3U Maths Question Thread

[3.14159265359]

Two questions.

1. Are you asking for an induction? Because this cannot be proved algebraically
2. Did you just mean >=? The >/= symbol is taken to mean NOT greater than or equal to

sorry for the late reply, my internet was down.
1. yes it is, sorry I should've specified
2. I meant "n greater than or equal to 4"

[RuiAce]

[itssona]

a) is pretty common: You're just trying to use the formula \( m_1m_2 = -1 \) for perpendicular lines. Have a go at that and come back if you're stuck

Ah thank you:D
but im not sure what lines equations to use for the m1m2= -1 question
we us the tangent equation on the formula sheet and sub in y=-a?

[RuiAce]

Ah thank you:D
but im not sure what lines equations to use for the m1m2= -1 question
we us the tangent equation on the formula sheet and sub in y=-a?

[Claudiaa]

So this is a multiple choice question in my 3u exam..and it has been fiercly debated by myself and the 3u math teachers....hoping someone could clarify this.

The number, N , of kangaroos in a certain area was studied over a period of time, t .
The initial number was 2500.
During this period,  y' <0  and  y''>0   

What does this say about the number of kangaroos over this time?
The answer in the exam was 'decreasing at a decreasing rate' however, it has been debated that the answer SHOULD be 'decreasing at an increasing rate'

Hope someone can confirm what the answer is!

[RuiAce]

So this is a multiple choice question in my 3u exam..and it has been fiercly debated by myself and the 3u math teachers....hoping someone could clarify this.

The number, N , of kangaroos in a certain area was studied over a period of time, t .
The initial number was 2500.
During this period,  y' <0  and  y''>0   

What does this say about the number of kangaroos over this time?
The answer in the exam was 'decreasing at a decreasing rate' however, it has been debated that the answer SHOULD be 'decreasing at an increasing rate'

Hope someone can confirm what the answer is!

You’re decreasing AND you’re concave up, so you’re behaving like the left branch of \(y=x^2\). Hence you’re decreasing at a decreasing rate.

The positive second derivative just tells us about the concave up nature of your model. You always need to combine it with the first derivative to know what’s going on.
y’ > 0 and y” > 0 is increasing at an increasing rate, but y’ < 0 and y” > 0 is decreasing at a decreasing rate.


Edit: I just realised that there was a typo - not sure how this has happened. Potentially, the confusion might've been caused by not correctly linking concave up/concave down to the sign of y".

If y' < 0 and y" > 0 then we decrease at a decreasing rate. But if they're BOTH negative, i.e. y' < 0 and y" < 0, we most certainly decrease at an increasing rate.

So essentially, decreasing but concave up means we decrease at a decreasing rate. Whereas decreasing but concave DOWN means we decrease at an increasing rate.

[Claudiaa]

You’re decreasing AND you’re concave up, so you’re behaving like the left branch of \(y=x^2\). Hence you’re decreasing at a decreasing rate.

The positive second derivative just tells us about the concave up nature of your model. You always need to combine it with the first derivative to know what’s going on.
y’ > 0 and y” > 0 is increasing at an increasing rate, but y’ < 0 and y” < 0 is decreasing at a decreasing rate.

That's what I thought! But then someone brought up this video
https://youtu.be/BAfpYgzDWo8?t=276   (at 4:36 he states that it is decreasing at an increasing rate)

[RuiAce]

That's what I thought! But then someone brought up this video
https://youtu.be/BAfpYgzDWo8?t=276   (at 4:36 he states that it is decreasing at an increasing rate)

I'm in China rn and I think YouTube is blocked here. Mind screenshotting and sending some attachments?

