differential calculus -...
TRANSCRIPT
DIFFERENTIAL_CALCULUS
Definition:
Let P be a given point on a given curve, and Q be any other point on it. Let
the normal at P and Q intersects at N. If N tends to definite position C as Q
tends P. Then C is called the centre of curvature of the curve at P.
The reciprocal of the distance CP is called the Curvature of the curve at P.
The circle with its centre at C and radius CP is called the circle of curvature of the
curve at P.
The distance CP is called the radius of curvature of the curve at P. The radius of
Curvature is usually denoted by ρ.
Note:
(dψ/ds) is the rate of bending of the curve.
Write to the curvature of the curve at point P.
CARTESIAN FORMULA FOR RADIUS OF CURVATURE
We know that (dy/dx) represents the slope of the tangent to the curve y = f
(x) at ( x,y ) .
Hence (dy/dx) = tan ψ.
Differentiate both sides write to ‘x’ we get
d²y / dx² = sec²ψ.(dψ/dx) ds/dψ = ρ
dψ/ds = 1/ρ
= sec²ψ.(dψ/ds).(ds/dx) dx/ds = cosψ
dy/ds = sinψ.
= sec²ψ.(1/ρ).(1/cosψ)
= sec³ψ / ρ
= (sec ψ)³ / ρ sec²ψ = 1+ tan²ψ
d²y / dx² = (1+ tan²ψ)3.1/2 secψ = (√1+tan² ψ)
____________
ρ sec ψ = (1+tan²ψ)½
Therefore, ρ = (1+ tan²ψ)3.1/2 tan ψ = dy / dx
______________
d²y / dx²
ρ = (1+y1²) 3/2
________
Y2
Note:
The following formula is very useful, when the tangent is parallel to the y–
axis.
[ 1 + (dx / dy)²]3/2
ρ = _________________
d²x / dy²
1) Find the radius of curvature for the curve √x + √y = 1. at (1/4,1/4) .
Soln:
Given √x+√y = 1
Differentiating write to ‘x’
x½+y½ = 1.
(½ x -1/2 )+ (1/2 y -1/2 ). dy/dx = 0.
1/(2√x) + 1/(2√y) . dy/dx = 0.
dy/dx = (- 1/(2√x)) . 2√y
dy/dx = - (√y/x) ---------------------I
Therefore (dy/dx)(1/4,1/4) = - √(1/4) / (1/4)
= -1. ---------------------II
Differentiate equation I write to ‘x’
dy/dx = -(√y) / (√x)
-√x(1/(2√y))(dy/dx ) + √y(1/(2√x))
d²y / dx² = _______________________
x
(-1/4)(1/(2√1/4))(-1) + (√1/4)(1/(2√1/4))
(d²y / dx²)(1/4,1/4) = _______________________________
¼
= 1 / ( 1/4) = 4.
ρ = (1+y1²) 3/2 ) / y2
= ((1+1)3/2 ) / 4 =(2√2) / 4 =√2/2
Therefore ρ = 1/√2
2) Find the radius of curvature at (x,y) for the curve a²y = x³-a³
Soln:
Given a²y = x³-a³.
Differentiate write to ‘x’ We get,
a²(dy/dx) = 3x²
dy/dx = 3x²/a²
d²y/dx² = 6x/a²
ρ = ((1+y1²)3/2) / y2
(1+(9x4/a4))3/2 ((a4+9x4)/a4)) 3/2
= ___________ = ____________
6x / a² 6x / a²
(a4+9x4)3/2 a²
= ________ . _________
a6 6x
= ((a4+9x4)3/2 ) / (6a4x)
3) Find ρ for the curve y = c cosh (x/c) at the point (0,c) (or)
y =( a/2)(ex/a + e -x/a)
Soln:
Given y = c cosh (x/c) (or) y = a/2(ex/a + e -x/a)
dy/dx = c sinh (x/c).(1/c)
= sinh (x/c)
(dy/dx)(0,c) = sinh (0/c) sinh0 = 0.
= 0.
(dy/dx)(0,c) = (1/c) cosh 0 cosh0 = 1.
= 1/c
Therefore ρ = (1+0)3/2 / (1/c)
= c
4) Find the radius of curvature of the curve xy² = a³-x³ at (a,0)
Soln:
Given xy² = a³-x³ -----------------------------
--I
Differentiate I write to ‘x’
(x.2y.(dy/dx)) +y² = -3x²
dy/dx = -(3x²+y²) / 2xy --------------------------------II
(dy/dx)(a,0) = -(3a²+0) / (2(a)(0)) = ∞
Hear dy/dx = ∞
Therefore we use the formula
(1+(dx/dy)²)3/2
ρ = _________________ ------------III
d²x / dy²
From the equation II (dx/dy)(a,0) = 0 -----------------------------IV
dx/dy = (-2xy) / (3x²+y²)
(3x²+y²)(-2x-(2y.(dx/dy))) + 2xy((6x.(dx/dy))+2y)
d²x/dy² = ______________________________________
(3x²+y²)²
(d²x/dy²)(a,0) = (-6a³+0) / 9a4 = -2/3a
Therefore ρ = 3a/2
5) Find the radius of curvature at x=c on the curve xy = c².
Soln:
Given xy = c² ------------------------ I
Since x = c
From yc = c²
Therefore y = c
We have to find the radius of curvature at (c,c) on xy = c²
Differentiate I write to ‘x’
x.(dy/dx)+y = 0
dy/dx = -y/x ------------------- II
Therefore (dy/dx)(c,c) = -c/c = -1
Differentiate II write to ‘x’
[x.((dy/dx)) – y.1]
d²y/dx² = - ______________
x²
[x(-y/x)-y]
= - ________
x²
d²y/dx² = 2y/x²
(d²y/dx²)(c,c) = 2c/c² = 2/c
Therefore the radius of curvature at (c,c) is
(1+y1²)3/2 (1+1)3/2
ρ = ____________ = _______
y2 2/c
= (2√2c ) / 2
Therefore ρ = √2 . c
6) Find the radius of curvature at any point (x,y) on the curve
y =c log sec(x/c).
Soln:
Given y=c log sec(x/c)
(dy/dx) = c . (1/(sec(x/c))) . sec(x/c) . tan(x/c) . (1/c)
(dy/dx) = tan(x/c) ----------------------------------- I
Differentiate I write to ‘x’
d²y / dx² = sec²(x/c) . (1/c)
Therefore ρ = ((1+y1²)3/2) / y2 at (x,y)
(1+ tan²(x/c))3/2 (sec²(x/c))3/2
= ____________ = __________
sec²(x/c) . (1/c) (1/c) sec²(x/c)
__________
= c √ sec ²(x/c)
Therefore ρ = c sec(x/c)
7) Find the radius of curvature of y =2a on the curve y² =4ax.
Soln:
Given curve y²=4ax
2y(dy/dx) = 4a
(dy/dx) = 2a /y
(dy/dx)(0,2a) = 2a / 2a =1 Therefore y1 =1
y.0 - (2a(dy/dx))
(d²y/dx²) = ______________
y²
(d²y/dx²)(0,2a) = (-2a) / 4a²
= -1 / 2a
Therefore ρ = (1+1)3/2 / (-1/2a)
= 2√2 / (-1/2a)
= - 4√2 . a
ρ = 4 √2 . a
8) For the curve y= (ax / a+x) if ρ is the radius of curvature at any point
(x,y) , Show that (2ρ /a)2/3 = (y/x)² + (x/y)² .
