Differential Calculus—I

UNIT I

1.1 INTRODUCTION

In many practical situations engineers and scientists come across problems which involve quantities of varying nature. Calculus in general, and differential calculus in particular, provide the analyst with several mathematical tools and techniques in studying how the functions involved in the problem behave. The student may recall at this stage that the derivative, obtained through the basic operation of calculus, called differentiation, measures the rate of change of the functions (dependent variable) with respect to the independent variable. In this chapter we examine how the concept of the derivative can be adopted in the study of curvedness or bending of curves.

1.2 RADIUS OF CURVATURE

Let P be any point on the curve C. Draw the tangent at P to the Y C circle. The circle having the same curvature as the curve at P touching the curve at P, is called the circle of curvature. It is also O called the osculating circle. The centre of the circle of the curvature is called the centre of curvature. The radius of the circle P of curvature is called the radius of curvature and is denoted by ‘ρ’.

Note : 1. If k (> 0) is the curvature of a curve at P, then the radius O X 1 of curvature of the curve of ρ is . This follows from the definition k Fig. 1.1 of radius of curvature and the result that the curvature of a circle is the reciprocal of its radius. dψ Note : 2. If for an arc of a curve, ψ decreases as s increases, then is negative, i.e., k is negative. ds 1 ds But the radius of a circle is non-negative. So to take ρ = = some authors regard k also as non-negative k dψ dψ i.e., k = . ds 1 2 ENGINEERING MATHEMATICS—II

dψ The sign of indicates the convexity and concavity of the curve in the neighbourhood of ds ds the point. Many authors take ρ = and discard negative sign if computed value is negative. dψ 1 ∴ Radius of curvature ρ = · k

1.2.1 Radius of Curvature in Cartesian Form Suppose the Cartesian equation of the curve C is given by y = f (x) and A be a fixed point on it. Let P(x, y) be a given point on C such that arc AP = s. Then we know that dy = tan ψ ...(1) dx where ψ is the angle made by the tangent to the curve C at P with the x-axis and

1 ds |R F dyI 2 |U 2 = S1 + G J V ...(2) dx T| H dxK W| Differentiating (1) w.r.t x, we get 2 ψ dy 2 d = sec ψ ⋅ dx2 dx dψ ds = di1 +⋅tan2 ψ ds dx

1 L F dyI 22O 1 L F dyI O 2 = M1 + G J P M1 + G J P [By using the (1) and (2)] NM H dxK QP ρ NM H dxK QP

3 1 |R F dyI 2 |U 2 = S1 + G J V ρ T| H dxK W|

3 |R F dyI 2 |U 2 S1 + G J V T| H dxK W| Therefore, ρ = ...(3) dy2 dx2

dy dy2 where y = and y = . 1 dx 2 dx2 DIFFERENTIAL CALCULUS—I 3

Equation (3) becomes, 3 + 2 2 ot1 y1 ρ = y2 This is the Cartesian form of the radius of curvature of the curve y = f (x) at P (x, y) on it.

1.2.2 Radius of Curvature in Parametric Form Let x = f (t) and y = g (t) be the Parametric equations of a curve C and P (x, y) be a given point on it. dy dy dt Then = ...(4) dx dx dt dy2 d Rdy/ dt U dt and = S V ⋅ dx2 dt Tdx/ dt W dx

2 2 dx ⋅−⋅dy dy dx dt dt 2 dt dt 2 1 = ⋅ GF dxJI 2 dx H dt K dt dx dy2 dy dx2 2 ⋅⋅– dy dt dt 2 dt dt 2 = ...(5) dx2 GF dxJI 3 H dt K

3 |RF I 22F I |U 2 SG dxJ + G dyJ V T|H dt K H dt K W| ∴ρ = 2 2 ...(6) dx ⋅−⋅dy dy dx dt dt 2 dt dt 2 4 ENGINEERING MATHEMATICS—II

dx dy dx2 dy2 where x′ = , y′ = , x″ = , y″ = dt dt dt 2 dt 2 3 otxy′+′222 ρ = xy′″– yx ′″ This is the cartesian form of the radius of curvature in parametric form.

