- Messages
- 165
- Reaction score
- 109
- Points
- 43
Q6, June 2008.
xy(x+y) = 2a^3
x^2y + xy^2 = 2a^3
Differentiating.
2xy + (x^2)(dx/dy) + y^2 + (2xy)(dy/dx) = 0
(x^2 + 2xy)(dy/dx) = - y^2 - 2xy
dy/dx = ( - y^2 - 2xy )/(x^2 + 2xy)
As the tangent is parallel to the x-axis, dy/dx=0.
dy/dx = ( - y^2 - 2xy )/(x^2 + 2xy)
0 = ( - y^2 - 2xy )/(x^2 + 2xy)
0 = - y^2 - 2xy
0 = -y ( y - 2x )
0 = y - 2x
y = -2x
Substitute '2x' in the place of 'y' in the equation of the curve.
x^2y + xy^2 = 2a^3
x^2(-2x) + x(-2x)^2 = 2a^3
-2x^3 + 4x^3 = 2a^3
2x^3 = 2a^3
x = a
Find the coordinate of 'y' now.
y = -2x
y = -2a
Therefore, the coordinates of the point on the curve at which the tangent is parallel to the x-axis are 'a, -2a'.
Q8(i), June 2008.
We've been given two hints in the question which'll help us prove that 'dy/dx = (1/2)(y^2)(cot x)'.
The first hint is that the area of the triangle PTN is equal to tan x.
Area of the triangle PTN = tan x
(1/2) x (TN) x (PN) = tan x
TN = (2 tan x)/PN
Second hint, the gradient of the curve at P is PN/TN.
dy/dx = PN/TN
dy / dx = PN/[(2 tan x)/PN]
dy / dx = (1/2)(PN^2)(1/tanx)
Recall the identity '(1/tan x) = cot x' and replace PN with 'y'.
dy / dx = (1/2)(PN^2)(1/tanx)
dy / dx = (1/2)(y^2)(cot x)
Therefore, dy / dx = (1/2)(y^2)(cot x).
Q8(ii), June 2008.
dy/dx = (1/2)(y^2)(cot x)
1/(y^2) dy = (1/2)(cot x) dx
(y^-2) dy = (1/2)(cos x / sin x) dx
Integrate both the sides.
(y^-2) dy = (1/2)(cos x / sin x) dx
-1/y = (1/2)(ln sin x) + c
Substitute 'π/6' and '2' in places of 'x' and 'y' respectively to calculate the value of the constant 'c'.
-1/2 = (1/2)(ln sin π/6) + c
-1/2 = (1/2)(ln 1/2) + c
-1/2 - (1/2)(ln 1/2) = c
Put this value of 'c' back into the integrated equation and obtain an expression of ';y' in terms of 'x'.
-1/y = (1/2)(ln sin x) + c
-1/y = (1/2)(ln sin x) -1/2 - (1/2)(ln 1/2)
-1/y = (1/2)(ln sin x) - (1/2)(ln 1/2) -1/2
-1/y = (1/2)(ln 2 sin x) - 1/2
-1/y = ln (2 sin x)^(1/2) - 1/2
-2/y = 2 ln (2 sin x)^(1/2) - 1
-2/y = 2 ln (2 sin x)^(1/2) - 1
-2/y = ln (2 sin x) - 1
-2/[ln (2 sin x) - 1] = y
Therefore, the expression of 'y' in terms of 'x' is 'y = -2/[ln (2 sin x) - 1]'
Q7(i), November 2008.
2x - y - 3z = 7
x + 2y + 2z = 0
(2).(1)
(-1).(2) = [(14)^(1/2)][3] cos x
(-3).(2)
2-2-6 = [(14)^(1/2)][3] cos x
(-6)/[3(14)^(1/2)] = cos x
122.3 = x
180 - 122.3 = 57.7
Therefore, the acute angle between the 2 planes is '57.7'.
Q7(ii), November 2008.
First of all, we'll find the common perpendicular of the two planes in question.
i j k
2 -1 -3
1 2 2
i ( -2 + 6 ) + j ( -3 -4 ) + k ( 4 + 1 )
4i - 7j + 5k
'4i - 7j + 5k'; This is the common perpendicular of the two planes or more precisely, the direction vector of their line of intersection.
Next, we'll be needing to find a point on this line.
(x).(2)
( y ).(-1) = 7
(z).(-3)
2x - y -3z = 7
(x).(1)
( y ).(2) = 0
(z).(2)
x + 2y + 2z = 0
As it's quiet difficult/impossible to find 3 variables by solving 2 simultaneous equations, we'll take any one of the 3 values ( x, y or z) to be equal to '0'. I have assumed that 'y=0', so we are left with:
2x -3z = 7 and x + 2z = 0
Solve simultaneously to find the values of 'x' and 'z'.
x + 2z = 0
x = -2z
2x -3z = 7
2(-2z) - 3z = 7
-7z = 7
z = -1
x = -2z
x= -2(-1)
z=2
Therefore, the coordinates of a point on this line are:
(2)
(0)
(-1)
Once we've found the coordinates, we can write the line's vector equation.
r = 2i – 1k + t(4i – 7j + 5k)
