SOS: detailed working out and comments needed (1 Viewer)

[ameh]

Prove x²+y²>=2xy, where x and y are integers

Consider the geometric series: 1 + lnx + (lnx)2 + ...
Find the values of x for which the limiting sum of this series exists.

Show that d/dx 3^(3x+1)=3^(3x+2)ln3

By using the substitution u = lnx or otherwize,

Integrate : 1/(xlnx) dx

Differentiate sin^3(2x-3)

Find, to the nearest degree, the acute angle between the curves y=4-x2 and y=4x-x2 at their point of intersection.

A 5m fence stands 4m from the wall of a house. A farmer wishes to reach a point A on the wall by the use of a ladder L that can reach from the ground outside the fence to the wall. Find the length of the shortest ladder that can reach from the ground outside the fence to the wall


Any help woud be appreciated. I've got the answers, just not sure why :?
[they're from the maths marathon ]

 

[ShaunSmith]

Prove x²+y²>=2xy, where x and y are integers

first check true for x=1, y=1 and x=2, y=2 i guess
2 = 2 and 8 = 8
then assue true for x = y = k
then prove for x = y = k+1
something like that

not sure.

 

[icycloud]

Prove x^2+y^2>=2xy, where x and y are integers
(x-y)^2 >= 0 {squares are always >= 0}
x^2 - 2xy + y^2 >= 0
x^2 + y^2 >= 2xy #

Consider the geometric series: 1 + lnx + (lnx)2 + ...
Find the values of x for which the limiting sum of this series exists.
a = 1
r = ln(x)

For limiting sum to exist, |r| < 1
|ln(x)| < 1
-1 < ln(x) < 1
e^(-1) < e^ln(x) < e^1
{x: 1/e < x < e} #

Show that d/dx 3^(3x+1)=3^(3x+2)ln3
Let y = 3^(3x+1)
ln(y) = ln(3^[3x+1])
= (3x+1) ln(3)
= 3x ln(3) + ln(3)

(1/y) (dy/dx) = 3ln(3)

LHS = d/dx [3^(3x+1)]
= dy/dx
= y * 3ln(3)
= 3^(3x+1) * 3 ln(3)
= 3^(3x+2) ln(3)
= RHS #

By using the substitution u = lnx or otherwise, Integrate : 1/(xlnx) dx
Let u = ln(x)
du = dx/x

Let I = ∫dx/{xln(x)}
= ∫du/u
= ln(u) + C
= ln[ln(x)] + C #

Differentiate sin^3(2x-3)
Let u = 2x - 3
du = 2 dx

y = sin^3 (2x-3)
= [sin(u)]^3

Let v = sin(u)
dv = cos(u) du

y = v^3
dy = 3v^2 dv
= 3[sin(u)]^2 cos(u) du
= 3[sin(2x-3)]^2 cos(2x-3) 2 dx

Thus, dy/dx = 6[sin(2x-3)]^2 cos(2x-3) #

Find, to the nearest degree, the acute angle between the curves y=4-x^2 and y=4x-x^2 at their point of intersection.
I believe this is 3U work. But,

Equation 1:
y = 4-x^2
y' = -2x

Equation 2:
y = 4x-x^2
y' = 4-2x

Intersection:
4-x^2 = 4x-x^2
x = 1
y = 3
Curves intersect at (1,3)

m₁ = -2(1) = -2
m₂ = 4-2(1) = 2

Using the angle between two lines formula:

Let x be the acute angle between the two tangents,
tan(x) = (m₁ - m₂) / (1+m₁m₂)
= (-2-2)/(1+(-2)(2))
= -4 / -3
= 4/3

x = 53 degrees 8 minutes (nearest minute) #

 

[ameh]

thanks for that icy cloud

Appreciate it

 

[icycloud]

thanks for that icy cloud

Appreciate it

No probs . And now the final question,

A 5m fence stands 4m from the wall of a house. A farmer wishes to reach a point A on the wall by the use of a ladder L that can reach from the ground outside the fence to the wall. Find the length of the shortest ladder that can reach from the ground outside the fence to the wall

This is hard to explain without a diagram, but...

Let x be the angle that the ladder makes with the horizontal (the ground). Now,

L = 4sec(x) + 5cosec(x)

We must minimize L,

dL/dx = 4sec(x)tan(x) - 5cosec(x)cot(x)

L is minimized/maximized when dL/dx = 0

Thus, we solve,

4sec(x)tan(x) = 5cosec(x)cot(x)
4sin(x) / cos^2(x) = 5cos(x)/sin^2(x)
4sin^3(x) = 5cos^3(x)
tan^3(x) = 5/4
tan(x) = CubeRoot(5/4)
x = arctan(CubeRoot(5/4))

(Don't forget to check that this is a minimum using a table of values or the second derivative. I won't put this step here.).

Substitute x into the original equation L, we get:

L = 4/cos(x) + 5/sin(x)
= 12.7m (1dp) #

 

[Riviet]

Wow, nice job icycloud. Answered pretty much all of them =D

 

[icycloud]

Wow, nice job icycloud. Answered pretty much all of them =D

Heh Maths is one of my favourite subjects .