]

So you've got \(3^{n-5}\) on the LHS, and \(3^{n-4}\) on the RHS. If you think about these as just a whole bunch of 3's, there is one extra 3 on the RHS compared to the LHS, because the power is one higher. So, Rui divided by \(3^{n-5}\), which removed the 3 on the LHS but left a single 3 on the RHS, if that makes sense

OHH! I get it! Thank you so much Jamon + Rui for your fast and helpful responses! So thankful to have found this resource. 

[jamonwindeyer]

OHH! I get it! Thank you so much Jamon + Rui for your fast and helpful responses! So thankful to have found this resource.

Happy to help! Glad to have you around the forums

[anotherworld2b]

Hi i was wondering if i was doing these questions correct

[RuiAce]

Hi i was wondering if i was doing these questions correct

a) is correct


c) is almost correct. Why did 2^1/2 turn into 1? 2^1/2 is approximately 1,41
d) Same problem as b). You don't expand the powers in like this.

(Note: These are 2U difficulty.)


In the next set

b) is ok. Probably not the easiest approach but still correct.

Same problem with c) and d) in expanding the powers.

[anotherworld2b]

thank you for your help

a) is correct


c) is almost correct. Why did 2^1/2 turn into 1? 2^1/2 is approximately 1,41
d) Same problem as b). You don't expand the powers in like this.

(Note: These are 2U difficulty.)


In the next set

b) is ok. Probably not the easiest approach but still correct.

Same problem with c) and d) in expanding the powers.

[bsdfjnlkasn]

I READ THE QUESTION WRONG.
Its between y=sinx, y=cosx and Y-AXIS

sorry about this mistake


Mod edit: Hi everyone, just letting you know that I (Rui) have resolved this with Rathin through external means.

Hey sorry to bring this questions up again, but how would you make the trig ratios in terms of y so as to integrate in regards to the y-axis? Also Rui, i'm not sure what you meant when you said that there wasn't a common area between y = sinx and y = cosx (in the post with the screenshot) I thought there was..


Thank you!!

[bsdfjnlkasn]

Could someone please help me with question 18? I have no idea how to even start - thank you

[jamonwindeyer]

(Image removed from quote.)

Could someone please help me with question 18? I have no idea how to even start - thank you

Cool question! Just perform the definite integral as normal:



Note that the \(\cos{2\pi n}\) term is equal to 1, because the cosine of any integer multiple of \(2\pi\) is 1 (and we know \(n\) is an integer this is a really common trick for recurrence relationships and other similar stuff, which I think is done in 4U, but pretty rare for 3U (VERY common. for electrical engineers performing Fourier Transformations though )

The next one, if you draw the curve, you'll get a sine curve that looks similar to the one below. To interpret the result geometrically, think about what that integral above represents: An area. So, what the result says is, the area underneath that curve from 0 to n (any integer), will be equal to n. So the area underneath the curve from 0 to 1 is 1 unit squared, from 0 to 5 is 5 units squared, and so on

[RuiAce]

Hey sorry to bring this questions up again, but how would you make the trig ratios in terms of y so as to integrate in regards to the y-axis? Also Rui, i'm not sure what you meant when you said that there wasn't a common area between y = sinx and y = cosx (in the post with the screenshot) I thought there was..


Thank you!!

There is most certainly a region bound by y=sin(x), y=cos(x) and the y-axis.

However, between just y=sin(x) and y=cos(x) there is not. Because we're talking about 0≤x≤π/4, if the y-axis isn't bounding it then the region isn't bounded; it's opened up. An opened up region isn't useful to us because it causes a lot of confusion as to which area we're interested in.

In fact, if we had something like -3π/4 ≤ x ≤ π/4, then we'd have a region bound only between sin and cos.

[bsdfjnlkasn]

There is most certainly a region bound by y=sin(x), y=cos(x) and the y-axis.

However, between just y=sin(x) and y=cos(x) there is not. Because we're talking about 0≤x≤π/4, if the y-axis isn't bounding it then the region isn't bounded; it's opened up. An opened up region isn't useful to us because it causes a lot of confusion as to which area we're interested in.

In fact, if we had something like -3π/4 ≤ x ≤ π/4, then we'd have a region bound only between sin and cos.

Perfect, that makes sense now. How would approach integrating with respect to y?

EDIT: I just attempted the question, I know it's messy! But could someone please just check to see if it's right? Thank you

[bsdfjnlkasn]

Cool question! Just perform the definite integral as normal:



Note that the \(\cos{2\pi n}\) term is equal to 1, because the cosine of any integer multiple of \(2\pi\) is 1 (and we know \(n\) is an integer this is a really common trick for recurrence relationships and other similar stuff, which I think is done in 4U, but pretty rare for 3U (VERY common. for electrical engineers performing Fourier Transformations though )

The next one, if you draw the curve, you'll get a sine curve that looks similar to the one below. To interpret the result geometrically, think about what that integral above represents: An area. So, what the result says is, the area underneath that curve from 0 to n (any integer), will be equal to n. So the area underneath the curve from 0 to 1 is 1 unit squared, from 0 to 5 is 5 units squared, and so on

(Image removed from quote.)

That was actually amazing. Thank you for the swift and succinct reply - it's greatly appreciated 

[jamonwindeyer]

Perfect, that makes sense now. How would approach integrating with respect to y?

EDIT: I just attempted the question, I know it's messy! But could someone please just check to see if it's right? Thank you

(Image removed from quote.)

Looks right to me!

Integrating those functions to find an area with respect to the y-axis would require the integration of inverse trig functions, which we don't learn in this course. If you want an area with respect to the y-axis for something like \(y=\sin{x}\), you would rearrange to \(x=\sin^{-1}{y}\), which we can't integrate. So, the trick would be to form some sort of rectangle and subtract an area with respect to the x-axis to find an area with respect to the y-axis; have you seen that technique before?

[bsdfjnlkasn]

Looks right to me!

Integrating those functions to find an area with respect to the y-axis would require the integration of inverse trig functions, which we don't learn in this course. If you want an area with respect to the y-axis for something like \(y=\sin{x}\), you would rearrange to \(x=\sin^{-1}{y}\), which we can't integrate. So, the trick would be to form some sort of rectangle and subtract an area with respect to the x-axis to find an area with respect to the y-axis; have you seen that technique before?

Yes I have! Also it's good to know that you can't perform integrations with respect to the y-axis because I was starting to worry that we hadn't been taught that yet (and we've finished the topic  )

Thanks for checking that through for me, could you help me with integrating tan²x? I've just tried to rearrange the tan2x formula to put in as the integrand but am getting very very confused