Author Topic: 3U Maths Question Thread (Read 1506351 times) Tweet Share

Dragomistress · « **Reply #3330 on:** March 23, 2018, 10:21:39 pm »

What is happening? Why is the question's differentiation formula different from the given one? Isn't a=0.5 and just substitute it in?

RuiAce · « **Reply #3331 on:** March 23, 2018, 10:54:01 pm »

Quote from: Dragomistress on March 23, 2018, 10:21:39 pm

What is happening? Why is the question's differentiation formula different from the given one? Isn't a=0.5 and just substitute it in?

Both their sample method and your suggested method will lead to the same answer. They chose to ignore the formula involving $a$’s

Dragomistress · « **Reply #3332 on:** March 24, 2018, 12:14:07 pm »

Which formula would you suggest using?

Also, how do I do part b?

dermite · « **Reply #3333 on:** March 24, 2018, 12:23:07 pm »

seems pretty basic but im having a mental block so

how would you factorise y = x^3 - 3x + 2?

thanks

RuiAce · « **Reply #3334 on:** March 24, 2018, 01:43:30 pm »

Quote from: Dragomistress on March 24, 2018, 12:14:07 pm

Which formula would you suggest using?

Also, how do I do part b?

Whichever you prefer.
$\text{Although, having said that, if you want to use the }a = \frac12\text{ method,}\\ \text{you end up getting an answer that you should simplify}\\ \begin{align*}\frac{d}{dx} \cos^{-1}2x &= \frac{1}{\sqrt{\frac14 - x^2}}\\ &= \frac{1}{\sqrt{\frac14 - \frac{4x^2}{4}}}\\ &= \frac{1}{\sqrt{\frac{1}{4} (1-4x^2)}}\\ &= \frac{1}{\frac12 \sqrt{1-4x^2}}\\ &= \frac{2}{\sqrt{1-4x^2}}\end{align*}$
Note, however, that you don't have to set out all that working. If you can jump from line 1 straight to the bottom line, that's also ok.
_____________________________________
$\theta = \sin^{-1}\frac{h}{6} \implies \frac{d\theta}{dh} = \frac{1}{\sqrt{36-h^2}}$
$\text{Also we're told that }\frac{dh}{dt} = 0.05\\ \text{So by using the chain rule,}\\ \begin{align*}\frac{d\theta}{dt}&= \frac{d\theta}{dh} \times \frac{dh}{dt}\\ &= \frac{1}{\sqrt{36-h^2}} \times 0.05\end{align*}\\ \text{From here, simply sub in }h=2.5\text{ and multiply by }\frac{180^\circ}{\pi}\\ \text{to convert from radians to degrees.}$
Remark: Technically both $ \frac{d\theta}{dt} $ and $ \frac{dh}{dt}$ should be negative. But because of the wording of the question it doesn't matter too much here.

RuiAce · « **Reply #3335 on:** March 24, 2018, 01:44:34 pm »

Quote from: dermite on March 24, 2018, 12:23:07 pm

seems pretty basic but im having a mental block so

how would you factorise y = x^3 - 3x + 2?

thanks

You don't. Unless you first guess and check that $x=1$ is a root and then do polynomial long division.

Dragomistress · « **Reply #3336 on:** March 25, 2018, 11:26:08 am »

For some reason I can't find a way to do b. :/

Mate2425 · « **Reply #3337 on:** March 25, 2018, 11:48:49 am »

Hi could i please have some help in finding the integral of Cos(x)Sin^4(x) + Sin(x)Cos^2(x).

Thank you

Sine · « **Reply #3338 on:** March 25, 2018, 11:52:28 am »

Quote from: Mate2425 on March 25, 2018, 11:48:49 am

Hi could i please have some help in finding the integral of Cos(x)Sin^4(x) + Sin(x)Cos^2(x).

Thank you

Split the integral into 2

For 1) Cos(x)Sin^4(x) set u = sin(x)
For 2) Sin(x)Cos^2(x) set u = cos(x)

Mate2425 · « **Reply #3339 on:** March 25, 2018, 12:04:23 pm »

Quote from: Sine on March 25, 2018, 11:52:28 am

Split the integral into 2

For 1) Cos(x)Sin^4(x) set u = sin(x)
For 2) Sin(x)Cos^2(x) set u = cos(x)

THANK YOU

legorgo18 · « **Reply #3340 on:** March 25, 2018, 12:09:36 pm »

Quote from: Dragomistress on March 25, 2018, 11:26:08 am

For some reason I can't find a way to do b. :/

Hey! Solutions in the link as usual, please ask if you have any questions: https://imgur.com/a/Am8GM

Dragomistress · « **Reply #3341 on:** March 25, 2018, 01:24:27 pm »

I don't quite get what you did from the first to the second line.

mathquestioner · « **Reply #3342 on:** March 25, 2018, 04:22:37 pm »

Hi guys, bit of an easy question but I've forgotten whether or not there's a formula for working out the second part? Thanks!

RuiAce · « **Reply #3343 on:** March 25, 2018, 04:34:49 pm »

Quote from: mathquestioner on March 25, 2018, 04:22:37 pm

Hi guys, bit of an easy question but I've forgotten whether or not there's a formula for working out the second part? Thanks!

You're using the theorem that tangents to an external point are equal. This tells you that $ BE = BD = 4$ and $CF = CE = 2$, and note that $BC = BE + CE$.

After you find $BC$, you'll need to note that $AD = AF = x$ before using the cosine rule, noting that $ \angle A = 60^\circ$. Also note that $BA = BD + AD$ and similarly for $ AC$

RuiAce · « **Reply #3344 on:** March 25, 2018, 04:52:00 pm »

Quote from: Dragomistress on March 25, 2018, 01:24:27 pm

I don't quite get what you did from the first to the second line.

$\text{He bashed the formula}\\ \int \frac{dx}{a^2x^2+b^2} = \frac{1}{ab} \tan^{-1} \frac{bx}{a}.$
$\text{If you're uncomfortable with the formula, do this instead.}\\ \begin{align*}\int \frac{3}{1+7x^2} dx &= 3 \int \frac{1}{7 \left( \frac{1}{7}+x^2 \right)}dx\\ &= \frac{3}{7} \int \frac{1}{\frac17 + x^2}\,dx\\ &= \frac37 \times \sqrt{7} \tan^{-1}( \sqrt7 x)\\ &= \frac{3}{\sqrt7} \tan^{-1} (\sqrt7 x) \end{align*}$

Search the forums now!

We have moved!

Author Topic: 3U Maths Question Thread (Read 1506351 times) Tweet Share

Dragomistress

Re: 3U Maths Question Thread

RuiAce

Re: 3U Maths Question Thread

Dragomistress

Re: 3U Maths Question Thread

dermite

Re: 3U Maths Question Thread

RuiAce

Re: 3U Maths Question Thread

RuiAce

Re: 3U Maths Question Thread

Dragomistress

Re: 3U Maths Question Thread

Mate2425

Re: 3U Maths Question Thread

Sine

Re: 3U Maths Question Thread

Mate2425

Re: 3U Maths Question Thread

legorgo18

Re: 3U Maths Question Thread

Dragomistress

Re: 3U Maths Question Thread

mathquestioner

Re: 3U Maths Question Thread

RuiAce

Re: 3U Maths Question Thread

RuiAce

Re: 3U Maths Question Thread

Recent Posts

User:
Password: