Author Topic: 3U Maths Question Thread (Read 1497822 times) Tweet Share

Never.Give.Up · « **Reply #3270 on:** February 13, 2018, 10:38:24 pm »

Quote from: RuiAce on February 13, 2018, 10:20:47 pm

$\text{If we assume that}\\ \boxed{9+14+19+\dots + (5k+4) = \frac{5k^2}{2} + \frac{13k}{2}}$
$\text{Then we can consider how}\\ \begin{align*}9+14+19+\dots + (5k+4) + (5k+9)&= \frac{5k^2}{2} + \frac{13k}{2} + 5k+9\\ &= \frac{5}{2}(k^2+2k+1) + \frac{13}{2}(k+1)\\ &= \frac{5(k+1)^2}{2} + \frac{13k}{2}.\end{align*}$
Making the transition from line 2 to line 3 a bit clearer..
\begin{align*}\frac{5k^2}{2} + \frac{13k}{2} + 5k + 9 &= \frac{5k^2}{2} + 5k + \frac{5}{2} + \frac{13k}{2} + \frac{13}{2}\\ &= \frac{5}{2}(k^2+2k+1) + \frac{13}{2}(k+1)\end{align*}
________________________________________________________________________
$\text{If we assume that}\\ \boxed{-4 + 8 - 16 + \dots -4(-2)^{k-1} = \frac{4\left((-2)^k-1 \right)}{3}}$
$\text{Then we can consider how}\\ \begin{align*}-4+8-16+\dots -4(-2)^{k-1} - 4(-2)^k &=\frac{4\left((-2)^k-1 \right)}{3} - 4(-2)^k\\ &= \frac{4(-2)^k - 4 - 12(-2)^k}{3} \\ &= \frac{-8(-2)^k - 4}{3}\\ &= \frac{4 \times (-2)(-2)^k - 4}{3}\\ &= \frac{4(-2)^{k+1}-4}{3}.\end{align*}$

wowow!!!!

thanks so much...ur a legend!!!

arii · « **Reply #3271 on:** February 16, 2018, 06:15:04 pm »

Can someone teach me this? I'm a little confused on how I do this without a given angle.

RuiAce · « **Reply #3272 on:** February 16, 2018, 06:18:14 pm »

Quote from: arii on February 16, 2018, 06:15:04 pm

Can someone teach me this? I'm a little confused on how I do this without a given angle.

Hint: Take advantage of the fact that the greatest height is 0.6 to solve for $\theta$.

arii · « **Reply #3273 on:** February 16, 2018, 06:28:10 pm »

Quote from: RuiAce on February 16, 2018, 06:18:14 pm

Hint: Take advantage of the fact that the greatest height is 0.6 to solve for $\theta$.

I'm thinking to do y'=0..

y'' = -10
dy'/dt = -10
dy' = -10dt
integrate both sides..
y' = -10t + C.
When t=0, y'=8sin theta
y' = -10t + 8 sin theta
dy/dt = -10t + 8 sin theta
dy = (-10t + 8 sin theta)dt
integrate both sides..
y = -5t^2 + 8t sin theta
Set y = 0.6..
0.6 = -5t^2 + 8t sin theta

Definitely did something wrong somewhere... i have 2 variables here t and theta...

RuiAce · « **Reply #3274 on:** February 16, 2018, 06:37:00 pm »

$\text{So you have correctly idenfitied}\\ \begin{align*}\ddot{y} &= -10\\ \dot{y} &= -10t +8\sin\theta \\ y&= -5t^2+8t\sin\theta \end{align*}$
$\text{The greatest height occurs when }\dot{y} = 0\\ \text{Setting }-10t+8\sin\theta = 0\text{ gives }\boxed{t = \frac{4\sin\theta}{5}}$
$\text{Since the greatest height itself is 0.6,}\\ \frac{3}{5} = -5\left(\frac{4\sin\theta}{5}\right) ^2+ 8\left(\frac{4\sin\theta}{5}\right)\sin \theta$
The fact that the greatest height occurs when $ \dot{y}=0 $ is quite important and fundamental to the course. Your intuition was correct, but you had to follow through with it.

KT Nyunt · « **Reply #3275 on:** February 18, 2018, 12:59:26 pm »

Question: Find the point of inflexion on the curve y=xlnx - x^2

I found the derivative to be: 1 + lnx - 2x
I believe it's right but I don't know what to do with it now.

