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May 16, 2022, 02:33:03 pm

Author Topic: VCE Methods Question Thread!  (Read 4151521 times)  Share 

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beep boop

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Re: VCE Methods Question Thread!
« Reply #19290 on: January 18, 2022, 09:55:43 pm »
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Can someone please help me w/ this?

What I've done so far :
b.i. I've tried deriving it. I know the amount of stationary points are 2 or 1.

b.ii I know for inverses must have be one to one in order to be an inverse. Which means there is only 1 turning point for the cubic.
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fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #19291 on: January 18, 2022, 10:06:19 pm »
+2
A question for you: is there anything that prevents \(p\) from having zero stationary points?

The picture more or less has the correct answer for ii).

Not particularly sure what else to say given it's not abundantly clear what you're confused by
 
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beep boop

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Re: VCE Methods Question Thread!
« Reply #19292 on: January 19, 2022, 06:14:36 pm »
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A question for you: is there anything that prevents \(p\) from having zero stationary points?

The picture more or less has the correct answer for ii).

Not particularly sure what else to say given it's not abundantly clear what you're confused by

I'm not sure what to put for values of m for both.
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Rose34

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Re: VCE Methods Question Thread!
« Reply #19293 on: January 20, 2022, 10:58:09 am »
+1
Can someone please help me w/ this?

What I've done so far :
b.i. I've tried deriving it. I know the amount of stationary points are 2 or 1.

b.ii I know for inverses must have be one to one in order to be an inverse. Which means there is only 1 turning point for the cubic.

What is part "a" of the question?is it related to part b? To find the "exact" values m can be, we must find the values of a, b and c, because the question is asking for actual value not "how many" stationary points there are. So for bi) after you derive the function and set it into zero(with the presence of a,b,c which we need to find) so we can solve for x and then sub the x values in the derived function so we can find m. Same with the second part, we should find the values of a,b,c,d first.

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Re: VCE Methods Question Thread!
« Reply #19294 on: January 20, 2022, 11:34:04 am »
+2
I'm not sure what to put for values of m for both.

i) m = 0, 1, or 2 since a cubic has at most 2 turning points.
ii) m = 0 or 1 since as you say, the function must be one to one (which isn't possible with two turning points)

What is part "a" of the question?is it related to part b? To find the "exact" values m can be, we must find the values of a, b and c, because the question is asking for actual value not "how many" stationary points there are. So for bi) after you derive the function and set it into zero(with the presence of a,b,c which we need to find) so we can solve for x and then sub the x values in the derived function so we can find m. Same with the second part, we should find the values of a,b,c,d first.

Unless I'm mistaken, m is just the number of stationary points - there's no indication that m is a particular stationary point. a, b, c, k are just arbitrary real numbers.
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beep boop

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Re: VCE Methods Question Thread!
« Reply #19295 on: January 20, 2022, 01:12:04 pm »
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What is part "a" of the question?is it related to part b? To find the "exact" values m can be, we must find the values of a, b and c, because the question is asking for actual value not "how many" stationary points there are. So for bi) after you derive the function and set it into zero(with the presence of a,b,c which we need to find) so we can solve for x and then sub the x values in the derived function so we can find m. Same with the second part, we should find the values of a,b,c,d first.

The part a of the question was a completely separate part and wasn't related to part b. I don't think it was asking to find values of a,b, c and k. I understand now, thank you.

i) m = 0, 1, or 2 since a cubic has at most 2 turning points.
ii) m = 0 or 1 since as you say, the function must be one to one (which isn't possible with two turning points)

Unless I'm mistaken, m is just the number of stationary points - there's no indication that m is a particular stationary point. a, b, c, k are just arbitrary real numbers.
I took a lot at it again and its most likely used to represent the real numbers in front of coefficients. Thanks a lot for your input fun_jirachi. It really helped!  ;D 8)

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Re: VCE Methods Question Thread!
« Reply #19296 on: January 25, 2022, 08:59:32 pm »
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hey, could someone please explain how to solve this?

find the equation of the straight line which passes through the point (2, 3) and is inclined at 30◦ to the positive direction of the x-axis

thanks in advance!

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Re: VCE Methods Question Thread!
« Reply #19297 on: January 25, 2022, 09:08:13 pm »
+3
hey, could someone please explain how to solve this?

find the equation of the straight line which passes through the point (2, 3) and is inclined at 30◦ to the positive direction of the x-axis

thanks in advance!

