List of Topics (M 2)

Unit 1 Differential Equations

Unit 2 Applications of Differential Equations

Unit 3 i Fourier Series

Unit 3 ii Reduction Formulae

Unit 3 iii {\beta} and {\gamma} Functions

Unit 4 i DUIS

Unit 4 ii Error Function

Unit 4 iii Curve Tracing and Rectification

Unit 5 i Three Dimensional Geometry

Unit 5 ii Spheres, Cones and Cylinders

Unit 6 i Double Integrals

Unit 6 ii Triple Integrals

Unit 6 iii Applications of Multiple Integration



M II May 2016

Q.5 a) Show that the plane {2x-2y+ z+12=0} touches the sphere {x^2+y^2+z^2-2x-4y+2z-3=0}. Also find the point of contact.

Solution : The sphere has center at the point {(1,2,-1)} and its radius is equal to

{\sqrt {(1)^2 + (2)^2 + (-1)^2 - (-3)} = \sqrt {1+4+1+3} = \sqrt {9} = 3 \ units}

Length of perpendicular from the center to the plane will be

{\begin {vmatrix} \frac {2(1) -2 (2)+ 1(-1) =12}{\sqrt {2^2 + (-2)^2 + 1(1)^2}} \end {vmatrix} = \begin {vmatrix} \frac {2-4-1+12}{\sqrt {9}} \end {vmatrix} = \frac {9}{3} = 3 \ units}

Thus the length of perpendicular from the center to the plane is equal to the radius of sphere. This proves that the plane touches the sphere.

The equation of line through the center and normal to the plane is given by

{\frac {x-1}{2} = \frac {y-2}{-2} = \frac {z+1}{1} = \lambda}


{x = 2 \lambda + 1, \ y= -2 \lambda +2, \ z= \lambda-1}

If this point lies on the plane, its coordinates must satisfy the equation of the plane. Therefore,

{2(2 \lambda+1)-2 (-2 \lambda+2) + 1 (\lambda-1)+12 =0}


{4 \lambda +2+4 \lambda -4+ \lambda -1+12=0, \ i.e. \ 9 \lambda+9 =0, \ i.e. \ \lambda =-1}

Substituting {\lambda} in the coordinates, we get the point of contact as

{(x,y,z) = \big ( 2(-1)+1, \ -2(-1)+2 , \ (-1)-1 \big ) = \big (-1,4,-2 \big )}

Q.5 b) Find the equation of right circular cone passing through {(2,-2,1)} with vertex at origin and axis parallel to the line

{\frac {x-2}{5} = \frac {y-1}{1} = \frac {z+2}{1}}

Solution : Let the vertex {(0,0,0)} be denoted by {V} and the point {(2,-2,1)} by {A}. So,

{\vec {VA} = 2 \hat i - 2 \hat j + \hat k}

Since parallel lines have same direction ratios, the axis of the cone will also have direction ratios {5,1,1}. Let {\vec d} be the vector along the axis, then

{\vec d = 5 \hat i + \hat j + \hat k}

Let {\alpha} be the semi-vertical angle of cone. By the definition of dot product of vectors,

{\vec {VA} \cdot \vec {d} = |\vec {VA}| |\vec d| cos (\alpha)}

This gives

{(2 \hat i - 2 \hat j + \hat k) \cdot (5 \hat i + \hat j + \hat k) = \sqrt {2^2 + (-2)^2 + (1)^2} \sqrt {(5)^2 +(1)^2 + (1)^2} \ cos (\alpha)}

{10-2+1 = \sqrt {9} \sqrt {27} cos (\alpha)}

{cos (\alpha) = \frac {10-2+1}{ \sqrt {9} \sqrt {27}} = \frac {9}{3 \times 3 \sqrt 3} = \frac {1}{\sqrt {3}}}

Having obtained {cos (\alpha)}, we can go for the equation of right circular cone. Note that knowing {cosine} of the semi-vertical angle is sufficient.

