Author Topic: Mathematics Question Thread (Read 1308081 times) Tweet Share

cocopops201 · « **Reply #3270 on:** February 14, 2018, 05:09:47 pm »

I need help with finding the radius of a circle: x^2+y^2-4x+8y+11=0
I totally forgot how to.
Thank you for anyone who replies!

kauac · « **Reply #3271 on:** February 14, 2018, 05:33:03 pm »

Quote from: cocopops201 on February 14, 2018, 05:09:47 pm

I need help with finding the radius of a circle: x^2+y^2-4x+8y+11=0
I totally forgot how to.
Thank you for anyone who replies!

Hi...
I think this might be how you do it...

Rearrange formula:

$x^2 -4x + y^2 + 8y = -11$

Then complete the square:

$x^2-4x+4+y^2+8y+16= -11+4+16$

Plug into circle formula:

$(x - a)^2 + (y - b)^2 = r^2$

$(x - 2)^2 + (y + 4)^2 = 9$

Therefore:
$r = \sqrt{9}$

$= 3$

Hope this is helpful (and correct, of course)!

StupidProdigy · « **Reply #3272 on:** February 14, 2018, 05:47:10 pm »

Quote from: kauac on February 14, 2018, 05:33:03 pm

Hi...
I think this might be how you do it...

Rearrange formula:

x^2 -4x + y^2 + 8y = -11

Then complete the square:

x^2 - 4x + 4 +y^2 +8y +16 = -11

Plug into circle formula:

(x - a)^2 + (y - b)^2 = r^2

(x - 2)^2 + (y + 4)^2 = -11

Therefore r = the square root of -11

Hope this is helpful (and correct, of course)!

Don't forget to add 4 and 16 to the RHS when completing the square. R=sqrt(-11) doesn't sound right for this question. Instead we should have R=sqrt(-11+20)=3

kauac · « **Reply #3273 on:** February 14, 2018, 06:23:40 pm »

Quote from: StupidProdigy on February 14, 2018, 05:47:10 pm

Don't forget to add 4 and 16 to the RHS when completing the square. R=sqrt(-11) doesn't sound right for this question. Instead we should have R=sqrt(-11+20)=3

Sorry, it's all fixed now. I had a feeling that I did something wrong, but couldn't work out what. Thanks for reminding me!

RuiAce · « **Reply #3274 on:** February 15, 2018, 10:02:13 am »

Quote from: kauac on February 12, 2018, 06:24:17 pm

Hi...
I have spent wayyyyy too long on question 24 and haven't really got anywhere... Any help greatly appreciated.

Edit: I think the image is attached now

Your thing's definitely up indeed

sorry for the delay, I've only had enough time to properly concentrate on this question just now.

Deleted materials: Spotted the problem - the textbook rounded (gross) (again thanks Shadowxo)

What I've found is that ultimately you're trying to minimise $ t = \frac{x}{7} + \frac{80 - \sqrt{x^2-400}}{11} $ (thanks Shadowxo for confirmation). But when I checked the textbook, the final answer was completely different to what WolframAlpha gave me. Wolfram claims that the minimum is at $x = \frac{55\sqrt2}{3}$ with a minimum value of $ \frac{40}{77}(14+3\sqrt2) $

$\text{The total distance that Grant travels will be the distance}\\ \text{he swims, plus the distance he runs.}$
$\text{We are told he swims a distance of }x\\ \text{so since his speed when swimming is 7}\\ \text{via }\boxed{t = \frac{D}{S}}\text{ the time taken to do this will be }\frac{x}{7}.$
$\text{For the distance he runs, essentially we must note}\\ \text{that via Pythagoras' theorem, the distance he }\textbf{does not run}\\ \text{will be }\sqrt{x^2-400}$
$\text{As a consequence, the distance he }\textbf{does}\text{ run}\\ \text{will be }80 - \sqrt{x^2-400}\\ \text{and consequently, at a speed of 11, the time taken to accomplish this is}\\ \frac{80}{11} - \frac{\sqrt{x^2-400}}{11}$
$\text{So the total time taken is }\boxed{T = \frac{x}{7} + \frac{80}{11} - \frac{\sqrt{400-x^2}}{11}}$
$\text{Now we just do the usual thing.}\\ \frac{dT}{dx} = \frac{1}{7} - \frac{x}{11\sqrt{400-x^2}}$
If you set this equal to 0 you'll get $x = \frac{55\sqrt2}{3}$ as stated in the spoiler. Which rounds to 26.

gilliesb18 · « **Reply #3275 on:** February 15, 2018, 10:22:44 am »

Hello again... just need help on another integration one... this time using simpson's rule.
\int_1^3 x^(4)dx using 3 function values

Thanks heaps..