Soln:
Given y = ax / a+x = a (x/(a+x)) ----------------------- I
Differentiate with respect to ‘x’
a[(a+x).1 – (x.1) a²+0 1 a²x² y²
(dy/dx) = _______________ = ________ = ____ . ____ = ___
(a+x)² (a+x)² x² (a+x)² x²
Differentiate I with respect to ‘x’
(d²y/dx²) = d/dx [a²(a+x)-2] = - 2a²(a+x) -3 . 1 = (-2a²) / (a+x)³
1 -2a³x³ -2y³
= ______ . __________ = ______________
ax³ (a+x)³ ax³
(1+(y4/x4))3/2
Therefore ρ = ________________
(-2y³ / ax³)
2ρ (1+(y4/x4))3/2
_____ = _____________
a (-y³ / x³) Raise the power 2/3 on both sides
(1+(y4/x4))
(2ρ/a) 2/3 = _______________
(y²/x²)
(x4+y4) x²
= _________ . ____
x4 y²
= (x4+y4) / (x²y²)
= (x²/y²) + (y²/x²)
Therefore (2ρ/a)3/2 = (x/y)² + (y/x)²
9) Find the radius of curvature for the curve √x+√y = 1 at (1/4,1/4).
Soln:
Given √x=√y = 1.
Differentiate with respect to ‘x’
1 1 dy
____ + ( _____ . ____ ) = 0
2√x 2√y dx
dy/dx = -(√y/√x) ________________ I
Differentiate equation I with respect to ‘x’
d²y - [√x.(1/(2√y)).(dy/dx) - (√y.(1/(2√x))) ]
_______ = _________________________________
dx² x
- [√x.(1/(2√y)).(-√y/√x) - (√y.(1/(2√x))) ]
= _________________________________
x
= (1/2 + 1/2) / (1/4)
= 4
[1+(-1)²]3/2
Therefore ρ = _______
4
= (2√2) / 4
ρ = 1/√2
10) Find the radius of curvature for y=ex at (0,1)
Soln:
Given y = ex
y1 = ex
y2 = ex
Hence at (0,1) y1 = 1
y2 = 1
ρ = ((1+1)3/2) / 1 = 2√2
x----------------------------------------------------x
PARAMETRIC FORM
Note 1:
In parametric form radius of curvature
(x’ ² + y’ ² )3/2
ρ = ______________
(x’y’’ – x’’y’)
1) Find ρ at any point P(at²,2at) on the parabola y² =4ax .
Soln:
x=at² y=2at
x’=2at y’=2a
x’’=2a y’’=0
(x’ ² + y ’ ²)3/2 (4a²t² + 4a²)3/2 -2³a³(t²+1)3/2
Therefore ρ = ________________ = __________________ =
___________________
X’y’’ – x’’y’ 0 – 4a² 4a²
= - 2a(1+t²)3/2
i.e) | ρ | = 2a(1+t²)3/2
2) S.T The radius of curvature at any point of the cycloid x=a(θ+sin θ),
y=a(1-cos θ) is 4acos(θ/2).
Soln:
Given x=a(θ+sin θ) y=a(1-cos θ)
dx/dθ = a(1+cos θ) dy/dθ=asin θ
dy dy/dθ asin θ 2sin θ/2 cos θ/2
____ = __________ = _____________ = _________________ = tan θ/2
dx dx/dθ a(1+cos θ) 2cos² θ/2
i.e) (dy/dx) = tan θ/2
(d²y/dx²) = (d/dx) (dy/dx) = (d/dθ) (dy/dx) (dθ/dx)
= (d/dθ) (tan θ/2) (1/(a(1+cos θ)))
= (sec² θ/2) (1/2) (1/(a2cos² θ/2))
(d²y/dx²) = (1/(4acos4 θ/2))
[1+(dy/dx)²]3/2 (1+tan² θ/2) 3/2
ρ = _____________________ = _____________________
(d²y/dx²) (1 / (4acos4 θ/2)
= (sec² θ/2)3/2 (4acos4 θ/2) = (4acos4 θ/2) (1/cos³ θ/2)
ρ = 4acos θ/2
3) Find ρ for the curve x=a(cos t+tsin t), y=a(sin t-tcos t).
Sonl:
Given x=a(cost+tsint) y=a(sint-tcos t)
(dx/dt)= a(-sin t+tcos t+sin t) dy/dt=a(cos t+tsin t-cos t)
= atcos t = atsin t
(dy/dx)=(atsin t / atcos t) = tan t
(d²y/dx²)=(d/dt) (dy/dx) (dt/dx)
=(d/dt) (tan t) (1 / atcos t)
=(sec² t) (1 / atcos t)
= 1 / atcos³ t
Therefore,
[1+(dy/dx)²]3/2 (1+tan² t)3/2
ρ = ____________ = ___________________________
(d²y/dx²) 1 / atcos³ t
ρ = (sec³ t) (atcos³ t) = at
ρ = at
4) Find ρ for the curve x =alog (sec θ) , y =a(tan θ- θ) at 0.
Soln:
Given x=alog (sec θ) y=a(tan θ-θ)
(dx/dθ) =a (1/sec θ) sec θ .tan θ
=atant θ
(dy/dθ)=a(sec² θ-1)
=atan² θ
(dy/dx)= (atan² θ) / (atan θ)
(dy/dx)= tan θ
(d²y/dx²) = (d/dθ) (dy/dx) (dθ/dx)
= (d/dθ) (tan θ) (1 / atan θ)
= (sec² θ) (1 / atan θ)
= (1 / cos² θ) (1 /a(sin θ/cosθ))
= (2 / asin 2θ)
= (2 / a) cosec 2θ
Therefore,
(1+tan² θ)3/2 asec ³ θ
ρ = ________________ = ______________
2/a . cosec 2θ 2cosec 2θ
1 1
= a.______________ . _______________
Cos³ θ 2 .(1 / sin 2θ)
= (a 2sinθ cosθ) / (2cos³ θ)
= a tanθ secθ
Therefore, ρ = a tanθ secθ
x-----------------------------------------------------x
POLAR FORM
1) Find ρ for the curve r =aeθcot α .
Soln:
Given r = aeθcot α
(dr/dθ) = acotα eθcot α
= rcotα
(d²r/dθ²) = (d/dθ) (dr/dθ) = (d/dθ)(rcotα
= (d/dr) (rcot α) (dr/dθ)
= cot α. rcotα
i.e) (d²r/dθ²) = rcot²α
(r²+r’ ²)3/2 = [r²+r²cot²α]3/2
= r³cosec³α
And
r² + 2(dr/dθ)2 – r(d²r/dθ²) = r² + 2r²cot² α - r²cot²α
= r² + r²cot²α = r²(1 + cot²α)
= r²cosec²α
Therefore,
ρ = r³cosec³ α / r²cosec² α
= rcosecα
2) For the curve rn =ancosnθ, the radius of curvature is (an.r - n+1 ) / (n+1) .
Soln:
Given rn=ancosnθ
Taking log on both sides
log rn = log ancosnθ
nlog r = log an+ log cos nθ
(n/r)(dr/dθ) = (-nsin nθ) / (con nθ)
(dr/dθ) = - rtan nθ
(d²r/dθ²) = (d/dθ) (dr/dθ)
= (d/dθ) (-rtan nθ)
= (-dr/dθ)tan nθ - nrsec² nθ
= rtan² nθ - nrsec² nθ
[r²+(dr/dθ)²]3/2
ρ = _____________________
(r² + 2(dr/dθ)² - r(d²r/dθ²)
(r² + r²tan² nθ)3/2
= __________________________________
(r² + 2r²tan² nθ - r²tan² nθ + nr²sec² nθ)
(r³ + r³tan³ nθ)
= __________________________________
(r² + r²tan² nθ + nr²sec² nθ)
(r³sec³ nθ)
= ___________________________________
(r²sec² nθ + nr²sec² nθ)
(rsec nθ . r²sec² nθ)
= ____________________________________
(r²sec² nθ(1+n)
= (r.an) / (1+n)rn
ρ = (an.r – n+1) / (1+n)
3) Find the radius of curvature of the curve r²cos 2θ = a² .
