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March 29, 2024, 05:40:00 am

Author Topic: Mathematics Question Thread  (Read 1296833 times)  Share 

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RuiAce

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Re: 96 in 2U Maths: Ask me Anything!
« Reply #30 on: February 27, 2016, 08:43:13 am »
+1
cos(2x+pie/6)=1/root2     -pie<=x<=pie

how do i find the angles in a simple way without getting the angles confused?




Edit: Holy crap the mistakes.
« Last Edit: February 27, 2016, 02:04:09 pm by RuiAce »

Phillorsm

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Re: 96 in 2U Maths: Ask me Anything!
« Reply #31 on: February 27, 2016, 05:03:16 pm »
+1
Thank you Jake and HPL, I really value all your help and time you spend here sharing your wisdom. HPL, what you said about not coming first giving you the motivation to work harder for the following exams was especially helpful :) Thanks again guys, now I need to get stuck into hardcore study mode for an ext1 exam on wednesday. Binomial, inverse, induction and harder 2u... Wish me luck!  ;D

amandali

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Re: 96 in 2U Maths: Ask me Anything!
« Reply #32 on: February 27, 2016, 07:55:06 pm »
0
how do you do this ques thanks  ;D
a man in a rowing boat is presently 6km from the nearest point A on the shore. He wants to reach as soon as possible a point B that is further 20km down the shore from A. If he can row at 8km/hr and run at 10km/hr , how far from A should he land?  ans: 8km

RuiAce

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Re: 96 in 2U Maths: Ask me Anything!
« Reply #33 on: February 27, 2016, 09:09:39 pm »
0
how do you do this ques thanks  ;D
a man in a rowing boat is presently 6km from the nearest point A on the shore. He wants to reach as soon as possible a point B that is further 20km down the shore from A. If he can row at 8km/hr and run at 10km/hr , how far from A should he land?  ans: 8km

Weakly lacking information. There are three cases:
1. We want to minimise the time of the entire journey (this sounds extremely preferable to the others)
2. We want to minimise the distance of the entire journey (sounds very unfeasible as there is actually a very obvious answer)
3. We want to minimise only a part of the journey (e.g. reduce some walking

If the earlier answer didn't make sense, all the hints were given.
a) Rewrite boundaries because otherwise you don't know what you're trying to get at
b) Always use ASTC
c) Negative angles work in reverse (clockwise around ASTC)

Exact values are different. Use the formula sheet for exact values.

jakesilove

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Re: 96 in 2U Maths: Ask me Anything!
« Reply #34 on: February 27, 2016, 09:12:34 pm »
+2
how do you do this ques thanks  ;D
a man in a rowing boat is presently 6km from the nearest point A on the shore. He wants to reach as soon as possible a point B that is further 20km down the shore from A. If he can row at 8km/hr and run at 10km/hr , how far from A should he land?  ans: 8km

Hey!

This was actually quite a difficult question! First, you need to construct the relevant equation. Then, the differentiation part is quite tricky for a 2U question. Finally, even solving the equation is a bit of work! Below is my answer, I hope you find it helpful! In the case of questions like this, just think through what is actually happening in the example. Then, always create some sort of equation that you can differentiate!




Jake
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RuiAce

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Re: 96 in 2U Maths: Ask me Anything!
« Reply #35 on: February 27, 2016, 09:20:05 pm »
+4
Assumed we wanted to minimise time:


P.S. Woah Jake what are you doing with the trig!
« Last Edit: February 27, 2016, 09:21:55 pm by RuiAce »

Happy Physics Land

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Re: 96 in 2U Maths: Ask me Anything!
« Reply #36 on: February 27, 2016, 09:22:05 pm »
+3
how do you do this ques thanks  ;D
a man in a rowing boat is presently 6km from the nearest point A on the shore. He wants to reach as soon as possible a point B that is further 20km down the shore from A. If he can row at 8km/hr and run at 10km/hr , how far from A should he land?  ans: 8km

Hey Amanda:

Another great question from you amanda! Ok, so when I first looked at the question, my instinct tells me I will have to draw a diagram, because it involves distance and speed and even worse theres also time. And I will have to admit, despite my distaste towards drawing diagrams, it is essential for you to draw one because graphics help you to visualise stuff. So next time when you see these types of questions, definitely draw a diagram, its worth the time.

Ok and then I manipulated pythagorus theorem, made an algebraic expression for  the time to travel distance MS, made another algebraic expression for the time to travel distance BS. The time taken to travel the total distance now becomes T(MS) + T(BS), where T =time. So after establishing such a relationship, the rest is straight forward: simply repeat the conventional process of simplifying, differentiating, make derivative = 0, and then find stationary point. Afterwards you test a value on both sides of stationary point to determine its nature (I.e. maximum or minimum) and then make a concluding statement for your answer. Actually it is quite crucial to include a concluding statement, because that makes it clearer for the marker what your final result is and you would less likely to be deducted a mark on not stating the result clearly.