Let {P(x,y,z)} be any point on the cone. Then,

{\vec {VP} = x \hat i + y\hat j + z \hat k}

Using the dot product of vectors, we can write

{\vec {VP} \cdot \vec d = |\vec {VP}| |\vec d| cos (\alpha)}

{(x \hat i + y \hat j + z \hat k) \cdot (5 \hat i + \hat j + \hat k) = \sqrt {x^2 + y^2 + z^2} \sqrt {(5)^2 +(1)^2 + (1)^2} \times \frac {1}{\sqrt 3}}

Taking the dot product and squaring both sides, we get

{(5x+y+z)^2 = (x^2+y^2+z^2) \times 27 \times \frac 1 3 = (x^2+y^2+z^2) \times 9}

Expanding the bracket,

{25x^2+y^2+z^2+10xy+2yz+10zx = 9x^2+9y^2+9z^2}

On further simplifying, the required equation of cone is

{16x^2-8y^2-8z^2+10xy + 2yz + 10zx =0}

Q.5 c) Find the equation of right circular cylinder, whose axis is {x=2y=-z} and radius is {4} units.

Solution : The axis of cylinder is {x=2y=-z}. Making the adjustments to bring these equations into general equations of a line in 3 dimensions, we get,

{\frac {x}{1} = \frac {y}{1/2} = \frac {z}{-1}}

The direction ratios of the axis are {1, \frac 1 2} and {-1}. Now, we need to find a point on the axis. (We can arbitrarily choose any value of {x} and get {y} and {z} from above equations. The most trivial value of {x} is {0}. This gives {y=0} and {z=0}.) So, we got a point on the axis, which is the origin itself, {O(0,0,0)}.

Let {P(x,y,z)} be any point on the cylinder. We drop a perpendicular from {P} on the axis. Let the foot be {M}. {PM} has length {4} units, because it is the radius of the cylinder.

{\Delta OPM} is a right angled triangle, with {\angle M = 90^o}. Let {\angle POM} be {\theta}. By property of right angled triangles,

{OP^2 = OM^2 + PM^2}

{\therefore x^2+y^2+z^2 = OM^2 + 4^2}

{cos (\theta) = \frac {OM}{OP}}, so {OM = OP cos (\theta)}. Once we get {OM} in terms of {x,y,z}, we can go ahead with above equation.

Let {\vec d} be the vector along the axis of the cylinder (which has direction ratios {1, \frac 1 2, -1}). Multiplying each direction ratio by same non-zero number will still give us direction ratios. So, {\vec d = 2 \hat i + \hat j - \hat k} (We multiplied by {2}, so the fraction {1/2} is gone, which could have possibly made the task tougher.)

The unit vector along the axis,

{\hat d = \frac {\vec d}{|\vec d|} = \frac {2 \hat i + \hat j - \hat k}{\sqrt {2^2 + 1^2 + (-1)^2}} = \frac 2 3 \hat i + \frac 1 3 \hat j - \frac 2 3 \hat k}.

We now consider the dot product {\vec {OP} \cdot \hat d}.

{\vec {OP} \cdot \hat d = |\vec {OP}| |\hat d| cos (\theta) = OP \times 1 \times cos (\theta) = OP \ cos (\theta) = OM}

{\therefore \ OM = \vec {OP} \cdot {\hat d} = (x \hat i + y \hat j + z \hat k) \cdot \left (\frac 2 3 \hat i + \frac 1 3 \hat j - \frac 2 3 \hat k \right ) = \frac {2x}{3} + \frac {y}{3} - \frac {2z}{3}}

Using this in {OP^2 = OM^2 + PM^2},

{x^2+y^2+z^2 = \left (\frac {2x}{3} + \frac {y}{3} - \frac {2z}{3} \right)^2 + 16}

On expanding and simplifying,

{x^2 + y^2 + z^2 -16 = \frac 1 9 \left (2x+y-2z \right)^2}

{\therefore 9x^2+9y^2 + 9z^2 - 144 = 4x^2+y^2+4z^2 + 4xy - 4yz - 8zx}

{\therefore 5x^2+8y^2+5z^2 - 4xy + 4yz + 8zx - 144 =0}

This is the required equation.

Q.6 a) Find the equation of the sphere which has its center at {(2,3,-1)} and touches the line

{\frac {x+1}{-5}= \frac {y-8}{3} = \frac {z-4}{4}}

Solution : The tangent line is

{\frac {x+1}{-5}= \frac {y-8}{3} = \frac {z-4}{4} = \lambda, \ \lambda \in \mathbb R}

Any point on this line can be represented by the coordinates {(-5 \lambda -1, \ 3 \lambda +8, \ 4 \lambda + 4)}.