Soln:
Given r²cos 2θ = a² ----------------------------------- I
Differential equation I w.r. to ‘θ’
r²(-sin 2θ.2) + (cos 2θ . 2r . (dr/dθ)) = 0
(dr/dθ) = (2r²sin 2θ) / (2rcos 2θ) = rtan 2θ
Now
(d²r/dθ²)= (2rsec² 2θ )+ (tan 2θ(dr/dθ))
= (2rsec² 2θ) + (rtan² 2θ)
(r²+r’ ²)3/2
ρ = ______________
r²+2r1²-rr’’
(r²+r²tan² 2θ) 3/2
= _______________________________
(r²+2r²tan² 2θ) – (r(2rsec² 2θ+rtan² 2θ))
(r³sec³ 2θ)
= ________________________________________
(r² + r²tan² 2θ - 2r²sec² 2θ)
(r³sec³ 2θ)
=_________________________________
r²(1 + tan² 2θ – 2sec² 2θ)
(r³sec³ 2θ)
= ________________________________ = -rsec θ
-r²sec² 2θ
Therefore,
ρ = rsec θ
x----------------------------------------------------x
CIRCLE OF CURVATURE
Let APB be a curve and P(x,y) be any point on it , and PT be a tangent at P.
Let K be a circle, with its centre at C having same curvature as the curve at the
point p.
This circle is called the circle of curvature of the given curve at P. And C is
known as centre of curvature . Any chord of the circle of curvature through the
point of contact P is called the chord of curvature of the given curve at P.
Definition:
The circle which touches the curve at P and whose radius is equal to
the radius of curvature is known as circle of curvature and its centre is known as
centre of curvature.
Centre of Curvature:
_
x = x – (y1/y2)(1+y1²)
_
y = y + (1/y2)(1+ y1²)
Equation of circle of curvature is,
_ _
(x – x)² + (y – y)² = ρ²
1) Find the centre of curvature of the curve y = 3x³ + 2x² - 3 at (0,- 3).
Soln:
Given y = 3x³ + 2x² - 3
(dy/dx) = 9x² + 4x
(dy/dx)(0,-3) = 0.
(d²y/dx²) = 18x + 4
(d²y/dx²)(0,-3) = +4
_ _
The centre of curvature (x , y) is given by
_
X = x - (y1/y2)(1+y1²)
= 0 – (0/4)(1+0)
= 0
_
y = y + (1/y2)(1+ y1²)
= -3 + (1/4)(1)
= -11/4
Hence the centre of curvature is (0 , -11/4)
2) Find the equation of the circle of curvature at (c,c) on xy =c² .
Soln:
The equation of circle of curvature is
_ _
(x-x)² + (y-y)² = ρ²
_ _
Where (x,y) is the centre of curvature and ρ is radius of curvature
Given xy = c² ---------------------------- I
Differentiate equation I w.r. to ‘x’
y + x(dy/dx) = 0.
(dy/dx) = -y/x
(dy/dx)(c,c) = -c/c = -1
(d²y/dx²)(c,c) = 2/c
Therefore
ρ =√2 . c
_
x = x – (y1/y2)(1+y1²)
= c + ((c/2)(1+1))
_
x = 2c
_
y = y + (1/y2)(1+ y1²)
= c + ((c/2)(1+1))
_
y = 2c
The circle of curvature is
_ _
(x-x)² + (y-y)² = ρ²
(x-2c)² + (y-2c)² = ρ²
(x-2c)² + (y-2c)² = 2c²
3) S.T The circle of curvature of √x+√y = √a at (a/4 , a/4) is
(x – (3a/4))² + (y – (3a/4))² = a²/2
Soln:
Given
√x+√y = √a
Differentiate the above eqn w.r. to ‘x’
(1 / 2√x) + (1 / 2√y)(dy/dx) = 0.
(dy/dx) = (-√y / √x) ------------------ I
i.e) (dy/dx)(a/4,a/4) = -1
Differentiate the equation I w.r. to ‘x’
d²y -(√x . (1/(2√y)) . (dy/dx)) - √y . (1/(2√x))
_____ = __________________________________
dx² (√x)²
___ __ __ ___
- [(√a/4).(1/(2√a/4)) - (√a/4).(1/(2√a/4))
(d²y/dx²) (a/4,a/4) = ________________________________
(a/4)
= (1 / a/4) = 4/a
Therefore
The radius of curvature is
ρ = ((1+y1²)3/2) / y2
= (1+(-1)²) 3/2 / (4/a)
_
= (2√2 a ) / 4
= a / √2
_
x = x – (y1/y2)(1+y1²)
= (a/4) + (a/4)(1+1)
_
x = 3a / 4
_
y = y + (1/y2)(1+ y1²)
=(a/4) + (a/4)(1+1)
_
y = 3a / 4
Circle of curvature is
_ _
(x-x)² + (y-y)² = ρ²
(x-(3a/4))² + (y-(3a/4))² = a²/2
4) Find the circle of curvature for √x + √y = 1 at (1/4 , 1/4).
5) Find the equation of the circle of curvature of the parabola y² = 12x at
(3,6).
Soln:
Given y² = 12x
(2y)(dy/dx) = 12
(dy/dx) = (6/y)
(dy/dx)(3,6) = (6/6) = 1
(d²y/dx²) = (-6/y²)
(d²y/d²x)(3,6) = -6/36
= -1/6
_ _
y = -6, x = 15, ρ = -12√2
Circle of curvature is (x-15)² + (y+6)² = (-12√2)²
6) Find the circle of curvature of xy = 12 at (3,4).
Soln:
Given curve is
xy=12
y + x(dy/dx) = 0 i.e) (dy/dx) =(-y/x)
(dy/dx)(3,4) = -4/3
x(dy/dx) - y
(d²y/dx²) = -{ ________________ }
x²
3(-4/3) - 4
(d²y/dx²)(3,4) = - { _________________ }
9
= 8/9
Centre of curvature
_
x = x – (y1/y2)(1+y1²)
4 9 16
= 3 + __ . __ ( 1 + ___ )
3 8 9
= 43/6
_
y = y + (1/y2)(1+ y1²)
9 16
= 4 + ___ (1 + ___ )
8 9
= 57 / 8
Radius of curvature is ρ = 125 / 24
Circle of curvature is
_ _
(x-x)² + (y-y)² = ρ²
(x – 46/3)² + (y – 57/8)² = (125 / 24)²
x--------------------------------------------------------x
INVOLUTES AND EVOLUTES
Let A1, A2, A3, A4, …….. be any points on the curve y=f(x).
Let B1,B2,B3,B4,………... be the centre of curvature corresponding to the points
A1,A2,A3,A4,…………….. respectively.
As we move along the curve y=f(x) from A1 to A2, A3 …… the centre of
curvature will also moves along a curve B1,B2,B3,B4…..
EVOLUTE:
The locus of centre of curvature of a curve (B1, B2, B3, B4,…….) is called
evolute of the given curve.