WORKED OUT EXAMPLES

F x I 1. Find the radius of curvature at any point on the curve y = a log sec G J . H aK Solution 3 + 2 2 ot1 y1 Radius of curvature ρ = y2 F x I Here, y = a log sec G J H aK 11F xI F xI y = a × ⋅ secG J tan G J ⋅ 1 F xI H aK H aaK sec G J H aK F x I y = tan G J 1 H aK

2 F xI 1 y = sec G J ⋅ 2 H aaK

3 R F xIU 2 S1 + tan2 G JV T H aKW Hence ρ = 1 F xI sec2 G J a H aK

32 R F xIU F xI Ssec2 G JV a sec3G J T H aKW H aK = = 1 F xI F xI sec2 G J sec2 G J a H aK H aK F x I = a sec G J H aK F xI ∴ Radius of curvature = a sec G J H aK DIFFERENTIAL CALCULUS—I 5

F xI y2 2. For the curve y = c cos h G J , show that ρ = · H cK c 3 + 2 2 di1 y1 Solution ρ = y2 F xI Here, y = chcos G J H cK F xI 1 F x I y = chsin G J × = sin h G J 1 H ccK H c K F xI 1 and y = cos h G J × 2 H ccK R U 3 3 2 F xI 2 F xI 2 S1 + sin h G JV chGcos 2 J T H cKW H cK ρ = = 1 F xI x cosh G J cosh c H cK c

2 F xI 1 F F xII = c cos h2 G J = Gchcos G JJ H c K c H H cKK

1 2 = ⋅ y c y 2 ∴ρ= · Hence proved. c 3. Find the radius of curvature at (1, –1) on the curve y = x2 – 3x + 1.

3 + 2 2 di1 y1 Solution. Where ρ = at (1, – 1) y2 Here, y = x2 – 3x + 1

y1 =2x – 3, y2 = 2

Now, ( y1)(1, –1) =– 1

( y2)(1, –1) =2 3 bg11+ 2 22 ∴ρ= = (1, –1) 2 2 = 2 4. Find the radius of curvature at (a, 0) on y = x3 (x – a).

3 + 2 2 di1 y1 Solution. We have ρ = at (a, 0) y2 6 ENGINEERING MATHEMATICS—II

Here, y = x3(x – a) = x4 – x3a 3 2 y1 =4x – 3ax 2 and y2 =12x – 6ax 3 3 3 Now ( y1)(a, 0) =4a – 3a = a 2 2 2 ( y2)(a, 0) =12a – 6a = 6a 3 R 2 U TS1 + dia 3 WV 2 ∴ρ= (a, 0) 6a 2 3 ot1 + a6 2 = · 6a 2 πa F xI 5. Find the radius of curvature at x = on y = a sec G J . 4 H aK

3 + 2 2 di1 y1 πa Solution. We have ρ = at x = y2 4 F xI Here y = a sec G J H aK F xI F x I 1 ∴ y = a secG J ⋅ tan G J × 1 H aK H aaK F xI F xI y = secG J tan G J 1 H aK H aK

32x 11F xI F xI and y = sec×+ secG J ⋅ tan G J ⋅ 2 aa H aK H aaK 1 L F xI F xI F xIO = Msec32G J + secG J tan G JP a N H aK H aK H aKQ πa ππ At x = , y = sec⋅= tan 2 4 1 44 1 32 and y = ej22+= 2 2 aa

3 R 2 U 2 S12+ ejV ρ T W 33 ∴ πa =⋅a x = = 4 32 32 a 3 = a . 2 DIFFERENTIAL CALCULUS—I 7

π F xI 6. Find ρ at x = on y = 2 log sin G J . 3 H 2K

3 + 2 2 di1 y1 π Solution. We have ρ = at x = y2 3 F xI The curve is y = 2 log sin G J H 2K

1 F xI 1 y = 2 ⋅ × cos G J × 1 F xI H 2K 2 sin G J H 2K F x I = cot G J H 2K F xI 1 and y = –cosec2 G J × 2 H 2K 2 π F πI At x = , y = cot G J = 3 3 1 H 6K –1 π and y = cosec2 = – 2 2 26 3 R 2 U 2 TS13+ ejWV ∴ ρ π x = = 3 – 2

3 bg13+ 2 42× = = = – 4. ––2 2

F 3a 3aI 7. Find the radius of curvature at G , J on x3 + y3 = 3axy. H 2 2 K

3 + 2 2 ot1 y1 F 3aa3 I Solution. We have ρ = at HG , KJ . y2 2 2 Here, x3 + y3 =3axy Differentiating with respect to x 2 2 3x + 3y y1 =3a (xy1 + y) 2 2 3 ( y – ax) y1 =3 (ay – x ) ay– x2 ⇒ y = ...(1) 1 yax2 − 8 ENGINEERING MATHEMATICS—II

Again differentiating w.r.t x. 2 −⋅bg − −−2 bg − diyaxayxayxyya1 22di1 ⇒ y2 = 2 ...(2) diyax2 −