Let {C (2,3,-1)} be the center of sphere and {M} be the point of contact. Then,

{\vec {CM} = (-5 \lambda -1 -2)\hat i + (3 \lambda + 8 - 3) \hat j + (4 \lambda + 4 - \ -1) \hat k }

{=(-5 \lambda - 3) \hat i + (3 \lambda + 5) \hat j + (4 \lambda + 5) \hat k}

Let {\vec d} be the vector along the line. Its components are the direction ratios of the line, i.e. {\vec d = -5 \hat i + 3 \hat j + 4 \hat k}.

Since {CM} is perpendicular to the line,

{\vec {CM} \cdot \vec d =0}

{\therefore \left [(-5 \lambda - 3) \hat i + (3 \lambda + 5) \hat j + (4 \lambda + 5) \hat k \right ] \cdot (-5 \hat i + 4 \hat j + 3 \hat k) = 0}

{\therefore 25 \lambda + 15 + 9 \lambda + 15 + 16 \lambda + 20 = 0}

{\therefore 50 \lambda + 50 = 0}

{\therefore \lambda = -1}

So, coordinates of {M} are {(-5 \times -1 -1, \ 3 \times -1 + 8, \ 4 \times -1 + 4) = (4,5,0)}.

Distance {CM} is the radius of the sphere and using distance formula,

{CM^2 = (4-2)^2 + (5-3)^2 + (0+1)^2 = 2^2+2^2+1^2 = 9}

The required equation of sphere in center-radius form is

{(x-2)^2 + (y-3)^2 + (z+1)^2 = 9}

Q.6 b) Find the equation of the cone with vertex at {(1,2,-3)}, semi-vertical angle {cos^{-1} \frac {1}{\sqrt 3}} and the line

{\frac {x-1}{1}= \frac {y-2}{2} + \frac {z+1}{-1}}

as the axis of the cone.

Solution : Let {V (1,2,-3)} be the vertex of cone and {\alpha} be the semi-vertical angle. Let {P (x,y,z)} be any point on the cone. Since the axis of the cone is {\frac {x-1}{1}= \frac {y-2}{2} + \frac {z+1}{-1}}, the direction ratios are {1,2,-1}. Let {\vec d} be the vector along the axis. Then,

{\vec d = \hat i + 2 \hat j - \hat k}

Using the definition of dot product,

{\vec {VX} \cdot \vec d = |\vec {VX}| |\vec d| cos (\alpha)}

{\therefore \left ( (x-1)\hat i + (y-2) \hat j + (z+3) \hat k \right ) \cdot (\hat i + 2 \hat j - \hat k)}

{= \sqrt {(x-1)^2 + (y-2)^2 + (z+3)^2} \times \sqrt {1^2 + 2^2 + (-3)^2} \times \frac {1}{\sqrt 3}}

{\therefore x-1 + 2 (y-2) + (-1) (z+3) = \sqrt {(x-1)^2 + (y-2)^2 + (z+3)^2} \times \sqrt {1+4+1} \times \frac {1}{\sqrt 3}}

Squaring both the sides,

{\left ( x-1+2y-4-z-3 \right)^2 = \left (x^2-2x+1+y^2-4y+4+z^2+6z+9 \right ) \times 6 \times \frac 1 3}

LHS is {(x+2y-z-8)^2}. Let’s expand it separately :

{LHS = (x+2y)^2 + (-z-8)^2 + 2 (x+2y)(-z-8)}

{= x^2 + 4xy + 4y^2 + z^2 +16z + 64 - 2zx - 16x-2yz -16y}

{RHS = 2x^2 + 2y^2 + 2z^2 - 4x - 8y + 12z + 28}

Equating 2 sides and taking all terms on one side,


This is the required equation.

Q.7 a)  Evaluate the following over the region {y \ge x^2, x \ge 1}.

{\iint \frac {1}{x^4+y^2} dxdy}

Solution The region of integration is plotted below :


Clearly, to the left of line {x=1}, the region is absent, hence the lower limit on {x} will be {1}. The upper limit will be {\infty}. No restriction on the highest value of {x}.

