• Congratulations to the Class of 2024 on your results!
    Let us know how you went here
    Got a question about your uni preferences? Ask us here

HSC 2016 Maths Marathon (archive) (4 Viewers)

Status
Not open for further replies.

trecex1

Active Member
Joined
Mar 19, 2016
Messages
196
Gender
Male
HSC
2017
Re: HSC 2016 2U Marathon

Correct! I'll leave the algebraic exercise as an exercise for 2U students.
a/b < c/d so
ad < bc
ad + ab < ab + bc
a(b+d ) < b(a+c)
a/b < (a+c)/(b+d)

similarly bc +cd > cd + ad
c(b+c) > d(a+c)
c/d > (a+c)/(b+d)

 
Last edited:

leehuan

Well-Known Member
Joined
May 31, 2014
Messages
5,805
Gender
Male
HSC
2015
Re: HSC 2016 2U Marathon

a/b < c/d so
ad < bc
ad + ab < ab + bc
a(b+d ) < b(a+c)
a/b < (b+d)/(a+c)

similarly bc +cd > cd + ad
c(b+c) > d(a+c)
c/d > (a+c)/(b+d)

Nicely done, however the assumption that a/b < c/d isn't necessarily true. It may be possible that c/d < a/b.

This is a type of assumption that's made because in essence, the other case is proved in the exact same manner. Should state that "without loss of generality, assume that a/b < c/d"
 

Paradoxica

-insert title here-
Joined
Jun 19, 2014
Messages
2,556
Location
Outside reality
Gender
Male
HSC
2016
Re: HSC 2016 2U Marathon

a/b < c/d so
ad < bc
ad + ab < ab + bc
a(b+d ) < b(a+c)
a/b < (a+c)/(b+d)

similarly bc +cd > cd + ad
c(b+c) > d(a+c)
c/d > (a+c)/(b+d)

Consider the equivalent equation (logx)/x³ = k

The curve attains a global maximum with stationary point at x = ∛e, so the value of k must be less than 1/(3e)

However, the curve also never drops below 0 for all sufficiently large values of x, so k must be greater than 0.

Thus:

0 < k < 1/(3e)

 

AfroNerd

New Member
Joined
Feb 21, 2016
Messages
24
Gender
Male
HSC
2017
Re: HSC 2016 2U Marathon

Differentiate y=x^2 + bx + c and hence find values of b and c if the line 3x+y-5=0 is a normal to the curve at the point X(3,-1).
 
Last edited:

davidgoes4wce

Well-Known Member
Joined
Jun 29, 2014
Messages
1,877
Location
Sydney, New South Wales
Gender
Male
HSC
N/A
Re: HSC 2016 2U Marathon

Differentiate y=x^2 + bx + c and hence find values of b and c if the line 3x+y-5=0 is a normal to the curve at the point X(3,-1).
Have you got an answer for b and c?

Just want to check that you mean the equation 3x+y-5=0 is the normal equation?



The reason I ask is because if you substitute x=3



If the wording of the question is right, the gradient of the normal is -3, and the gradient of the tangent is

In which case,

 

AfroNerd

New Member
Joined
Feb 21, 2016
Messages
24
Gender
Male
HSC
2017
Re: HSC 2016 2U Marathon

Yep Im pretty sure 3x+y-5=0 is the normal.

Ty and also would that mean c=7
 
Last edited:

duxxx

New Member
Joined
Jan 19, 2015
Messages
13
Gender
Male
HSC
2015
Re: HSC 2016 2U Marathon

The tidal heights over the next five days for the Yangtze River mouth in Shanghai are shown in screenshot below.
Screen Shot 2016-10-14 at 2.13.05 PM.png
The heights in the tidal prediction chart show the height above the chart datum which is 7m. For example, a low tide of 0.5m is actually a depth of 7.5m.

a)Use the tidal data to synthesise a model for the depth of water at the Yangtze River mouth for all the data you have captured.
A horizontal translation is required in your model. Use the following information to assist you
The general trigonometric function
f(x) = asinb)x+c) + d
has an amplitude of a, a period of 2pi/b, a horizontal translation of c units to the left and a vertical translation of d units up.

b) Produce a single graph containing the actual data given in the tidal prediction chart and your model. Comment on the strength and limitations of the model.
 

pikachu975

Premium Member
Joined
May 31, 2015
Messages
2,739
Location
NSW
Gender
Male
HSC
2017
Re: HSC 2016 2U Marathon

The tidal heights over the next five days for the Yangtze River mouth in Shanghai are shown in screenshot below.
View attachment 33502
The heights in the tidal prediction chart show the height above the chart datum which is 7m. For example, a low tide of 0.5m is actually a depth of 7.5m.

a)Use the tidal data to synthesise a model for the depth of water at the Yangtze River mouth for all the data you have captured.
A horizontal translation is required in your model. Use the following information to assist you
The general trigonometric function
f(x) = asinb)x+c) + d
has an amplitude of a, a period of 2pi/b, a horizontal translation of c units to the left and a vertical translation of d units up.

b) Produce a single graph containing the actual data given in the tidal prediction chart and your model. Comment on the strength and limitations of the model.
Are you sure this is even 2 unit
 

duxxx

New Member
Joined
Jan 19, 2015
Messages
13
Gender
Male
HSC
2015
Re: HSC 2016 2U Marathon

Yeah it actually is, soooo hard.
 

duxxx

New Member
Joined
Jan 19, 2015
Messages
13
Gender
Male
HSC
2015
Re: HSC 2016 2U Marathon

Help me, I know you can you do extension 2!
 

InteGrand

Well-Known Member
Joined
Dec 11, 2014
Messages
6,109
Gender
Male
HSC
N/A
Re: HSC 2016 2U Marathon

The tidal heights over the next five days for the Yangtze River mouth in Shanghai are shown in screenshot below.
View attachment 33502
The heights in the tidal prediction chart show the height above the chart datum which is 7m. For example, a low tide of 0.5m is actually a depth of 7.5m.

a)Use the tidal data to synthesise a model for the depth of water at the Yangtze River mouth for all the data you have captured.
A horizontal translation is required in your model. Use the following information to assist you
The general trigonometric function
f(x) = asinb)x+c) + d
has an amplitude of a, a period of 2pi/b, a horizontal translation of c units to the left and a vertical translation of d units up.

b) Produce a single graph containing the actual data given in the tidal prediction chart and your model. Comment on the strength and limitations of the model.
Since you have the actual data, you can get a computer package to fit the data to that model, as well as produce the graphs.
 
Status
Not open for further replies.

Users Who Are Viewing This Thread (Users: 0, Guests: 4)

Top