F 3aa3 I Now, from (1), at G , J H 2 2 K

F 3aaI F 3 I 2 a G J − G J H 2 K H 2 K y = 1 F 3aI 2 F 3aI G J − a G J H 2 K H 2 K 69aa22− = 96aa22−

–9diaa22− 6 = = –1 di96aa22−

F 3aa3 I From (2), at G , J H 2 2 K F 9aa223 I F 3aa 229 I G − J bg––aa−−3 G − J bg3aa− H 2 2 K H 2 4 K y = 2 F 22I 2 G 9aa− 3 J H 4 2 K

3 3a2 ––aa2 ××4 4a 4 4 = F 2 I 2 G 3a J H 4 K

––6a 3 32 = = 9a 4 3a 16 Using these 3 2 2 ρ {}11+ bg– GF 3aa3 JI H , K = F I 2 2 G− 32J H 3a K 22× 3aa− 3 = – = 32 82 F 3aa3 I 3a ∴ Radius of curvature at G , J is · H 2 2 K 82 DIFFERENTIAL CALCULUS—I 9

8. Find the radius of curvature of b2x2 + a2y2 = a2b2 at its point of intersection with the y-axis.

3 + 2 2 ot1 y1 Solution. We have ρ = at x = 0 y2 Here, b2x2 + a2y2 = a2b2 When x = 0, a2y2 = a2b2 y2 = b2 ⇒ y = ± b i.e., the point is (0, b) or (0, – b) The curve is b2x2 + a2y2 = a2b2. Differentiating w.r. to x 2 2 2b x + 2a yy1 =0 bx2 y = – 1 ay2 Differentiating again w.r. to x 2 F − I – b yxy1 y = G J 2 a 2 H y2 K

– b2bg0 Now at (0, b), y = = 0 1 ab2bg

– b2 F b − 0I and y = G J 2 a22H b K – b = a2 i.e., Radius of curvature at (0, b) is 3 bg10+ 2 – a 2 ∴ρ= = (0, b) GF – bJI b H a 2 K a 2 ∴ Radius of curvature is b Next consider (0, – b), – b2 0 y = × = 0 1 a 2 – b –––b2 F b 0I a 2 y = G J = 2 a 22H b K b 10 ENGINEERING MATHEMATICS—II

3 bg10+ 2 a2 ρ = = (0, – b) GF b JI b H a 2 K a2 ∴ Radius of curvature of (0, – b) is . b r 3 9. Show that at any point P on the rectangular hyperbola xy = c2, ρ = where r is the 2c2 distance of the point from the origin. Solution. The curve is xy = c2 Differentiating w.r. to x

xy1 + y =0 y y = – 1 x Again differentiating w.r.t. x R xy− y U y = – S 1 V 2 T x2 W R – xy U | − y | x 2y = – S V = T| x22W| x

3 |R F yI 2 |U 2 3 S1 + G J V + 2 2 H K di1 y1 T| x W| ρ = = 2y y2 x2 3 dixy22+ 2 = 2y x3 × x2 3 dixy22+ 2 = 2xy where x2 + y2 = r2 and xy = c2. r3 ∴ρ= . 2c2 x2 y2 ab22 10. Show that, for the ellipse + = 1, ρ = where p is the length of the perpen- a 2 b2 p3 dicular from the centre upon the tangent at (x, y) to the ellipse. DIFFERENTIAL CALCULUS—I 11

x2 y2 Solution. The ellipse is + = 1 a 2 b2 Differentiating w.r.t. x 22x yy + 1 =0 a 2 b2 b2 x ⇒ y = – 1 a 2 y Again Differentiating w.r. to x – b2 L yxy− O y = M 1 P 2 a 2 N y2 Q

L b2 x2 O M y +⋅P – b2 M a 2 y P = 2 M 2 P a NM y QP

b4 L y2 x2 O = – M + P ay23Nb2 a 2 Q

– b4 L x2 y2 O y = MQ +=1P 2 ay23 N a2 b2 Q

3 42 3 F bx I 2 1 + + 2 2 G 42J ot1 y1 H axK Now, ρ = = y F b4 I 2 G– J H ay23K

3 42 42 diay+ bx 2 ay23 = − × ay63 b4

3 diay42+ bx 422 ρ = – ab44 3 diay42+ bx 422 Taking magnitude ρ = ...(1) ab44 x2 y2 The tangent at (x , y ) to the ellipse + = 1 is 0 0 a 2 b2 xx yy 0 + 0 =1 a2 b2 12 ENGINEERING MATHEMATICS—II