INVOLUTE:
If a curve C2 is the evolute of a curve C1. then C1 is said to be an Involute of a
curve C2.
1) Find the equation of the evolute of the parabola y²=4ax
Soln:
The equation of the given curve is y² = 4ax --------------------------------- I
The parametric form of equation I is x = at² , y = 2at
(dx/dt) = 2at , (dy/dt) = 2a
(dy/dx) = (dy/dt) / (dx/dt) = (2a) / (2at)
y1 = 1/t
y2 = (d²y/dx²) = (d/dx)(dy/dx) = (d/dx)(1/t) = (d/dt)(1/t)(dt/dx) = (-1/t²)(1 /2at)
y2 = (-1/ 2at³) _ _
The coordinates of the centre of curvature is (x , y)
_
Where x = x – (y1/y 2)(1+y1²)
(1/t)
= at² - ________ . (1+ 1/t²)
(-1/ 2at³)
= at² + (2at³/t)((t²+1) /t²)
= at² + 2a(t²+1)
= at² + 2at² + 2a
_
x = 3at² + 2a --------------------------------------------- II
_
y = y + (1/y2)(1+ y1²)
= 2at – 2at³((1+t²) /t²)
= 2at - 2at - 2at³
_
y = -2at³ ---------------------------------------------- III
Now eliminating t between II and III we get
_
x = 3at² + 2a
_
3at² = x – 2a
_
t² = (x - 2a) /3a
_
t = ((x – 2a) /3a)½ ---------------------------------------------IV
Sub. t value in III
_ _
y = -2a . ( (x-2a) /3a)3/2
Squaring on both sides, we get
_ ² _
y = 4a² . ( (x-2a) /3a)3
_ ² _
y = 4a² . (x-2a)³ /27a³)
_ ² _
y = 4 (x-2a)³
___________
27a
_ _
27ay 2 = 4(x – 2a)³
_ _
Since the evolute is the locus of centre of curvature. Replace x by x and y by y.
27ay² = 4(x-2a)³ is the evolute of the parabola y² = 4ax.
2) Find the equation of the evolute of the curve x 2/3+ y 2/3= a2/3 .
Soln:
The equation of the given curve is x 2/3+ y 2/3= a2/3 -------------------------- I
The parametric form of equation I is
x = acos³ θ y = asin³ θ
(dx/dθ ) = -3acos² θ sin θ (dy/dθ) = 3asin² θcos θ
y1 = (dy/dx) = (3asin² θcos θ) / (-3acos² θ sin θ ) = - tan θ
y2 = ((d²y/dx²) = (d/dx)(dy/dx) = (d/dx)(-tan θ) = (d/dθ)(-tan θ).(dθ/dx)
= (-sec² θ) / (-3acos² θsin θ)
y2 = (sec4 θcosec θ) / 3a _ _
The co-ordinates of the centre of curvature is ( x , y )
_
Where x = x – (y1/y 2)(1+y1²)
- tan θ
= acos³ θ - { ___________________ } (1+tan² θ)
(sec4 θcosec θ) / 3a
3a tan θ
= acos³ θ +{ ___________________ } sec² θ
(sec4 θcosec θ)
3atan θ
= acos³ θ + { ___________________ }
(sec² θcosec θ)
_
x = acos³ θ + (3a(sin θ/cos θ)cos² θsin θ)
_
x = acos³ θ + (3asin² θcos θ) ---------------------- II
_
y = y + (1/y2)(1+ y1²)
3a
= asin³ θ +{ ___________________ } sec² θ
(sec4 θcosec θ)
3a
= asin³ θ +{ ___________________ }
_ (sec² θcosec θ)
y = asin³ θ+3acos² θsinθ ------------------------ III
Now eliminate θ between the equations II and III
_ _
II + III ==> x + y = acos³ θ + 3asin² θcos θ + asin³ θ+3acos² θsinθ
= a(cos³ θ +3cos² θsin θ + 3cos θsin² θ + sin³ θ)
x +y =a(sin θ + cos θ)³ ---------------------------- IV
_ _
Similarly x – y = a(cos θ – sin θ)³ ----------------------------- V
_ _ _ _
( x + y )2/3 + ( x - y )2/3 = {a(cos θ+sin θ)³} 2/3 + { a(cos θ-sin θ)³} 2/3
= a2/3 {(cos θ+sin θ)²+(cos θ-sin θ)²}
= a2/3 {2(cos² θ+sin² θ)}
= 2a2/3
(x+y)2/3+ (x-y) 2/3 = 2a2/3
3) Find the equation of the evolute of the ellipse (x²/a²) + (y²/b²) = 1 .
Soln:
The equation of given curve is (x²/a²) + (y²/b²) = 1 ---------------------------- I
The parametric equation of I is
x = acos θ y = bsinθ
(dx/dθ) = -asin θ (dy/dθ) = bcos θ
y1 = (dy/dx) = (bcos θ) / (-asin θ) = (-b/a)cot θ
y2 = (d²y/dx²) = (d/dx)(dy/dx) = (d/dx)(-b/a cotθ)
= (d/dθ) ((-b/a) cot θ)) (dθ/dx)
= (b/a)(cosec² θ)(1/-asin θ)
y2 = (-b/a²)cosec³ θ
_ _
The co-ordinate of the centre of curvature is ( x , y )
_
Where x = x – (y1/y 2)(1+y1²)
(-b/a)cot θ b²
= acos θ - ________________(1 + __ cot² θ)
(-b/a²)cosec³ θ a²
= acos θ - acot θsin³ θ(1+ (b²/a²)cot² θ)
= acos θ - acot θsin³ θ - a(b²/a²)cot³ θsin³ θ
= acos θ - a(cos θ/sin θ)sin³ θ - a(b²/a²)(cos³ θ/sin³ θ)sin³ θ)
= acos θ - acos θsin² θ - (b²/a)cos ³ θ
= acos θ.(1-sin² θ) - (b²/a)cos³ θ
_ = (a²cos³ θ - b²cos³ θ) /a
x = (cos³ θ(a²-b²)) /a ------------- II
_
y = y + (1/y2)(1+ y1²)
(1+(b²/a²)cot² θ
= bsin θ + ________________
(-b/a²)cosec³ θ
= bsin θ – (a²/b)sin³ θ(1+ (b²/a²)cot² θ)
= bsin θ – (a²/b)sin³ θ - b(cos2 θ/sin² θ)sin³ θ
= bsin θ – (a²/b)sin³ θ - bcos² θsin θ
= bsin θ(1-cos² θ) - (a²/b)sin³ θ
= bsin ³ θ - (a²/b)sin³ θ
_ = (sin³ θ(b²-a² )) /b
y = - (sin³ θ(a²-b² )) /b ------------- III
To eliminate θ between the equations II and III
From II
_
ax = (a²-b²)cos³ θ
_
(ax)2/3 = (a²-b²)2/3cos² θ
_
cos² θ = (a x)2/3 / (a²-b²)2/3 -------------- IV
From III
_
by = -(a²-b²)sin³ θ
_
(by)2/3 = (a²-b²)2/3sin² θ
_
sin² θ = ( by)2/3 / (a²-b²)2/3 _ _ ---------------- V
(ax)2/3 (by)2/3
IV + V ===> sin² θ+cos² θ = _________ + __________
(a²-b²)2/3 (a²-b²)2/3
_ _
(ax)2/3 (by)2/3
1 =_________ + __________
(a²-b²)2/3 (a²-b²)2/3
i.e) (ax)2/3 + (by)2/3 = (a²-b²)2/3
4) Find the evolute of the rectangular hyperbola xy= c² .
