Author Topic: 3U Maths Question Thread (Read 1506344 times) Tweet Share

RuiAce · « **Reply #3405 on:** April 30, 2018, 03:34:35 pm »

Quote from: vikasarkalgud on April 30, 2018, 02:55:06 pm

Hi, I want help for this motion in a straight line question. I don't seem to understand when u take the positive/negative case for velocity so I end up with +- not knowing which to take

The acceleration of a particle is 2x-5/ms^2, where x is the distance in metres from the origin
a) Find an expression for the velocity of this particle in terms of x, given that the particle is at rest one metre to the left of the origin initially

Answer: -sqrt(2x^2-10x-12). (Why negative???)

ty for help

For this particular question, we can indirectly take advantage of the "initially" fact. The idea is to think about the acceleration.
$\text{Whe }t=0\text{ we have }x=-1.\\ \text{Let's see what happens when we plug }-1\text{ into the acceleration.}$
$\text{At }t=0, \, \ddot{x} = -9\text{ which is negative.}\\ \text{Hence, the particle is initially accelerating, }\textbf{in the negative direction.}$
$\text{This means, that immediately after time }t=0\text{ (where it is at rest),}\\ \text{the particle will start heading in the negative direction }\textbf{as well}.$
$\text{Thus, immediately after }t=0\text{, }v < 0.\\ \text{Hence, we take the negative square root: }v = -\sqrt{2x^2-10x-12}$

vikasarkalgud · « **Reply #3406 on:** May 02, 2018, 03:21:56 pm »

Ty for help with previous question Rui

I have another question in same topic,

3. When x m from an origin on a straight line, the velocity v m/s, of a particle is given by v = 1 + e^-x. Find the acceleration when at the origin.
a) What is the position of the particle when t = 2?

I found a = -2 at origin if im not mistaken. in part a) i got up to t = ln(e^x+1) + C via integration, and then I don't know how to get rid of variables and find when t = 2. Not sure what I'm missing. Also, are there any general tips for motion topics?

Thankyou for help

RuiAce · « **Reply #3407 on:** May 02, 2018, 03:33:59 pm »

Quote from: vikasarkalgud on May 02, 2018, 03:21:56 pm

Ty for help with previous question Rui

I have another question in same topic,

3. When x m from an origin on a straight line, the velocity v m/s, of a particle is given by v = 1 + e^-x. Find the acceleration when at the origin.
a) What is the position of the particle when t = 2?

I found a = -2 at origin if im not mistaken. in part a) i got up to t = ln(e^x+1) + C via integration, and then I don't know how to get rid of variables and find when t = 2. Not sure what I'm missing. Also, are there any general tips for motion topics?

Thankyou for help

They don't give you enough information for this particular question. Because they didn't say anything about $t=$ something else, nor did they say anything about "initially".

"General tips" is very vague and in general the only general one is to not get mixed up between time, displacement, velocity and acceleration. (That and not overlooking key bits of information.) Please elaborate

beeangkah · « **Reply #3408 on:** May 03, 2018, 03:51:37 pm »

Hii could I please get help for this projectile motion Q:

A ball is thrown from 2 m above the ground with a velocity of 22 ms−1 and hits a 10 m high target 35 m from where the ball was thrown. Find two possible angles at which the ball was thrown.

Does the 10m mean from the ground or from the 2m above the ground?? I've tried both but probs my calculations have been messed up somewhere else bc both don't work for me.

Correct answer is 63 deg 26 min and 39 deg 27 min
Thanks!

RuiAce · « **Reply #3409 on:** May 03, 2018, 07:19:35 pm »

Quote from: beeangkah on May 03, 2018, 03:51:37 pm

Hii could I please get help for this projectile motion Q:

A ball is thrown from 2 m above the ground with a velocity of 22 ms−1 and hits a 10 m high target 35 m from where the ball was thrown. Find two possible angles at which the ball was thrown.

Does the 10m mean from the ground or from the 2m above the ground?? I've tried both but probs my calculations have been messed up somewhere else bc both don't work for me.

Correct answer is 63 deg 26 min and 39 deg 27 min
Thanks!

What's the acceleration from gravity?

Also, it should be $10$ metres high. The $35$ metres is only used to describe the range.

fireives1967 · « **Reply #3410 on:** May 03, 2018, 07:39:46 pm »

Please help! $:-\$

RuiAce · « **Reply #3411 on:** May 03, 2018, 07:43:09 pm »

Quote from: fireives1967 on May 03, 2018, 07:39:46 pm

Please help! $:-\$

$\frac{d}{dx} \left(\frac{v^2}{2}\right) = \frac{1}{1+x^2} \implies v^2 = 2 \tan^{-1}x + C$
$\text{Initially }x=0\text{ and }v=0\text{, so upon subbing we have }C=0.$
$\text{Thus }v^2 = 2\tan^{-1}x\text{, i.e. }v = \sqrt{2\tan^{-1}x}.$
Note that the acceleration was written in terms of the displacement, not time.

beeangkah · « **Reply #3412 on:** May 03, 2018, 08:50:57 pm »

Quote from: RuiAce on May 03, 2018, 07:19:35 pm

What's the acceleration from gravity?