Anyways, my solution as below:




Sorry for the messy working btw, if you have any further questions dont hesitate to ask! :)

Best Regards
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jakesilove

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Re: 96 in 2U Maths: Ask me Anything!
« Reply #37 on: February 27, 2016, 09:26:50 pm »
+1
Assumed we wanted to minimise time:
(Image removed from quote.)

P.S. Woah Jake what are you doing with the trig!

So, this is a wayyyy easier solution. There are so many ways to answer questions like these, so whichever makes more sense to you! Still, thanks everyone for participating and answering questions on the forums!

My method is definitely more tricky, because I have a tendency to over-use trig. If you can use algebra, that's usually a better and simpler solution.

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16ebond

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Re: 96 in 2U Maths: Ask me Anything!
« Reply #38 on: March 07, 2016, 06:09:18 pm »
0
Hey Jake
So we have been doing e in class and I am a bit stuck on how to answer this question.
I have attached the question, so hopefully the file will open.
Thanks heaps
Em  :)

RuiAce

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96 in 2U Maths: Ask me Anything!
« Reply #39 on: March 16, 2016, 04:50:39 pm »
+3
I'm not sure if this was meant to be an unanswered question but here's a solution


Pick an arbitrary point (a,0) to be the first vertex of the rectangle. By default (a, e^(-a^2)) is also on the rectangle. But because it HAS TO BE A RECTANGLE and we have an even function
(-a,0) and (-a, e^(-a^2)) must lie on the rectangle. This is most easily shown with a diagram.




So the breadth and length of the triangle are:




Hence we can combine these to give an expression for area:



Set dA/dx=0 to maximise:




Reject x=0 though, because if x=0 we have no rectangle (we can't have a width of 0).

So just show that the remaining value gives a maxima using a table of values or the second derivative. Note that because the curve is an even function, the negative term can be ignored.

I would've used latex but I'm thoroughly lazy to on a phone (Hello, Jamon here, I went through and added LaTex  ;D)
« Last Edit: March 17, 2016, 01:04:15 pm by jamonwindeyer »

jamonwindeyer

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Re: 96 in 2U Maths: Ask me Anything!
« Reply #40 on: March 17, 2016, 01:05:32 pm »
+2
I'm not sure if this was meant to be an unanswered question but here's a solution


Pick an arbitrary point (a,0) to be the first vertex of the rectangle. By default (a, e^(-a^2)) is also on the rectangle. But because it HAS TO BE A RECTANGLE and we have an even function
(-a,0) and (-a, e^(-a^2)) must lie on the rectangle. This is most easily shown with a diagram.

(Image removed from quote.)


So the breadth and length of the triangle are:




Hence we can combine these to give an expression for area:



Set dA/dx=0 to maximise:




Reject x=0 though, because if x=0 we have no rectangle (we can't have a width of 0).

So just show that the remaining value gives a maxima using a table of values or the second derivative. Note that because the curve is an even function, the negative term can be ignored.

I would've used latex but I'm thoroughly lazy to on a phone (Hello, Jamon here, I went through and added LaTex  ;D)

LEGEND! Thanks RuiAce! I went through and added LaTex and a diagram for you, cheers for the solution! Jake has been on holiday and I've been trying to keep an eye on all his forums, I must have missed this, you are a champion  ;D

kal.123

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Re: Mathematics Question Thread
« Reply #41 on: March 19, 2016, 11:43:17 am »
+1
I was wondering how to draw the x-intercepts and vertex??? For Y= 3-2xsquared

jamonwindeyer

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Re: Mathematics Question Thread
« Reply #42 on: March 19, 2016, 12:00:28 pm »
+1
I was wondering how to draw the x-intercepts and vertex??? For Y= 3-2xsquared

Sure! So we have:



The x-intercepts occur when y=0, so we just substitute and solve. This is a quadratic, so we can expect anywhere from o to 2 answers:



Now, for the vertex, we are actually taught that the x-coordinate of the vertex is given by:



Now for your quadratic, a=2, b=0 and c=-3 (when rearranged), so the x-coordinate of the vertex will actually be zero. The y-coordinate will then be y=3 by substitution of x=0.

So, we have intercepts and a vertex, we can now draw a diagram. Hope this helps!


RuiAce

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Re: 96 in 2U Maths: Ask me Anything!
« Reply #43 on: March 19, 2016, 05:53:28 pm »
+1
LEGEND! Thanks RuiAce! I went through and added LaTex and a diagram for you, cheers for the solution! Jake has been on holiday and I've been trying to keep an eye on all his forums, I must have missed this, you are a champion  ;D

Ah sweet. Thanks for the aid too Jamon!

imtrying

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Re: Mathematics Question Thread
« Reply #44 on: March 21, 2016, 04:26:50 pm »
0
Hey :)
This is a Geometrical Applications of Calculus question I'm stuck on (I've attached a photo of the question and answer). I know the quotient and product rules, but for some reason I keep ending up with some ridiculous answer. A step by step explanation would be hugely helpful :)
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