DIFFERENTIAL_CALCULUS

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1 DIFFERENTIALCALCULUS 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.

2 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+y 1 ²) 3/2 Y 2

3 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). 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) = II

4 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+y 1 ²) 3/2 ) / y 2 = ((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 Given a²y = x³-a³. Differentiate write to x We get, a²(dy/dx) = 3x² dy/dx = 3x²/a² d²y/dx² = 6x/a²

5 ρ = ((1+y 1 ²) 3/2 ) / y 2 (1+(9x 4 /a 4 )) 3/2 ((a 4 +9x 4 )/a 4 )) 3/2 = = 6x / a² 6x / a² (a 4 +9x 4 ) 3/2 a² =. a 6 = ((a 4 +9x 4 ) 3/2 ) / (6a 4 x) 6x 3) Find ρ for the curve y = c cosh (x/c) at the point (0,c) (or) y =( a/2)(e x/a + e -x/a ) Given y = c cosh (x/c) (or) y = a/2(e x/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

6 4) Find the radius of curvature of the curve xy² = a³-x³ at (a,0) 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) = 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) / 9a 4 = -2/3a Therefore ρ = 3a/2 5) Find the radius of curvature at x=c on the curve xy = c².

7 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 d²y/dx² [x.((dy/dx)) y.1] = - x² [x(-y/x)-y] = - x² d²y/dx² = 2y/x² (d²y/dx²) (c,c) = 2c/c² = 2/c

8 Therefore the radius of curvature at (c,c) is ρ = (1+y 1 ²) 3/2 (1+1) 3/2 y 2 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). 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+y 1 ²) 3/2 ) / y 2 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)

9 7) Find the radius of curvature of y =2a on the curve y² =4ax. Given curve y²=4ax 2y(dy/dx) = 4a (dy/dx) = 2a /y (dy/dx) (0,2a) = 2a / 2a =1 Therefore y 1 =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) = 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. Given y = ax / a+x = a (x/(a+x)) I Differentiate with respect to x

10 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) = (-2a²) / (a+x)³ 1-2a³x³ -2y³ =. = ax³ (a+x)³ ax³ (1+(y 4 /x 4 )) 3/2 Therefore ρ = (-2y³ / ax³) 2ρ (1+(y 4 /x 4 )) 3/2 = a (-y³ / x³) Raise the power 2/3 on both sides (2ρ/a) 2/3 (1+(y 4 /x 4 )) = (y²/x²) (x 4 +y 4 ) x² =. x 4 y²

11 = (x 4 +y 4 ) / (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). 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

12 Therefore [1+(-1)²] 3/2 ρ = 4 = (2 2) / 4 ρ = 1/ 2 10) Find the radius of curvature for y=e x at (0,1) Given y = e x y 1 = e x y 2 = e x Hence at (0,1) y 1 = 1 y 2 = 1 ρ = ((1+1) 3/2 ) / 1 = 2 2 x x

13 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. x=at² x =2at x =2a y=2at y =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). Given x=a(θ+sin θ) y=a(1-cos θ)

14 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/(4acos 4 θ/2)) [1+(dy/dx)²] 3/2 (1+tan² θ/2) 3/2 ρ = = (d²y/dx²) (1 / (4acos 4 θ/2) = (sec² θ/2) 3/2 (4acos 4 θ/2) = (4acos 4 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)

15 (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. 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

16 (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

17 POLAR FORM 1) Find ρ for the curve r =ae θcot α. Given r = ae θcot α (dr/dθ) = acotα e θcot 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 r n =a n cosnθ, the radius of curvature is (a n.r - n+1 ) / (n+1). Given r n =a n cosnθ Taking log on both sides

18 log r n = log a n cosnθ nlog r = log a n + 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)

19 = (r.a n ) / (1+n)r n ρ = (a n.r n+1 ) / (1+n) 3) Find the radius of curvature of the curve r²cos 2θ = a². Given r²cos 2θ = a² Differential equation I w.r. to θ I 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+2r 1 ²-rr (r²+r²tan² 2θ) 3/2 = (r²+2r²tan² 2θ) (r(2rsec² 2θ+rtan² 2θ)) (r³sec³ 2θ) = (r² + r²tan² 2θ - 2r²sec² 2θ)

20 (r³sec³ 2θ) = r²(1 + tan² 2θ 2sec² 2θ) (r³sec³ 2θ) = = -rsec θ -r²sec² 2θ Therefore, ρ = rsec θ x x

21 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 (y 1 /y 2 )(1+y 1 ²) y = y + (1/y 2 )(1+ y 1 ²) Equation of circle of curvature is, (x x)² + (y y)² = ρ²

22 1) Find the centre of curvature of the curve y = 3x³ + 2x² - 3 at (0,- 3). 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 - (y 1 /y 2 )(1+y 1 ²) = 0 (0/4)(1+0) = 0 y = y + (1/y 2 )(1+ y 1 ²) = -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. 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