(Or alternatively if someone could check it out for me that'd be really helpful)

[sayma]

Hi, could someone help me with this question please,
Sketch, showing any stationary points, points of inflextion or asymptotes.
y=(x)^2/2x+3

I'm unsure quite unsure as to how i should approach this question. How do i find the Horizontal Asymptote and also, when finding the stationary points how do i differentiate it?
Thanks

[Opengangs]

That's what I thought! But then someone brought up this video
https://youtu.be/BAfpYgzDWo8?t=276   (at 4:36 he states that it is decreasing at an increasing rate)

It's fairly difficult to grasp at first, because it "defies" how we think. However, consider the nature of the graph. We can clearly see that the graph is decreasing; we can tell because the displacement of f, the y-axis, is decreasing for all of x, the x-axis. To determine the nature of this decreasing function, consider the behaviour of the tangent in relation to the x-axis; for the sake of simplicity, we will call it t for time.

As t increases, what happens to the behaviour of the tangent? It is increasingly steeper and stepper, albeit in the negative direction. This implies that the tangent's rate of change is increasing, which demonstrates why the f'(x) < 0, f''(x) < 0 is decreasing at an increasing rate.

In the same way, consider what happens to the tangent as t increases for the other graph.

[itssona]

heey
so for binomial, should I know both the T_k and T_k+1 formula?
also, if I want to find greatest term or greatest coefficient, I use T_k>= T_k+1?
how would I use this?
e.g. Find the term in the expansion of (3+5x)^20 with the greatest coefficient.

thank you

[Opengangs]

See if you can find a value for k, and find the greatest coefficient. Remember that k is an integer.

[Natasha.97]

Hi, could someone help me with this question please,
Sketch, showing any stationary points, points of inflextion or asymptotes.
y=(x)^2/2x+3

I'm unsure quite unsure as to how i should approach this question. How do i find the Horizontal Asymptote and also, when finding the stationary points how do i differentiate it?
Thanks

Hi!

Not completely sure if this is correct:

I don't think that this graph has any horizontal asymptotes, as the highest power of x in the numerator is larger than the highest power of x in the denominator. However, it does have both vertical and oblique asymptotes.

To find the vertical asymptote, let the denominator be 'not' equal to 0, resulting in the asymptote x = -1.5

To find the oblique asymptote, divide both the numerator and the denominator by the highest power of the denominator (in this case, x) which should result in x/(2 + 3/x). Use the fact that when x tends to infinity, the limit of 1/x is 0. This results in an oblique asymptote of x/2.

Use the quotient rule to differentiate this equation: y' = (vu'-uv')/v². Differentiate once to find stationary points, and twice to find points of inflexion.

Hope this helps

[Opengangs]

Hi, could someone help me with this question please,
Sketch, showing any stationary points, points of inflextion or asymptotes.
y=(x)^2/2x+3

I'm unsure quite unsure as to how i should approach this question. How do i find the Horizontal Asymptote and also, when finding the stationary points how do i differentiate it?
Thanks

To find the oblique asymptote, use long division of polynomials to find that x^2/(2x + 3) = x/2 - 3/4 + 9/(4(2x + 3)), this implies that there is an oblique asymptote at x/2 - 3/4.

[RuiAce]

Hi!

Not completely sure if this is correct:

I don't think that this graph has any horizontal asymptotes, as the highest power of x in the numerator is larger than the highest power of x in the denominator. However, it does have both vertical and oblique asymptotes.

To find the vertical asymptote, let the denominator be 'not' equal to 0, resulting in the asymptote x = 1.5.

To find the oblique asymptote, divide both the numerator and the denominator by the highest power of the denominator (in this case, x) which should result in x/(2 + 3/x). Use the fact that when x tends to infinity, the limit of 1/x is 0. This results in an oblique asymptote of x/2.

Use the quotient rule to differentiate this equation: y' = (vu'-uv')/v². Differentiate once to find stationary points, and twice to find points of inflexion.

Hope this helps

Of course, the most important thing to note is that there will certainly be an oblique asymptote, by comparing the powers of x in the numerator and denominator. But the 1/x limit needs to be used more cautiously.

It's weird how this happens, but we can't treat the case where the numerator has a higher power the same way as though the denominator has the higher power (or the leading power is the same). It's partly because we know that the limit is ultimately \( \infty \), but what we're interested in is how fast it reaches infinity, not the obvious fact that it does.