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March 29, 2024, 08:20:51 am

Author Topic: VCE Methods Question Thread!  (Read 4802785 times)  Share 

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SmartWorker

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Re: VCE Methods Question Thread!
« Reply #19050 on: January 17, 2021, 05:28:56 pm »
+6
Thanks so much makes heaps more sense!
Theres 2 more questions i've tried to work out but can't get to an answer.

1. Find the value of m for which the following simultaneous equations have infinitely many solutions
mx−4y=6
3x−(m−1)y=2m

2. A function has rule y=ae^kt. Given that y=3 when t=2 and that y=6 when t=3,  find the values of a and  k.



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bluebird

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Re: VCE Methods Question Thread!
« Reply #19051 on: January 18, 2021, 11:44:23 am »
0
Hi, I hope this is the right place but I'm kinda stumped on these distance speed time questions. Any help is appreciated.
Thanks
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SmartWorker

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Re: VCE Methods Question Thread!
« Reply #19052 on: January 19, 2021, 05:09:30 pm »
+5
Hi, I hope this is the right place but I'm kinda stumped on these distance speed time questions. Any help is appreciated.
Thanks
-bluebird

Hey Bluebird!

To do these type of questions i would recommend making diagrams and logically drawing conclusions.

Qu 4:


Qu 5:

« Last Edit: January 19, 2021, 05:11:28 pm by SmartWorker »
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fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #19053 on: January 19, 2021, 11:13:00 pm »
+4
Hey is someone able to help me out with these q's.
1. The volume, Vcm3, of water in a tank at time t seconds is given by V(t)=5t^3+5t−8. What is the average rate of change of volume between t=1 and t=4.
-I first differentiated the equation to be 5t^2+5.
-Then I tried to sub t=1 into the equation which gave me 20. Then I subbed 4 into the equation and I got 245. So I subtracted 20 from 245 and it gave me 225cm3. But its wrong? Where've I gone wrong?

A golf ball is hit so that its height h(t) metres above the ground t seconds after it is hit is given by h(t)=10t(4−t). What is the maximum height reached.
I graphed the equation and found the max height to be 40 on the y-axis? idk if thats right?



For the first question, you're pretty damn close :D - the question asks for the average rate of change, which you haven't done. I'm not going to give anything away because you're literally that close.

For the second question, you're definitely right :)
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alexandra.vo

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Re: VCE Methods Question Thread!
« Reply #19054 on: January 21, 2021, 01:34:00 pm »
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Is someone able to check my answers for me plz!

fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #19055 on: January 21, 2021, 02:15:01 pm »
+4
Last two seem correct, first one is wrong.

When checking things like these, there are a few checks you can do to definitively check if your answer is wrong.
- Test a few points (not boundary points, points either clearly outside or inside the zone defined by the equation) and check if they satisfy the equation. Common points you might like to try are points on the x-axis, y-axis or most commonly the origin.
- In the case of multiple equations defining a composite area, do a bit of a simple sanity check and check that your answer actually defines an enclosed area that makes enough sense

Hope this helps :)
« Last Edit: January 21, 2021, 02:16:51 pm by fun_jirachi »
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alexandra.vo

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Re: VCE Methods Question Thread!
« Reply #19056 on: January 21, 2021, 02:21:06 pm »
0
Last two seem correct, first one is wrong.

When checking things like these, there are a few checks you can do to definitively check if your answer is wrong.
- Test a few points (not boundary points, points either clearly outside or inside the zone defined by the equation) and check if they satisfy the equation. Common points you might like to try are points on the x-axis, y-axis or most commonly the origin.
- In the case of multiple equations defining a composite area, do a bit of a simple sanity check and check that your answer actually defines an enclosed area that makes enough sense

Hope this helps :)

Thanks so much! Makes more sense, I've re-looked at the graph and I'm thinking B now for the first graph?

Jinju-san

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Re: VCE Methods Question Thread!
« Reply #19057 on: January 22, 2021, 01:05:30 pm »
0
Hey guys!
I was wondering if anyone could help me out with this question (I’m a bit confused as to how I’m supposed to go about solving it)…

Its question 15 from the picture below..

Thanks!

fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #19058 on: January 22, 2021, 01:11:30 pm »
+6
Thanks so much! Makes more sense, I've re-looked at the graph and I'm thinking B now for the first graph?

See, the thing is with B, x can't be greater than or equal to 2 because the point (3, 0) lies outside the shaded area. You're closer though - 2/3 equations satisfied as opposed to 1, but give it another shot :)

Hey guys!
I was wondering if anyone could help me out with this question (I’m a bit confused as to how I’m supposed to go about solving it)…

Its question 15 from the picture below..

Thanks!

They basically tell you that \(\int_2^a \frac{1}{2} \ln \left(\frac{x}{2}\right) \ dx = 1\) but with extra steps. Integrating the left hand side with respect to x will give you an equation in terms of a that you can solve.

Hope this helps both of you :D
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Jinju-san

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Re: VCE Methods Question Thread!
« Reply #19059 on: January 23, 2021, 02:22:05 pm »
0
They basically tell you that \(\int_2^a \frac{1}{2} \ln \left(\frac{x}{2}\right) \ dx = 1\) but with extra steps. Integrating the left hand side with respect to x will give you an equation in terms of a that you can solve.

Hope this helps both of you :D

Thanks fun_jirachi!
Also, I had another question… Sorry, I’m probably being really annoying bringing this up  :P
But I am confused as to how I am supposed to solve the question below…

Thanks!

p0kem0n21

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Re: VCE Methods Question Thread!
« Reply #19060 on: January 23, 2021, 02:55:49 pm »
+5
Thanks fun_jirachi!
Also, I had another question… Sorry, I’m probably being really annoying bringing this up  :P
But I am confused as to how I am supposed to solve the question below…

Thanks!