Let’s now consider a strip parallel to {Y} axis. It will first intersect the region at {y=x^2}. So, the lower limit on {y} is {x^2}. The upper limit will be {\infty} because no curve bounds the region in that direction. Hence, given integral will be

{\int \limits_{x=1}^{\infty} \int \limits_{y=x^2}^{\infty} \frac {1}{x^4+y^2} dydx = I \ say}

Since inner limits are of {y}, let’s first integrate w.r.t. {y}.

{I = \int \limits_{x=1}^{\infty} \left [\frac {1}{x^2} tan^{-1} \left (\frac {y}{x^2} \right) \right]_{x^2}^{\infty}dx}

{= \int \limits_{x=1}^{\infty} \frac {1}{x^2} \left[ \frac {\pi}{2} - \frac {\pi}{4} \right] dx}

{= \frac {\pi}{4} \int \limits_1^{\infty} x^{-2}dx}

{= \frac {\pi}{4} \left [\frac {x^{-2+1}}{-2+1} \right ]_{1}^{\infty}}

{= \frac {\pi}{4} \times \frac {1}{-1} \times \left[\frac 1 x \right]_1^{\infty}}

{= \frac {\pi}{4} \times -1 \times [0 - 1]}

{= \frac {\pi}{4}}

Curve Tracing II

In the previous section, we learnt how to trace the curves, when equations are in Cartesian form. In this section, we will learn to trace polar curves.

  • General Observations (For Equations in Polar Coordinates)

I) The equation in polar form is generally r=f(\theta), where r^2 = x^2+y^2 and tan (\theta) = \frac y x.

II) The polar coordinate system has a pole and an initial line. The directed distance r is measured from pole and the angle \theta is measured w.r.t. the initial line.

So, initial line is analogous to X axis and the pole is analogous to the origin.

III) Symmetry


IV) If for {\theta=0}, {r=0}, then the pole lies on the curve.

V) Slope of tangent at a point {(r_1, \theta_1)} is equal to { r \frac {d \theta}{dr}}

V) Polar Curves Nomenclature:

Cardioid : {r = a (1 \pm cos \ \theta)}, {r = a (1 \pm sin \ \theta)}

Parabola : {r (1 \pm cos \ \theta) = 2a}, {r (1 \pm sin \ \theta) = 2a}

Rose Curves : {r = a sin (n \theta)}, {r = a cos (n \theta)}

Lemniscate of Bernoulli : {r^2 = a^2 cos (2 \theta)}

Spirals : {r = a \theta}, {r = ae^{m \theta}, \ a,m > 0}

  • Illustrative Examples

Q.1) Trace the curve {r = a cos (3 \theta)} (June 2016)

Remarks :

1) {Cosine} is an even function, so function will have same value for {\theta} and {- \theta}.

2) {Cosine} is bounded between {-1} and {1}, hence {r} will be bounded between {-a} and {a}.

3) Intersection with the initial line will be at {r=a}, because at {\theta = 0}, {cos \ \theta = 1}.

4) When {3 \theta} becomes {\frac {\pi}{2}}, the curve will pass through the pole, because {cos \frac \pi 2 =0}. Hence, at {\theta = \frac {pi}{6}}, {r=0}.

5) The next value of {\theta} when {r} becomes {0} will be {\frac \pi 2}, because at {\theta = \frac {\pi}{2}}, {3 \theta = \frac {3 \pi}{2}} and {cos \ \frac {3 \pi}{2}= 0}.

6) As value of {\theta} varies from {0} to {\frac \pi 6}, value of {r} will reduce from {a} to {0}. Since the curve is symmetric about the initial line, we will get a loop (or a petal) of the curve.

7) Considering the periodicity of {cosine} function, same values of {r} will be obtained in the intervals {(0, \frac {2 \pi} {3})}, {(\frac {2 \pi}{3}, \frac {4 \pi}{3})} and {(\frac {4 \pi}{3}, 2 \pi)}.

8) Since at {\theta = \frac {\pi}{2}}, {r=0} and again at {\theta = \frac {2 \pi} {3}}, {r} becomes maximum, we’ve a good idea about the curve in the interval {0 \le \theta \le \frac {2 \pi}{3}}.

9) Let’s repeat this in the intervals {\frac {2 \pi}{3} \le \theta \le \frac {4 \pi}{3}} and {\frac {4 \pi}{3} \le \theta \le \frac {6 \pi}{3}= 2 {\pi}}.

