3U Maths Question Thread

[bsdfjnlkasn]

Hey!

I was just wondering what will be covered in the ATAR Notes 3u lecture?

Thank you 

[jakesilove]

Hey!

I was just wondering what will be covered in the ATAR Notes 3u lecture?

Thank you 

Hey! Jamon said that he'll be going through integration by substitution, exponential growth and decay and chain rule!

Jake

[anotherworld2b]

Thank you so much for you help
I have a quick question how do you know that a rectangle and a right angle triangle would be formed?

I've posted my solution (insomnia at 3am), let me know if it's the correct solution and if it's clear enough (bit messy sorry)

[Shadowxo]

Thank you so much for you help
I have a quick question how do you know that a rectangle and a right angle triangle would be formed?

No worries
the lines connecting the two circles are tangents to the circles, meaning those angles are right angles. From there you can create a line parallel to the tangent that goes from the centre of the smaller circle until it touches the line connecting the centre of the bigger circle and the point the tangent touches. Since the line's parallel, it'll make it a rectangle shape, and since all those angles is 90º, we know the triangle will have a right angle there
Does this help?

[anotherworld2b]

Thank you for your help

I was wondering if my approach to q29 is right.
I ask wanted to ask would you fillow the same appraoch for q30?

No worries
the lines connecting the two circles are tangents to the circles, meaning those angles are right angles. From there you can create a line parallel to the tangent that goes from the centre of the smaller circle until it touches the line connecting the centre of the bigger circle and the point the tangent touches. Since the line's parallel, it'll make it a rectangle shape, and since all those angles is 90º, we know the triangle will have a right angle there
Does this help?

[bsdfjnlkasn]

Thank you for your help

I was wondering if my approach to q29 is right.
I ask wanted to ask would you fillow the same appraoch for q30?

Hey sorry, this might be irrelevant but which textbook are you getting these questions from? 

[RuiAce]

Hey sorry, this might be irrelevant but which textbook are you getting these questions from? 

He's from Western Australia, just keep that in mind

[Shadowxo]

Thank you for your help

I was wondering if my approach to q29 is right.
I ask wanted to ask would you fillow the same appraoch for q30?

Hi,
For Q29, it is except the DC part.You're looking for the area of the segment, not the length of the arc, so it's 1/2*r²*(x-sinx) instead of rx, and if it were rx you need to remember the radius is 107.7 not 100 . Also for finding the angle I'd use tan instead of cos so you can use 40 and 100 as they're exact. My answer for this was 413.87 + 16000cm² (segment + rectangle) if that helps (let me know if I made an error / you got something different)

For 30, I'd use the same approach as the similar question earlier. You should find the length of the ropes connecting the two are exact values, I got 36 and 24 and go from there using the same rectangle/triangle approach

[anotherworld2b]

the answer in the book for Q29 is 16410 cm^2 (how accurate should the answer be?)
I will attempt Q30 again

Hi,
For Q29, it is except the DC part.You're looking for the area of the segment, not the length of the arc, so it's 1/2*r²*(x-sinx) instead of rx, and if it were rx you need to remember the radius is 107.7 not 100 . Also for finding the angle I'd use tan instead of cos so you can use 40 and 100 as they're exact. My answer for this was 413.87 + 16000cm² (segment + rectangle) if that helps (let me know if I made an error / you got something different)

For 30, I'd use the same approach as the similar question earlier. You should find the length of the ropes connecting the two are exact values, I got 36 and 24 and go from there using the same rectangle/triangle approach

[Shadowxo]

the answer in the book for Q29 is 16410 cm^2 (how accurate should the answer be?)
I will attempt Q30 again

Yep my answer was the same - I separated it into parts so it would be easier to see just the value of the segment. And I believe in the question it says to the nearest 10cm², my answer was 413.87 + 16000 = 16414 = 16410 rounded
And good luck!

[anotherworld2b]

Thank for your help
I was also wondering if i could get help to solve this question please?
I'm sorry for asking so many questions 

Yep my answer was the same - I separated it into parts so it would be easier to see just the value of the segment. And I believe in the question it says to the nearest 10cm², my answer was 413.87 + 16000 = 16414 = 16410 rounded
And good luck!

[RuiAce]

Thank for your help
I was also wondering if i could get help to solve this question please?
I'm sorry for asking so many questions 

Hint: This is just f(x)=√(2x) plugged RIGHT into the definition of the derivative.

(Aka differentiation from first principles)


i.e. You're basically finding f'(x)

[Shadowxo]

Thank for your help
I was also wondering if i could get help to solve this question please?
I'm sorry for asking so many questions 

No worries
I'm not entirely sure how they want you to answer it. You could say that that equation = lim_h->0 (f(x+h)-f(x))/h = dy/dx where y = √(2x)
therefore it = 1/√(2x). If they say use an appropriate derivative maybe it's that (anyone else know?) Edit: RuiAce did this way so probably this
Another way is just working through it.
Multiply both top and bottom by (√(x+h)+√(x)) so you end up (including the lim with all of these) with √2*(x+h-x)/(h(√(x+h)+√(x)))
= √2*h/(h*(√(x+h)+√(x))) = √(2)/(√(x+h)+√(x))
then remove the lim and sub h = 0
=√(2)/(√(x) + √(x)) = 1/√(2x)

[anotherworld2b]

Liks this?

No worries
I'm not entirely sure how they want you to answer it. You could say that that equation = lim_h->0 (f(x+h)-f(x))/h = dy/dx where y = √(2x)
therefore it = 1/√(2x). If they say use an appropriate derivative maybe it's that (anyone else know?) Edit: RuiAce did this way so probably this
Another way is just working through it.
Multiply both top and bottom by (√(x+h)+√(x)) so you end up (including the lim with all of these) with √2*(x+h-x)/(h(√(x+h)+√(x)))
= √2*h/(h*(√(x+h)+√(x))) = √(2)/(√(x+h)+√(x))
then remove the lim and sub h = 0
=√(2)/(√(x) + √(x)) = 1/√(2x)

[kiwiberry]

Liks this?

When you factorise the √2 out, remember to keep the square root sign on the x+h and x as well

and then from here you can multiply top and bottom by √(x+h)+√x to cancel the h on the bottom!