3U Maths Question Thread

[anotherworld2b]

I've been working on q30 and it is quite tedious. The answer is 177cm.
Could i have help with this question please?

[anotherworld2b]

Question 30 for ^
I was also wondering is possible to do q31? I dont think there is enough information provided

[RuiAce]

I was also wondering is possible to do q31? I dont think there is enough information provided

Noting that the arc length is l=rθ
(where θ is the angle subtended in radians)

Perimeter:
14 = 2r + rθ

Area:
10 = r²θ/2

Simultaneous equations.

[anotherworld2b]

I see now
Thank you for your help

Noting that the arc length is l=rθ
(where θ is the angle subtended in radians)

Perimeter:
14 = 2r + rθ

Area:
10 = r²θ/2

Simultaneous equations.

[anotherworld2b]

I am a bit confused on how Q31 part b works
I put solve (10=1.5x r^2 x 2 x pi- 4/5, r)
And I got 1.85

[RuiAce]

I am a bit confused on how Q31 part b works
I put solve (10=1.5x r^2 x 2 x pi- 4/5, r)
And I got 1.85

Try figuring out why there are TWO solutions to the WolframAlpha output

Not sure what calculator you use, but I know some calculators only output one solution

[jamonwindeyer]

I am a bit confused on how Q31 part b works
I put solve (10=1.5x r^2 x 2 x pi- 4/5, r)
And I got 1.85

See my working here:

WORKING

So the working up until I check \(\theta\) is common for parts A and B. The two solutions to the simultaneous equations represent the two answers to A and B; you need to choose which is which.

Use \(r\) to check the value of \(\theta\). One, 5 radians, is bigger than \(\pi\), so it must correspond to the larger or major segment. The other is smaller, so it must correspond to the minor segment.

So, r=5 for A, r=2 for B

Let me know if you need clarification on the actual difference between major/minor segments or how this working comes about

[anotherworld2b]

thank you for your help

See my working here:

WORKING

So the working up until I check \(\theta\) is common for parts A and B. The two solutions to the simultaneous equations represent the two answers to A and B; you need to choose which is which.

Use \(r\) to check the value of \(\theta\). One, 5 radians, is bigger than \(\pi\), so it must correspond to the larger or major segment. The other is smaller, so it must correspond to the minor segment.

So, r=5 for A, r=2 for B

Let me know if you need clarification on the actual difference between major/minor segments or how this working comes about

[anotherworld2b]

I was working on this question but i got the wrong answer

[RuiAce]

I was working on this question but i got the wrong answer

I had a look through your working. Shouldn't \( x=\sqrt{15^2-5^2}\)?

Assuming everything else is right, come back if that above correction fails to work. Edit: Looked further, not sure what "sector ACD" is

Also, be careful with your labeling. By making two points (the centers AND the ends of the line) BOTH A and B, you're kinda doing incorrect notation. Maybe label the centers E and F instead. (Compliment: That's one really NEAT diagram though)

[jamonwindeyer]

Compliment: That's one really NEAT diagram though

Such nice circles omg

[Rathin]

Our of curiosity anotherworld2b do you do HSC maths?

[RuiAce]

Our of curiosity anotherworld2b do you do HSC maths?

He's from WA

[legorgo18]

Hi, im new to the forum. Dont understand what i am doing wrong in the d part of this question

A particle is moving in a straight line, starting from the origin. At time t seconds the particle has a displacement of x metres from the origin and a velocity v m/s. The displacement is given by x= 2t- 3ln(t+1).

d) Find the distance travelled by the particle in the first three seconds, to 2 dp.

Any help would be appreciated!

[Syndicate]

Hi, im new to the forum. Dont understand what i am doing wrong in the d part of this question

A particle is moving in a straight line, starting from the origin. At time t seconds the particle has a displacement of x metres from the origin and a velocity v m/s. The displacement is given by x= 2t- 3ln(t+1).

d) Find the distance travelled by the particle in the first three seconds, to 2 dp.

Any help would be appreciated!

If you substitute in 3 directly, you will get the displacement of the particle, which is not what the question is asking for.
2t - 3ln(x+1) is a curve. Therefore, you will need to measure the arc length of the curve in the domain [0,3] (0_<x_<3)


Can anyone confirm if I am right?