In the figure below, I’ve plotted 2 such curves, one with {a=1}, the violet curve and second with {a=2}, the orange curve. The red line is {\theta = \frac \pi 6}, the green line is {\theta = \frac {2 \pi}{3}} and the blue line is {\theta = \frac \pi 2}.


Q.2) Trace the curve {r = a \ sin (2 \theta)}. (December 2015)

This is similar to Q.1.

Q.3) Trace the curve {r = a (1 + cos \theta)} (June 2015)

Remarks : (Assuming {a>0})

1) {Cosine} is an even function, so the curve will be symmetric about the initial line {\theta = 0}.

2) The curve will not have multiple loops, because the argument of {cosine} is {\theta} and not {n \theta, n \ne 1} as it was in Q.1.

3) Intersection with initial line will be at the point {(2a,0)} and intersection with {\theta = \pi} will be at the point {(0, \pi)}, i.e. the pole. Intersection with the line {\theta = \frac \pi 2} will be the point {(a,0)}.

4) The equation of tangent at the pole is obtained by equating {r} to {0}, which gives {1 + cos (\theta)= 0}, or {\theta = \pi}. Since the curve is symmetrical about {\theta =0} and value of {r} reduces from {\theta = 0} to {\theta = \pi}, there will be a cusp at the pole.

5) Slope of tangent at the point {(2a,0)} is obtained by {r \frac {d \theta}{dr}}. From the equation, {\frac {dr}{d \theta} = -a \ sin \ \theta}. So, {\frac {d \theta}{dr} = \frac {-1}{a \ sin \ \theta}} = {- \infty} for {\theta = 0}. Hence the tangent will be parallel to {\theta = \frac {\pi}{2}} at {(2a,0)}.

With this information, the curve will look like this :

Value of {a} that I took is {5}. Tangent at {(2a,0)} is the orange line. The curve is in green colour.






  • Definition of Plane

Let there be three non-collinear points. There exist 3 distinct lines, which pass through these points taken 2 at a time. The triangle so formed is a planar surface.

A line joining any 2 points on a plane always lies on the plane.

  • Equation of Plane : Normal Form

Consider a plane. Let {N} be a point on this plane, such that. Position vector of {N}, {\vec n} would be {p \hat n}, if {p} is the magnitude of {\vec {ON}}. {\hat n} is the unique vector along {\vec {ON}}. Let {P (\vec r)} be any point on the plane. Clearly, {\vec {ON} \perp \vec {NP} \ i.e. \ \hat n \perp \vec {NP}}.


{(\vec r - p \hat n) \cdot \hat n =0 \ i.e. \ \vec r \cdot \hat n = p}

Cartesian Equivalent of normal form is {ax+by+cz+d=0}, where {a,b} and {c} are the direction ratios of any line normal to the plane.

  • Equation of Plane through a Point and Normal to a Vector

Let {A (\vec a)} be the point and {\vec b} be the vector. If {\vec b} is perpendicular to the plane, it will be perpendicular to any vector in that plane. Let {P (\vec r)} be a point on the plane. Then,

{\vec {NP} \perp \vec b \ i.e. \ (\vec {NP}) \cdot \vec b =0 \ i.e. \ \vec r \cdot \vec n - \vec a \cdot n = 0}

Cartesian Equivalent of above equation is

{(x-a_x)b_1 + (y-a_y)b_2 + (z-a_a)b_3 =0,}

if {\vec a = a_x \hat i + a_y \hat j + a_z \hat k} and {\vec b = b_1 \hat i + b_2 \hat j + b_3 \hat k}.

  • Equation of Plane through a Point and Parallel to 2 Non-parallel Non-zero Vectors

Let {A (\vec a)} be the point and {\vec b} and {\vec c} be the vectors. {\vec b \times \vec c} will perpendicular to the plane, hence,

{(\vec r - \vec a) \cdot (\vec b \times \vec c) = [\vec r- \vec a \ \vec b \ \vec c] = 0}

  • Equation of Plane through 3 Non-collinear Points

Let {A (\vec a), \ B (\vec b)} and {C (\vec c)} be the points and {P (\vec r)} be any point on the plane.Thus, {\vec {AP}}, {\vec {AB}} and {\vec {AC}} will be coplanar. Thus,

{(\vec {AP}) \cdot (\vec {AB} \times \vec {AC}) = 0}

This is the required equation.