There is a formula, gradient of line = tan(x˚) where x is the angle between the line and the +ve direction of the x-axis.

So now, we know the equation is y = tan(30˚) x + c, and from your knowledge of circular function exact values, tan(30˚) = sqrt(3)/3.

Sub in the point (2,3) to find your c value (y-int)
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beep boop

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Re: VCE Methods Question Thread!
« Reply #19298 on: January 26, 2022, 08:12:15 pm »
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I've done part a.

Now I'm currently stuck on part b. Do I make a transformation matrix first?
Thanks in advance.
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Re: VCE Methods Question Thread!
« Reply #19299 on: January 26, 2022, 09:35:40 pm »
+6
I've done part a.

Now I'm currently stuck on part b. Do I make a transformation matrix first?
Thanks in advance.

The following transformations are occuring;
- A reflection in the \(y\) axis
- Translation of 2 units to the right (in the positive direction of the \(x\) axis.

Now can you apply these transformations to the function? In this case you would need to adjust the domain accordingly to the transformation and apply the transformation on both sub-functions of the function.

So knowing that the transformation maps \(f(x)\) to \(f(-(x-2))\).

\(f(-(x-2))=\begin{cases} 1-2\left(-\left(x-2\right)\right),\ -\left(x-2\right)\le2 \\ \frac{3}{4}\left(-\left(x-2\right)-2\right)^{2}-3,\ -\left(x-2\right)>2\end{cases}\)

Simplification will yield to our final answer

\(g(x)=\begin{cases} 2x-3,\ x\ge0 \\ \frac{3}{4}x^{2}-3,\ x\le0\end{cases}\)

beep boop

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Re: VCE Methods Question Thread!
« Reply #19300 on: February 03, 2022, 09:17:15 pm »
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Hi all.

I've done pre much everything but part b.ii.

Any help appreciated.

Ty in advance.

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Re: VCE Methods Question Thread!
« Reply #19301 on: February 03, 2022, 09:45:54 pm »
+5
Hi all.

I've done pre much everything but part b.ii.

Any help appreciated.

Ty in advance.

Hi! I don't have time to solve the entire question, but I can give you some pointers on where to start.
 Firstly, find the coordinates of that top corner of the truck, you know the height and width, and that the truck drives directly through the arcs middle. Therefore, that top right corner will be (1/2*2.7, 3.7).

You also know that any point along the parabola will have the coordinates (x, x^2+whatever number you found in the previous part).

Then, you need to substitute these coordinates into the distance formula, which will give you some distance function with x in it. Since it asks for minimum distance, you can diff the function, let it equal 0 and solve for the x value (use CAS). Finally, put this x value in back in the distance formula to get back a distance.

I hope this helps a bit! Let me know if you need me to clarify anything.

beep boop

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Re: VCE Methods Question Thread!
« Reply #19302 on: February 03, 2022, 10:07:59 pm »
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Then, you need to substitute these coordinates into the distance formula, which will give you some distance function with x in it. Since it asks for minimum distance, you can diff the function, let it equal 0 and solve for the x value (use CAS). Finally, put this x value in back in the distance formula to get back a distance.

I hope this helps a bit! Let me know if you need me to clarify anything.

So with the x value I found, do I sub this back into the original to find the y-coordinate?

Then use the corner of the truck coordinates and coordinates I just found into the distance formula?

Also tysm ArtyDreams, you've helped me a lot so far!
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Re: VCE Methods Question Thread!
« Reply #19303 on: February 03, 2022, 10:48:14 pm »
+1
So with the x value I found, do I sub this back into the original to find the y-coordinate?

Then use the corner of the truck coordinates and coordinates I just found into the distance formula?

Also tysm ArtyDreams, you've helped me a lot so far!

Yep, that will work! You can also just sub the x value you found straight into the distance formula you had before you differentiated it. (You'll the exact same outcome)

No problem, glad it helped!

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Re: VCE Methods Question Thread!
« Reply #19304 on: February 16, 2022, 10:00:14 pm »
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The cost of a taxi trip in a particular city is $4.00 up to and including 2 km. After 2 km the passenger pays an additional $2.00 per kilometre. Find the function f which describes this method of payment and sketch the graph of y=f(x), where x is the number of kilometres travelled. (Use a continuous model.) How do you do this?