Soln:
The given equation is xy= c² ,The parametric eqn is x=ct ,y=(c/t)
----------------- I
(dx/dt) = c (dy/dt) = (-c/t²)
(dy/dx) = (-c/t² /c) = (-1/t²) ----------------- II
(d²y/dx²) = (d/dx)(dy/dx) = (d/dx)(-1/t²) = (d/dt)(-1/t²)(dt/dx)
= (-d/dt)(1/t²)(dt/dx) = (-d/dt)(t -2)(1/c)
y2= (d²y/dx²) = (2/ ct³) ---------------- III
_
Where x = x – (y1/y 2)(1+y1²)
-1/t²
= ct – ( ________ )(1+ 1/t4)
2/ct³
= ct + (ct/2)(1+ 1/t4)
_ = ct + (ct/2) + (c/2t³)
x = (3ct/2) + (c/2t³) -------------------- IV
_
y = y + (1/y2)(1+ y1²)
= (c/t) + (ct³/2)(1+ 1/t4) = (c/t ) + (ct³/2) + (c/2t)
_ = (2c/2t) + (c/2t) + (ct³/2)
y = (3c/2t) + (ct³/2) ---------------------- V
Now IV + V ===>
= (3ct/2) + (c/2t³) + (3c/2t) + (ct³/2) (or) = (3c/2)(t+ 1/t) + (c/2)(t³+ 1/t³)
= c/ 2t³ [ 1 + 3t4 + 3t² + t6 ] = c/2 [ t³ +3t + 3/t + 1/t³ ]
_ _ = c/ 2t³ [ (t² +1)³ ]
x + y = c/2 [ (t²+(1 /t))³ ] = c/2 [ (t + (1/t))³ ] ------- VI
Now IV – V ===>
_ _ = (3ct/2) + (c /2t³) – (3c/2t) – (ct³/2) = (-c/2) [ t³- 3t + 3/t – 1/t³ ]
x –y = (-c/2)(t – 1/t)³ ----------------------------- VII
_ _ _ _
( x + y )2/3 + ( x - y )2/3 = (c/2)2/3 [ (t + 1/t)² - (t – 1/t)² ]
= (c/2)2/3 (4)
= (c/2)2/3 (2²)
= (c2/3 / 22/3)(2²)
= (c2/3)(22- 2/3)
= (c2/3)(24/3) = (c2/3)(42/3)
_ _ _ _
( x + y )2/3 + ( x - y )2/3 = (4c)2/3.
5) Find the evolute of the hyperbola (x²/a² - y²/b²) = 1 .
Soln:
The given equation is (x²/a² - y²/b²) = 1 --------------------------- I
The parametric equation of the equation I is
x = asec θ y = btan θ
Now (dx/dθ) = asec θtan θ (dy/dθ) = bsec² θ
y1 = (bsec² θ) / (asec θtan θ) = (b/a)(sec θ / tan θ) =(b/a)((1/cosθ)/(sinθ/ cosθ))
= (b/a)cosec θ -------------------------- II
y2 = (d/dx)(dy/dx) = (d/dθ)(dy/dx)(dθ/dx) = (d/dθ) ((b/a)cosec θ) (1/ asecθ
tanθ)
= (-b/a) cosec θ cot θ (1/ asec θtan θ)
= (-b/a2) (1/sin θ) (cos θ/sin θ) (cos θ) (cos θ/sin θ)
= (-b/a)cot³ θ
_
Where x = x – (y1/y 2)(1+y1²)
= asec θ - [((b/a)cosec θ) / ((-b/a)cot³ θ)](1+ (b²/a²)cosec² θ)
= asec θ + (b/a) [((cosec θ(a²+ b²cosec² θ))/ (b/a²)a²cot³ θ ]
= asec θ + (1/a) (1/sin θ) (a²+ b²/sin² θ) (sin³ θ/cos3 θ)
= asec θ + (asin³ θ/sin θcos³ θ) + (b²/a)(1/sin³ θ)(sin ³ θ/cos³ θ)
= asec θ + (asin² θ/cos³ θ) + (b²/acos³ θ)
= asec θ + (a(1-cos² θ) / cos³ θ) + (b²sec³ θ/a)
_ = asec θ + asec³ θ - asec θ + (b²/a)sec³ θ
a x = a²sec θ + a²sec³ θ - a²sec θ + b²sec³ θ
= (a²+b²)sec³ θ
_
sec³ θ = ax/(a²+b²)
_
sec θ = [ ax/ (a²+b²) ]1/3 ------------------------ III
_
y = y + (1/y2)(1+ y1²)
= btan θ + [ (1+ (b²/a²)cosec² θ) / ( (-b/a²)cot³ θ) ]
= btan θ - [ (a²+b²cosec² θ) / (bcot³ θ) ]
= btan θ - [(a²+b²(1/sin² θ) ) / (b(cos³θ /sin³ θ)) ]
= btan θ - (a²/b)(sin³ θ/cos³ θ) - (bsin θ/cos³ θ)
= btan θ - (a²/b)tan³ θ - btan θsec² θ
_
by = b²tan θ - a²tan³ θ - b²tan θ(1+tan² θ)
_ = b²tan θ - a²tan³ θ - b²tan θ - b²tan³ θ
by = - (a²+b²)tan³ θ
Therefore, _
tan³ θ = (-by)/(a²+b²)
_
= [ (-by/(a²+b²) ]1/3 -------------------------- IV
WKT. (sec² θ - tan² θ) = 1 ------------------------ V
Sub. the equations III and IV in V _ _
{ [ ax/ (a²+b²) ]2/3 - [ -by/(a²+b²)]2/3 } =1
6) S.T The evolute of the cycloid x=a(θ-sin θ), y=a(1-cos θ) is another
equal cycloid.
Soln:
x = a(θ-sin θ) y = a(1-cos θ)
(dx/dθ) = a(1-cos θ) (dy/dθ) = asin θ
y1=(dy/dx) = (asin θ/ a(1-cos θ))= (2sin (θ/2) cos( θ/2)/(2sin²( θ/2)) =cot θ/2 ----- I
y2 = (d²y/dx²) = (d/dx)(dy/dx) = (d/dx)(cot θ/2) = (d/dθ)(cot θ/2)(dθ/dx)
= (- cosec² θ/2) (1/2) (1/ a(1-cos θ)
= ( -1 ) 1
____ { ________________ }
2a (sin² θ/2 . 2sin² θ/2)
= (-1 / 4asin4 θ/2)
--------------------------II
x = x - (y1/y2)(1+y1²)
= a(θ-sin θ) - [ (cos θ/2 / sin θ/2)(1+ cos² (θ/2)/sin² (θ/2) ] /
[-1/ 4asin4( θ/2)]
= a(θ-sin θ) + [(4asin4 θ/2)(cos( θ/2) / sin θ/2)] +
[(4asin4 θ/2)(cos θ/2 / sin θ/2)(cos² θ/2 / sin²
θ/2)]
= a(θ-sin θ) + 4asin³ θ/2 cos θ/2 + 4asin θ/2 cos³ θ/2
= a(θ-sin θ) + 4asin θ/2 cos θ/2 (sin² θ/2 + cos² θ/2)
= a(θ-sin θ) + 4asin θ/2 cos θ/2(1)
= a(θ-sin θ) + 2a.2sin θ/2 cos θ/2
= a(θ-sin θ) + 2asin2 θ/2
= a(θ-sin θ) + 2asin θ
= aθ - asin θ+2asin θ
_ = aθ + asin θ
x = a(θ+sin θ) -------------------------- III
_
y = y - (1/y2)(1+ y1²)
= a(1-cos θ) - 4asin4 θ/2(1+ cot² θ/2)
= a - acos θ - 4asin4 θ/2(sec² θ/2)
= a - acos θ - 4asin4 θ/2(1/sin² θ/2)
= a - acos θ - 4asin² θ/2
_ = a(1-cos θ) - 2a(1-cos θ)
y = -a(1-cos θ) ----------------------------- IV
From the equations III and IV we get the locus of x and y is
x = a(θ+sin θ)
y = -a(1-cos θ)
7) Find the evolute of the curve x = a(cos θ+θsin θ), y = a(sin θ-θcos θ).