23 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 (y 1 /y 2 )(1+y 1 ²) = c + ((c/2)(1+1)) x = 2c y = y + (1/y 2 )(1+ y 1 ²) = 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

24 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) Therefore = (1 / a/4) = 4/a The radius of curvature is ρ = ((1+y 1 ²) 3/2 ) / y 2 = (1+(-1)²) 3/2 / (4/a) = (2 2 a ) / 4 = a / 2

25 x = x (y 1 /y 2 )(1+y 1 ²) = (a/4) + (a/4)(1+1) x = 3a / 4 y = y + (1/y 2 )(1+ y 1 ²) =(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). 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

26 = -1/6 y = -6, x = 15, ρ = 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). 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 (y 1 /y 2 )(1+y 1 ²) = 3 +. ( 1 + ) = 43/6

27 y = y + (1/y 2 )(1+ y 1 ²) 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

28 INVOLUTES AND EVOLUTES Let A 1, A 2, A 3, A 4,.. be any points on the curve y=f(x). Let B 1,B 2,B 3,B 4,... be the centre of curvature corresponding to the points A 1,A 2,A 3,A 4,.. respectively. As we move along the curve y=f(x) from A 1 to A 2, A 3 the centre of curvature will also moves along a curve B 1,B 2,B 3,B 4.. EVOLUTE: The locus of centre of curvature of a curve (B 1, B 2, B 3, B 4,.) is called evolute of the given curve. INVOLUTE: If a curve C 2 is the evolute of a curve C 1. then C 1 is said to be an Involute of a curve C 2.

29 1) Find the equation of the evolute of the parabola y²=4ax 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) y 1 = 1/t y 2 = (d²y/dx²) = (d/dx)(dy/dx) = (d/dx)(1/t) = (d/dt)(1/t)(dt/dx) = (-1/t²)(1 /2at) y 2 = (-1/ 2at³) The coordinates of the centre of curvature is (x, y) Where x = x (y 1 /y 2)(1+y 1 ²) (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/y 2 )(1+ y 1 ²) = 2at 2at³((1+t²) /t²)

30 = 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

31 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 = a 2/3. The equation of the given curve is x 2/3 + y 2/3 = a 2/ 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 y 1 = (dy/dx) = (3asin² θcos θ) / (-3acos² θ sin θ ) = - tan θ y 2 = ((d²y/dx²) = (d/dx)(dy/dx) = (d/dx)(-tan θ) = (d/dθ)(-tan θ).(dθ/dx) = (-sec² θ) / (-3acos² θsin θ) y 2 = (sec 4 θcosec θ) / 3a The co-ordinates of the centre of curvature is ( x, y ) Where x = x (y 1 /y 2)(1+y 1 ²) - tan θ = acos³ θ - { } (1+tan² θ) (sec 4 θcosec θ) / 3a 3a tan θ = acos³ θ +{ } sec² θ (sec 4 θcosec θ)

32 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/y 2 )(1+ y 1 ²) 3a = asin³ θ +{ } sec² θ (sec 4 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

33 ( x + y ) 2/3 + ( x - y ) 2/3 = {a(cos θ+sin θ)³} 2/3 + { a(cos θ-sin θ)³} 2/3 = a 2/3 {(cos θ+sin θ)²+(cos θ-sin θ)²} = a 2/3 {2(cos² θ+sin² θ)} = 2a 2/3 (x+y) 2/3 + (x-y) 2/3 = 2a 2/3 3) Find the equation of the evolute of the ellipse (x²/a²) + (y²/b²) = 1. The equation of given curve is (x²/a²) + (y²/b²) = I The parametric equation of I is x = acos θ y = bsinθ (dx/dθ) = -asin θ (dy/dθ) = bcos θ y 1 = (dy/dx) = (bcos θ) / (-asin θ) = (-b/a)cot θ y 2 = (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) y 2 = (-b/a²)cosec³ θ The co-ordinate of the centre of curvature is ( x, y ) Where x = x (y 1 /y 2)(1+y 1 ²) (-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)

34 = 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/y 2 )(1+ y 1 ²) (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(cos 2 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) 2/3 cos² θ cos² θ = (a x) 2/3 / (a²-b²) 2/ IV

35 From III by = -(a²-b²)sin³ θ (by) 2/3 = (a²-b²) 2/3 sin² θ sin² θ = ( by) 2/3 / (a²-b²) 2/ 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. 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) y 2 = (d²y/dx²) = (2/ ct³) III

