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March 29, 2024, 06:19:24 pm

Author Topic: Maths 3A/3B  (Read 54317 times)  Share 

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jamonwindeyer

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Re: Maths 3A/3B
« Reply #75 on: September 25, 2016, 04:29:47 pm »
+1
Thank you for you help.
Could i please get help with part e and f?

For Part E, just find when the derivative of the function is equal to zero! That should tell you when deflation has stopped.

For Part F, \(a=0\). This is by definition, we can't have negative times! The upper limit \(b\) occurs when the turning point you find in Part E occurs, because after this, the deflation is negative (which doesn't make sense) ;D

anotherworld2b

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Re: Maths 3A/3B
« Reply #76 on: September 25, 2016, 05:30:12 pm »
0
how exactly can I find when the derivative of the function is equal to zero?
I am not sure how to do it
For Part E, just find when the derivative of the function is equal to zero! That should tell you when deflation has stopped.

For Part F, \(a=0\). This is by definition, we can't have negative times! The upper limit \(b\) occurs when the turning point you find in Part E occurs, because after this, the deflation is negative (which doesn't make sense) ;D

RuiAce

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Re: Maths 3A/3B
« Reply #77 on: September 25, 2016, 05:31:04 pm »
0
how exactly can I find when the derivative of the function is equal to zero?
I am not sure how to do it
You find dV/dt and let it equal to 0

jamonwindeyer

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Re: Maths 3A/3B
« Reply #78 on: September 25, 2016, 06:47:14 pm »
+1
how exactly can I find when the derivative of the function is equal to zero?
I am not sure how to do it

Pretty much!! So we find the derivative:



Then we set that equal to zero:



And there's your answer ;D
« Last Edit: September 25, 2016, 06:48:59 pm by jamonwindeyer »

anotherworld2b

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Re: Maths 3A/3B
« Reply #79 on: September 25, 2016, 08:42:58 pm »
0
Thank you very much for your help.
I also tried this question but I still get how to do it the secind part
« Last Edit: September 25, 2016, 08:44:53 pm by anotherworld2b »

RuiAce

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Re: Maths 3A/3B
« Reply #80 on: September 25, 2016, 08:46:09 pm »
+1
Thank you very much for your help.
I also tried this question but I still get how to do it



anotherworld2b

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Re: Maths 3A/3B
« Reply #81 on: September 29, 2016, 12:58:53 am »
0
Im sorry but im still confused how to do it?  :-\ could i also get help with this question?




« Last Edit: September 29, 2016, 10:33:33 am by anotherworld2b »

ml125

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Re: Maths 3A/3B
« Reply #82 on: September 29, 2016, 03:36:19 am »
+1
Im sorry but im still confused how to do it?  :-\


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anotherworld2b

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Re: Maths 3A/3B
« Reply #83 on: September 30, 2016, 11:47:25 am »
0
Oh i get it now :)
Also how would you do this question?




Syndicate

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Re: Maths 3A/3B
« Reply #84 on: September 30, 2016, 12:34:21 pm »
0
Oh i get it now :)
Also how would you do this question?

In order to locate the turning points of a cubic, you need to differentiate it's given equation.



- Now you basically need to solve for x



a) Since part a of the question wants you to justify that (5,-195) is a turning point of the given cubic, just sub in x = 5 into the equation.



Therefore (5, -195) is a minimum T.P. of the given cubic.

b) Well since only two x-values were discovered (and that there are two T.P.s), it means that the turning points displayed on the graphs are the only existing turning points of the function.

c) Using the same method utilised in part a of the question, you can determine the exact location of the maximum T.P (by subbing-in x= 3 into the equation).



Therefore, the exact location of the local maximum point is (-3, 61). Which seems correct, since the x coordinate has to be negative, and the y coordinate has to be positive (as seen on the graph).
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RuiAce

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Re: Maths 3A/3B
« Reply #85 on: September 30, 2016, 01:50:42 pm »
+3

anotherworld2b

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Re: Maths 3A/3B
« Reply #86 on: September 30, 2016, 04:58:29 pm »
0
Thank you for your help
I was wondering how to do this question as well
In order to locate the turning points of a cubic, you need to differentiate it's given equation.



- Now you basically need to solve for x



a) Since part a of the question wants you to justify that (5,-195) is a turning point of the given cubic, just sub in x = 5 into the equation.



Therefore (5, -195) is a minimum T.P. of the given cubic.

b) Well since only two x-values were discovered (and that there are two T.P.s), it means that the turning points displayed on the graphs are the only existing turning points of the function.

c) Using the same method utilised in part a of the question, you can determine the exact location of the maximum T.P (by subbing-in x= 3 into the equation).



Therefore, the exact location of the local maximum point is (-3, 61). Which seems correct, since the x coordinate has to be negative, and the y coordinate has to be positive (as seen on the graph).

Syndicate

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Re: Maths 3A/3B
« Reply #87 on: September 30, 2016, 06:47:28 pm »
+1
Thank you for your help
I was wondering how to do this question as well

Please excuse my handwriting  :P
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« Last Edit: September 30, 2016, 06:49:41 pm by Aaron »
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anotherworld2b

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Re: Maths 3A/3B
« Reply #88 on: October 04, 2016, 06:56:44 pm »
0
Thank you
Could i aldo get help with this question as well please?

Please excuse my handwriting  :P
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ml125

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Re: Maths 3A/3B
« Reply #89 on: October 04, 2016, 11:47:37 pm »
+1
Thank you
Could i aldo get help with this question as well please?
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« Last Edit: October 04, 2016, 11:54:51 pm by ml125 »
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