Length of perpendicular from (0, 0) upon this tangent 1 = F x I 2 F y I 2 G 0 J + G 0 J H a2 K H b2 K

ab22 = 4 24+ 2 ay0 bx0 So, the length of perpendicular from the origin upon the tangent at (x, y) is ab22 p = 4 24+ 2 ay0 bx0

By replacing x0 by x and y0 by y ab22 p = ay42+ bx 42 Reciprocal and cube on both sides, we get, 3 42+ 422 1 diay bx ⇒ = p3 ab66 3 42+ 422 diay bx 1 = × ab44 ab 22 By using eq. (1), we get 1 ρ = p3 ab22 ab22 ⇒ρ= · p3

2 2 2 ax F 2ρI 3 F xI F yI 11. Show that, for the curve y = , G J = G J + G J · ax+ H a K H yK H xK

ax Solution. Here, y = ax+ Differentiating w.r.t. x L O bgbgax+−1 x a 2 y = a M P = 1 NM bgax+ 2 QP bgax+ 2 Again Differentiating w.r.t. x ––22a 2 y = a 2 = 2 bgbgax+ 3 ax+ 3 DIFFERENTIAL CALCULUS—I 13

3 + 2 2 di1 y1 Now, ρ = y2

Substituting y1 and y2, we get R U 3 | a4 | 2 S1 + V T| bgax+ 4 W| = |R 2 |U S − 2a V T| bgax+ 3 W|

3 {}bgax++4 a4 2 ρ = – ...(1) 2aax2 bg+ 3

23 2 F 2ρI F xI F yI 2 To show that G J = G J + G J H a K H yK H xK

2 R 3 U 3 2 |2 {}bgax++4 a4 2 | F 2ρI 3 | | L.H.S. G J = S V using (1) H a K | 3bg+ 3 | T| 2aa x W|

4 bgax++ a4 = aa2 bg+ x2

2 bgax+ a2 = + ...(2) a2 bgax+ 2

2 bgax+ a2 = + ...(3) a2 bgax+ 2 ∴ L.H.S. = R.H.S. using (2) and (3). 14 ENGINEERING MATHEMATICS—II

12. Find ρ at any point on x = a (θ + sinθ) and y = a (1 – cosθ). Solution. Here x = a (θ + sinθ), y = a (1 – cosθ) Differentiating w.r.t. θ dx dy = a (1 + cos θ), = a sin θ dθ dθ dy dy dθ a sinθ y = = = 1 dx dx a bg1 + cosθ dθ θθ 2 sin cos 22 = θ 2 cos2 2 θ y = tan 1 2 Again differentiating w.r.t. θ d F θI y = G tan J 2 dx H 2K d F θθI d = G tan J × dθ H 2K dx F θI 1 1 = sec2 G J ×× H 2K 2 a bg1 + cos θ θ sec2 2 = θ 22a × cos2 2 1 y2 = θ 4a cos4 2 3 + 2 2 ot1 y1 ρ = y2

3 R θU 2 S1 + tan2 V T 2W = R U | 1 | S θ V |4a cos4 | T 2 W DIFFERENTIAL CALCULUS—I 15

3 R F θθIU 2 F I = Ssec2 G JV × 4a cos4 G J T H 2KW H 2K 1 F θI = × 4a cos4 G J F θI H 2K cos3 G J H 2K F θI ρ = 4a cos G J · H 2K 13. Find the radius of curvature at the point ′θ′ on the curve x = a log sec θ, y = a (tan θ –θ). Solution x = a log sec θ, y = a (tan θ – θ) Differentiating w.r.t. θ dx 1 dy = a ⋅⋅secθθ tan , = a (sec2 θ – 1) dθ secθ dθ = a tan θ = a tan2 θ dy dy θ ∴ y = = d 1 dx dx dθ a tan2 θ = a tanθ θ y1 = tan dy2 d y = = bgtanθ 2 dx2 dx d dθ = bgtanθ ⋅ dθ dx 1 = sec2 θ × a tan θ sec2 θ = a tan θ

