Author Topic: 4U Maths Question Thread (Read 664143 times) Tweet Share

RuiAce · « **Reply #1290 on:** June 25, 2017, 07:31:46 pm »

Quote from: bluecookie on June 25, 2017, 07:29:44 pm

Thanks! Can I've some help with this question?

A particle P is attached by a light string of length l to a fixed point A, and P describes a horizontal circle with uniform angular velocity w around the vertical AC. Prove that if theta is the inclination of the string to the vertical, then h does not depend on l, and find a expression for h in terms of w.

Find the tension in the string when the particle P is rotating at 1 rev/s, the mass of P is 1kg and the length of the string is 35cm. Take pi^2=10.

What's h in this question? Is there a diagram that goes with the question that should be provided?

bluecookie · « **Reply #1291 on:** June 25, 2017, 07:48:30 pm »

Yep, sorry!

RuiAce · « **Reply #1292 on:** June 25, 2017, 08:00:58 pm »

itssona · « **Reply #1293 on:** June 25, 2017, 08:15:46 pm »

Ooh also please explain this thank youuu :
Solve z^2 + (1+3i)z -8 -i =0

RuiAce · « **Reply #1294 on:** June 25, 2017, 08:22:56 pm »

Quote from: sssona09 on June 25, 2017, 08:15:46 pm

Ooh also please explain this thank youuu :
Solve z^2 + (1+3i)z -8 -i =0

$\text{When you substitute into the quadratic formula you arrive at}\\ z = \frac{-1-3i \pm \sqrt{24+10i}}{2}$
$\text{You should be very familiar with the method of dealing with complex square roots}\\ \text{Let }(x+iy)^2 = 24+10i\text{ to get}\\\begin{align*}x^2-y^2&=24\\ 2xy&=10\end{align*}$
$\text{And then we may use the trick}\\ |x+iy|^2 = |24+10i| \implies \boxed{x^2+y^2=26}$
$\text{Everything has been set up and I leave the remainder as your exercise}\\ \text{Post any further progress if you get stuck}$

Edit: Typo, forgot the square root

itssona · « **Reply #1295 on:** June 25, 2017, 09:06:37 pm »

Quote from: RuiAce on June 25, 2017, 08:22:56 pm

$\text{When you substitute into the quadratic formula you arrive at}\\ z = \frac{-1-3i \pm {24+10i}}{2}$
$\text{You should be very familiar with the method of dealing with complex square roots}\\ \text{Let }(x+iy)^2 = 24+10i\text{ to get}\\\begin{align*}x^2-y^2&=24\\ 2xy&=10\end{align*}$
$\text{And then we may use the trick}\\ |x+iy|^2 = |24+10i| \implies \boxed{x^2+y^2=26}$
$\text{Everything has been set up and I leave the remainder as your exercise}\\ \text{Post any further progress if you get stuck}$

ahh i get it now, thank u

ProfLayton2000 · « **Reply #1296 on:** June 29, 2017, 03:08:32 pm »

Hey for complex numbers how do we indicate the locus if it one of the axes? (Eg locus of Im(z) = |z|)

RuiAce · « **Reply #1297 on:** June 29, 2017, 03:15:31 pm »

Quote from: ProfLayton2000 on June 29, 2017, 03:08:32 pm

Hey for complex numbers how do we indicate the locus if it one of the axes? (Eg locus of Im(z) = |z|)

You can just call it the y-axis (or imaginary axis) surely

chloe9756 · « **Reply #1298 on:** June 30, 2017, 03:41:26 pm »

The combined air and road resistance of a car in motion is proportional to v2, where v is its speed. When the engine is disengaged the car moves down an incline making an angle arcsin(1/30) with the horizontal, with a velocity of 30m/s. Find the force required to drive the car up the incline with a steady speed of 24m/s, given that the mass of the car is 1200kg.

Answer = 656N

RuiAce · « **Reply #1299 on:** June 30, 2017, 04:36:29 pm »

Quote from: chloe9756 on June 30, 2017, 03:41:26 pm

The combined air and road resistance of a car in motion is proportional to v2, where v is its speed. When the engine is disengaged the car moves down an incline making an angle arcsin(1/30) with the horizontal, with a velocity of 30m/s. Find the force required to drive the car up the incline with a steady speed of 24m/s, given that the mass of the car is 1200kg.

Answer = 656N

An assumption had to be made that g=10 to make this work

beau77bro · « **Reply #1300 on:** July 03, 2017, 03:37:13 am »

Ok so the answer is b. But it doesn't consider two yellows being selected - or somethings idk. I got 10/41 so yea I'm way off - help much appreciated

RuiAce · « **Reply #1301 on:** July 03, 2017, 08:53:02 am »

Quote from: beau77bro on July 03, 2017, 03:37:13 am

(Image removed from quote.)