RuiAce · « **Reply #3276 on:** February 15, 2018, 10:35:42 am »

Quote from: gilliesb18 on February 15, 2018, 10:22:44 am

Hello again... just need help on another integration one... this time using simpson's rule.
\int_1^3 x^(4)dx using 3 function values

Thanks heaps..

\begin{align*}\int_1^3 x^4\,dx &\approx \frac{3-1}{6} \left(1^4 + 4(2^4) + 3^4 \right)\\ &= 38\end{align*}

gilliesb18 · « **Reply #3277 on:** February 15, 2018, 10:42:11 am »

That's awesome! Thanks once again....

kauac · « **Reply #3278 on:** February 15, 2018, 04:32:26 pm »

Quote from: RuiAce on February 15, 2018, 10:02:13 am

Your thing's definitely up indeed sorry for the delay, I've only had enough time to properly concentrate on this question just now.

$\text{The total distance that Grant travels will be the distance}\\ \text{he swims, plus the distance he runs.}$
$\text{We are told he swims a distance of }x\\ \text{so since his speed when swimming is 7}\\ \text{via }\boxed{t = \frac{D}{S}}\text{ the time taken to do this will be }\frac{x}{7}.$
$\text{For the distance he runs, essentially we must note}\\ \text{that via Pythagoras' theorem, the distance he }\textbf{does not run}\\ \text{will be }\sqrt{x^2-400}$
$\text{As a consequence, the distance he }\textbf{does}\text{ run}\\ \text{will be }80 - \sqrt{x^2-400}\\ \text{and consequently, at a speed of 11, the time taken to accomplish this is}\\ \frac{80}{11} - \frac{\sqrt{x^2-400}}{11}$
$\text{So the total time taken is }\boxed{T = \frac{x}{7} + \frac{80}{11} - \frac{\sqrt{400-x^2}}{11}}$
$\text{Now we just do the usual thing.}\\ \frac{dT}{dx} = \frac{1}{7} - \frac{x}{11\sqrt{400-x^2}}$
If you set this equal to 0 you'll get $x = \frac{55\sqrt2}{3}$ as stated in the spoiler. Which rounds to 26.

Thanks so much!! The wording of the question had me stumped, but your explanation is really clear.

Calley123 · « **Reply #3279 on:** February 15, 2018, 06:21:32 pm »

HEy,
How do I show that y=x^3-3x^2+27x-3 is monotonically increasing for all values of x?
THanks

Opengangs · « **Reply #3280 on:** February 15, 2018, 06:55:48 pm »

Quote from: Calley123 on February 15, 2018, 06:21:32 pm

HEy,
How do I show that y=x^3-3x^2+27x-3 is monotonically increasing for all values of x?
THanks

Hey, Calley123!

To show that a function is monotonically increasing, we need to show that the derivative at any point for x is positive.
So, we need to show that f'(x) > 0 for all x
$\begin{align*}f'(x) = 3x^2-6x+27 &= 3(x^2-2x+9) \\ \text{Completing the square, we get: } f'(x) &= 3(x^2 - 2x + 1 + 8) \\ &= 3(x-1)^2 + 24 > 0 \end{align*}$

Since our derivative is always positive, then we have a monotonically increasing function.

Calley123 · « **Reply #3281 on:** February 15, 2018, 07:38:47 pm »

Quote from: Opengangs on February 15, 2018, 06:55:48 pm

Hey, Calley123!

To show that a function is monotonically increasing, we need to show that the derivative at any point for x is positive.
So, we need to show that f'(x) > 0 for all x
$\begin{align*}f'(x) = 3x^2-6x+27 &= 3(x^2-2x+9) \\ \text{Completing the square, we get: } f'(x) &= 3(x^2 - 2x + 1 + 8) \\ &= 3(x-1)^2 + 24 > 0 \end{align*}$

Since our derivative is always positive, then we have a monotonically increasing function.

Thank youu!

owidjaja · « **Reply #3282 on:** February 16, 2018, 10:17:30 pm »

Hey guys,
I need help with the following question.

Thanks in advance.

RuiAce · « **Reply #3283 on:** February 16, 2018, 10:38:41 pm »

Quote from: owidjaja on February 16, 2018, 10:17:30 pm

Hey guys,
I need help with the following question.

Thanks in advance.

$\text{This is just }V = \pi \int_1^2 (e^{-x}+1)^2\, dx$
$\text{Having said that, take care when evaluating the integral}\\ \text{to not forget the extra 1 hanging about after expanding brackets.}\\\begin{align*}V&= \pi \int_1^2 (e^{-2x} + 2e^{-x} + 1)\,dx\\ &= \pi \left[-\frac{e^{-2x}}{2} - 2e^{-x} + x \right]_1^2\\ &= \pi\left(1 - \frac{e^{-4}}{2} - 2e^{-2} + \frac{e^{-2}}{2} + 2e^{-1}\right)\end{align*} \\ \text{You may simplify this however you wish to.}$

brooksykait · « **Reply #3284 on:** February 19, 2018, 12:25:57 pm »

Hey

. This question is from the Magaret Grove Extension Math Book, (chapter 2). I've been stuck on in for ages.
Grant is at point A on one side
of a 20 m wide river and needs to
get to point B on the other side
80 m along the bank.

Grant swims to any point on the
other bank and then runs along
the side of the river to point B. If
he can swim at 7 km/h and run
at 11 km/h, find the distance he
swims (x) to minimise the time
taken to reach point B. Answer to
the nearest metre.

If somebody could please help me with this solution that would be great! thank you

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