3U Maths Question Thread

[Rathin]

Hey Jamon!

I don't understand how you got the first line arrow, it could be because i'm a bit rusty on my substitution method but would love an explanation regardless!

Thank you 

Edit: I also don't understand why the derivative expression needed to be flipped (next to the third therefore sign) 

u=1-r
du/dr=-1
du=-dr


and the derivative needs to be flipped so we can introduce theta into the equation by differentiating theta with respect to r instead of the other way around which didn't initially have theta introduced.   

[jamonwindeyer]

Hey Jamon!

I don't understand how you got the first line arrow, it could be because i'm a bit rusty on my substitution method but would love an explanation regardless!

Thank you 

Edit: I also don't understand why the derivative expression needed to be flipped (next to the third therefore sign) 

Great questions! Rathin has stepped you through, let me explain the rationale a tad. So if we differentiate \(u=1-r\), we get \(\frac{du}{dr}=-1\). We sort of 'multiply' the dr to obtain the relationship \(du=-dr\), or alternatively, \(dr=-du\). It is this second version which lets us rewrite \(\frac{dr}{d\theta}\) as \(\frac{-du}{d\theta}\). Just a quick replacement there

The derivative also needs to be flipped because we need to integrate with respect to \(u\) - Since the expression is in terms of \(u\). We can't integrate with respect to anything else, because there isn't anything else there! So, we can't leave it as \(\frac{du}{d\theta}\), because we can't integrate with respect to \(\theta\). We need to flip to proceed

[anotherworld2b]

Could i please get help with these 2 questions? I attempted the first question but i got the wrong answer

[jamonwindeyer]

I think the calculation for the area of the minor segment is wrong - Is your calculator in the right mode? You might not have swapped it from degrees from the working you did to find \(\theta\)

The second question is literally the same as the first one! It is just finding the area of both minor segments (they'll have the same area), and subtracting both from the circle area. The way you find the area of the segment is otherwise identical to what you did earlier!

[Shadowxo]

Could i please get help with these 2 questions? I attempted the first question but i got the wrong answer

Hi, you were on the right track. First of all, for finding the angle I'd just do inverse cos (6/10). Saves you some trouble and gets you the same answer, and I'd set your calculator to radian mode for these questions. With the segment, looks like you had the right formula but your calculator must have been in degree mode, so sin(1.85...) was calculated to be 0.03236 instead of 0.96.
Just a simple calculator problem, set your calculator to radian mode and you should be set
Next question is pretty much the same, if you can do the first you should be able to do the second

[anotherworld2b]

Oh  you very much for your help Jamon and Shadowxo

I was also wondering how would you do q22 b and q28?

[jakesilove]

Oh  you very much for your help Jamon and Shadowxo

I was also wondering how would you do q22 b and q28?

Have you not finished 22b)? Just take the length of the chord (which you found), and subtract it from the length of the arc (which you found in 22a)).

Now, the belt goes around half of each wheel. So, those parts of the belt will be



Now, draw a line from the centre of the small circle to the point of departure of the rope from the big wheel. This forms a triangle; one side is 16cm, one side is 26cm. Using pythag, we know that the third side (hypotenuse) is 30.5cm. Now, use the same line but the triangle above (such that the length of half the rope is part of the triangle). One side is 30.5cm, one side is 6cm, and the third side is unknown. Using pythag (30.5 is still the hypotenuse) we find that the third side is 29.9cm. Doubling this, and adding it to the original perimeters, will get us the final answer.

[Shadowxo]

Oh  you very much for your help Jamon and Shadowxo

I was also wondering how would you do q22 b and q28?

Hi, just like to make some edits to jake's post
First of all, for 22b, you had it right just in degree not radian mode again Typing it in radian mode should give you 58.4cm, so difference between them is 1.6cm or 16mm

For 28 it was a bit of a tough one. Way I tried to do it was continuing the tangents until they met at a point P. As the two triangles made with P and one point of contact with each of the circles and the tangents and their centres were similar (hard to do without diagrams, I can post if you want) the ratios of the sides were the same. So 16/6 = (16+4+6+a)/a and solved for a, where a is the distance between p and the centre of the smallest circle

An easier way though was to draw a rectangle like they did here https://www.illustrativemathematics.org/content-standards/tasks/621
So this results in a rectangle of width 6 and length B
It also results in a triangle of hypotenuse length 26 and one side length 10. Using pythag we know 5,12,13 and 10,24,26, So the other length (B) is 24. Now we know 24 is the length of the rectangle too
The angle between the two tangents on the larger circle is 2 * angle in the triangle
Angle in triangle = cos^-1(10/26).
Now that we can figure out the angles, you can find out the length of the arc of each of the circles and add 2* length of rectangle to it to find the answer

It's hard to describe without images, let me know if you want me to create / upload some
And I think jake mistook the tangents to be halfway through each circle, but it's not (unfortunately) and pythag wouldn't work for his triangle as it wouldn't have a right angle.
that question's a tough one and I probably could have explained more clearly but hope I helped a little 

[jakesilove]

Hi, just like to make some edits to jake's post
First of all, for 22b, you had it right just in degree not radian mode again Typing it in radian mode should give you 58.4cm, so difference between them is 1.6cm or 16mm

For 28 it was a bit of a tough one. Way I tried to do it was continuing the tangents until they met at a point P. As the two triangles made with P and one point of contact with each of the circles and the tangents and their centres were similar (hard to do without diagrams, I can post if you want) the ratios of the sides were the same. So 16/6 = (16+4+6+a)/a and solved for a, where a is the distance between p and the centre of the smallest circle

An easier way though was to draw a rectangle like they did here https://www.illustrativemathematics.org/content-standards/tasks/621
So this results in a rectangle of width 6 and length B
It also results in a triangle of hypotenuse length 26 and one side length 10. Using pythag we know 5,12,13 and 10,24,26, So the other length (B) is 24. Now we know 24 is the length of the rectangle too
The angle between the two tangents on the larger circle is 2 * angle in the triangle
Angle in triangle = cos^-1(10/26).
Now that we can figure out the angles, you can find out the length of the arc of each of the circles and add 2* length of rectangle to it to find the answer

It's hard to describe without images, let me know if you want me to create / upload some
And I think jake mistook the tangents to be halfway through each circle, but it's not (unfortunately) and pythag wouldn't work for his triangle as it wouldn't have a right angle.
that question's a tough one and I probably could have explained more clearly but hope I helped a little 

Hey! Thanks so much for your response, seriously great to have you answering questions (especially tough ones like this!). I did a bit of research before answering the question, because I had the same thought; this is easy if it's halfway through the circle, but hard if it isn't. All similar questions I found online assumed/said that this was true, and so I stuck with that. Maybe there's a physics reason, maybe it's just not true, but I think you're way is just more comprehensive anyway. If you think about ACTUALLY setting up the experiment, I do tend to believe that there is a physics reason for the rope to be exactly halfway around each circle, if the radius are concentric. I guess I assumed this last bit as well. Either way, thanks so much for your response!

[Shadowxo]

Hey! Thanks so much for your response, seriously great to have you answering questions (especially tough ones like this!). I did a bit of research before answering the question, because I had the same thought; this is easy if it's halfway through the circle, but hard if it isn't. All similar questions I found online assumed/said that this was true, and so I stuck with that. Maybe there's a physics reason, maybe it's just not true, but I think you're way is just more comprehensive anyway. If you think about ACTUALLY setting up the experiment, I do tend to believe that there is a physics reason for the rope to be exactly halfway around each circle, if the radius are concentric. I guess I assumed this last bit as well. Either way, thanks so much for your response!

No worries. When I looked it up I could mostly only find problems where the circles were the same size, meaning the rope/ tangent would leave the circle exactly halfway, but this wouldn't apply to different sized circles. At different sizes, if the rope were to leave halfway around the circle, it would never reach the other circle as the line would be horizontal, not tilted down.

[stephsteph_xx]

does anyone know where to get practice 4U papers?

[kiwiberry]

does anyone know where to get practice 4U papers?

Heaps of papers on this site for every subject: https://thsconline.github.io/s/yr12/index2.html
And these ones specifically for 4U
http://www.angelfire.com/ab7/fourunit/
http://4unitmaths.com/exams.html

[anotherworld2b]

For q22 b i ended up with 2. 47 cm 
Could i please get some diagrams for q28 please ?

Hi, just like to make some edits to jake's post
First of all, for 22b, you had it right just in degree not radian mode again Typing it in radian mode should give you 58.4cm, so difference between them is 1.6cm or 16mm

For 28 it was a bit of a tough one. Way I tried to do it was continuing the tangents until they met at a point P. As the two triangles made with P and one point of contact with each of the circles and the tangents and their centres were similar (hard to do without diagrams, I can post if you want) the ratios of the sides were the same. So 16/6 = (16+4+6+a)/a and solved for a, where a is the distance between p and the centre of the smallest circle

An easier way though was to draw a rectangle like they did here https://www.illustrativemathematics.org/content-standards/tasks/621
So this results in a rectangle of width 6 and length B
It also results in a triangle of hypotenuse length 26 and one side length 10. Using pythag we know 5,12,13 and 10,24,26, So the other length (B) is 24. Now we know 24 is the length of the rectangle too
The angle between the two tangents on the larger circle is 2 * angle in the triangle
Angle in triangle = cos^-1(10/26).
Now that we can figure out the angles, you can find out the length of the arc of each of the circles and add 2* length of rectangle to it to find the answer

It's hard to describe without images, let me know if you want me to create / upload some
And I think jake mistook the tangents to be halfway through each circle, but it's not (unfortunately) and pythag wouldn't work for his triangle as it wouldn't have a right angle.
that question's a tough one and I probably could have explained more clearly but hope I helped a little 

[Shadowxo]

For 22b I'm not sure how you got that, I typed in 2*75*sin (0.4) and got 58.4

For 28, sure I'll do it tomorrow (if I forget just send me a pm)

[Shadowxo]

For q22 b i ended up with 2. 47 cm 
Could i please get some diagrams for q28 please ?

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)