Soln:
Given
x = a(cos θ+θsin θ)
(dx/dθ) = a(-sinθ+sin θ+θcos θ) = aθcos θ
y = a(sin θ-θcos θ)
(dy/dθ) = a(cos θ-cos θ+θsinθ = aθsin θ
y1 = (dy/dx) = (aθsin θ/aθcos θ) = tan θ ------------ I
y2 = (d²y/dx²) = (d/dx)tan θ = (d/dθ)tan θ(dθ/dx)
= sec² θ(1/ aθcos θ)
y2 = 1/ aθcos³ θ --------------------- II
_ _
Let the centre of curvature be( x , y )
_
x = x - (y1/y2)(1+y²)
= acos θ + aθsin θ - (tan θ/(1/ aθcos³ θ).(1+tan² θ)
= acos θ + aθsin θ - aθcos³ θ(sin θ/cos θ)(sec² θ)
_ = acos θ + aθsin θ - aθcos² θ sin θ(1/cos² θ)
x = acos θ -------------------- III
_
y = y + (1/y2)(1+ y1²) = asin θ - aθcos θ + aθ cos³ θsec² θ
_ = asin θ - aθcos θ + aθcos θ
y = asin θ --------------------- IV
From the equation III ====>
_
cos θ = x/a -------------------- V
_
sin θ = y/a -------------------- VI
(V)² +(VI)²==>
_ _
cos² θ+sin² θ = (x/a)² + (y/a)²
x²+y² = a².
x---------------------------------------------------------x
ENVELOPE
Family of Curves:
Let us consider the equation y=mx+c. It represents a straight line with slope
m. By giving different values of m we get different straight lines. The straight lines
are called family of straight lines and m is known as the parameter of the family.
The family of Curves is represented by f(x,y,m)=0. Where m is a parameter.
The Curve given by f(x,y,m+sm) is termed as neighbouring curve.
Definition:
Envelope:
A Curve which touches all the curves of a family is called the envelope of the
family.
1) Find the envelope of the family of straight lines y=mx + 1/m.
Soln:
Given y = mx + (1/m). -----------------------------
- I
Here ‘m’ is the parameter. Differentiate I partially w.r. to ‘m’
0=x – (1/m²) (or) x=1/m² (or) m²=1/m (or) m=1/√m -------- II
Now to find envelope of the equation I we have to eliminate m between the
equations I and II. Sub. the equation II in I we get,
y = √x+√x = 2√x = 2x1/2
y = 2x1/2
y² = 4x
Which is the envelope of the given curve.
2) Find the envelope of the family of lines y=mx + a/m .
( Result y² = 4ax)
3) Find the envelope of the family of line xcos³α + ysin³ α = a.The
parameter being α .
Soln:
Given xcos³α + ysin³ α = a --------------------------- I
Differentiating the equation I partially w.r. to α ,we get
-3xcos² α sin α + 3ysin² α cos α = 0.
3xcos²α sin α = 3ysin²α cos α
sin α /cos α = x/y
tan α = x/y
Therefore, _____ _____
sin α = x/√x²+y² , cos α = y/√x²+y²
Substitute these values in the equation I,
[x (y³/ (x²+y²)3/2)] + [y (x³/ (x²+y²)3/2)] = a
xy³+yx³ = a(x²+y²)3/2
xy(y²+x²) = a(x²+y²)(x²+y²)1/2
xy = a(x²+y²)1/2
x2y2 = a(x²+y²)
Which is the equation of the envelope of the given family of straight lines.
4) Find the envelope of xcos α /a+ysin α /b = 1, (α is the parameter).
Soln:
Given ( xcos α /a) + (ysin α /b) = 1 -------------------------- I
Differentiate partially w.r. to α
(-xsin α /a) + (ycos α /b) = 0.
x sin α /a = y cos α /b
bxsin α = aycos α
sin α / cos α = (ay / bx)
i.e) tan α =(ay / bx)
_______
Therefore sin α = (ay / √a²y²+b²x² ) -------------------------- II
________
cos α = (bx / √a²y²+b²x² ) -------------------------- III
Sub. the equations II and III in I
_______ _______
(x/a)(bx / √a²y²+b²x²) + (y/b)(ay / √a²y²+b²x² ) = 1
_______
(b²x²+a²y²) / (ab√a²y²+b²x²) = 1
_______
√a²y²+b²x² = ab.
a²y² + b²x² = a²b²
(y²/b²)+(x²/a²) = 1
Which is the envelope of the equation I.
5) Find the envelope of the family of straight lines ycos α - xsin α = acos
2α, (α being the parameter [ MU ’00 ] .
Soln:
Given ycos α - xsin α = acos 2 α ----------------------- I
Differentiate partially w.R. to α
-ysin α - xcos α = -2asin 2 α
ysin α + xcos α = 2asin 2 α ----------------------- II
I X cos α ===> ycos² α - xsin α cos α = acos 2α cos α
II X sin α ===> ysin² α + xsin α cos α = 2asin 2 α sin α
y(cos² α + sin² α) = a(cos2α cos α + 2sin2α sinα)
y = a[(cos² α - sin² α )cos α + 2.2sin α cos α sin α]
y = a[(cos³ α - cos α sin² α + 4sin² α cos α]
y = a[cos³ α + 3sin² α cos α] ----------------------- IV
I X sin α ===> ysin α cos α - xsin² α = asin α cos 2 α ------------------------ V
II X cos ( ===> ysin α cos α + xcos² α = 2asin2α cos α ------------------------ VI
V – VI ====>
-x = a(sin α cos2α - 2sin 2 α cos α)
x = a(2sin 2 α cos α - cos 2 α sin α)
x = a(4sin α cos² α - cos² α sin α + sin³ α
x = a(3sin α cos² α + sin³ α) ----------------------- III
Adding III and IV
(x+y) = a [3sin α cos² α + sin³ α + cos³ α + 3sin² α cos α]
= a(cos α +sin α)³
(x-y) = a(cos α -sin α)³
(x+y)2/3+(x-y)2/3 = a2/3(cos α +sin α)2+ a2/3(cos α -sin α)2
= a2/3 (2) = 2a2/3
(x+y)2/3+(x-y)2/3 = 2a2/3 is the required envelope.
6) Find the envelope of the family of circles x²+y²-2axcos α -2aysin α =c²,
(α is the parameter).
Result: (x²+y²-c²) = 4a²(x²+y²)
7) Find the envelope of the family of curves. (a²cos θ/x) – (b²sin θ/y) = c for
different values of θ.