  • Equation of Plane through Intersection of 2 Planes

This is similar to the equation of family of straight lines through a point (in 2D).

Note that intersection of 2 planes gives us a straight line and infinitely many planes can pass through the intersection of 2 planes. Let the planes be {\vec r \cdot \vec n_1 = p_1} and {\vec r \cdot \vec n_2 = p_2}. The vector equation of a plane through the intersection of these 2 is given by

{\vec r \cdot (\vec n_1 + \lambda \vec n_2) = p_1 + \lambda p_2}

  • Angle between Two Planes

Angle between 2 planes is the angle between their normals. So, if the normals are {\vec n_1} and {\vec {n_2}}, then the angle {\theta} will be

{cos (\theta) = \frac {\vec n_1 \cdot \vec n_2}{|\vec n_1| |\vec n_2|}}

For acute angle, we take the modulus of the RHS.

If {cos (\theta)} is {0}, planes are perpendicular. If {cos (\theta)} is {\pm 1}, planes are either parallel or coincident.

  • Angle between a Line and a Plane

Let the line be {\vec r = \vec a + \lambda \vec b}. Let {\vec n} be a vector normal to the plane. Clearly, angle between the line and the plane will be equal to {90^o} minus the acute angle made by the line (or {\vec b}) and {\vec n}.

  • When are 2 Lines Coplanar?

We know that in 3 dimensions, lines can be skew or coplanar. Let the lines be {\vec r = \vec a_1 + \lambda \vec b_1} and {\vec r = \vec a_2 + \mu \lambda b_2} be the lines. The lines will be coplanar if and only if the scalar triple product of the following vectors is zero :

{(\vec a_2 - \vec a_1), \ \vec b_1, \ \vec b_2}

  • Distance of a Point from a Plane

This is similar to distance of a point from a line. We find the value of parameter {lambda}. Let {A (\vec a)} be the point and {\vec r \cdot \vec n = p} be the plane. Let the foot of the perpendicular from {A} on the plane be {M}. Vector equation of line {AM} will be

{\vec r = \vec a + \lambda \vec n}

Since {M} lies on the plane, we have

 {(\vec a + \lambda \vec n) \cdot \vec n = p}

From this, we get the value of {\lambda}. Having obtained {\lambda}, we can easily get coordinates of {M} and eventually the distance {AM}.

Differentiation Under Integral Sign and Error Function

  • Differentiation under Integral Sign, DUIS

  •  Introduction

Not all integrals can be evaluated using analytical techniques, such as integration by substitution, by parts or by partial fractions. People come up with different ways of solving the integrals and DUIS is one of them.

The class of definite integrals can be treated as integrals of function of a variable x and a parameter t and can be evaluated using the DUIS.

  • Theorem I

Let f be a function of single variable x and a parameter t. Then,

{\frac {d}{dt} \int \limits_a^b f(x,t) dx = \int \limits_a^b \frac {\partial }{\partial t} f(x,t) dx}

Note that, sometimes, we might require to integrate again w.r.t. t. (To be done, when there are 2 parameters.)

  • Theorem II (Leibniz Rule)

Theorem I is a special case of theorem II. In this case, the limits are not constants, but are functions of the parameter t.

{\frac {d}{dt} \int \limits_{h(t)}^{g(t)} f(x,t) dx = \int \limits_{h(t)}^{g(t)} \frac {\partial }{\partial t} f(x,t) dx + f[g(t),t] \frac {d \ g(t)}{dt} - f[h(t),t] \frac {d \ h(t)}{dt}}

  • Error Function

  • Introduction

The error function is in the form a definite integral, whose limits contain the independent variable. It is defined as follows :

{erf(x) = \frac {1}{\sqrt {\pi}} \int \limits_{-x}^{x} e^{-u^2}du}

Using the property of definite integrals,

{erf(x) = \frac {2}{\sqrt {\pi}} \int \limits_{0}^{x} e^{-u^2}du}

How the function got the name error function is an interesting thing.

The function appears in probability, statistics and partial differential equations describing diffusion. In statistics, when any of population parameters (mean \mu or variance \sigma^2) are not known, they’re estimated by some techniques. The distribution of the difference between the actual and the estimated parameter is normal. (Normal distribution is the most general probability distribution). Now, the error function gives the probability that the error lies in [-x,x].