Ok so the answer is b. But it doesn't consider two yellows being selected - or somethings idk. I got 10/41 so yea I'm way off - help much appreciated

Of course we can't have two yellows because we know that one jellybean has to be black
$\text{Edit: Later on I figured out what they did}\\ \text{and I personally believe that this question is unfair to 4U}\\ \text{because it requires }\textbf{conditional probability}$
$\text{Given two events }A\text{ and }B,\text{ the probability of }B\text{ happening}\\ \textbf{given that }A\text{ has happened is}\\ \mathbb{P}(B \mid A)=\frac{\mathbb{P}(B \cap A)}{\mathbb{P}(A)}\\ \text{where, for our purposes only, the numerator is the pr. of both }A\text{ and }B$
$\text{In our case}\\ \mathbb{P}(\text{Two black} \mid \text{One black}) = \frac{\mathbb{P}(\text{Two black, One black})}{\mathbb{P}(\text{One black})}$
$\text{Of course, the probability of getting two black AND one black}\\ \text{is just the probability of getting two black}\\ \text{which is }\frac{2}{9}$
$\text{For the probability of getting one black, it's easier to consider the complement}\\ \text{Note that saying one black implicitly means at least one black}\\ \text{So the complement is no black}\\ \text{Which has a probability of }\frac{3}{10}\times \frac{2}{9} + 2\times \frac{3}{10}+\frac{2}{9}+\frac{2}{10}\times \frac{1}{9} = \frac{2}{9}\\ \text{Everything should puzzle up from here}$

limtou · « **Reply #1302 on:** July 04, 2017, 01:57:20 pm »

Hello! Could you please explain the solutions to these questions?
all of Q8 and part ii) for c)
Thanks

RuiAce · « **Reply #1303 on:** July 04, 2017, 04:09:30 pm »

Quote from: limtou on July 04, 2017, 01:57:20 pm

Hello! Could you please explain the solutions to these questions?
all of Q8 and part ii) for c)
Thanks

(Briefly recall that three lines being concurrent means that they intersect at a common point)
$\text{As none of the said lines are parallel, any two lines chosen will ensure a point of intersection}\\ \text{as unless we do 3D plane geometry, every pair of non-parallel lines intersect}$
$\text{The fact that no three or more lines are concurrent means that the points of intersection are }\textbf{distinct}\\ \text{Hence, any choice of 2 out of the 6 lines guarantees a } \textbf{unique}\text{ point of intersection}\\ \text{Therefore there are }\binom62\text{ points of intersection}$
_____________________________
$\text{You should be well aware of the definition of probability: }\frac{\text{Favourable outcomes}}{\text{Total outcomes}}\\ \text{Total outcomes being the same as the sample space.}$
$\text{The total possible outcomes is the number of ways we may choose 3 points}\\ \text{Out of the 15 above}\\ \text{This is trivially }\binom{15}3$
For the number of favourable outcomes, it helps to draw a diagram with 6 non-parallel non-concurrent lines.
$\text{The favourable outcomes occur when the points chosen are collinear}\\ \text{Recalling that we had 6 lines above, to ensure collinearity}\\ \text{We must choose one of the 6 lines, and then 3 points on that line}$
$\text{Also carefully observe that each line involves }\textbf{five}\text{ such points of intersection}\\ \textit{Exercise: Why?}\\ \text{So we must choose 3 out of these 5 points}\\ \text{giving us the number of favourable outcomes: }6\times \binom{5}{3}\\ \text{So the probability is just }\frac{6\binom53}{\binom{15}3}$
_____________________________
$\text{The sample space is obviously going to be the same as for the previous part}\\ \text{If we want }\textbf{none}\text{ of these points to be on the lines}\\ \text{We have to be careful about our choice of lines}$
Again, you should draw a diagram to visualise what's going on
$\text{Upon the choice of a point, we identify two lines that the point passes through}\\ \textbf{No other points can lie on these two lines.}\\ \text{So after choosing 2 lines out of 6, we only have 4 lines left.}\\ \text{The number of ways of choosing 2 lines out of 6 is }\binom62$
$\text{The second point must lie on any of the remaining 4 lines.}\\ \text{And the number of ways we choose 2 lines out of 4 is }\binom42\\ \text{In a similar fashion, the number of ways to choose 2 lines from the only 2 lines left is }\binom22$
$\text{Hence the number of favourable outcomes may be }\binom62\binom42\binom22\\ \text{But we have repetition. Recall that the }\textbf{order}\text{ in which they're chosen is irrelevant}\\ \text{Hence the division by }3!\\ \text{so our favourable outcomes is }\binom62\binom42\binom22\frac{1}{6!}$

I am not sure where the confusion is with the inequality proof. The solutions are quite self-explanatory. Please indicate where the problem is. Also please confirm that the stationary point obtained in the previous part is at x=c/2.

limtou · « **Reply #1304 on:** July 04, 2017, 05:03:53 pm »

Quote from: RuiAce on July 04, 2017, 04:09:30 pm

(Briefly recall that three lines being concurrent means that they intersect at a common point)
$\text{As none of the said lines are parallel, any two lines chosen will ensure a point of intersection}\\ \text{as unless we do 3D plane geometry, every pair of non-parallel lines intersect}$ ...

Thanks for the super long explanation for this question!
And don't worry about the inequality proof, I was momentarily confused at $x^k + (c-x)^k < 2\left(\frac{c}{2}\right)^k$ , but I see it now

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