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July 21, 2017, 06:48:16 am

AuthorTopic: 3U Maths Question Thread  (Read 123371 times) Tweet Share

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RuiAce

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Re: 3U Maths Question Thread
« Reply #1170 on: January 10, 2017, 09:15:44 pm »
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Hi I am back  .
I've been working on this question back i am unable to get the answer for parts c, d and e
Part e is REALLY interesting, and can be done geometrically.

I've made a GeoGebra simulation for you. Try playing around with the sliders and see what exactly is going on...

With c), by only considering f(3) = f(-5)

k(-5+a)^2 + 16 = k(3+a)^2 + 16

(-5+a)^2 = (3+a)^2

That could prove useful

Edit: Yeah the GeoGebra link wasn't working. Fixed.
« Last Edit: January 10, 2017, 10:30:45 pm by RuiAce »
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anotherworld2b

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Re: 3U Maths Question Thread
« Reply #1171 on: January 10, 2017, 10:25:59 pm »
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a geogebra simulation? It doesn't seem to work for me :/
I solved c)
But I'm not sure how to find k for part d). I tried to expand the given equation then substituted x = 0. I ended up with -16/a^2 - k. I'm not sure what to do from this point onwards.
Also for part e how can it be done geometrically?

Part e is REALLY interesting, and can be done geometrically.

I've made a GeoGebra simulation for you. Try playing around with the sliders and see what exactly is going on...

With c), by only considering f(3) = f(-5)

k(-5+a)^2 + 16 = k(3+a)^2 + 16

(-5+a)^2 = (3+a)^2

That could prove useful

RuiAce

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Re: 3U Maths Question Thread
« Reply #1172 on: January 10, 2017, 10:31:58 pm »
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a geogebra simulation? It doesn't seem to work for me :/
I solved c)
But I'm not sure how to find k for part d). I tried to expand the given equation then substituted x = 0. I ended up with -16/a^2 - k. I'm not sure what to do from this point onwards.
Also for part e how can it be done geometrically?
Both d) and e) are done geometrically it seems.

Fixed the link. Sorry about that; try it again.
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anotherworld2b

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Re: 3U Maths Question Thread
« Reply #1173 on: January 11, 2017, 03:51:13 pm »
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Oh thank you XD
I was also wondering could I please get help with these two questions? I just started learning about the chain rule but I'm still unsure how to use it.

Both d) and e) are done geometrically it seems.

Fixed the link. Sorry about that; try it again.

jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #1174 on: January 11, 2017, 03:59:20 pm »
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Oh thank you XD
I was also wondering could I please get help with these two questions? I just started learning about the chain rule but I'm still unsure how to use it.

Let me show you the first one as an example, then have another tackle of the second! They are identical in approach

So we know by the chain rule that:

$\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}$

The $du$'s sort of cancel each other out like they would regular numbers! So we just substitute the derivatives:

$\frac{dy}{dx}=7\times(4x+5)=28x+35$

The only difference with the second one is that you'll end up with both an $s$ and a $t$ in your answer - Anywhere you see $s$, replace it with $2t+1$ according to the equation you are given. Give it a try!

Rathin

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Re: 3U Maths Question Thread
« Reply #1175 on: January 11, 2017, 04:12:05 pm »
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I have noticed a pattern here...when integrating a cubic function using Simpson's rule it yields the same answer as the exact area. Why so?
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RuiAce

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Re: 3U Maths Question Thread
« Reply #1176 on: January 11, 2017, 04:23:38 pm »
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I have noticed a pattern here...when integrating a cubic function using Simpson's rule it yields the same answer as the exact area. Why so?
This is pretty hard to prove with high-school level mathematics.

Given first-year university mathematics, the proof becomes trivialised.