36 Where x = x (y 1 /y 2)(1+y 1 ²) -1/t² = ct ( )(1+ 1/t 4 ) 2/ct³ = ct + (ct/2)(1+ 1/t 4 ) = ct + (ct/2) + (c/2t³) x = (3ct/2) + (c/2t³) IV y = y + (1/y 2 )(1+ y 1 ²) = (c/t) + (ct³/2)(1+ 1/t 4 ) = (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 + 3t 4 + 3t² + t 6 ] = 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)

37 = (c/2) 2/3 (2²) = (c 2/3 / 2 2/3 )(2²) = (c 2/3 )(2 2-2/3 ) = (c 2/3 )(2 4/3 ) = (c 2/3 )(4 2/3 ) ( x + y ) 2/3 + ( x - y ) 2/3 = (4c) 2/3. 5) Find the evolute of the hyperbola (x²/a² - y²/b²) = 1. The given equation is (x²/a² - y²/b²) = 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 y 1 = (bsec² θ) / (asec θtan θ) = (b/a)(sec θ / tan θ) =(b/a)((1/cosθ)/(sinθ/ cosθ)) = (b/a)cosec θ II y 2 = (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/a 2 ) (1/sin θ) (cos θ/sin θ) (cos θ) (cos θ/sin θ) = (-b/a)cot³ θ Where x = x (y 1 /y 2)(1+y 1 ²) = 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/cos 3 θ)

38 = 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/ III y = y + (1/y 2 )(1+ y 1 ²) = 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/ IV WKT. (sec² θ - tan² θ) = V Sub. the equations III and IV in V { [ ax/ (a²+b²) ] 2/3 - [ -by/(a²+b²)] 2/3 } =1

39 6) S.T The evolute of the cycloid x=a(θ-sin θ), y=a(1-cos θ) is another equal cycloid. x = a(θ-sin θ) y = a(1-cos θ) (dx/dθ) = a(1-cos θ) (dy/dθ) = asin θ y 1 =(dy/dx) = (asin θ/ a(1-cos θ))= (2sin (θ/2) cos( θ/2)/(2sin²( θ/2)) =cot θ/ I y 2 = (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) θ/2)] = (-1 / 4asin 4 θ/2) II x = x - (y 1 /y 2 )(1+y 1 ²) = a(θ-sin θ) - [ (cos θ/2 / sin θ/2)(1+ cos² (θ/2)/sin² (θ/2) ] / [-1/ 4asin 4( θ/2)] = a(θ-sin θ) + [(4asin 4 θ/2)(cos( θ/2) / sin θ/2)] + [(4asin 4 θ/2)(cos θ/2 / sin θ/2)(cos² θ/2 / sin² = 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

40 y = y - (1/y 2 )(1+ y 1 ²) = a(1-cos θ) - 4asin 4 θ/2(1+ cot² θ/2) = a - acos θ - 4asin 4 θ/2(sec² θ/2) = a - acos θ - 4asin 4 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). 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 y 1 = (dy/dx) = (aθsin θ/aθcos θ) = tan θ I y 2 = (d²y/dx²) = (d/dx)tan θ = (d/dθ)tan θ(dθ/dx) = sec² θ(1/ aθcos θ) y 2 = 1/ aθcos³ θ II Let the centre of curvature be( x, y )

41 x = x - (y 1 /y 2 )(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/y 2 )(1+ y 1 ²) = 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 sin θ = y/a V VI (V)² +(VI)²==> cos² θ+sin² θ = (x/a)² + (y/a)² x²+y² = a². x x

42 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. 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 = 2x 1/2 y = 2x 1/2 y² = 4x

43 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. 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 x 2 y 2 = a(x²+y²) Which is the equation of the envelope of the given family of straight lines.

44 4) Find the envelope of xcos α /a+ysin α /b = 1, (α is the parameter). Given ( xcos α /a) + (ysin α /b) = 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 ]. Given ycos α - xsin α = acos 2 α I

45 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 = a 2/3 (cos α +sin α) 2 + a 2/3 (cos α -sin α) 2 = a 2/3 (2) = 2a 2/3 (x+y) 2/3 +(x-y) 2/3 = 2a 2/3 is the required envelope.

46 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. Differentiate partially w.r. to θ (-a²sin θ/x) (b²cos θ/y) = 0 tan θ = (-b²x /a²y) sin θ = (-b²x / a 4 y²+b 4 x² ) cos θ = (a²y / a 4 y²+b 4 x²) Result: (a 4 y²+b 4 x²) = 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.

