February 25, 2020, 05:50:03 pm

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#### Sine

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##### Re: VCE Methods Question Thread!
« Reply #18360 on: January 23, 2020, 12:30:30 am »
+3
Hey,
Does anyone know how to integrate xcos(x) by hand? Is there a method to do it straight away? I found it on an exam where they have it in parts with part a) asking us to diff 2xsin(x) and then use that to evaluate the integral which u can do by ∫(0.5*dy/dx)dx - ∫sin(x) dx. (note: dy/dx is the derivative of 2xsin(x)) (the limits are from 3pi/4 to 4pi/3). I get that but, is there a method to do it straight away from x cos (x)?

thanks
Yeah it is possible. It is called integration by parts which allows you to integrate expressions that are in the form f(x)g'(x). The formula for integration by parts is derived from the product rule.

However, this technique is not part of the methods nor spesh syllabus. It is most commonly taught in 1st/2nd year university maths. Since in methods it will be a multipart question to integrate this if you do it in one step you won't get all the marks.

#### cotangent

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##### Re: VCE Methods Question Thread!
« Reply #18361 on: January 23, 2020, 12:41:44 am »
0
Yeah it is possible. It is called integration by parts which allows you to integrate expressions that are in the form f(x)g'(x). The formula for integration by parts is derived from the product rule.

However, this technique is not part of the methods nor spesh syllabus. It is most commonly taught in 1st/2nd year university maths. Since in methods it will be a multipart question to integrate this if you do it in one step you won't get all the marks.

Thanks
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#### SmartWorker

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##### Re: VCE Methods Question Thread!
« Reply #18362 on: January 24, 2020, 10:35:59 am »
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Hey,
How do you do this question, it has something to do with quadratics.
Thanks.
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#### sk2000

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##### Re: VCE Methods Question Thread!
« Reply #18363 on: January 24, 2020, 11:00:25 am »
+5
Hey,
How do you do this question, it has something to do with quadratics.
Thanks.

Here, as we have speed, distance and time involved, we can come to the logical conclusion that the speed distance time formula must come into play somewhere.

So first let's set out what we know. We have the same distance (210km) and a normal speed and time, and a new speed and time.
Let's define the normal speed as s and the normal time as t.
We know that the new speed is 5km/h faster, so the new speed will be s + 5.
For this new speed, it takes one hour less. So the new time will be t - 1.

Now we have two sets of variables, so we can set up two equations and solve simultaneously.
We know that our normal distance of 210km = s * t.
We also know that, to get to the same distance of 210km, we can use our new time and speed. So instead of using s and t, our normal time and speed, we can use the new ones. Thus, 210km  = (s+5)*(t-1)

So we have two equations:
210 = s*t
210 = (s+5)*(t-1)

Solving simultaneously, you can simply plug it into your CAS or do it manually. You can do it manually but solving with the CAS gives you the variable s = 30, thus we can conclude that the normal speed is 30km/h.

Whenever you get a question like this, set out what you know, and what you have to figure out. This is usually the easiest way to guide you to the answer, and most of the time questions like these end up being simultaneous equations.

Let me know if you have any more questions
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#### SmartWorker

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##### Re: VCE Methods Question Thread!
« Reply #18364 on: January 24, 2020, 12:19:18 pm »
0
Here, as we have speed, distance and time involved, we can come to the logical conclusion that the speed distance time formula must come into play somewhere.

So first let's set out what we know. We have the same distance (210km) and a normal speed and time, and a new speed and time.
Let's define the normal speed as s and the normal time as t.
We know that the new speed is 5km/h faster, so the new speed will be s + 5.
For this new speed, it takes one hour less. So the new time will be t - 1.

Now we have two sets of variables, so we can set up two equations and solve simultaneously.
We know that our normal distance of 210km = s * t.
We also know that, to get to the same distance of 210km, we can use our new time and speed. So instead of using s and t, our normal time and speed, we can use the new ones. Thus, 210km  = (s+5)*(t-1)

So we have two equations:
210 = s*t
210 = (s+5)*(t-1)

Solving simultaneously, you can simply plug it into your CAS or do it manually. You can do it manually but solving with the CAS gives you the variable s = 30, thus we can conclude that the normal speed is 30km/h.

Whenever you get a question like this, set out what you know, and what you have to figure out. This is usually the easiest way to guide you to the answer, and most of the time questions like these end up being simultaneous equations.

Let me know if you have any more questions

Hey thanks mate! very well explained thank you.

I did have a few other questions: have attached them below

Questions 5c, 7, 9
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#### sk2000

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##### Re: VCE Methods Question Thread!
« Reply #18365 on: January 24, 2020, 12:56:18 pm »
+3
Hey thanks mate! very well explained thank you.

I did have a few other questions: have attached them below

5c. Whenever we see similar triangles, we know that the ratio of one side to the other of one of the triangles is going to be exactly the same for the other triangle. Using this rule, we know that AB:BC = DE:EF.

Once we know this rule, it becomes clear what we have to do. We only have x as a pronumeral and 3 as a constant, so there will be no need for simultaneous equations here. If AB:BC = DE:EF, just plug in these numbers as fractions. In other words, we set up an equation that shows the ratios are equivalent, using fractions.

So, x/(x+2) i.e AB/BC is equal to 3/x i.e DE/EF.

Now that we know x/(x+2) = 3/x, solve manually or plug it into the CAS (seems like most of these are CAS questions) to get x = (√33+3)/2

7. This questions may look complex from that first equation but it is actually pretty simple. We just need to set up an equation that links everything together. We know that AB = 8, so then we can assume that BP = the entire line (AP) - AB (8 ), or BP = AP - 8.
Now, it gives us AB * AP = BP2, which may look harder to solve because we only know AB, which is 8. However, if we can eliminate one of those pronumerals completely, we will only have one variable to solve making it much simpler.

