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March 19, 2024, 03:59:47 pm

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2160799 times)  Share 

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tinagranger

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Re: Specialist 3/4 Question Thread!
« Reply #8790 on: August 13, 2017, 10:51:28 pm »
0
Thanks so much!!

I have two more questions (sorry have been catching up on set work all day) - questions 14 and 15 in the screenshot.

14) The diagram I drew didn't give me the right answer when I used Lami's theorem. Could you talk through your thought process when drawing the diagram from the explanation?
I got P = 4.295 but the ans said 42.09, and I got T = 5.19 but the ans is 50.82N.

15) I drew the diagram but don't know how to find the tensions given that we only have one force, and 2 side lengths. (I prefer Lami's theorem/the triangle method rather than resolving)

Thanks in advance  :D
« Last Edit: August 13, 2017, 11:01:52 pm by tinagranger »
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Shadowxo

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Re: Specialist 3/4 Question Thread!
« Reply #8791 on: August 14, 2017, 08:20:39 am »
+4
Thanks so much!!

I have two more questions (sorry have been catching up on set work all day) - questions 14 and 15 in the screenshot.

14) The diagram I drew didn't give me the right answer when I used Lami's theorem. Could you talk through your thought process when drawing the diagram from the explanation?
I got P = 4.295 but the ans said 42.09, and I got T = 5.19 but the ans is 50.82N.

15) I drew the diagram but don't know how to find the tensions given that we only have one force, and 2 side lengths. (I prefer Lami's theorem/the triangle method rather than resolving)

Thanks in advance  :D
14.
You know the force is at an angle of 75° with the downwards vertical, so that's a bit below horizontal
The length of the string is 2.5m and it is a horizontal distance of 2m from the starting point. This means it's a vertical distance of 1.5m from the top (3,4,5 triangle)
You also have the weight directly downwards of 2g N
From there you should have all the information needed - could you post your working so I can see where you went wrong? (Don't have time to do a full solution atm) :)
Also for Q15, I can't see it attached. What question are you referring to? :)
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humblepie

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Re: Specialist 3/4 Question Thread!
« Reply #8792 on: August 14, 2017, 09:37:22 pm »
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(please refer to attachments) - I get that the answer is y=3cosx, but why is the domain from -pi to 0? And is it necessary to state the domain?

Thanks in advance :)
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VanillaRice

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Re: Specialist 3/4 Question Thread!
« Reply #8793 on: August 14, 2017, 09:59:28 pm »
+5
(please refer to attachments) - I get that the answer is y=3cosx, but why is the domain from -pi to 0? And is it necessary to state the domain?

Thanks in advance :)
An intermediate step you would get while solving the differential equation should be:

Since this is an inverse trig function, you need to take note of the possible values of x (remember that inverse trig functions must be one to one). This is what they've used to determine the domain in the second image.
For a solution such as this, I would say stating the domain is relatively important, as it is indeed restricted - due to the inverse sine function. This detail may have been lost/missed when you took the sine of both sides in order to cancel out the arcsin (i.e. when you tried to get y as the subject).

Hope this helps  :)
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humblepie

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Re: Specialist 3/4 Question Thread!
« Reply #8794 on: August 14, 2017, 10:05:55 pm »
+1
Thank-you so much VanillaRice! :D
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sonnyangel

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Re: Specialist 3/4 Question Thread!
« Reply #8795 on: August 17, 2017, 08:23:09 pm »
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I got the maximal height reached but I don't know how to go about finding the speed when next at the point of projection. I thought it would be when x=0 but that doesn't give the right answer (if you could explain why, that would really help)
Thanks in advance!
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Shadowxo

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Re: Specialist 3/4 Question Thread!
« Reply #8796 on: August 17, 2017, 09:41:00 pm »
+5
I got the maximal height reached but I don't know how to go about finding the speed when next at the point of projection. I thought it would be when x=0 but that doesn't give the right answer (if you could explain why, that would really help)
Thanks in advance!
You'll need to make a separate equation to the one you used to get the maximal height, as air resistance will be in the opposite direction (upwards instead of downards). From this you should be able to solve. If you still have problems post your working and we can see where you went wrong!  :)
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humblepie

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Re: Specialist 3/4 Question Thread!
« Reply #8797 on: August 18, 2017, 12:17:23 pm »
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Hi! For the differential equation dy/dx=y2 given y=1 when x=1, I understand that the solution is 1/(2-x), but could someone explain to me why x<2?