Normal distribution in theory of probability is given by the probability density function

{f(x) = \mathcal {N} (\mu, \sigma) = \frac {1}{\sigma \sqrt {2 \pi}}\int e^{-\frac {(x - \mu)^2}{2 \sigma^2}} dx}

(Try tracing this curve f(x)! The article on curve tracing might be useful.)

  • Properties

I) erf(0)= 0

II) erf ( \infty) = 1

III) By substituting u^2 =t, an alternative form of erf(x) is obtained, given by

{erf (x) = \frac {1}{\sqrt \pi} \int \limits_0^{x^2} e^{-t} \frac {1}{\sqrt t} dt}

IV) Complementary error function is given by

{erf_c(x) = \frac {2}{\sqrt \pi} \int \limits_0^{\infty} e^{-u^2} du}

Thus, {erf(x) + erf_c(x)= 1}

V) {erf (-x) = - erf (x)}. It is an odd function of x.

NOTE : Since x is the parameter in the definite integral defining erf (x), the rule of DUIS is used to solve problems on definite integration.


Related : Curve Tracing

The process of rectification involves computation of lengths of curves. The logic behind this is based on right triangle geometry and calculus.

Consider a curve in XY plane. Let the curve be divided into infinitesimally smaller parts. Let dS be the length of a part. As we take the limit, the arc length approximates to a straight line and we have,

{(ds)^2 = (dx)^2 + (dy)^2}

On integrating,

{S = \int \limits_{x_1}^{x_2} \sqrt {1 + \Big (\frac {dy}{dx} \Big)^2}dx}

This gives the length of curve between the lines x=x_1 and x=x_2.

For parametric curves x=f(t),y=g(t),

{S = \int \limits_{t_1}^{t_2} \sqrt { \Big ( \frac {dx}{dt} \Big )^2 + \Big ( \frac {dy}{dt} \Big )^2}dt}

For polar curves r= f(\theta),

{S = \int \limits_{\theta_1}^{\theta_2} \sqrt {r^2 + \Big (\frac {dr}{d \theta} \Big )^2}d \theta}

For polar curves \theta = f(r),

{S = \int \limits_{r_1}^{r_2} \sqrt {1 + r^2 \Big (\frac {d \theta}{dr} \Big )^2}dr}

Curve Tracing I

  • Introduction

Given an equation of a curve, say y=f(x), the standard process of plotting involves obtaining many pairs of coordinates (x,y), which satisfy the equation of curve. Having plotted a sufficiently large number of points, one gets a good picture of the curve.

In curve tracing (or curve sketching), we do not plot the points (generally) and get the exact shape of curve. However, using some mathematical facts and conclusions, we try to find an approximate shape of the curve. Thus, the figure so obtained may not always be to the scale and hence, the actual curve may look slightly different than the traced one. Nonetheless, we get an idea of how the curve looks like.

  • Types of Equations

I) Equations in Cartesian Coordinates

{y^2 = 4ax, \ x= (y-1)(y-2)(y-3), \ x^2y + y^x = 6x}

II) Equations in Polar Coordinates

{r = sin (a \theta), \ r^2 = a^2 cos (2 \theta), \ r = \frac {2a}{1+ cos (\theta)}}

III) Parametric Equations

{x=at^2, y = 2at, \ x= acos^3 (\theta) , y= bsin^3 (\theta)}

  • General Observations (For equations in Cartesian Coordinates)

I) Points of Intersection with X and Y axes

By putting x=0, we get the point of intersection of the curve with Y-axis. (Similar procedure for X axis and curve)

If there is no constant term in the equation of curve, it passes through the origin (0,0).

II) Symmetry


III) Tangents

If the curve passes through the origin, the equation of tangent is obtaining by equating the lowest degree term to 0; e.g. y^2 =4ax passes through origin. The lowest degree term is 4ax. Equating to 0, 4ax =0 or x=0, which is Y axis.

To get the slope of tangent at a point (x_1,y_1), we find \frac {dy}{dx} at that point.

Recall : Parallel lines have same slope, slope of X axis is 0, slope of Y axis is not defined. A positive (negative) value of slope implies an acute (obtuse) angle with X axis.