Also, it should be $10$ metres high. The $35$ metres is only used to describe the range.

g = 9.8

arii · « **Reply #3413 on:** May 03, 2018, 08:55:31 pm »

How would you do this algebraically? I know you can put it in a graphing program but I don't feel like it's a reliable method.

RuiAce · « **Reply #3414 on:** May 03, 2018, 09:12:25 pm »

Quote from: beeangkah on May 03, 2018, 08:50:57 pm

g = 9.8

\begin{align*}\ddot{x}&=0\\ \dot{x} &= 22\cos\theta\\ \dot{x}&=22t\cos\theta\\ \ddot{y}&=-9.8\\ \dot{y} &= -9.8t + 22\sin \theta\\ y&= -4.9t^2 + 22t\sin \theta + 2\end{align*}
$\text{Using all of this, the relevant Cartesian equation of motion is}\\ \boxed{y = -\frac{49x^2}{4840}(1+\tan^2\theta) + x \tan \theta + 2}$
$\text{If the ball passes through }(35,10)\text{, then after subbing we have}\\ \boxed{10 = -\frac{343}{968}(1+\tan^2\theta) + 35 \tan \theta +2}.\\ \text{Now we fight the algebra.}$
\begin{align*}- \frac{12005}{968}(1 + \tan^2 \theta) + 35 \tan \theta - 8 &= 0\\ -\frac{12005}{968} - \frac{12005}{968} \tan^2\theta + 35 \tan \theta - 8 &= 0\\ -\frac{12005}{968}\tan^2\theta + 35 \tan \theta - \frac{19749}{968} &= 0\end{align*}
\begin{align*}\therefore \tan \theta &= \frac{-35 \pm \sqrt{35^2 - 4 \times\frac{12005}{968} \times \frac{19749}{968} }}{2\times- \frac{12005}{968}} \\ \theta &= \tan^{-1}\left( \frac{-35 \pm \sqrt{35^2 - 4 \times\frac{12005}{968} \times \frac{19749}{968} }}{2\times- \frac{12005}{968}} \right)\\ &\approx 63^\circ 26', \quad 39^\circ 27'\end{align*}

RuiAce · « **Reply #3415 on:** May 03, 2018, 09:25:24 pm »

Quote from: arii on May 03, 2018, 08:55:31 pm

How would you do this algebraically? I know you can put it in a graphing program but I don't feel like it's a reliable method.

$\text{Note that the domain of }3\sqrt{x(1-x)}\text{ is just }x (1-x) \ge 0\\ \text{which solves to give }\boxed{0 \le x \le 1}.\\ \text{This means we only need to narrow our focus to this little strip.}$

Something to think about, but we won't use.

$\text{It's helpful to recall the definition of the absolute value:}\\ |x| = \begin{cases} x & \text{if }x\geq 0\\ -x & \text{if }x < 0 \end{cases}.\\ \text{In our case, we're dealing with }|x-2|\text{, and its corresponding definition is}\\ |x-2| = \begin{cases} x-2 &\text{if }x\geq 2\\ -(x-2) &\text{if }x < 2\end{cases}.$
$\text{Recall that we're }\textbf{only}\text{ interested in the interval }0 \leq x \leq 1.\\ \text{This interval is contained }\textbf{entirely}\text{ in }x-2.$
$\text{So if }0\leq x \leq 1, \text{ we actually arrive at }|x-2| = -(x-2).\\ \text{So the equation reduces to just solving }3\sqrt{x(1-x)} \leq 2-x$

A small remark in that we may square both sides safely. This is because the quantities on the $LHS$ and the $RHS$ are both positive.
$\text{Starting with }3\sqrt{x(1-x)} < |x-2|\text{, upon squaring both sides we have}\\ \begin{align*}9x(1-x) &< (x-2)^2\\ 9x - 9x^2 &< x^2 - 4x + 4\\ 0 &< 10x^2 - 13x + 4\\ 0 &< (5x-4)(2x-1)\end{align*}$
$\text{Which solves to give}\\ x < \frac12, x > \frac54$
$\text{But recall that we can only ever have }\boxed{0\leq x \leq 1}.\\ \text{So the solutions we obtain from the inequality are}\\ \boxed{0 \leq x < \frac12}\text{ and }\boxed{\frac54 < x \leq 1}.$

arii · « **Reply #3416 on:** May 04, 2018, 08:09:41 pm »

Got another question. I started it but there's parts of (a) and (b) I'm stuck on.

For (a), I can see that if I can somehow prove the relationship, then that's all I need.
For (b), I got the inequality relationship but I'm not sure how to actually show "How large..."

