4U Maths Question Thread

[3.14159265359]

That's cool, but like was looking for just the final **answer**. Anyway here is a sketch solution.

why did you introduce another R?

[RuiAce]

why did you introduce another R?

Because the radius of the circular motion isn’t necessarily the same as the radius of the actual hemisphere that makes up the bowl.

r is the radius of the hemisphere. But the particle would typically be at a point further below the rim of the hemisphere. It’ll be undergoing circular motion with a smaller radius because of the fact it’s at a lower depth.

[Jeeffffffffffffff]

Hey people, so just as I thought mechanics was going alright, good ol' fitzpatrick threw me a bit of a curveball.
A body, projected vertically upwards with speed U, returns with a speed V. Assuming constant gravity and air resistance proportional to the square of the speed, find the total time taken.
I just find it very vague and I'm not really sure how I'm meant to go about doing this, any help would be appreciated, thanks.

[RuiAce]

Hey people, so just as I thought mechanics was going alright, good ol' fitzpatrick threw me a bit of a curveball.
A body, projected vertically upwards with speed U, returns with a speed V. Assuming constant gravity and air resistance proportional to the square of the speed, find the total time taken.
I just find it very vague and I'm not really sure how I'm meant to go about doing this, any help would be appreciated, thanks.

Your final answer should be in terms of \(U\) and \(V\). You should consider the cases where the particle is going up and going down separately. In the case of the particle going up you will have \( \ddot{x} = -g - kv^2\) and the boundary cases:
- Initially, \(t = 0\) and \(v = U\)
- At the max height, \(t = \text{what you want to find}\) and \( v = 0\).

Then when the particle is going down you will have \( \ddot{x} = g - kv^2\).

[Jeeffffffffffffff]

Your final answer should be in terms of \(U\) and \(V\). You should consider the cases where the particle is going up and going down separately. In the case of the particle going up you will have \( \ddot{x} = -g - kv^2\) and the boundary cases:
- Initially, \(t = 0\) and \(v = U\)
- At the max height, \(t = \text{what you want to find}\) and \( v = 0\).

Then when the particle is going down you will have \( \ddot{x} = g - kv^2\).

Aaaaahhh yes that makes a lot of sense, thanks for that Rui. Just out of curiosity though, I have two different formulas for integrating something with 1/(a²-x²) and the way to tell which one to use is that for one |x|<a and the other you use when |x|>a,but in this question a ended up being the square root of g/k and x was the velocity, which was just v, so how would I go about figuring out if |v| is larger than or less than the square root of g/k??

[RuiAce]

Aaaaahhh yes that makes a lot of sense, thanks for that Rui. Just out of curiosity though, I have two different formulas for integrating something with 1/(a²-x²) and the way to tell which one to use is that for one |x|<a and the other you use when |x|>a,but in this question a ended up being the square root of g/k and x was the velocity, which was just v, so how would I go about figuring out if |v| is larger than or less than the square root of g/k??

You really should be using partial fractions to integrate that expression instead of jump to \( \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right| \). But to compare \(v\) to something like \(\sqrt{\frac{g}{k}} \), you can exploit the fact that \( \ddot{x} > 0\) after you change the orientation to be the more convenient one.

(Equivalently, just set \( \ddot{x} = 0\) because \( \sqrt{\frac{g}{k}}\) can be shown to be the terminal velocity, which is always a cap for the velocity.)

[aadharmg]

Are we allowed to use the standard integral formulas in the HSC exam or not because of its omission from the paper? For example the ln(x^2 + sqrroot(x^2 +/- a^2) formula.

[RuiAce]

Are we allowed to use the standard integral formulas in the HSC exam or not because of its omission from the paper? For example the ln(x^2 + sqrroot(x^2 +/- a^2) formula.

Not anymore in your actual exam. But the odds that they will examine that now are also extremely low.

[3.14159265359]

hey can someone please help me with this question?
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let a, b be positive real numbers so that a+b=1. prove: a/(1+a) + b/(1+b) is less that or equal to 2/3

[RuiAce]

hey can someone please help me with this question?
-----

let a, b be positive real numbers so that a+b=1. prove: a/(1+a) + b/(1+b) is less that or equal to 2/3

[envisagator]

Hi Rui, just a question on volumes of solids of revolution by slices.



Thank You!!!

[3.14159265359]

thank you so much. but shouldn't the second equation you boxed be 1>=4ab not < ?

is it ok if you can try part ii of this question? I can't seem to get the volume equation they have.

[RuiAce]

Hi Rui, just a question on volumes of solids of revolution by slices.



Thank You!!!

Fairly sure if you're using slicing the big radius is just \(1\). Not \(1 - \sin y\). (Which will give you \(\delta V = \pi (1 - \sin^2 y)\delta y \))

thank you so much. but shouldn't the second equation you boxed be 1>=4ab not < ?

is it ok if you can try part ii of this question? I can't seem to get the volume equation they have.

(Image removed from quote.)

Yeah I'll fix the typo shortly



Edit: Pictures

[3.14159265359]

I don't get how u got x =10 + (10-h)

[RuiAce]

I don't get how u got x =10 + (10-h)

See the edit. It's the exact same method I use in the book