Solved IGNOU Assignments and Question papers: 2020

Sunday, January 26, 2020

CURVE TRACING EXAMPLE 7

Given equation of the curve is

${x^3} + {y^3} = 3a{x^2}$

Symmetry

1.Replacing -y by y in the equation of the given curve.

$\Rightarrow {x^3} - {y^3} = 3a{x^2}$

which is not same as the equation of the given curve.
So the curve is not symmetric about the x-axis.

2.Replacing -x by x in the equation of the given curve/

$\begin{array}{l} {\left( { - x} \right)^3} + {y^3} = 3a{\left( { - x} \right)^2}\\ - {x^3} + {y^3} = 3a{x^2} \end{array}$

which is not same as the equation of the given curve.
So the curve is not symmetric about the y-axis neither in opposite quadrants.

3.Interchange y by x and x by y in the equation of the given curve.

${y^3} + {x^3} = 3a{y^2}$

which is not same as the equation of the given curve.
So, the curve is not symmetric about the line y=x.

4. Interchange y by -x and -y and x in the equation of the given curve.

$\begin{array}{l} {\left( { - y} \right)^3} + {\left( { - x} \right)^3} = 3a{\left( { - y} \right)^2}\\ \Rightarrow - \left( {{x^3} + {y^3}} \right) = 3a{y^2} \end{array}$

which is not same as the equation of the given curve.
So, the curve is not symmetric about the line y=-x.

Check for Origin

Replace (x,y) by (0,0) in the equation

${x^3} + {y^3} = 3a{x^2}$

${0^3} + {0^3} = 3a{\left( 0 \right)^2}$

$0 = 0$ .

The curve passes through the origin.

${x^3} + {y^3} - 3a{x^2} = 0$

Equating the lowest degree term of x to zero we get,

x=0 is the tangent at the origin and origin is the node there.

Points of intersection with x-axis and the y-axis

Put y=0 in

${x^3} + {y^3} - 3a{x^2} = 0$

we get the points of intersection with the x-axis as

(3a,0) and (0,0).

The curve intersects y-axis at the origin only as if we put x=0 in,

${x^3} + {y^3} - 3a{x^2} = 0$

we get,

y=0.

Asymptotes

Given equation of the curve is

${x^3} + {y^3} - 3a{x^2} = 0$ ._____________________(3)

Equating the highest degree term of y to zero.

1=0.

which is not true.
Thus there are no asymptotes parallel to y-axis.

Equating the highest degree term of x to zero.

1=0.

which is not true.
Thus there are no asymptotes parallel to x-axis.

Put y=mx+c in (3) to find the oblique asymptote.

${x^3} + {\left( {mx + c} \right)^3} - 3a{x^2} = 0$

${x^3} + {m^3}{x^3} + 3{m^2}{x^2}c + 3mc{x^2} - 3a{x^2} = 0$ ___________(4)

Equating the coefficient of highest power of x to zero,

${m^3} + 1 = 0$

m=-1.

Equating the coefficient of next highest power of x to zero,

3c-3a=0

c=a.

So the oblique asymptote is,

y=-x+a.

Domain

The given curve is defined $\forall \,\mathbb{R}$ .

Intervals of increase and decrease

Given equation of the curve is

${y^3} = 3a{x^2} - {x^3}$

$2{y^2}\frac{{dy}}{{dx}} = 6ax - 3a{x^2}$

The critical points are x=0 and x=2

y is increasing in the interval (0,2) and decreasing in the interval $\left( { - \infty ,0} \right)\, \cup \,\left( {2,\infty } \right)$ .

CURVE TRACING EXAMPLE 6

Given equation of the curve is

${y^2} = {x^2}\left( {\frac{{a - x}}{{a + x}}} \right)$

Symmetry

1.Replace y by -y in the equation of the given curve.

$\begin{array}{l} {\left( { - y} \right)^2} = {x^2}\left( {\frac{{a - x}}{{a + x}}} \right)\\ \Rightarrow {y^2} = {x^2}\left( {\frac{{a - x}}{{a + x}}} \right) \end{array}$

which is same as the equation of the given curve.
So, the curve is symmetric about the x-axis.