Soln:
Differentiate partially w.r. to ‘θ’
(-a²sin θ/x) – (b²cos θ/y) = 0 tan θ = (-b²x /a²y)
________ _______
sin θ = (-b²x / √a4y²+b4x² ) cos θ = (a²y / √a4y²+b4x²)
Result: (a4y²+b4x²) = c²
Note:
If the envelope of the family of curves A α²+Bα+c = 0. Where A,B,C are
functions of x and y.
B² - 4ac = 0.
1) Find the envelope of the family of straight lines y=mx+ 1/m , where m is
the parameter.
Soln:
The given equation can be written as m²x-my+1 = 0
Here A=x, B= -y, C=1.
B²-4ac = 0, y²-4x = 0 y² = 4x
2) Find the envelope of the family of straight lines represented by the
equation xcos α +ysin α = asec α( where α is a parameter).
Soln:
Given xcos α +ysin α = asec α -------------------- I
Dividing the equation I by cos α is
x+y tan α = a(sec α /cos α)
x+y tan α = a sec2 α
x+y tan α = a(1+tan² α)
x+y tan α = a+a tan² α
i.e) a tan² α – y tan α +(a-x) = 0
B²-4ac = 0, y²-4a(a-x) = 0
y²-4a²+4ax = 0.
3) Find the envelope of the family of straight lines y=mx+√a²m²+b² , where
m is the parameter.
Soln: ______
Given y=mx+√a²m²+b²
______
y-mx = √a²m²+b²
(y-mx)² = a²m²+b²
y² - 2mxy + m²x² - a²m² - b² = 0
m²(x²-a²) – 2yxm + (y²-b²) = 0
Here, A=x²-a², B= -2yx, C=y²-b²
Therefore,
The envelope is
(-2yx)² - 4(x²-a²)(y²-b²) = 0
4x²y² - 4x²y² + 4x²b² + 4a²y² - 4a²b² = 0
b²x² + a²y² = a²b²
i.e) (x²/a²) + (y²/b²) = 1 is the envelope of the given curve.
4) Find the envelope of (x- α)² + (y- α)² = 2α
Soln:
A=2, B= -2(x+y+1), C=x²+y²
Result: (x+y+1)2 = 2(x²+y²)
5) Find the envelope of the family of straight lines (x/a) + (y/b) = 1. where a
and b are connected by the relation (i) a+b = c, (ii) ab = c², where c is a
constant.
Soln:
Given (x/a) + (y/b) = 1 ------------------------- I
And a+b = c ------------------------ II
From the equation II we get b=c-a ------------------------- III
Sub. the equation III in I,
(x/a) + (y/ c-a) = 1 --------------------------IV
Here a is the only parameter.
(c-a)x +ay = a(c-a)
(cx-ax) + ay = ac-a²
i.e) a² + a(y-x-c) + cx = 0
Here A=1, B=y-x-c, C=cx
Therefore
The envelope is
(y-x-c)² - 4cx = 0 using B²=4AC
(y-x-c)² = 4cx
(y-x-c) = 2√c√x
y= x+c+2√c√x
y= (√x + √c)²
i.e) y = +(√x+√x) (or) y= -(√x+√c)
i.e) √x+√y = √c
Which gives the envelope of the given family of curves.
ii) Given
(x/a + y/b) = 1 ----------------- I
ab = c² ----------------- II
From II b = c²/a ---------------- III
Sub. the equation III in I we get,
(x/a + ay/c²) = 1
(c²x + a²y) = ac²
(a²y - ac² + c²x) = 0
(c4 – 4yc²x) = 0
c²c2 = 4xyc²
i.e) 4xy = c²
6) Find the envelope of the straight lines (x/a +y/b) = 1. Where the
parameters ‘a’ and ‘b’ are related by the equation an+bn = cn c being a
constant.
Soln:
Given (x/a +y/b) = 1 --------------------- I
an+bn = cn --------------------- II
Differentiate the equations I and II w.r. to ‘a’
(-x/a²) + (-y/b²)(db/da) = 0 --------------------- III
nan-1+ nbn-1(db/da) = 0
i.e) (db/da) = -an-1/bn-1 --------------------- IV
Sub. the equation IV in III we get,
(-x/a²) + (y/b²)(an-1/bn-1) = 0
(x/a²) = (y/b²)(an-1/bn-1)
(x/an+1) = (y/bn+1)
i.e) (x/a)/an = (y/b)/bn
(x/a + y/b) / (an+bn) = 1/cn
i.e) (x/an+1) = 1/cn
an+1 = cnx
a = (cnx)(1/ 1+n)
an = (cnx)(n/ n+1) ---------------------- V
(y/bn+1) = 1/cn
bn+1 = cny
bn = (cny)(n/ 1+n) ---------------------- VI
Sub. the equations V and VI in II
(cnx)(n/ n+1) + (cny)(n/ n+1) = cn
i.e) x(n/ n+1) + y(n/ n+1) = c(n/ n+1)
x(n/ n+1) + y(n/ n+1) = (cn)(c(-n²/ n+1))
= c(n²+n-n²)/ n+1
x(n/ n+1) + y(n/ n+1) = cn/ n+1
7) Find the envelope of the family of ellipses (x²/a² + y²/b²) = 1, where
a+b=c.
Soln:
Given (x²/a² + y²/b²) = 1 ----------------------- I
a+b = c ----------------------- II
Differentiate the equations I and II w.r. to ‘b’
(-2x²/ a³)(da/db) + (-2y²/b³) = 0
(da/db) = (-2y²/b³)(a³/ 2x²) = (-a³y²) / (b³x²) ---- III
From II we get,
(da/db)+1 = 0 (da/db) = -1 ----------------------- IV
From (III + IV) we get,
(-a³y²/ b³x²) = -1
i.e) (x²/a³) = (y²/b³)
(x²/a²) /a = (y²/b²) /b = (x²/a² + y²/b²) /(a+b) = 1/c
i.e) (x²/a³) = 1/c (y²/b³) = 1/c
x²c = a³ cy² = b³
a = (x²c)1/3 b = (cy²)1/3
Sub. a and b in the equation II we get,
(x²c)1/3 + (y²c)1/3 = c
c1/3 (x2/3 + y2/3) = c
i.e) (x2/3 + y2/3) = c2/3
Which is the required envelope.
x------------------------------------------------------------x
EVOLUTE AS THE ENVELOPE OF NORMALS
As the centre of curvature of a curve for a given point ‘P’ on it is the limiting
position of the intersection of the normal at P with the normal at any point
Q as Q implies P and evolute is the locus of the centre of curvature, the evolute
of a curve is the envelope of the normals of that curve.
1) Find the evolute of the parabola y² = 4ax as the envelope of normals.
Soln:
Equation of normals to the parabola is
y = mx -2am - am³ ------------------------------ I
m=parameter.
Differentiate the equation I w.r. to ‘m’
0 = x – 2a – 3m²a
i.e) 3m²a = x-2a
m² = x-2a /3a
m = (x-2a /3a)1/2 ------------------------------- II
Substitute the equation II in I we get,
y = [(x-2a) /3a]1/2 x - 2a[(x-2a) /3a]1/2 - a[(x-2a) /3a]3/2
y = [(x-2a) /3a]1/2 [x – 2a – a(x-2a) /3a]
y = [(x-2a) /3a]1/2 [(x-2a)(1- a/3a)]
y = [(x-2a) /3a]1/2 [(x-2a)²(4/9)]
y² = 4(x-2a)³ /27a
27ay² = 4(x-2a)³
2) Considering the evolute as the envelope of the normals, find the evolute
of the astroid x2/3 + y2/3 = a2/3 .