RuiAce · « **Reply #3417 on:** May 04, 2018, 08:52:56 pm »

Quote from: arii on May 04, 2018, 08:09:41 pm

Got another question. I started it but there's parts of (a) and (b) I'm stuck on.

For (a), I can see that if I can somehow prove the relationship, then that's all I need.
For (b), I got the inequality relationship but I'm not sure how to actually show "How large..."

In the future, please post questions of the harder 3U topic in the relevant 4U question thread. Otherwise I may move the question there myself.

Note that a bit of working backwards was done behind the scenes to deduce this proof.
$\text{We're given that }\boxed{\frac{x+y}{2} \ge \sqrt{n}} \text{ which is just a really common 4U proof}\\ \text{and for a) we assume that }\boxed{x > \sqrt{n}}$
Recall that all quantities are positive here.
$\text{To say that }\frac{x+y}{2}\text{ approximates }\sqrt{n}\text{ better than }x\text{ approximates }n\\ \text{means that the distance between }\frac{x+y}{2}\text{ and }\sqrt{n}\text{, i.e. }\boxed{\frac{x+y}{2} - \sqrt{n}}\\ \text{must be }\textbf{less than}\text{ the distance between }x\text{ and }\sqrt{n}\text{, i.e. }\boxed{x - \sqrt{n}}$
$\text{i.e. we wish to prove that}\boxed{\frac{x+y}{2} - \sqrt{n} < x - \sqrt{n}}$
Note: The distance between $x$ and $\sqrt{n}$ is technically $ |x - \sqrt{n}| $. The reason why we were allowed to take off the absolute values was because $ x > \sqrt{n}$, and from the definition of the absolute value, $ |x-n| = \begin{cases}x-n & \text{if }x\ge n\\ -(x-n) &\text{if }x < n\end{cases} $. A similar thing happens for (x+y)/2.
$\text{To do this proof, we start from our assumption }\boxed{x > \sqrt{n}}.\\ \text{Squaring both sides gives }x^2 > n.\text{ But we know that }n = xy,\text{ so we have }\boxed{x^2 > xy}$
$\text{Since }x\text{ is positive, we can divide out }x\text{ to obtain }\boxed{x > y}.\\ \text{From here, things are easy.}\\ \begin{align*}y &< x\\ x+y &< 2x\\ \frac{x+y}{2} &< x\\ \frac{x+y}{2} - \sqrt{n} &< x - \sqrt{n}\end{align*}\\ \text{as required.}$
------------------------------------------------------------------------------------------------------------------------------------------------
$\text{Now suppose instead that }x < \sqrt{n}.\\ \text{Note that the distance between }x\text{ and }\sqrt{n}\text{ is now }\boxed{- (x-\sqrt{n})}.\\ \text{You should think about why this is the case.}$
$\text{If we want }\frac{x+y}{2}\text{ to estimate }\sqrt{n},\textbf{ better }\text{than }x\text{ does,}\\ \text{then the condition we require is }\boxed{\frac{x+y}{2} - \sqrt{n} < \sqrt{n} - x}.$
$\text{Recalling that }\boxed{y = \frac{n}{x}}\text{, we begin to solve this inequality.}\\ \begin{align*}\frac{3x+y}{2} - 2\sqrt{n} &< 0\\ 3x + y - 4\sqrt{n} &< 0\\ 3x - 4\sqrt{n} + \frac{n}{x} &< 0\\ 3x^2 - 4\sqrt{n}x + n &< 0\\ (3x-\sqrt{n}) (x-\sqrt{n}) &< 0\end{align*}$
$\text{Which solves to give }\frac{\sqrt{n}}{3} < x < \sqrt{n}\\ \text{so }x\text{ has to be greater than }\boxed{\frac{\sqrt{n}}3}$
A side remark regarding your thoughts - It seemed like the main cause of your problem was not really understanding what the question meant. Whilst you had written a bunch of very useful ideas, that was all I could salvage because the method wasn't appropriate for the question. But the fact that they asked for approximations in this regard (as opposed to something usual like Simpson's rule), as well as the introduction of all of these variables, were immediately hinting to me that this is not the typical 3U question. This question actually required you to understand what it meant for an approximation to be "better". An approximation is better, essentially when it approximates the value better. So if you have two values $x_1$ and $x_2$ to approximate $x$, the only way $x_1$ can be the "better" estimator is when $x_1$ is closer to $x$, than $x_2$ is. That is to say, $ |x_1 - x| < |x_2 - x| $.

K98100 · « **Reply #3418 on:** May 10, 2018, 09:55:06 pm »

Hey,
For what values is y=1/(x^2+2x-2) concave up and concave down?
Thanks

ravioli · « **Reply #3419 on:** May 10, 2018, 10:03:42 pm »

how would I go about finding when a function has a horizontal asymptote?

so my questions asking me to find the values of a that cause f(x) to have a horizontal asymptote when f(x)=1/(x^2+2x+a), and I'm confused because I thought that it would just always have an asymptote at y=0?

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