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Differential Calculus—I

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Differential Calculus—I UNIT I Differential Calculus—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 cur- vature 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 dy dy2 Substituting the values of and in the Cartesian form of the radius of curvature of the dx dx2 curve y = f (x) [Eqn. (3)] 3 |R F dyI 2 |U 2 3 S1 + G J V + 2 2 H K ot1 y1 T| dx W| ∴ρ= = 2 y2 dy dx2 3 |R F dy/ dt I 2 |U 2 S1 + G J V T| H dx/ dt K W| = Rdx dy2 dy dx2 U F dxI 3 S ⋅⋅–/V G J T dt dt 2 dt dt 2 W 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 3 4. Find the radius of curvature at (a, 0) on y = x (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.
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