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March 29, 2024, 10:46:23 am

Author Topic: VCE Specialist 3/4 Question Thread!  (Read 2164425 times)  Share 

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Special At Specialist

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Re: Specialist 3/4 Question Thread!
« Reply #120 on: January 01, 2012, 08:15:32 pm »
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The equation of a curve C is where k is a constant.

C does have a tangent parallel to the y axis.
Show that the y coordinate at the point of contact satisfies

I was thinking that since it is parallel to the y axis, it makes a vertical line, so I could make the substitution let dy/dx = infinity.
« Last Edit: January 01, 2012, 08:21:01 pm by Special At Specialist »
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TrueTears

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Re: Specialist 3/4 Question Thread!
« Reply #121 on: January 01, 2012, 08:19:50 pm »
+2
The equation of a curve C is where k is a constant.

C does have a tangent parallel to the y axis.
Show that the y coordinate at the point of contact satisfies
a few hints:

implicit differentiate.

now parallel to y axis implies "infinite" gradient (think about why?)

from then you can easily manipulate to satisfied the required expression
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Special At Specialist

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Re: Specialist 3/4 Question Thread!
« Reply #122 on: January 01, 2012, 08:22:55 pm »
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I think I get it. Can I let the denominator equal 0?
So 6y^2 + x = 0.
But I haven't got any y^6 terms...
Wait I think I know:
x = -6y^2
Now substitute into the original equation:
x^3 + xy + 2y^3 = k
(-6y^2)^3 + (-6y^2)y + 2y^3 = k
-216y^6 - 6y^3 + 2y^3 = k
-216y^6 - 4y^3 = k
-216y^6 - 4y^3 - k = 0
216y^6 + 4y^3 + k = 0

Got it. Thanks TrueTears and Paulsterio :)
« Last Edit: January 01, 2012, 08:27:29 pm by Special At Specialist »
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paulsterio

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Re: Specialist 3/4 Question Thread!
« Reply #123 on: January 01, 2012, 08:26:45 pm »
+1
notice that it says at the point of contact.

x = -6y^2 from what you had

sub that back into the eqn for the curve, you get the answer

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Re: Specialist 3/4 Question Thread!
« Reply #124 on: January 01, 2012, 08:29:09 pm »
+2




Now for the gradient to be parallel to the y-axis the gradient will be undefined, i.e. the bottom of the fraction will be equal to 0.


Sub this back into the original equation.




EDIT: Beaten, dam latex is slower.
« Last Edit: January 01, 2012, 08:33:18 pm by b^3 »
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paulsterio

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Re: Specialist 3/4 Question Thread!
« Reply #125 on: January 01, 2012, 08:31:44 pm »
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EDIT: Beaten, dam latex is slower.

1) - You made an error in the last line with y^6

2) - You always seem to get beaten cause of extravagant formatting/LATEX :P

b^3

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Re: Specialist 3/4 Question Thread!
« Reply #126 on: January 01, 2012, 08:33:02 pm »
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EDIT: Beaten, dam latex is slower.

1) - You made an error in the last line with y^6

2) - You always seem to get beaten cause of extravagant formatting/LATEX :P
Thanks Paul, and not always beaten, its just there was a lot of fractions in that one, too much {} and so on. It's not always slower, just depends on when you pick up the question. Plus LaTeX looks nicer :P
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Re: Specialist 3/4 Question Thread!
« Reply #127 on: January 01, 2012, 08:33:55 pm »
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Hence show that k≤(1/54)
I have no idea how to do this part either...
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Re: Specialist 3/4 Question Thread!
« Reply #128 on: January 01, 2012, 08:39:04 pm »
+1
let y^3 = Y, then 216Y^2+4Y+k=0, this must have one or two solutions for the case when there are two points of vertical tangents
So the discriminant must be greater than 0, do this and you should get k<=1/54
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paulsterio

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Re: Specialist 3/4 Question Thread!
« Reply #129 on: January 01, 2012, 09:02:25 pm »
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damn, that wasn't too tough, it just shows how much I've gotten rusty just two months after exams! damn!!

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Re: Specialist 3/4 Question Thread!
« Reply #130 on: January 01, 2012, 09:05:38 pm »
+1
wow 2 months? I forget everything after 1 week after uni exams.
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paulsterio

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Re: Specialist 3/4 Question Thread!
« Reply #131 on: January 01, 2012, 09:09:05 pm »
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wow 2 months? I forget everything after 1 week after uni exams.

hmm, I guess Year 12 is a little harder to forget ;P

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Re: Specialist 3/4 Question Thread!
« Reply #132 on: January 01, 2012, 09:15:46 pm »
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Find the possible values of k in the case where x = -6 is a tangent to C.

I spent so long on this question and I give up because I keep going around in circles and I don't know what to do :(
If it was any equation that didn't lead to a vertical line with infinite gradient, then I could've solved it so much quicker.

I think I know!
After literally 30 minutes of thinking, the idea occured to me:
Rather than letting dy/dx = infinity, I can simply let dx/dy = 0.
Actually I don't know if that helps me with this specific question :( It would've helped me in the last one though.

OMFG yes I finally found the answer after almost an hour!
Let dx/dy = 0
(6y^2 + x) / (-3x^2 - y) = 0
6y^2 + x = 0
y^2 = -x/6
y = +or- sqrt(-x/6)
when x = -6
y = 1 or y = -1
If y = 1:
216 + 4 + k = 0 so k = -220
If y = -1:
216 - 4 + k = 0 so k = -212
So k = -220 or k = -212
« Last Edit: January 01, 2012, 09:46:47 pm by Special At Specialist »
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Special At Specialist

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Re: Specialist 3/4 Question Thread!
« Reply #133 on: January 02, 2012, 10:25:15 pm »
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Find an antiderivative of cos(x) / (sin(x))^3

My first method was:
cos(x) / (sin(x))^3 = cot(x)*csc^2(x)
Let u = cot(x) and du/dx = -csc^2(x)
= Integral of -u du
= -1/2 u^2 + C
= -1/2 cot^2(x) + C

My second method was:
Let u = sin(x), du/dx = cos(x)
= Integral of u^(-3) du
= -1/2 u^(-2) + C
= -1/2 sin^(-2) + C
= -1 / (2*sin^2(x)) + C

-1/2 cot^2(x) is not the same as -1 / (2*sin^2(x)).

Why do I get a different answer???
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paulsterio

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Re: Specialist 3/4 Question Thread!
« Reply #134 on: January 02, 2012, 10:29:55 pm »
+1
I think they might actually be the same, sub something like pi/6 in and see if it comes to the same answer, if it does, try and manipulate it with theorems...it's got me stumped!