3 + 2 2 ot1 y1 Now, ρ = y2

3 di1 + tan2 θ 2 = F 2 I G sec θJ H tan θ K 16 ENGINEERING MATHEMATICS—II

sec3 θ = × a tan θ sec2 θ ρ = a sec θ tan θ. 14. For the curve x = a (cos θ + θ sin θ), y = a (sin θ – θ cos θ), show that the radius of curvature at ′θ′ varies as θ. Solution x = a (cos θ + θ sin θ) dx ⇒ = a (– sin θ + θ cos θ + sin θ) = a θ cos θ dθ y = a (sin θ – θ cos θ) dy ⇒ = a (cos θ + θ sin θ – cos θ) = a θ sin θ dθ dy a θθsin y = ==tan θ 1 dx a θθcos d F dyI d y = G J = bgtan θ 2 dx H dxK dx d dθ = bgtan θ ⋅ dθ dx 1 = sec2 θ × a θθcos 1 = a θθcos3

3 + 2 2 di1 y1 Now, ρ = y2

3 di1 + tan2 θ 2 = F I G 1 J H a θθcos3 K = sec3 θ × a θ cos3 θ = a θ i.e., ρ∝θ. ρ ρ 15. If 1 and 2 are the radii of curvatures at the extremities of a focal chord of the parabola –2 –2 −2 2 ρρ3 + 3 y = 4ax, then show that 1 2 = bg2a 3 .

2 2 Solution. If diat1 , 2 at1 and diat2 , 2 at2 are the extremities of a focal chord of the parabola y2 = 4ax.

Then t1 · t2 =– 1 DIFFERENTIAL CALCULUS—I 17

The parametric equations to the parabola are x = at2, y = 2at x′ =2at y′= 2a x″ =2ay″ = 0

3 otxy′+′222 ∴ρ= xy′″– x ″′ y

3 {}bgbg22at22+ a 2 = di– 4a2

3 81at32di+ 2 = – 4a 2 3 ρ = −+21atdi2 2

–2 – 2 −1 2 ρ 3 = bg21at3 di+

–2 1 1 ρ 3 = × 2 di1 + t 2 bg2a 3 1 – Let t1 and t2 be extremities of a focal chord. Then t2 = · t1 – 2 – 2 1 1 ρ 3 ρ 3 =× Now, = tt= 2 1 1 di1 + t 2 bg2a 3 1

– 2 – 2 1 1 ρ 3 ρ 3 =× = – 1 2 2 t = di+ 2 t bg 1 t1 1 2a 3

1 t 2 = × 1 2 di1 + t 2 bg2a 3 1

––2 2 1 R 1 t 2 U Adding ρρ3 + 3 = S + 1 V 1 2 2 T11+ t 2 + t 2 W bg2a 3 1 1

2 – 2 + di1 t1 = bg2a 3 × + 2 di1 t1

––2 2 – 2 3 3 ρρ+ bg3 i.e., 1 2 = 2a · Hence proved. 18 ENGINEERING MATHEMATICS—II

EXERCISE 1.1

1. Find ρ at any point on y = log sin x. Ans. cosec x

log x L 3 O 2. Find ρ at x = 1 on y = . MAns. P x N 22Q L 3 O π F xI M F 3I 2 P 3. Find the radius of curvature at x = on y = log tan HG KJ · Ans. 2 × G J 4 2 NM H 2K QP xy22 L 125O 4. Find the radius of curvature of (3, 4) on + = 2. MAns. P 916 N 12 Q

L 3 O M 2 24+ 2 2 P 2 2 2 2 2 2 ejbx1 ay1 5. Find the radius of curvature at (x1, y1) on b x – a y = a b . MAns. P N ab44 Q

6. Find ρ at (4, 2) on y2 = 4 (x – 3). Ans. 42

3 di4xa44+ 2 7. Show that ρ at any point on 2xy = a2 is · 8ax23 8. Show that ρ at (0, 0) on y2 = 12x is 6. xxbg− 2 L 1O 9. Find radius of curvature at x = 2 on y2 = · MAns. P x − 5 N 3Q

L a O 10. Find ρ at (a, a) on x3 + y3 = 2a3. MAns. P N 2 Q

L 55a O 2 2 2 MAns. P 11. Find the radius of curvature at (a, 2a) on x y = a(x + a ). NM 6 QP

12. Find the radius of curvature at (–2a, 2a) on x2y = a (x2 + y2). Ans. 2a

2 2 2 13. Show that ρ at (a cos3θ, a sin3θ) on xy3 + 3 = a 3 is 3a sin θ cos θ.