I like the proof given here, but it is a bit cumbersome to read due to the font.
___________________________

Brief note, however, it all stems down to the fact that Simpson's rule MUST exactly approximate the area for any quadratic. This is because Simpson's rule IS using a parabola to approximate the area.
« Last Edit: January 11, 2017, 04:29:06 pm by RuiAce »
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anotherworld2b

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Re: 3U Maths Question Thread
« Reply #1177 on: January 11, 2017, 05:18:08 pm »
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Would this be the correct way to do it jamon?
Let me show you the first one as an example, then have another tackle of the second! They are identical in approach

So we know by the chain rule that:

$\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{dx}$

The $du$'s sort of cancel each other out like they would regular numbers! So we just substitute the derivatives:

$\frac{dy}{dx}=7\times(4x+5)=28x+35$

The only difference with the second one is that you'll end up with both an $s$ and a $t$ in your answer - Anywhere you see $s$, replace it with $2t+1$ according to the equation you are given. Give it a try!

RuiAce

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Re: 3U Maths Question Thread
« Reply #1178 on: January 11, 2017, 05:20:11 pm »
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Would this be the correct way to do it jamon?
Word of advice. If you're using the chain rule, avoid being lazy with the dash (prime symbol '). It makes it unclear what you're differentiating with respect to.

But the final answer looks correct (Edited)
« Last Edit: January 11, 2017, 05:22:47 pm by RuiAce »
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anotherworld2b

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Re: 3U Maths Question Thread
« Reply #1179 on: January 11, 2017, 05:38:20 pm »
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Thanks for the advice RuiAce

Word of advice. If you're using the chain rule, avoid being lazy with the dash (prime symbol '). It makes it unclear what you're differentiating with respect to.

But the final answer looks correct (Edited)

anotherworld2b

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Re: 3U Maths Question Thread
« Reply #1180 on: January 11, 2017, 06:08:42 pm »
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Could i please get my working checked? I just want to make sure that I've written it out properly

bsdfjnlkasn

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Re: 3U Maths Question Thread
« Reply #1181 on: January 11, 2017, 07:10:40 pm »
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How would I go about solving for R in the following?

3cosx+4sinx ≡ Rsin(x+a)

Thank you!

RuiAce

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Re: 3U Maths Question Thread
« Reply #1182 on: January 11, 2017, 07:19:30 pm »
+1
How would I go about solving for R in the following?

3cosx+4sinx ≡ Rsin(x+a)

Thank you!
$\text{By formula}\\ a\sin x + b \cos x = R \sin (x+\alpha)\\ \text{where }R=\sqrt{a^2+b^2}\\ \tan \alpha=\frac{b}{a}\text{ and }0 <\alpha <\frac{\pi}{2}$
\text{To prove this, we form an identity:}\\ \begin{align*}a\sin x + b \cos x &\equiv R\sin (x+\alpha)\\ &\equiv R(\sin x \cos \alpha + \cos x \sin \alpha)\\ &\equiv (R\cos \alpha)\sin x + (R\sin \alpha)\cos x\end{align*}
\text{So by equating coefficients:}\\ \begin{align*}a&=R\cos \alpha \tag{1}\\ b&=R\sin \alpha \tag{2}\end{align*}
$(1)^2+(2)^2\text{ gives}\\ a^2+b^2=R^2(\cos^2\alpha+\sin^2\alpha)=R^2\\ (2)\div (1) \text{ gives}\\ \frac{b}{a}=\frac{\sin \alpha}{\cos \alpha}=\tan \alpha$
$\text{This is the usual way of considering what's known as the}\\ \textit{auxiliary angle transformation}$
$\text{So in your scenario, }R=5$
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RuiAce

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Re: 3U Maths Question Thread
« Reply #1183 on: January 11, 2017, 07:19:56 pm »
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Could i please get my working checked? I just want to make sure that I've written it out properly
At least the method is right. Haven't checked the actual computations
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jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #1184 on: January 11, 2017, 08:01:02 pm »
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Could i please get my working checked? I just want to make sure that I've written it out properly

All correct, nice job!