4U Maths Question Thread

[KT Nyunt]

Hey Rui, I'm a bit confused about something in the ATAR Math ext. 2 topic notes...
With your LIATE rule, you said when we're integrating e^xsinx dx that we would differentiate sinx and integrate e^x.
But then in your solution, you integrated sinx and then differentiate e^x.

I feel like I'm missing something, but I don't know what... Pls help!

[RuiAce]

Hey Rui, I'm a bit confused about something in the ATAR Math ext. 2 topic notes...
With your LIATE rule, you said when we're integrating e^xsinx dx that we would differentiate sinx and integrate e^x.
But then in your solution, you integrated sinx and then differentiate e^x.

I feel like I'm missing something, but I don't know what... Pls help!

My bad there, went against my own rule. I need to go fix that later.

Fortunately however, the order of the last two doesn't matter. I've found that with \( e^x\cos x\) I integrate cos, whereas with \(e^x \sin x\) I differentiate sin. But trying to say "LIAET" would be pretty hard lols

[envisagator]

Hei Rui, conics has been a topic I really struggle as i cant grasp what the question actually asks and the usefulness of sum of roots such as in this question: y=mx+1 is a chord of the hyperbola x^2 - 4y^2 = 16 , that passes through (0,1) for all real values of m. Find the locus of the midpoint of the chord as m varies. (This is a textbook question btw)

[RuiAce]

Hei Rui, conics has been a topic I really struggle as i cant grasp what the question actually asks and the usefulness of sum of roots such as in this question: y=mx+1 is a chord of the hyperbola x^2 - 4y^2 = 16 , that passes through (0,1) for all real values of m. Find the locus of the midpoint of the chord as m varies. (This is a textbook question btw)

These questions aren't hard because of the fact they're conics. What makes them hard is trying to find the shortcut that makes the problem a lot simpler, and actually tying the concepts together. This one you have here is one of the classic ones. Note that drawing a diagram is highly recommended.


Which, upon expanding and rearranging, becomes a quadratic. The roots are, again, the points of intersection.

At this point, I remark on what's significant. The points of intersection between the chord and the hyperbola are simply the endpoints of the chord itself. Therefore, the MIDPOINT of the chord is actually just the midpoint of the points of intersection. So the \(x\)-coordinate of the midpoint, actually turns out to be \( x= \frac{\alpha+\beta}{2} \).

We can now kill this problem by realising that rewriting \(x = \frac{\alpha+\beta}{2} \) gives us \(x = \frac{\text{sum of roots}}{2} \). And then boom, the \(x\)-coordinate of the midpoint of the chord has been found.

(Now plug into \(y=mx+1\) for the corresponding \(y\)-coordinate, before attempting to eliminate the parameter. Which is \(m\) in this case.)

A remark: We could've actually solved the quadratic to find what \(x_1\) and \(x_2\) explicitly are. But then we still need to compute \( \frac{x_1+x_2}{2} \) anyway. The sum of roots is just a more convenient shortcut for this purpose.

[clovvy]

Hey, I am doing a volumes question where the equation y=x^3+1 between -1 to 0 is rotated about the x-axis and I have to use the slicing method.

[RuiAce]

Hey, I am doing a volumes question where the equation y=x^3+1 between -1 to 0 is rotated about the x-axis and I have to use the slicing method.

[clovvy]

[Opengangs]

The volume is evaluated as:
\[ V = \pi \int_a^b y^2\,dx \]

[clovvy]

The volume is evaluated as:
\[ V = \pi \int_a^b y^2\,dx \]

ah ok so I don't even need slicing.. make sense...

[RuiAce]

The slicing formula is \( \delta V = \pi r^2 \times \text{thickness}\). The square is supposed to be there for the radius, which in your case is just \(r = x^3+1\). You should go back and revise the formula.

In general, slicing eventually gives you the same integral as if you did it the old way.

The volume is evaluated as:
\[ V = \pi \int_a^b y^2\,dx \]

They actually need to do a bit more work in MX2. I just skipped all the details in my earlier reply.

[vikasarkalgud]

Hey, Rui, for complex locus, I don't get when I'm meant to be putting in open circles for intersections and dotted lines...coz I thought dotted lines were only for when there is no =, hence it cannot exist there. So this question I've attached..

Thankyou for help

[RuiAce]

Hey, Rui, for complex locus, I don't get when I'm meant to be putting in open circles for intersections and dotted lines...coz I thought dotted lines were only for when there is no =, hence it cannot exist there. So this question I've attached..

Thankyou for help

Part i) shouldn't have any open circles for that region..? It doesn't look like there's any open circles there which should be fine, since nothing is dotted here.

[vikasarkalgud]

Hey Rui,

for another subject, I don't really know what area, perimeter and volume actually are categorised as. What I mean is, length, height, perpendicular height, radius are dimensions. Therefore, using these, u can get area, perimeter, volume, circumference.etc. But what are these things actually called? These results?

Thankyou in advance

[envisagator]

Hei Rui,can you help with this induction q, its a cambridge 4u textbook question. I'm not quite sure what exactly im supposed to assume. Thank You in advance!!!

[RuiAce]

Hei Rui,can you help with this induction q, its a cambridge 4u textbook question. I'm not quite sure what exactly im supposed to assume. Thank You in advance!!!

Fairly sure you can continue with this question once I give you what to assume.

Hey Rui,

for another subject, I don't really know what area, perimeter and volume actually are categorised as. What I mean is, length, height, perpendicular height, radius are dimensions. Therefore, using these, u can get area, perimeter, volume, circumference.etc. But what are these things actually called? These results?

Thankyou in advance

Length, height, perpendicular height and radius are 1-dimensional concepts.

Perimeter is also a 1-dimensional concept; it's just a special type of length (just like how perpendicular height and radius are). Area, however, is a 2-dimensional concept. Volume is a 3-dimensional concept.

The radian measure of an angle is a special case of a 0-dimensional concept.

Colloquially, the word "dimensions" typically just refers to lengths, especially when you're dealing with a volume. But in science (yes, this is actually more of a science question than a maths question), the dimensions refer to the SI base units that you require to build up a new concept. Volume involves 3 length-dimensions, because its SI unit is m^3, which is built from 3 SI base units. Force involves a mass dimension and a length dimension, PER 2 time-dimensions, because its SI unit is kg m s^-2.

(As an extra example, we can get a bit bizarre, and argue voltage involves a mass dimension and length dimension, PER 3 time-dimensions and current-dimension, because its SI unit is kg m s^-3 A^-1. But it gets a bit bizarre, because there's debate surrounding whether current is the dimension or electric charge is the dimension.)