L 3 O π M 22+ 2 P θ θ θ ejab12 14. Find radius of curvature at = on x = a sin , y = b cos 2 . MAns. P 3 N 4ab Q

15. Find ρ for x = t – sin ht cos ht, y = 2cos ht. Ans. 2cosht2 sin ht DIFFERENTIAL CALCULUS—I 19

1.2.3 Radius of Curvature in Pedal Form Let polar form of the equation of a curve be r = f (θ) and P(r, θ) be a given point on it. Let the tangent to the curve at P subtend an angle ψ with the initial side. If the angle q r = f ( ) P (r, q ) between the radius vector OP and the tangent at P is φ then ψ θ φ f we have = + (see figure). r Let p denote the length of the perpendicular from the pole O to the tangent at P. Then from the figure, q y O OM p X sin φ = = OP r Hence, p = r sin φ ...(1) p

1 dψ dθφθφd d d dr M ∴ = = +=+⋅ ...(2) ρ ds ds ds ds dr ds dθ Fig. 1.2 We know that tan φ = r ⋅ dr dθ r ⋅ sin φ ds i.e., = cos φ dr ds dθ Hence, sin φ = r ⋅ ds dr and cos φ = ds 1 sin φ dφ From (2), = +⋅cos φ ρ r dr 1 L d φ O = Msinφφ+⋅r cos P r N dr Q 1 d = ⋅ bgr sin φ r dr Since, r sin φ = p dr Therefore, ρ = r ⋅ ...(3) dp This is the Pedal form of the radius of curvature.

1.2.4 Radius of Curvature in Polar Form Let r = f (θ) be the equation of a curve in the polar form and p(r, θ) be a point on it. Then we know that 1 11F dr I 2 Differentiating w.r.t. r, we get= + G J ...(4) p2 rr24H dθK 20 ENGINEERING MATHEMATICS—II

–2 dp – 2 F dr I 2 1 dr d F dr I ⋅ = + G J di– 4r –5 +⋅⋅⋅2 G J p3 dr r 3 H dθθθK r 4 d dr H d K

– 24F dr I 2 2dr dr2 dθ = − . G J +⋅⋅ ⋅ rr35H dθθK r 4 d dθ2 dr

–2 4F dr I 2 2 dr2 = − G J +⋅ rr35H dθK r 4 dθ2

L 2 2 O dp 3 12F dr I 1dr Hence, = p M + G J – P dr NMrr35H dθK r 4 dθ2 QP dr r Now, ρ = r ⋅= dp |R 12F dr I 2 1dr2 |U p3 S + G J − V T|rr35H dθK r 4 dθ2 W| 1 r 6 ⋅ p3 = F dr I 2 dr2 r 2 + 2 G J −⋅r H dθK dθ2 By using equation (4),

3 |R 11F dr I 2 |U 2 r 6 ⋅+S G J V T|rr24H dθK W| = F dr I 2 dr2 r 2 + 2 G J – r ⋅ H dθK dθ2

3 |R F dr I 2 |U 2 Sr 2 + G J V T| H dθK W| ρ = ...(5) F dr I 2 dr2 r 2 + 2 G J −⋅r H dθK dθ2 dr dr2 where r = , r = · 1 dθ 2 dθ2 3 2 + 2 2 otrr1 ∴ρ= 2 +−2 rrrr2 1 2 This is the formula for the radius of curvature in the polar form. DIFFERENTIAL CALCULUS—I 21

WORKED OUT EXAMPLES

1. Find the radius of the curvature of each of the following curves: (i) r3 = 2ap2 (Cardiod) (ii) p2 = ar 1 1 1 r 2 (iii) =+− (Ellipse). p222a b ab22 Solution. (i) Here r3 =2ap2 Differentiating w.r.t. p, we get dr 3r 2 ⋅ =4ap dp dr 4ap ⇒ = dp 3r 2 dr 4ap 4ap Hence, ρ = r ⋅=⋅=r dp 3r 2 3r

1 F r 3 I 2 where p = G J H 2aK

1 F r 3 I 2 4a ⋅ G J H 2aK ρ = 3r 3 4ar2 22ar = = 32ra 3 (ii) Here p2 = ar Differentiating w.r.t. p, we get dr Then 2p = a ⋅ dp dr 2 p ⇒ = dp a where p = ar · 3 dr 22⋅ ar r 2 ρ = r =⋅r = dp a a 1 11r 2 (iii) Given = + – p2 ab22ab22 Differentiating w.r.t. p, we get –2 –1 dr = 2r ⋅ p3 ab22 dp 22 ENGINEERING MATHEMATICS—II

dr ab22 Hence = dp pr3 dr ab22 ab22 Therefore, ρ = r ..==r dp pr3 p3 2. Find the radius of curvature of the cardiod r = a (1 + cos θ) at any point (r, θ) on it. Also ρ2 prove that is a constant. r Solution. Given r = a (1 + cos θ) Differentiating w.r.t. θ dr r = = –sina θ 1 dθ dr2 and r = = –cosa θ 2 dθ2 ∴ The radius of curvature in the polar form 3 2 + 2 2 otrr1 ρ = 2 +−2 rrrr2 1 2 3 {}aa2 bg1 ++cosθθ2 22 sin 2 = aaaa2 bg12++cosθθθθ2 22 sin −+ bgbg 1 cos – cos 3 a 322ot12+++cosθθθ cos sin 2 = a 2222ot12+++++cosθθ cos 2 sin θθθ cos cos