Also, for the attached question, why does the absolute value sign appear in the sixth line (in attachment b)?

Thank-you in advance :)
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exit

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Re: Specialist 3/4 Question Thread!
« Reply #8798 on: August 18, 2017, 12:26:55 pm »
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Hi! For the differential equation dy/dx=y2 given y=1 when x=1, I understand that the solution is 1/(2-x), but could someone explain to me why x<2?

Also, for the attached question, why does the absolute value sign appear in the sixth line (in attachment b)?

Thank-you in advance :)

For the first question, x=1 when y=1. This means x<2 and as the function exists at x=1 and thus cant exist x>2

Second question, all logs you make in Spesh have a modulus sign as it could either 'go left' or 'go right' to the asymptote. You have to figure out whether you need to insert a negative into the log components or leave it be if you want to get rid of the modulus sign. This is depending on the domain in which the function is defined - they can give you an initial condition to assist with this

Good luck :)
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gnaf

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Re: Specialist 3/4 Question Thread!
« Reply #8799 on: August 18, 2017, 04:33:33 pm »
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In the vcaa 2006 exam 2, how do you do 4d) (slope field question)?

humblepie

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Re: Specialist 3/4 Question Thread!
« Reply #8800 on: August 18, 2017, 07:57:33 pm »
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For the first question, x=1 when y=1. This means x<2 and as the function exists at x=1 and thus cant exist x>2

Second question, all logs you make in Spesh have a modulus sign as it could either 'go left' or 'go right' to the asymptote. You have to figure out whether you need to insert a negative into the log components or leave it be if you want to get rid of the modulus sign. This is depending on the domain in which the function is defined - they can give you an initial condition to assist with this

Good luck :)

Thanks for the explanation! Hmmm... I still don't really get the second question; you don't usually put modulus signs when you simply rearrange the equation :/ Or do you have to in this case since there is a "c"? 
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Shadowxo

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Re: Specialist 3/4 Question Thread!
« Reply #8801 on: August 18, 2017, 08:29:12 pm »
+2
Thanks for the explanation! Hmmm... I still don't really get the second question; you don't usually put modulus signs when you simply rearrange the equation :/ Or do you have to in this case since there is a "c"? 

For your first question with dy/dx = y2, I think it should be x≠2. If you graph the function, the gradient is always positive (and y2) so I don't think there's any real need to only take x<2 unless there was some other condition such as it having to be continuous

For the second question, you shouldn't need to put in modulus signs, from previous working x-c is negative so c-x should be positive. Modulus signs should usually only appear in integration (for 1/x or similar). I think it's probably an error, as I don't believe c-x < 0 would work.
Also when I graphed the solution they gave, half the graph had a negative gradient which is not possible as e^anything > 0. Without the modulus signs it's always positive though.
« Last Edit: August 18, 2017, 08:35:33 pm by Shadowxo »
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humblepie

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Re: Specialist 3/4 Question Thread!
« Reply #8802 on: August 18, 2017, 09:33:22 pm »
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For your first question with dy/dx = y2, I think it should be x≠2. If you graph the function, the gradient is always positive (and y2) so I don't think there's any real need to only take x<2 unless there was some other condition such as it having to be continuous

For the second question, you shouldn't need to put in modulus signs, from previous working x-c is negative so c-x should be positive. Modulus signs should usually only appear in integration (for 1/x or similar). I think it's probably an error, as I don't believe c-x < 0 would work.
Also when I graphed the solution they gave, half the graph had a negative gradient which is not possible as e^anything > 0. Without the modulus signs it's always positive though.

Ah that makes sense :) Thank-you Shadowxo!
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Willba99

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Re: Specialist 3/4 Question Thread!
« Reply #8803 on: August 20, 2017, 11:36:52 am »
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Hey guys, is there any way to derive something in the form of 1/(a-bx^2) by hand? I've tried putting it into my CAS and i get an inverse tan with negative square roots so this seems unlikely.

Thanks in advance!!
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Shadowxo

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Re: Specialist 3/4 Question Thread!
« Reply #8804 on: August 20, 2017, 11:52:51 am »
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Hey guys, is there any way to derive something in the form of 1/(a-bx^2) by hand? I've tried putting it into my CAS and i get an inverse tan with negative square roots so this seems unlikely.

Thanks in advance!!
If you want to find its integral, it's not a standard integral so I believe you'd have to do partial fractions and integrate it that way.
If you want to find the derivative, just use the chain rule
« Last Edit: August 20, 2017, 12:28:48 pm by Shadowxo »
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