47 1) Find the envelope of the family of straight lines y=mx+ 1/m, where m is the parameter. 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). 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 sec 2 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. Given y=mx+ a²m²+b²

48 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 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. Given (x/a) + (y/b) = I And a+b = c II From the equation II we get b=c-a Sub. the equation III in I, III

49 (x/a) + (y/ c-a) = 1 Here a is the only parameter. (c-a)x +ay = a(c-a) (cx-ax) + ay = ac-a² IV 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 (y-x-c)² = 4cx (y-x-c) = 2 c x y= x+c+2 c x y= ( x + c)² using B²=4AC 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) = 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 (c 4 4yc²x) = 0

50 c²c 2 = 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 a n +b n = c n c being a constant. Given (x/a +y/b) = I a n +b n = c n II Differentiate the equations I and II w.r. to a (-x/a²) + (-y/b²)(db/da) = III na n-1 + nb n-1 (db/da) = 0 i.e) (db/da) = -a n-1 /b n IV Sub. the equation IV in III we get, (-x/a²) + (y/b²)(a n-1 /b n-1 ) = 0 (x/a²) = (y/b²)(a n-1 /b n-1 ) (x/a n+1 ) = (y/b n+1 ) i.e) (x/a)/a n = (y/b)/b n (x/a + y/b) / (a n +b n ) = 1/c n i.e) (x/a n+1 ) = 1/c n a n+1 = c n x a = (c n (1/ 1+n) x) a n = (c n (n/ n+1) x) V (y/b n+1 ) = 1/c n b n+1 = c n y

51 b n = (c n y) (n/ 1+n) VI Sub. the equations V and VI in II (c n x) (n/ n+1) + (c n y) (n/ n+1) = c n i.e) x (n/ n+1) + y (n/ n+1) (n/ n+1) = c x (n/ n+1) + y (n/ n+1) = (c n )(c (-n²/ n+1) ) (n²+n-n²)/ n+1 = c x (n/ n+1) + y (n/ n+1) n/ n+1 = c 7) Find the envelope of the family of ellipses (x²/a² + y²/b²) = 1, where a+b=c. Given (x²/a² + y²/b²) = 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) = 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

52 Sub. a and b in the equation II we get, (x²c) 1/3 + (y²c) 1/3 = c c 1/3 (x 2/3 + y 2/3 ) = c i.e) (x 2/3 + y 2/3 ) = c 2/3 Which is the required envelope. x x

53 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. Equation of normals to the parabola is y = mx -2am - am³ m=parameter I 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/ 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

54 2) Considering the evolute as the envelope of the normals, find the evolute of the astroid x 2/3 + y 2/3 = a 2/3. 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-y 1 = (-1/m)(x-x 1 ) y-asin³ θ = (1/tan θ)(x-acos³ θ) y-asin³ θ = (cos θ/sin θ)(x-acos³ θ) (ysinθ - asin 4 θ) = (xcos θ acos 4 θ) (ysinθ - xcos θ) = a(sin 4 θ - cos 4 θ) (ysinθ - xcos θ) = a(sin 2 θ cos 2 θ)(sin² θ + cos² θ) (ysinθ - xcos θ) = a(sin 2 θ cos 2 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

55 (V-VI)==> x-y = a(cosθ sinθ)³ VIII (x+y) 2/3 + (x-y) 2/3 (x+y) 2/3 + (x-y) 2/3 = 2a 2/3 = a 2/3 [(cosθ+sinθ)² + (cosθ sinθ)²] 3) Find the evolute of x²=4ay, considering it as the envelope of the normals. 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 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-y 1 = (-1/m)(x-x 1 ) 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)

56 (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 } = (a²b²) [ (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

57 OBJECTIVE QUESTIONS 1.The radius of curvature and curvature of (a)3, (b)3, (c)27, (d)9, 2.For the given curve y = f(x), if = = 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 y 2 =4ax is The normal at any point of a curve is tangent to its evolutes touching at the corresponding 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 u = r 9. A stationary point (a,b) is called as saddle point,then the value of f(x,y) at (a,b) is is PART-A 1.Find the radius of curvature at (3, 4) for the curve =25 ( r, θ ) 2.If x= r cos ө y= r sin ө find ( x, y) 3. If u = tan 1 y ( ) x du Find dx where x y = a 2 by treating u as a function of x & y only. 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 + y 2 = 4 α, α being the parameter. 7.State the properties of evolutes.

58 8.Obtain du when u = log (xy) if x 2 + y 2 = a 2 dx 9.Find the stationary points of x 2 - xy + y 2-2x + y. 10.State the necessary conditions for f(x,y) to have an extremum at (a,b). Are these conditions sufficient? 11.If u = y 2 and v = x 2, 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 5.Examine f(x,y)= e x cosy in powers of x & y as far as the terms of second degree. 3x 2 - y x for its extreme values. x 6.Find the envelope of the family of lines + a connected by relation a n + b n = c n y =1 where parameters a & b are b 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 x 3 + y 3 = 3axy. 3a, 3a on the curve 2 2

59 9.Find the centre of curvature at t = π on the curve x = 2 cos t + cos 2t, y = 2 sin t + sin 2t Find the evolute of the tractrix x = a cos θ + log tan θ it as the envelope of its normals., y = a sin θ, treating 2 11.Expand e x 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.