Hey! I'm going to assume that you're allowed to use CAS for this question. I'm also going to use 'u' to represent the mean (it looks like mu), and 's' to represent the standard deviation.

The first thing to note is that X is normally distributed. Whenever this happens, I always like to write X ~ N(u, s2). This reminds us that there are two variables which we can solve for, meaning that we probably need at least 2 equations.

We actually have these 2 equations! They are Pr(X<39.9161)=0.5789 [1] and Pr(X>38.2491)=0.4799 [2]. However, these are really hard to work with since we don't have the mean. This is where we have to convert these equations with z-scores, which provides us with a constant mean and standard deviation. We can then apply the inverse normal distribution CAS thingamajig to proceed. Hopefully these are enough hints to get you going, but I've provided the rest of the working below.

Spoiler

With [1], we can write Pr(Z< (39.9161-u)/s) = 0.5789. With [2], we can write Pr(Z > (38.2491-u)/s) = 0.4799. Note that the second equation should be written with a 'less than' sign because CAS (at least ti-nspire) likes using the 'less than' area. Rewritten, we have Pr(Z < (38.2491-u)/s) = 0.5201.

Now we can apply the CAS inverse normal distribution thingo. We keep the mean as 0 and the standard deviation as 1, but insert the 0.5789 and 0.5201 from above. This outputs two long decimal values, which I will just state as 0.20 (from 0.5789) and 0.05 (from 0.5201) (to 2 d.p.).

Recall how we calculated the z-scores from above in terms of u and s (these were (39.9161-u)/s and (38.2491-u)/s ). These are equivalent to the values obtained from InvNormal above (0.20 and 0.05 respectively). Since this now gives us two equations for two variables, we solve them as simultaneous equations. It would be much faster to solve these equations using CAS. Assuming I didn't mix up any values in my working, we should get s=11.21 (2 d.p.) and u=37.68 (also 2 d.p.).



Jinju-san

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Re: VCE Methods Question Thread!
« Reply #19061 on: January 23, 2021, 04:09:57 pm »
0
Hey! I'm going to assume that you're allowed to use CAS for this question. I'm also going to use 'u' to represent the mean (it looks like mu), and 's' to represent the standard deviation.

The first thing to note is that X is normally distributed. Whenever this happens, I always like to write X ~ N(u, s2). This reminds us that there are two variables which we can solve for, meaning that we probably need at least 2 equations.

We actually have these 2 equations! They are Pr(X<39.9161)=0.5789 [1] and Pr(X>38.2491)=0.4799 [2]. However, these are really hard to work with since we don't have the mean. This is where we have to convert these equations with z-scores, which provides us with a constant mean and standard deviation. We can then apply the inverse normal distribution CAS thingamajig to proceed. Hopefully these are enough hints to get you going, but I've provided the rest of the working below.

Spoiler

With [1], we can write Pr(Z< (39.9161-u)/s) = 0.5789. With [2], we can write Pr(Z > (38.2491-u)/s) = 0.4799. Note that the second equation should be written with a 'less than' sign because CAS (at least ti-nspire) likes using the 'less than' area. Rewritten, we have Pr(Z < (38.2491-u)/s) = 0.5201.

Now we can apply the CAS inverse normal distribution thingo. We keep the mean as 0 and the standard deviation as 1, but insert the 0.5789 and 0.5201 from above. This outputs two long decimal values, which I will just state as 0.20 (from 0.5789) and 0.05 (from 0.5201) (to 2 d.p.).

Recall how we calculated the z-scores from above in terms of u and s (these were (39.9161-u)/s and (38.2491-u)/s ). These are equivalent to the values obtained from InvNormal above (0.20 and 0.05 respectively). Since this now gives us two equations for two variables, we solve them as simultaneous equations. It would be much faster to solve these equations using CAS. Assuming I didn't mix up any values in my working, we should get s=11.21 (2 d.p.) and u=37.68 (also 2 d.p.).


Thank you so much p0kem0n21!!
Yep, I thought we were meant to use the z-score formulas.. But I didn’t know you could calculate them simply by putting mean as 0 and s as 1 in the invNorm function on the cas!!!
 

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Re: VCE Methods Question Thread!
« Reply #19062 on: January 26, 2021, 02:25:00 pm »
0
Hey would someone be able to help me with this question?:

"ABCD is a parallelogram with vertices A(2,2), B(1.5,4), and C(6,6), Find the equations of the diagonals AC and BD"


fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #19063 on: January 26, 2021, 02:32:03 pm »
+6
Hey :D

The position of C relative to B will be the same as the position of D relative to A ie. given that B and C have the coordinates \((x_1, y_1)\) and \((x_2, y_2)\) respectively, and that A has coordinates \((x_3, y_3)\) D will have the coordinates \((x_3 + (x_2-x_1), y_3 + (y_2 - y_1))\). Try adapting this to find the coordinates of D.

From here you can find the slope of lines AC and BD with \(m = \frac{y_2-y_1}{x_2-x_1}\), then use the point gradient form of a line to find the equations of the diagonals.

if you need further assistance please ask again! :D
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jasmine24

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Re: VCE Methods Question Thread!
« Reply #19064 on: January 27, 2021, 09:33:31 pm »
+1
Hi, i just wanted to clarify some things I'm not sure about. Is it that e^x cannot equal zero or must be greater than zero or both? Also, would anybody be able to explain why?
Thanks!