Please help!

RuiAce · « **Reply #3276 on:** February 18, 2018, 01:01:02 pm »

Quote from: KT Nyunt on February 18, 2018, 12:59:26 pm

Question: Find the point of inflexion on the curve y=xlnx - x^2

I found the derivative to be: 1 + lnx - 2x
I believe it's right but I don't know what to do with it now.

Please help!

$\text{The possible points of inflexion occur when}\\ \text{the }\textbf{second derivative}\text{ equals }0.\\ \text{Your first derivative is correct, but therefore you must consider}\\ \frac{1}{x} - 2 = 0.$
$\text{Which solves to give }x = \frac12.\\ \text{You can then follow the usual procedure from there.}$

KT Nyunt · « **Reply #3277 on:** February 18, 2018, 01:07:32 pm »

Quote from: RuiAce on February 18, 2018, 01:01:02 pm

$\text{The possible points of inflexion occur when}\\ \text{the }\textbf{second derivative}\text{ equals }0.\\ \text{Your first derivative is correct, but therefore you must consider}\\ \frac{1}{x} - 2 = 0.$
$\text{Which solves to give }x = \frac12.\\ \text{You can then follow the usual procedure from there.}$

Got it! thank you!

KT Nyunt · « **Reply #3278 on:** February 18, 2018, 01:57:34 pm »

Question: Find the indefinite integral (primitive function) of 3^(2x-1)

Could you please send the worked solutions? I got no clue how to do this!

RuiAce · « **Reply #3279 on:** February 18, 2018, 02:01:08 pm »

Quote from: KT Nyunt on February 18, 2018, 01:57:34 pm

Question: Find the indefinite integral (primitive function) of 3^(2x-1)

Could you please send the worked solutions? I got no clue how to do this!

$\text{Using the fact that }\boxed{x = e^{\ln x}}\text{ for all }x > 0$
\begin{align*}\int 3^{2x-1}\,dx &= \int e^{\ln (3^{2x-1})}\, dx \\ &= \int e^{(2x-1)\ln 3}\,dx \tag{log law}\\ &= \frac{1}{2\ln 3}e^{(2x-1)\ln 3}+C\\ &= \frac1{2\ln 3}e^{\ln (3^{2x-1})}+C\\ &= \frac1{2\ln 3}\times3^{2x-1}+C\end{align*}

asd987 · « **Reply #3280 on:** February 19, 2018, 10:20:55 pm »

can i get some help with this question:
Find the coefficient of x^3 in (5-x)^3 (2+x)

legorgo18 · « **Reply #3281 on:** February 19, 2018, 11:09:06 pm »

Quote from: asd987 on February 19, 2018, 10:20:55 pm

can i get some help with this question:
Find the coefficient of x^3 in (5-x)^3 (2+x)

Hey! Solutions in link, please ask if you dont understand a step!

https://imgur.com/a/C8ts5

jamonwindeyer · « **Reply #3282 on:** February 19, 2018, 11:16:50 pm »

Quote from: asd987 on February 19, 2018, 10:20:55 pm

can i get some help with this question:
Find the coefficient of x^3 in (5-x)^3 (2+x)

Hey! Let's just expand this step by step:

$(5-x)^3(2+x)=(5-x)(25-10x+x^2)(2+x)=(10+3x-x^2)(25-10x+x^2)\\=-x^4 + 13 x^3 - 45 x^2 - 25 x + 250$

You might need to do that last step slow, be careful! But once you're done, you'll see that $13$ is our answer

Edit: Beaten to the punch! But an alternative method nonetheless, thanks legorgo!

beeangkah · « **Reply #3283 on:** February 20, 2018, 08:40:54 pm »

Hii, can someone help me find the integral of:

sin²x^o ?

I'm assuming that little circle means degrees, not sure I would solve it though, it's from my textbook...

Thank you

RuiAce · « **Reply #3284 on:** February 20, 2018, 08:51:18 pm »

Quote from: beeangkah on February 20, 2018, 08:40:54 pm

Hii, can someone help me find the integral of:

sin²x^o ?

I'm assuming that little circle means degrees, not sure I would solve it though, it's from my textbook...

Thank you

Which textbook and exercise is this? Normally we never do trigonometric calculus in degrees, but I know that Cambridge throws in some weird oddballs.

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