Mathematics Question Thread

[RuiAce]

Hi, could you please help me with finding the intercepts for these 2 graphs. I know how to find the area by integrating, its just this bit that is confusing me.
Calculate area of the region bounded by x-axis and curves y= x^1/2 and y=6-x.
Thanks

[Zainbow]

A circle has a chord of 25 mm with an angle of  π/6  subtended at the centre. Find, to 1 decimal place, the length of the arc cut off by the chord.

I'm a bit confused, how can you find the length of the arc only from this information?

[RuiAce]

A circle has a chord of 25 mm with an angle of  π/6  subtended at the centre. Find, to 1 decimal place, the length of the arc cut off by the chord.

I'm a bit confused, how can you find the length of the arc only from this information?

Hint: Draw a diagram. You can use the cosine rule to find the radius of the sector you're interested in.

[jamonwindeyer]

A circle has a chord of 25 mm with an angle of  π/6  subtended at the centre. Find, to 1 decimal place, the length of the arc cut off by the chord.

I'm a bit confused, how can you find the length of the arc only from this information?

Hey! Bit of a tricky question, but what we need to do is form a triangle with the chord, and the two radii to the centre, and use the cosine rule to find the radii:



You can then use that, with the formula \(l=r\theta\), to find the answer!! Does that help? Happy to show more working if you need it

Edit: What Rui said

[Shadowxo]

Or, alternatively, use chord length = 2rsin(θ/2)
25 = 2r*sin(π/12)
r = 12.5/sin(π/12)
Then l=rθ

Or use trig sin(θ)=O/H
sin(π/12)=12.5/r
r=12.5/sin(π/12)

Few different ways 

[jamonwindeyer]

Or, alternatively, use chord length = 2rsin(θ/2)
25 = 2r*sin(π/12)
r = 12.5/sin(π/12)
Then l=rθ

Or use trig sin(θ)=O/H
sin(π/12)=12.5/r
r=12.5/sin(π/12)

Few different ways 

Oh wow that last way is wayyy cleaner, thanks Shadow ps - Is \(CL=2r\sin{\frac{\theta}{2}}\) a quotable result for VCE? Lucky devils

[Zainbow]

Hint: Draw a diagram. You can use the cosine rule to find the radius of the sector you're interested in.

Hey! Bit of a tricky question, but what we need to do is form a triangle with the chord, and the two radii to the centre, and use the cosine rule to find the radii:



You can then use that, with the formula \(l=r\theta\), to find the answer!! Does that help? Happy to show more working if you need it

Edit: What Rui said

Or, alternatively, use chord length = 2rsin(θ/2)
25 = 2r*sin(π/12)
r = 12.5/sin(π/12)
Then l=rθ

Or use trig sin(θ)=O/H
sin(π/12)=12.5/r
r=12.5/sin(π/12)

Few different ways 

Thanks everyone    this really helped

[Shadowxo]

Oh wow that last way is wayyy cleaner, thanks Shadow ps - Is \(CL=2r\sin{\frac{\theta}{2}}\) a quotable result for VCE? Lucky devils

Actually, I only had to do chords/arcs/circle equations in year 11, wasn't on the VCE year 12 course
\(CL=2r\sin{\frac{\theta}{2}}\) is basically a simplified version of using sin(θ)=O/H to find the radius. Pretty easy to use sin(θ)=O/H instead if the chord length formula isn't allowed

[RuiAce]

Uh. Ok. I will add that to the list of formulas I need to prove.

Edit: It's just a generalisation.

[Shadowxo]

Uh. Ok. I will add that to the list of formulas I need to prove.

Edit: It's just a generalisation.

I always worked it out a different way, drawing it then splitting it into two right-angled triangles

[RuiAce]

I always worked it out a different way, drawing it then splitting it into two right-angled triangles

Makes sense. I have a bias against splitting things up if unnecessary though.

[jamonwindeyer]

Makes sense. I have a bias against splitting things up if unnecessary though.

Rui's all about togetherness

[smile123]

Please help

[jamonwindeyer]

Please help

Hey! So the expression that gives the number of infected cells after \(n\) days is:



Basically, you are solving for \(n\) when \(A=0\) - The problem is that it has no solutions! What you are looking for is the lowest \(n\) value in that equation for which the result gives a number of cells less than 1 - Meaning, the first time all the cells could feasibly be dead. At least, that is how I interpret it I think I'd go A?

[smile123]

thanks