IV) Asymptotes

An asymptote is a line which is tangential to the curve at infinity. e.g. X and Y axes are asymptotic to the rectangular hyperbola xy = C.

Asymptote parallel to X (Y) axis is obtained by equating the highest degree term in y (x) to 0.

To get an oblique asymptote, we put y=mx+c in the equation of the curve and get the values of m and c.

  • Illustrative Examples :

Q.1) Sketch the curve : {y^2 = x^5 (2a-x)}. (December 2015)

Remarks : (Assuming {a>0})

1) Even powers of {y} implies symmetry about {X} axis.

2) Since {LHS} is never negative, the region of absence of the curve will correspond to those values of {x}, for which {RHS} is negative. When {x<0}, clearly {x^5} will be negative and {(2a-x)} will be positive. Hence to the left of the line {x=0}, the curve will not exist. (The Black Line)

3) Further, when {x > 2a}, {x^5} will be positive and {(2a-x)} will be negative. Hence to the right of the line {x=2a}, the curve will not exist. (The Blue Line)

4) The curve passes through the origin and intersects {X} axis again at the point {(2a,0)}.

5) In the interval {0 \le x \le 2a}, values of {y} are finite and hence there will be a peak (1st quadrant) and a trough (4th quadrant).

6) The tangent at origin will be obtained by equating the lowest degree term, i.e. {y^2} to {0}. This gives 2 coincident lines {y=0}. Hence at {(0,0)}, the curve has a cusp.

7) At the point {(2a,0)}, the tangent will be parallel to {Y} axis, since {\frac {dy}{dx}} is not defined.

With all this information, the sketch of the curve will look like this :

(I’ve taken the value of {a} as {2}).



Q.2)  Sketch the curve {ay^2 = x^2(a-x)} (June 2015)

Remarks : (Assuming {a > 0})

1) Even power of {y} implies symmetry about {X} axis.

2) Since {LHS} is never negative, the region of absence of the curve will correspond to those values of {x}, for which {RHS} is negative. For {x > a}, {(a-x)} and hence {RHS} will be negative. Hence, the curve will be absent to the right of the line {x=a}. (The Orange Line)

3) As {x \to - \infty}, {x^2 (a-x)} will tend to {\infty} and so will be {y^2}. Hence, in the second and third quadrants, the curve will extend to {\infty}.

4) The curve passes through origin. It intersects {X} axis at the point {(a,0)}.

5) The lowest degree term is {a (y^2-x^2)}, so the equation(s) of tangent(s) at origin will be {y-x=0} and {y+x=0} i.e. {y=x} and {y=-x}. (The Blue Line and The Green Line)

6) 2 tangents at origin imply that the curve passes through the origin twice.

With this information, the curve will look like this : (The Red Curve)


Q.3) Sketch the curve  {y^2=x^2 (1-x)} (December 2013)

This is similar to Q.2, the only difference is the number {1}. In the previous problem, it was {2}.

Q.4) Sketch the curve {x^2y^2 = a^2 (y^2-x^2)} (June 2014)

Remarks :

1) The curve passes through origin. It is symmetric about {X} and {Y} axes, because all powers are even.

2) {LHS} is always non-negative. Hence in the region where {y^2< x^2}, i.e. {y < x} or {y > -x}, the curve will be absent.

3) Equation of tangent to the curve at origin will be obtained by equating the lowest degree term to {0}, which in this case is {y^2-x^2}, which gives 2 tangents, {y+x=0} and {y-x = 0}, i.e. {y=- x} and {y=x}.

The line {y=x} is the violet line and the line {y=-x} is the black line.

4) From points 2 and 3, we can conclude that the region bounded by the lines which contains the {X} axis is the region of absence of the curve, because in that region, {y^2-x^2} is negative.

5) Rewriting the function, we get

{y^2 = \frac {-a^2 x^2}{x^2-a^2} = \frac {a^2 x^2}{a^2 - x^2}}

Taking the limit as {x \to a}, we get the following :

{\lim \limits_{x \to a} \left | \frac {a^2x^2}{a^2-x^2} \right | = \infty}

Hence the asymptotes parallel to {Y} axis are {x=a} and {x=-a}. (The Green Line and The Blue Line respectively)

Based on these points, the curve will look like this :


The value of {a} taken by me is {8}.