2.Replace x by -x in the equation of the given curve.

${y^2} = {\left( { - x} \right)^2}\left( {\frac{{a - \left( { - x} \right)}}{{a + \left( { - x} \right)}}} \right)$
$\Rightarrow {y^2} = {x^2}\left( {\frac{{a + x}}{{a - x}}} \right)$
which is same as the equation of the given curve.
So the curve is not symmetric about the y-axis and neither about the opposite quadrants.

3.Interchange y and x in the equation of the given curve.

${y^2} = {x^2}\left( {\frac{{a + x}}{{a - x}}} \right)$
$\Rightarrow {x^2} = {y^2}\left( {\frac{{a + y}}{{a - y}}} \right)$
which is not same as the equation of the given curve.
So the curve is not symmetric about the line y=x.

4.Interchange y and -x along with -y and x in the equation of the given curve.

${\left( { - x} \right)^2} = {\left( { - y} \right)^2}\left( {\frac{{a - y}}{{a + y}}} \right)$
$\Rightarrow {x^2} = {y^2}\left( {\frac{{a + y}}{{a - y}}} \right)$

which is not same as the equation of the given curve.
So the curve is not symmetric about the line y=-x.

Check for Origin

Replace (x,y) instead of (0,0) in the equation of the given curve.

${0^2} = {0^2}\left( {\frac{{a + 0}}{{a - 0}}} \right)$
$\Rightarrow 0 = 0$ .______________________(1)

The given equation of the curve can also be written as

$\begin{array}{l} {y^2}\left( {a + x} \right) = {x^2}\left( {a - x} \right)\\ {x^2}a - {y^2}a - {x^3} - {y^2}x = 0 \end{array}$

Equating the lowest degree terms to zero

${x^2} - {y^2} = 0$
$\left( {x - y} \right)\left( {x + y} \right) = 0$ _______________________(2)

From (1) and (2) the curve passes through the origin and (x-y=0) and (x+y=0) are the tangents thereat.

Point of Intersection with the x- axis and the y-axis

Put x=0 in the equation

${x^2}a - {y^2}a - {x^3} - {y^2}x = 0$ __________________________(3)

we get

y=0.

The curve meets y-axis at the origin only.

Putting y=0 in equation (3)

we get

${x^2}a - {x^3} = 0$

$x = 0,a$

The curve meets x-axis at (a,0) and at origin.

Asymptotes

Equate the highest degree term of y to zero in

${x^2}a - {y^2}a - {x^3} - {y^2}x = 0$

which leads us to

$\left( {x + a} \right) = 0$

is the asymptote parallel to y-axis.

Equate the highest degree term of x to zero in

${x^2}a - {y^2}a - {x^3} - {y^2}x = 0$

which leads us to

-1=0

which is not true.

So there are no asymptotes parallel to the x-axis.

Domain

The region in which the curve lies is defined as follows,

$y = x\sqrt {\frac{{a - x}}{{a + x}}}$

The curve is defined in the region

$\left( { - a,a} \right]$

Intervals of increase and decrease

Given equation of the curve is

${y^2} = {x^2}\left( {\frac{{a - x}}{{a + x}}} \right)$

Differentiating,

$2y\frac{{dy}}{{dx}} = \frac{{2{a^2}x - 2a{x^2} - 2{x^3}}}{{{{\left( {a + x} \right)}^2}}}$
$2y\frac{{dy}}{{dx}} = \frac{{ - 2x\left( {{x^2} + xa - {a^2}} \right)}}{{{{\left( {a + x} \right)}^2}}}$

In the domain the curve increases in the interval,

$\left( { - a,0.618a} \right)$

and decreases in the interval

$\left( {0.618a,a} \right)$ .

Saturday, January 25, 2020

CURVE TRACING EXAMPLE 5

Given equation of the curve is

${y^2} = \frac{{2x - 1}}{{{x^2} - 1}}$ ______________________(1)

Symmetry

1.Replace y by -y in the equation of the given curve.

$\begin{array}{l} {\left( { - y} \right)^2} = \frac{{2x - 1}}{{{x^2} - 1}}\\ \Rightarrow {y^2} = \frac{{2x - 1}}{{{x^2} - 1}} \end{array}$

which is same as the equation of the given curve.
So, the curve is symmetric about the x-axis.

