3U Maths Question Thread

[chelseam]

So like this?

Yep!

[RuiAce]

did u mean two 5c coins?

Lol, yes. Fixed.

[owidjaja]

Hey there,
I'm starting to understand substitution but could you please tell me where I'm going wrong? Here is my working out.

Thanks in advance.

[legorgo18]

Hey there,
I'm starting to understand substitution but could you please tell me where I'm going wrong? Here is my working out.

Thanks in advance.

Hey! It seems like you forgot to a step, remember you need to differentiate in order to obtain a substitution for dx in terms of du

Edit (didnt look thoroughly): index error in second step?

[Sine]

Hey there,
I'm starting to understand substitution but could you please tell me where I'm going wrong? Here is my working out.

Thanks in advance.

2nd line of integral working

[arii]

Can someone please teach me how to do this question?

A particle moves according to x=3-2cos²2t, in units of centimetres and seconds. Express the equation in the form x=x₀-Acoswt.

[RuiAce]

Can someone please teach me how to do this question?

A particle moves according to x=3-2cos²2t, in units of centimetres and seconds. Express the equation in the form x=x₀-Acoswt.

[owidjaja]

Hey,
I don't think I'm getting it. This is my working out for question I attached.

Thanks in advance.

[RuiAce]

Hey,
I don't think I'm getting it. This is my working out for question I attached.

Thanks in advance.

Careful. At quick glance you wrote \( \frac{du}{dx} = \frac12 \). But instead \( \frac{dx}{du} = \frac12 \) and we have \( \frac{du}{dx}=2 \)

(Still scanning it for any other errors.)
____________

Edit: For some reason, your \( \sqrt{u} \) isn't in the denominator anymore.
\begin{align*}\int \frac{x}{\sqrt{2x+5}}\,dx&= \int \frac{\frac{u+5}{2}}{\sqrt{2x-5}}\,dx\\ &= \int\frac{\frac{u+5}{2}}{\sqrt{u}}\times 2\,du\\ &= \int \frac{u+5}{\sqrt{u}}\,du\end{align*}

[arii]

Thank you so much. I was wondering, is there a way to figure out the range of motion directly from the equation that was just found?

[owidjaja]

Careful. At quick glance you wrote \( \frac{du}{dx} = \frac12 \). But instead \( \frac{dx}{du} = \frac12 \) and we have \( \frac{du}{dx}=2 \)

(Still scanning it for any other errors.)
____________

Edit: For some reason, your \( \sqrt{u} \) isn't in the denominator anymore.
\begin{align*}\int \frac{x}{\sqrt{2x+5}}\,dx&= \int \frac{\frac{u+5}{2}}{\sqrt{2x-5}}\,dx\\ &= \int\frac{\frac{u+5}{2}}{\sqrt{2}}\times 2\,du\\ &= \int \frac{u+5}{\sqrt{u}}\,du\end{align*}

In your second line of working, how did the denominator turn to sqrt2 ?

[chelseam]

Hey,
I don't think I'm getting it. This is my working out for question I attached.

Thanks in advance.

I think a way to make this process easier is making sure that the integral is in terms of one variable only, not two! You changed the numerator and dx to be in terms of u, but not the denominator - it'll be clearer if you do I've attached my working out below, but have a go again and let me know if you want to clarify any steps

[RuiAce]

In your second line of working, how did the denominator turn to sqrt2 ?

Typo. Fixed.

Thank you so much. I was wondering, is there a way to figure out the range of motion directly from the equation that was just found?

The centre of motion is at \(x_0 = 2\) and the amplitude is \(1\). Hence the particle is oscillating between \(x=1\) and \(x=3\)

[owidjaja]

I think a way to make this process easier is making sure that the integral is in terms of one variable only, not two! You changed the numerator and dx to be in terms of u, but not the denominator - it'll be clearer if you do I've attached my working out below, but have a go again and let me know if you want to clarify any steps

But the answers have a different answer. I attached the screenshot of the answer.

[RuiAce]

But the answers have a different answer. I attached the screenshot of the answer.

Look at her second last line. It's the same thing except rearranged - her answer is just tidier than what they provided.

Either would be marked correct.