2u Mathematics Marathon v1.0 (1 Viewer)

[Riviet]

I don't really know what a secant is

A secant is a line that intersects a curve at two or more points.

In the context of circles, a secant is a line that cuts the circle at two points.

P.S Got a new question SoulSearcher?

 

[SoulSearcher]

I wonder how I managed to get through 2 unit maths last year without knowing that

Anyways, new question:
Let triangle ABC be right-angled at C, so a² + b² = c². Find the ratio c : a if:
i) b is the arithmetic mean of a and c
ii) b is the geometric mean of a and c

 

[Mc_Meaney]

New questions!

A particle's displacement is given by: x=3+2cos2t. Find

a) Initial displacement, velocity and acceleration
b) When the particle is first at rest

 

[Riviet]

New questions!

A particle's displacement is given by: x=3+2cos2t. Find

a) Initial displacement, velocity and acceleration
b) When the particle is first at rest

Spoiler

a) v=dx/dt=-4sin2t
a=dv/dt=-8cos2t
when t=0,
x=5, v=0 and a=-8.

b) when v=0,
t=0, pi
.'. particle is first at rest after pi seconds.

The question before this one can become the next question.

 

[Mc_Meaney]

Make it hard for me then. thanks riv

 

[Riviet]

Make it hard for me then. thanks riv

The question's in SoulSearcher's last post.

 

[Mc_Meaney]

I know. It's hard because I can't do it

 

[Sober]

Let triangle ABC be right-angled at C, so a² + b² = c². Find the ratio c : a if:
i) b is the arithmetic mean of a and c
ii) b is the geometric mean of a and c

Question (i)
(1) b = (a+c)/2
(2) a² + b² = c²

Sub (1) into (2)

a² + (a² + c² + 2ac)/4 = c²
5a² + c² + 2ac = 4c²
5a² + 2ac - 3c²= 0
(5a - 3c)(a+c) = 0

c:a = 5/3 (discard negative solution)

Question (ii)

(1) b = (ac)^1/2
(2) a² + b² = c²

Sub (1) into (2)

a² + ac - c² = 0

a = (-c +√(c² + 4c²)) / 2
a = c(-1+√5)/2
c:a = -2 / (1-√5) (discard negative solution)
c:a = (1+√5)/2 = φ

 

[SoulSearcher]

I would like a question that I can do at the dead of night Sober

 

[Sober]

I would like a question that I can do at the dead of night Sober

Sorry, I didn't have any time, I have my trials today, doing maths was my break from English study (can English really be that bad?). I will post one later today but anybody may feel free to beat me to it.

 

[SoulSearcher]

Ahh ok then, understandable.

can English really be that bad?

Yes it can be

 

[Sober]

Three circles, each of radius r are placed so that the center of any circle lies on the circumference of the other two circles. What is the area where all three circles overlap?

For those interested the name of the resulting shape is a reuleaux triangle.

 

[SoulSearcher]

Three circles, each of radius r are placed so that the center of any circle lies on the circumference of the other two circles. What is the area where all three circles overlap?

For those interested the name of the resulting shape is a reuleaux triangle.

For this question, we will have to consider 2 areas; the area of the equilateral triangle formed by joining the centre of each circle to each other, and the area of the segments that remain in the shaded area.

Spoiler

Since the circles have equal radii, when the centre of the circles are joined together, they will create an equilateral triangle, with side length r.
To find the area of the triangle, use 2 sides and the angle inbetween the sides, which will be pi/3.
A_Triangle = 1/2 * r * r * sin (pi/3)

= r² * √3/4
Now we have to find the area of the segments that are left over by the triangle. For each circle, the area of the sector is calculated as
1/2 * r² * @, where @ = pi/3
= 1/2 * r² * pi/3
= pi/6 * r²
And the area of the segment is equal to the area of the sector - the area of the triangle made by joining the centres of the circles together, i.e.
A_Segmentpi/6 * r² - r² * √3/4

= r²(pi/6 - √3/4)
Now we can find the total area of the shaded triangle by simply adding those 2 areas together.
A_{Shaded Area} = 3 * A_Segment (as there are 3 segments of equal area in the shaded area) + A_Triangle

= 3r²(pi/6 - √3/4) + r² * √3/4
= r²[3pi/6 - 3√3/4 + √3/4]
= r²[pi/2 - 2√3/4]
= r²[pi/2 - √3/2]
= 1/2 * r² * (pi - √3)

I'll post a question later today.

 

[pLuvia]

Spoiler

(a)
x=(t³+1)⁶
v=18t²(t³+1)⁵
a=36t(t³+1)⁵+270t⁴(t³+1)⁴
Initial when t=0
x=1
v=0

(b)
When t=2
a=32600000 (3 sig fig)

(c)
When x=0
0=(t³+1)⁶
0=t³+1
-1=t³
t=-1
t has to be positive
Hence particle is never at origin

 

[SoulSearcher]

Here's my promised question:

Show that a rectangle with fixed perimeter has its shortest diagonal when it is a square

 

[airie]

Spoiler

Let the dimensions of the rectangle be a*b,
Then the diagonal is sqrt(a² + b²), which is the least when a² + b² is the least.
And since (a + b)² is constant ie. a² + b² + 2ab is constant,
a² + b² is the least when ab is at its maximum.

Since (a+b)/2 >= sqrt(ab) by AM-GM where equality occurs when a=b,
Therefore ab is at most when a = b ie. the rectangle is a sqaure.

 

[SoulSearcher]

I think you can post up a question now airie

 

[airie]

Oh yeah. Forgot about that. Eep, can't really think of one...Can someone else please post up a question instead? (note to self: need to remember not to answer a question when I don't have one ready as the next >.o)

 

[SoulSearcher]

I have one.
The temp T*Celsius of a compount in a science experiment is increasing exponentially over t minutes such that dT/dt = 0.03T. The temp was initially 23*C
(a) By what percentage is the tempterature increasing each minute?
(b) What is the temp after 10 minutes?
(c) When does the temp reach 100*C

Spoiler

General equation of exponential growth is
T = Ae^kt, and
dT/dt = kT
Compare the rate you have been given to the general equation:
dT/dt = 0.03T
Therefore,
0.03T = kT
.'. k = 0.03
Since T = 23 when t = 0,
23 = Ae⁰
.'. A = 23
Therefore,
T = 23e^0.03t
1) From dT/dt = 0.03T, we can say that the temperature is increasing at about 3% per minute.
2) After the first 10 minutes,
T = 23e^0.30
= 31.04675257 ....
= 31.05 degrees to 2 decimal places
3) Find t when T is 100 degrees
100 = 23e^0.03t
100/23 = e^0.03t
0.03t = ln(100/23)
t = ln(100/23) / 0.03
= 48.989199...
= 49 minutes, to the nearest minute.
I might be wrong though, haven't done these types of questions in a long time.

 

[SoulSearcher]

Ok:

Two circles of radii 2cm and 3cm touch externally at P. AB is a common tangent. Calculate, in cm² correct to two decimal places, the area of the region bounded by the tangent and the arcs AP and BP