3 a mr21bg+ cos θ 2 = 31bg+ cos θ 1 22a bg 1+ cos θ 2 = 3 1 F θI 2 22a G 2cos2 J H 2K = 3 4 θ ρ = a cos 32 Squaring on both sides, we get θ 16 θ L 2 1 O ρ2 = a 22cos Mwhere cos=+bg1 cos θ P 92 N 2 2 Q 8 a 2 L r O = bg1 + cos θ Mwhere 1 + cos θ= P 9 N a Q DIFFERENTIAL CALCULUS—I 23

8ar2 8ar ρ2 = ⋅= 9 a 9 ρ2 8a Hence, = which is constant. r 9 an 3. Show that for the curve rn = an cos nθ the radius of curvature is · bgnr+ 1 n – 1 Solution. Here rn = an cos nθ Taking logarithms on both sides, we get n log r = n log a + log cos nθ Differentiating w.r.t. θ, we have n dr nnsin θ = 0– r dθ cos nθ dr r = =– r tan nθ 1 dθ Differentiating w.r.t. θ again, we obtain dr2 R dr U r = = −+⋅Srnsec2 nθθ tan n V 2 dθ2 T dθ W = –sectanotnr22 nθθ− r n = r tan2 nθ – nr sec2 nθ Using the polar form of ρ, we get 3 2 + 2 2 otrr1 ρ = 2 +−2 rrrr2 1 2 3 otrr222+ tan nθ 2 = r2 +−−2 bg–tan r nθθθ2 rdi r tan22 n nr sec n

rn33sec θ = rnnnn222212+−+tanθθ tan sec θ

rnsec3 θ = bgnn+ 1 sec2 θ r = bgnn+ 1 cos θ

r L r n O = Mwhere cos nθ= P F r n I N a n Q bgn + 1 G J H a n K a n = · bgnr+ 1 n – 1 24 ENGINEERING MATHEMATICS—II

4. Find the radii of curvature of the following curves: (i) r = aeθ cot α (ii) r (1 + cos θ) = a

ra22− F aI (iii) θ = –cos–1 G J · a H r K Solution. (i) Here r = aeθ cot α Differentiating w.r.t θ dr = aeθ cot α · cot α dθ = r · cot α r So that, tan φ = dr dθ r = = tan α r cot α Hence, φ = α, since p = r sin φ We get, p = r sin α. This is the Pedal equation of the given curve. From which, we get dr 1 = dp sin α dr Hence, p = r ⋅ = r cosec α. dp (ii) Given equation of the curve is r (1 + cos θ)=a Differentiating w.r.t. θ, we get dr r (– sin θ) + (1 + cos θ) · = 0 dθ dr r sin θ or = dθ 1 + cos θ

1 11F dr I 2 We have, = + G J p2 rr24H dθK 11r 22sin θ = +⋅ rr24bg1 + cos θ 2 L O 1 sin2 θ = M1 + P r 2 NM bg1 + cos θ 2 QP

L 2 2 O 1 bg1 ++cosθθ sin = M P r 2 NM bg1 + cos θ 2 QP DIFFERENTIAL CALCULUS—I 25 L O 1 M 21bg+ cos θ P = 22 r NM bg1 + cos θ QP

= 2 r 2 bg1 + cos θ a where 1 + cos θ = r

1 22= 2 = a p r 2 ⋅ ar r ar Hence, p2 = which is the pedal equation of the curve. 2 Differentiating w.r.t. p, we get adr 2p = ⋅ 2 dp dr 4p ⇒ = dp a dr ∴ρ= r ⋅ dp

4 p ar = r ⋅ where p = a 2 4 ar = r . a 2

3 = 22 ar2 .

ra22− F aI (iii) Here, θ = − cos–1 G J a H r K θ F I d 2r + 1 G – aJ Then, = 1 H K dr ara⋅⋅2 22 − r F a 2 I 2 G1 − J H r 2 K r a = – ar22−− a rr 22 a

ra22− = ar r22− a 26 ENGINEERING MATHEMATICS—II

dθ ra22− = dr ar dr ar so that θ = d ra22− We have the Pedal equation, we get 1 11F dr I 2 = + G J p2 rr24H dθK 11 ar22 = +⋅ rr24dira22−