This becomes clear when considering our linking expression, BP=AP-8. If we can express BP in terms of AP, this completely eliminates the need for BP and slims down the given expression AB * AP = BP2 to only one variable. So, substituting BP with AP-8, we get AB * AP = (AP-8)2. We know that AB=8, so 8 * AP = (AP-8 )2. Switch out AP with something simple like a for the CAS and we get 8*a=(a-8)2. Solve it for a and we get AP = 4(√5+3).

Now for BP, we already have an expression for that so its much simpler to solve. Just sub AP (which is 4(√5+3)) into BP = AP - 8 and we get BP = 4√5+4.

9. If I'm interpreting this question right, it is asking us to form two equations which will then be solved simultaneously.
Let the first number be a, and the second number be b. We don't know a or b, the only thing we know is that b is going to be 5 more or less than a, so we can assume that b = a+5.

Then, we know the sum of their squares is 100. In other words, a2 + b2 = 100.

So we have two equations:
a+5=b
a2 + b2 = 100.

Solve simultaneously using CAS and we get a=4.11438 and b=9.11438.

Hope this helps!
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#### SmartWorker

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##### Re: VCE Methods Question Thread!
« Reply #18366 on: January 28, 2020, 02:32:00 pm »
0
Hey,

What is the compound interest and simple interest formula and how do you derive it?

Thank you.
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#### colline

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##### Re: VCE Methods Question Thread!
« Reply #18367 on: January 28, 2020, 06:34:31 pm »
+4
Hey,

What is the compound interest and simple interest formula and how do you derive it?

Thank you.
Just wondering, why would you need this for methods?

But to answer your question, this is the simple interest formula (A = final amount, P = initial amount, r = rate, t = time); and this is the compound interest formula. I think the best way to figure out how they're derived is to start with some actual numbers (for example, an initial amount of $100) and calculate the interests to find patterns. 99.75: Methods [50] Literature [50] Further [48] Chemistry [40] Resources!! . Current: Veterinary Medicine, University of Melbourne #### SmartWorker • Forum Regular • Posts: 92 • Wake determined. Sleep Satisfied. • Respect: +15 ##### Re: VCE Methods Question Thread! « Reply #18368 on: January 29, 2020, 11:18:29 am » 0 Just wondering, why would you need this for methods? But to answer your question, this is the simple interest formula (A = final amount, P = initial amount, r = rate, t = time); and this is the compound interest formula. I think the best way to figure out how they're derived is to start with some actual numbers (for example, an initial amount of$100) and calculate the interests to find patterns.

As part of our holiday homework, they got us to work through some yr 10 questions and I found this. Thanks for the help.

I had another question: attached it below.
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#### MoonChild1234

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##### Re: VCE Methods Question Thread!
« Reply #18369 on: January 29, 2020, 11:24:14 am »
0
is addition of ordinates present on the methods exam? like, would we ever be required to graph it?

thanks!

#### jnlfs2010

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##### Re: VCE Methods Question Thread!
« Reply #18370 on: January 29, 2020, 11:40:34 am »
+4
As part of our holiday homework, they got us to work through some yr 10 questions and I found this. Thanks for the help.

I had another question: attached it below.

Use Pythagoras theorem as you have a right-angled triangle.
a2+b2=c2

(5x)2= (3x+1)2+ (5x-2)2

If you used the theorem you have an equation with only x (as all the side lengths are in terms of x.
Thus move everything to one side and solve the quadratic equation.
« Last Edit: January 29, 2020, 11:43:08 am by jnlfs2010 »
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#### SmartWorker

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##### Re: VCE Methods Question Thread!
« Reply #18371 on: January 29, 2020, 11:51:16 am »
0
Use Pythagoras theorem as you have a right-angled triangle.
a2+b2=c2

(5x)2= (3x+1)2+ (5x-2)2

If you used the theorem you have an equation with only x (as all the side lengths are in terms of x.
Thus move everything to one side and solve the quadratic equation.

Hey,

thanks for the help,

I have another question:
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#### colline

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##### Re: VCE Methods Question Thread!
« Reply #18372 on: January 29, 2020, 11:54:52 am »
+4
is addition of ordinates present on the methods exam? like, would we ever be required to graph it?

thanks!
Yes. This doesn't get tested very often but it's on the study design and anything on the study design could potentially pop up.

Additional of ordinates (including graphing) was on the 2018 VCAA exam.
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#### jnlfs2010

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##### Re: VCE Methods Question Thread!
« Reply #18373 on: January 29, 2020, 12:54:44 pm »
+4
Hey,

thanks for the help,

I have another question:

if AP=BQ=CR=DS, therefore, set all these things as variable x

Other lengths such as PB= 16-x
and QC=12-x
DR=16-x
AS= 12-x

Therefore find all the areas of the unshaded triangles in terms of x. The total amount of unshaded + shaded area= 16 x 12

You have a quadratic and then solve.
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#### SmartWorker

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##### Re: VCE Methods Question Thread!
« Reply #18374 on: February 03, 2020, 08:00:02 pm »
0
Hey guys,

How do you do this questions, I am really stuck on it. 😩 Thanks.

Edit: Nvm I found out how to do it. Thanks anyway. But I was stuck another question --> Factorise:a^2 −2ax+x^2 −4b^2
« Last Edit: February 03, 2020, 08:22:44 pm by SmartWorker »
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