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March 28, 2024, 11:22:28 pm

Author Topic: VCE Methods Question Thread!  (Read 4802315 times)  Share 

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swagsxcboi

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Re: VCE Methods Question Thread!
« Reply #4650 on: May 01, 2014, 10:00:04 pm »
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...I have no idea where to even begin
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M_BONG

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Re: VCE Methods Question Thread!
« Reply #4651 on: May 01, 2014, 10:01:44 pm »
+3
...I have no idea where to even begin

You have two pairs of coordinates

(x , x^2) and (5,0)

Sub all of them into the distance formula - the one we learned in Year 11 Methods (first chapter)

Differentiate the distance formula, set derivative to 0, solve for x. THEN sub that x-value into the distance formula. That value is the shortest distance.

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Re: VCE Methods Question Thread!
« Reply #4652 on: May 01, 2014, 10:05:40 pm »
+5
...I have no idea where to even begin
We have the point and a point on the parabola will be represented by , which we can also write as . We also know that the distance between two points and is given by , so you can get that function in terms of only, then differentiate and equate the derivative to zero to minimise the function and substitute that value of back in to find the minimum value of

EDIT: Beaten.
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Zealous

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Re: VCE Methods Question Thread!
« Reply #4653 on: May 01, 2014, 10:54:00 pm »
+12
Ok here's a question I need help with.....


Can anyone help pls!!!!!! I won't get any sleep tonight if I don't get this tbh -.-

"A massive conical vat, 10 metres in height, is filled to its capacity of 120pi cubic metres of pure glacier water

A massive crane raises the conical vat above an empty swimming pool at an exclusive resort. (diagram of swimming pool provided) in attachment

The glacier water is siphoned into the swimming pool.

If the water level in the conical vat decreases at a constant rate of 1 meter per second, at what level does the water in the swimming pool rise at the point in time when the water level in the vat is 6 metres deep?"

Final answer: 0.39m/s

Srs spent 2 hours on this and still can't figure it out... Our teacher is really, really sadistic...


EDIT: Photo cut off; it's 2m (the cut-off part)

Thanks!

Also, this is the only diagram the question provided, if you were wondering..

I believe I have a solution to it. I got 0.38m/s instead of 0.39m/s,  I'm not sure what happened there.

For the very last section where I divide by 108 metres squared, you can go into more algebraic proof as to why that works by setting up even more related rates for the rectangular prism. I just took a shortcut to save time.



[EDIT] Explanation of last step, we want to find dh/dt for the swimming pool:







Sub dh/dv into dh/dt:



Substitude dV/dt=40.715



Hopefully that doesn't make it more confusing!
« Last Edit: May 04, 2014, 09:17:03 am by Zealous »
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Irving4Prez

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Re: VCE Methods Question Thread!
« Reply #4654 on: May 03, 2014, 04:53:19 pm »
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Hey AN,

Can someone shed some light on how to determine the exact max value of f(x), where f(x)=(cos(x)-3/4)(cos(x)-1/4) where x belongs to all real. Thanks

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Re: VCE Methods Question Thread!
« Reply #4655 on: May 03, 2014, 04:58:49 pm »
+1
Hey AN,

Can someone shed some light on how to determine the exact max value of f(x), where f(x)=(cos(x)-3/4)(cos(x)-1/4) where x belongs to all real. Thanks

Let u = cos x
f(x) = (u - 3/4) (u-1/4) = (u-1/2 + 1/4) (u-1/2 -1/4) = (u-1/2)^2 - 1/16
so f(x) = (cos x - 1/2)^2 - 1/16

The min value is at cos x = 1/2 but you don't want that. You want the max value.
It's the same as trying to maximise f(u) = (u-1/2)^2 - 1/16 for -1<= u <=1
If you draw that graph, you'll find that the maximum value will occur at when u = -1. If there is no local maximum, the function maximum must occur at an endpoint.
So if u = -1, f(u) = (-3/2)^2 - 1/16 = 9/4 - 1/16 = 35/16
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Irving4Prez

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Re: VCE Methods Question Thread!
« Reply #4656 on: May 03, 2014, 09:27:40 pm »
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The min value is at cos x = 1/2 but you don't want that. You want the max value.
It's the same as trying to maximise f(u) = (u-1/2)^2 - 1/16 for -1<= u <=1
If you draw that graph, you'll find that the maximum value will occur at when u = -1. If there is no local maximum, the function maximum must occur at an endpoint.
So if u = -1, f(u) = (-3/2)^2 - 1/16 = 9/4 - 1/16 = 35/16