1 R a2 U = S1 + V r 2 T ra22− W 1 1 = p2 ra22− Hence p 2 = r 2 – a 2 dr p From this we get = dp r p ∴ρ= r ⋅==pra22 − ⋅ r

EXERCISE 1.2

1. Find the radius curvature at the point ( p, r) on each of the following curves:

L r 3 O (i) pr = a2 (Hyperbola) MAns. P N a 2 Q

L a 2 O 3 2 MAns. P (ii) r = a p (Lemniscate) N 3r Q

L an O n n+1 MAns. P iii pa r n −1 ( ) = (Sine spiral) NM bgnr+ 1 QP

L 3 O M diar22+ 2 P r 4 MAns. P (iv) p = (Archimedian spiral) M ra22+ 2 P ra22+ NM QP DIFFERENTIAL CALCULUS—I 27

2. Find the radius of curvature at (r, θ) on each of the following curves:

L 3 O M radi22+ r 2 P a MAns. P L a O (i) r = M a 3 P (ii) r = a cos θ MAns. P θ NM QP N 2 Q

L a2 O L an O 2 2 M P n n MAns. P iii r a θ Ans. iv r = a nθ n −1 ( ) = cos 2 N 3 r Q ( ) sin NM bgnr+ 1 QP

L 3 O r a L 2 ar O (v) r2 cos 2θ = a2 MAns. P (vi) r = bg1 − cos θ MAns. P N a 2 Q 2 NM 3 QP

L r 4 O L na O (vii) r = a sec 2θ MAns. P (viii) r = a sin nθ MAns. P N 3p2 Q N 2 Q ρ ρ 3. If 1 and 2 are the radii of curvature at the extremities of any chord of the cardiode 2 2 2 16a r = a (1 + cos θ) which passes through the pole. Prove that ρρ+ = · 1 2 9

1.3 SOME FUNDAMENTAL THEOREM

1.3.1 Rolle’s Theorem If a function f (x) is 1. continuous in a closed interval [a, b], 2. differentiable in the open interval (a, b) and 3. f (a) = f (b). Then there exists at least one value c of x in (a, b) such that f ′ (c) = 0 (No proof).

1.3.2 Lagrange’s Mean Value Theorem Suppose a function f (x) satisfies the following two conditions. 1. f (x) is continuous in the closed interval [a, b]. 2. f (x) is differentiable in the open interval (a, b). Then there exists at least one value c of x in the open interval (a, b), such that fb()– fabg = f ′ (c) ba– 28 ENGINEERING MATHEMATICS—II

Proof. Let us define a new function φ(x)=f (x) – k·x ...(1) where k is a constant. Since f (x), kx and φ (x) is continuous in [a, b], differentiable in (a, b). From (1) we have, φ (a)=f (a) – k·a φ (b)=f (b) – k·b ∴φ(a)=φ (b) holds good if f (a) – k·a = f (b) = k·b i.e., k (b – a)=f (b) – f (a)

fbbg− fa bg or k = ...(2) ba− Hence, if k is chosen as given by (2), then φ (x) satisfy all the conditions of Rolle’s theorem. Therefore, by Rolle’s theorem there exists at least one point c in (a, b) such that φ′(c) = 0. Differentiating (1) w.r.t. x we have, φ′(x)=f ′(x) – k and φ′(c) = 0 gives f ′(c) – k = 0 i.e., k = f ′(c) ...(3) Equating the R.H.S. of (2) and (3) we have

fbbg− fa bg = f ′(c) ...(4) ba− This proves Lagrange’s mean value theorem.

1.3.3 Cauchy’s Mean Value Theorem If two functions f (x) and g (x) are such that 1. f (x) and g (x) are continuous in the closed interval [a, b]. 2. f (x) and g (x) are differentiable in the open interval (a, b). 3. g′ (x) ≠ 0 for all x ∈ (a, b). Then there exists at least one value c ∈ (a, b) such that fb()– fabg fc′ bg = · gbbg– ga bg gc′ bg Proof: Let us define a new function φ (x)=f (x) – kg (x) ...(1) where k is a constant. From the given conditions it is evident that φ (x) is also continuous in [a, b], differentiable in (a, b). Further (1), we have φ (a)=f (a) – k g (a); φ (b) = f (b) – k g (b)