Soln:
The parametric equation of astroid is
x=acos³ θ, y=asin³ θ
(dx/dθ) = -3acos² θsin θ (dy/dθ) = 3asin² θcos θ
(dy/dx) = (3asin² θcos θ) / (-3acos² θsin θ) = -tan θ
Equation of the normal to the given asteroid is
y-y1 = (-1/m)(x-x1)
y-asin³ θ = (1/tan θ)(x-acos³ θ)
y-asin³ θ = (cos θ/sin θ)(x-acos³ θ)
(ysinθ - asin4 θ) = (xcos θ – acos4 θ)
(ysinθ - xcos θ) = a(sin4 θ - cos4 θ)
(ysinθ - xcos θ) = a(sin2 θ – cos2 θ)(sin² θ + cos² θ)
(ysinθ - xcos θ) = a(sin2 θ – cos2 θ)
(ysinθ - xcos θ) = -acos 2θ ------------------- I
Partially Differentiate the equation I w.r. to ‘θ’
(ycos θ + xsin θ) = 2asin 2θ ------------------ II
I X sin θ ====> ysin² θ – xcos θsi θ = -acos 2θsinθ
II X cos θ ===> ycos² θ + xcos θsinθ = 2asin 2θcosθ
y(sin² θ + cos² θ) = a(2sin2θcos θ – cos2θsi θ) ;( since, cos 2θ = cos² θ-sin² θ)
y = a[4sin θcos² θ - cos² θsin θ + sin³ θ]
y = a[3cos² θsinθ + sin³ θ] ------------------ V
Similarly,
x = a[cos³ θ + 3cosθsin² θ] ------------------ VI
(V+VI)==>
x+y = a(cosθ+sinθ)³ ----------------- VII
(V-VI)==>
x-y = a(cosθ – sinθ)³ ---------------- VIII
(x+y)2/3 + (x-y)2/3 = a2/3[(cosθ+sinθ)² + (cosθ – sinθ)²]
(x+y)2/3 + (x-y)2/3 = 2a2/3
3) Find the evolute of x²=4ay, considering it as the envelope of the normals.
Soln:
Any point on x²=4ay is (2at,at²)
Therefore, m=t
Result: 27ax² = 4(y-2a)³
4) Find the evolute of the parabola y²=4x considering it as the envelope of
its normals.
5) Considering the evolute as the envelope of normals find the evolute of
(x²/a² )+( y²/b² ) = 1
Soln:
The parametric equations are
x=acos θ y=bsin θ
(dx/dθ) = -asin θ (dy/dθ) = bcos θ
(dy/dx) = (bcos θ)/(-asin θ) = (-b/a)cot θ = m
The equation of the normal is
y-y1 = (-1/m)(x-x1)
y-bsin θ = (a/bcot θ)(x-acos θ)
= (asin θ/bcosθ)(x-acos θ)
(ybcos θ - b²sin θ cos θ) = (xasin θ - a²cos θsin θ)
(yb/sin θ)–b² = (ax/cos θ)-a²
(ax/cos θ) – (by/sin θ) = a²-b² ------------------------- I
Differentiate the equation I partially w.r. to ‘θ’
(-ax/cos² θ)(-sin θ) + (by/sin² θ)(cos θ) = 0
(ax/cos² θ)(sin θ) = (-by/sin² θ)(cos θ)
(sin³ θ/cos³ θ) = (-by/ ax)
tan θ = (-by/ ax)1/3 ___________
sin θ = [(-by)1/3 / √(ax)2/3+(by)2/3 ]
___________
cos θ = [(ax)1/3 / √(ax)2/3+(by)2/3 ]
__________ __________
I ==> { ax /[(ax)1/3 / √(ax)2/3+(by)2/3 ] } - { (by /[(-by)1/3 / √(ax)2/3+(by)2/3 ] }
= (a²-b²)
__________ __________
{[ax√(ax)2/3+(by)2/3] / (ax)1/3 } + {[by√(ax)2/3+(by)2/3] /(by)1/3 } = (a²-
b²)
__________
[√(ax)2/3+(by)2/3 ] [(ax /(ax)1/3) + (by /(by)1/3] = (a²-b²)
__________
[√(ax)2/3+(by)2/3 ] [(ax)2/3+(by)2/3] = (a²-b²)
[ (ax)2/3+(by)2/3 ]3/2 = (a²-b²)
Multiplying both sides by power 2/3
[(ax)2/3+(by)2/3] = (a²-b²)2/3
Which is the required equation.
xx-------------------------------------------------------xx
OBJECTIVE QUESTIONS
1.The radius of curvature and curvature of is
(a)3, (b)3, (c)27, (d)9,
2.For the given curve y = f(x), if = = ------------------
3. Say true or false: (i)The circle is the only curve of constant curvature.
(ii)The radius of curvature of the curve s = 4a sin Ψ is
4.The parametric equation of the parabola 2y =4ax is-------------
5.The normal at any point of a curve is tangent to its evolutes touching at the
corresponding ----------
6.The envelope of family of curves not touching every member of the family of
curves. (T/F)
7.Euler’s theorem is valid only for homogeneous function. (T/F)
8.If u =Ø(x,y) and x =f(r,s),y = g (r,s) then r
u
∂
∂=
9. A stationary point (a,b) is called as saddle point ,then the value of f(x,y) at
(a,b) is ----------
PART-A 1.Find the radius of curvature at (3, 4) for the curve =25
2.If x= r cos ө y= r sin ө find ),(
),(
yx
r
∂
∂ θ
3. If u = )(1
tanx
y− where 2x +
2y = 2a by treating u as a function of x & y only.
Find dx
du
4. Find the envelope of the family of lines y= mx +m
a m being the parameter.
5.What is the curvature of (a)Straight line (b)Circle of curvature 2 units. 6.Find the envelope of the family of circles ( x - α ) 2 + y2 = 4 α , α being the parameter. 7.State the properties of evolutes.
8.Obtain du when u = log (xy) if x2 + y2 = a2 dx 9.Find the stationary points of x2 - xy + y2 - 2x + y . 10.State the necessary conditions for f(x,y) to have an extremum at (a,b). Are these conditions sufficient? 11.If u = y2 and v = x2 , find ∂(x,y) . x y ∂(u,v)
PART-B
1.P.T. the for the catenary y=c cosh( ) is equal to the portion of the normal
intercepted between the curve and the x-axis and that it varies as the square of the ordinate.
2.P.T. for the curve + =3axy, the radius of curvature at that point ( ) is
numerically equal to
3. A rectangular box open at the top is to have a volume of 32 cc. Find the
dimensions of the box that require the least material for its construction.
4.Expand xe cosy in powers of x & y as far as the terms of second degree.
5.Examine f(x,y)= 23x - 2y +
3x for its extreme values.
6.Find the envelope of the family of lines +a
x
b
y=1 where parameters a & b are
connected by relation na + n
b = nc
7.Show that evolute of the cycloid x=a(Ө-sin Ө) y=a(1-cos Ө) is another cycloid.
8.Find the radius of curvature at the point 3a , 3a on the curve x3 + y3 = 3axy. 2 2
9.Find the centre of curvature at t = π on the curve x = 2 cos t + cos 2t , y = 2 sin t + sin 2t. 2 10.Find the evolute of the tractrix x = a cos θ + log tan θ , y = a sin θ, treating it as the envelope of its 2 normals. 11.Expand ex cos y in powers of x and y as far as the terms of second degree. 12.Find the maximum and minimum values of f(x,y) = sin x sin y sin ( x + y ) ; 0 < x , y < π.