Hm, I've followed your steps but I'm struggling to grasp how you're able to replace cos(x) with u and find the maximum of f(x) = (cos x - 1/2)^2 - 1/16 based on f(u) = (u-1/2)^2 - 1/16 for -1<= u <=1

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Re: VCE Methods Question Thread!
« Reply #4657 on: May 03, 2014, 10:42:19 pm »
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For the love of anyone, please take me through this, especially HOW to break it down. It wording just seems that everything is everywhere and I can't seem to dissect it properly. I KNOW the concept of the stationary point and what not, but the question is just all bluh-bluh confusing.
I actually have no idea what I'm saying or talking about.

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Re: VCE Methods Question Thread!
« Reply #4658 on: May 03, 2014, 10:44:22 pm »
+4
Hm, I've followed your steps but I'm struggling to grasp how you're able to replace cos(x) with u and find the maximum of f(x) = (cos x - 1/2)^2 - 1/16 based on f(u) = (u-1/2)^2 - 1/16 for -1<= u <=1
We know that no matter what values of we put into , the smallest number we can get out is and the largest number out is . That is, . So we can replace the in our function with another variable that can take these same values, so that you're effectively getting the same values of back out. If you plot for those values you'll get the same values out if you were to plot for all values of . (see how the values are the same https://www.desmos.com/calculator/zmjvisslih). So you can then maximise your new function to maximise the old one. The only change will be that you'll be putting a different value into that function to get the maximum, so you'd find it at a different value of than , but if you were asked to find this value of , you know how the two are related (), and so can work it back.

Hope that helps :)
« Last Edit: May 04, 2014, 05:42:58 pm by b^3 »
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Re: VCE Methods Question Thread!
« Reply #4659 on: May 03, 2014, 11:03:12 pm »
+4
For the love of anyone, please take me through this, especially HOW to break it down. It wording just seems that everything is everywhere and I can't seem to dissect it properly. I KNOW the concept of the stationary point and what not, but the question is just all bluh-bluh confusing.
When it says 'touches', it means the line and the curve don't cross but have a point where they have the same value. So that is at that value of the line is tangential to the curve, that is the two have the same gradient at that value of . Rearranging the line gives a gradient of , so you can find the derivative and let it equal .

Since you have 4 unknowns, you need pieces of information. You have one above, two more will come from the points themselves (you know the value of for a given ). The last comes from the stationary point, as at that value of the gradient (so the derivative) will be zero. This should give you 4 equations which will allow you to solve the system.

« Last Edit: May 03, 2014, 11:19:48 pm by b^3 »
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Re: VCE Methods Question Thread!
« Reply #4660 on: May 04, 2014, 11:48:12 am »
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Could anyone tell me how this questions ends up as -1/6 ?
I got 6. Not too sure where I went wrong!
Thanks in advance.
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Re: VCE Methods Question Thread!
« Reply #4661 on: May 04, 2014, 11:53:38 am »
+3
Could anyone tell me how this questions ends up as -1/6 ?
I got 6. Not too sure where I went wrong!
Thanks in advance.
Factorise the denominator and take out the negative in the factor (x-5) and you should end up with -1/6 :)









Sub in x=5 and your answer is -1/6
« Last Edit: May 04, 2014, 11:55:52 am by IndefatigableLover »

Bluegirl

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Re: VCE Methods Question Thread!
« Reply #4662 on: May 04, 2014, 12:03:44 pm »
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Could someone please explain how to differentiate y= -4(3x-5)^2 +4 using the chain rule?
I'm doing something wrong.
Thanks

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Re: VCE Methods Question Thread!
« Reply #4663 on: May 04, 2014, 12:06:57 pm »
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Could someone please explain how to differentiate y= -4(3x-5)^2 +4 using the chain rule?
I'm doing something wrong.
Thanks

Let u=3x-5 du/dx=3
y=-4u^2+4 so dy/du=-8u

Hence: dy/dx= dy/du x du/dx
= -8u x 3 = -24u = -24(3x-5)
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Bluegirl

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Re: VCE Methods Question Thread!
« Reply #4664 on: May 04, 2014, 12:15:06 pm »
+1
Let u=3x-5 du/dx=3
y=-4u^2+4 so dy/du=-8u

Hence: dy/dx= dy/du x du/dx
= -8u x 3 = -24u = -24(3x-5)

Thankyou!