2.Replace x by -x in the equation of the given curve.

$\begin{array}{l} {y^2} = \frac{{\left( { - 2x - 1} \right)}}{{{{\left( { - x} \right)}^2} - 1}}\\ \Rightarrow {y^2} = \frac{{ - \left( {2x + 1} \right)}}{{{x^2} - 1}} \end{array}$

which is not same as the equation of the given curve.
So, the curve is not symmetric about the y-axis and neither in opposite quadrants.

3.Interchange y and x in the equation of the given curve

${y^2} = \frac{{2x - 1}}{{{x^2} - 1}}$
${x^2} = \frac{{2y - 1}}{{{y^2} - 1}}$

which is not same as the equation of the given curve.
So the curve is not symmetric about the line y=x.

4.Interchange y and -x and x by -y in the equation of the given curve.

$\begin{array}{l} {\left( { - y} \right)^2} = \frac{{2\left( { - x} \right) - 1}}{{{{\left( { - x} \right)}^2} - 1}}\\ \Rightarrow {y^2} = \frac{{ - \left( {2x + 1} \right)}}{{{x^2} - 1}} \end{array}$

which is not same as the equation of the given curve.
So the curve is not symmetric about the line y=-x.

Check for Origin

Put (x,y)=(0,0) in

${y^2} = \frac{{2x - 1}}{{{x^2} - 1}}$

${\left( 0 \right)^2} = \frac{{2 * \left( 0 \right) - 1}}{{{{\left( 0 \right)}^2} - 1}}$
$\Rightarrow 0 = 1$

which is not true.

The curve does not pass through the origin.

Points of intersection with the x-axis and the y-axis

Put x=0 in

${x^2}{y^2} - {y^2} - 2x + 1 = 0$

$- {y^2} + 1 = 0$

$\left( {y + 1} \right)\left( {y - 1} \right) = 0$

The curve intersects the y-axis at (0,1) and (0,-1).

Put y=0 in

${x^2}{y^2} - {y^2} - 2x + 1 = 0$

$- 2x + 1 = 0$

The curve intersects the x-axis at $\left( {\frac{1}{2},0} \right)$ .

Asymptotes

${x^2}{y^2} - {y^2} - 2x + 1 = 0$

Equating the highest degree term of y to zero,

$\Rightarrow \left( {x - 1} \right)\left( {x + 1} \right) = 0$

(x-1=0) and (x+1=0) are the two asymptotes parallel to the y-axis.

Equating the highest degree term of x to zero,
y=0, is the asymptote parallel to x-axis.

For oblique asymptote put y=mx+c in
$\left( {{x^2} - 1} \right){y^2} - 2x + 1 = 0$
$\left( {{x^2} - 1} \right){\left( {mx + c} \right)^2} - 2x + 1 = 0$
$\left( {{x^2} - 1} \right)\left( {{m^2}{x^2} + 2mxc + {c^2}} \right) - 2x + 1 = 0$
${m^2}{x^4} + 2m{x^3}c + {c^2}{x^2} - {m^2}{x^2} - 2mcx - {c^2} - 2x - 1 = 0$

So the value of m comes out to be 0 therefore there are no obique asymptotes of the given curve.

Domain

For domain of definition we proceed as follows,

${y^2} = \frac{{2x - 1}}{{{x^2} - 1}}$

$y = \sqrt {\frac{{2x - 1}}{{{x^2} - 1}}}$

y is defined for $\left( { - 1,\frac{1}{2}} \right]\, \cup \,\left( {1,\infty } \right)$

Intervals of increase and decrease

Let us find intervals of increase and decrease in the region for which the curve is defined.

${y^2} = \frac{{2x - 1}}{{{x^2} - 1}}$

$2y\frac{{dy}}{{dx}} = \frac{{ - 2\left( {{x^2} - x + 1} \right)}}{{{{\left( {{x^2} - 1} \right)}^2}}}$

In the domain $\frac{{dy}}{{dx}}$ is decreasing for $\left( { - 1,\frac{1}{2}} \right]$ and increasing for $\left( {1,\infty } \right)$ .