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Geometrical Applications of Differentiation Aaaaa 2 Geometrical Applications of Differentiation aaaaa 2.1 INTRODUCTION Though we have some algebraic results which give useful information about the graph of function and the function rate of change over most if not all of the functions. But to know the complete insight and details about the graph of the curve in space, we need to know first about certain other things like Maxima-Minima problems, Estimating approximation Errors, Intermediate forms, Role’s Theorem, mean value theorem, Taylor’s theorem, concavity, points of inflexion, sign of first derivatives, Asymptotes, etc. and which in turn can be known only with the differential co-efficient/derivative of the function at a point or over a certain change. Derivatives are interpreted as slope of curves and as instantaneous rate of change. We know that the first and second derivatives together tell how the graph of the function is shaped. Second derivative helps in estimating the linear approximation of the function. Collectively all above inference help in sketching the trace of the curve. 2.2 TANGENTS AND NORMALS I. Tangent and its Equation Let P(x, y) and Q(x + δx, y + δy) be neighbouring points on the curve y = f(x) with a supposition that the curve is continuous near P. Equation of any line through P(x, y) is Y – y = m(X – x) …(1) Y where X, Y are the current coordinates of any point on this B line (Fig. 2.1). y = f( x ) T´ Now, as Q → P, the straight line PQ tends in general to a (xxyy + δ, + δ) Q definite straight line TPT', which is called the tangent to the curve at P(x, y). (,x yP ) Whence the slope of PQ, which in other words known as A Ψ ΨδΨ + ()yyy+δ − δy X gradient of PQ becomes = and in the limiting O T T´´ ()xxx+δ − δ x Fig. 2.1 δydy case when Q → P, lttan===Ψm δ→x0δxdx 97 98 Engineering Mathematics through Applications Therefore, the eqn. (1) becomes equation of the tangent at P(x, y), and reduces to dy Yy−=() Xx − dx or Y – y = tanΨ(X– x) …(2) Cor. 1: Intercepts of the tangent on x-axis (i.e. OT): Putting Y = 0 in eqn (2), we get y OT()==− X x dy …(3) dx Intercepts of the tangent of y-axis (i.e. OT"): Putting X = 0 in eqn (2), we get dy OT"()==− Y y x … (4) dx ∂f dy f ∂ Cor. 2: =−x =− x If the given equation of the curve be f(x, y) = 0, then we know that ∂ dx fy f and hence eqn (2) reduces to ∂y −=−fx() − Yy Xx or (X – x)fx + (Y – y)fy = 0 …(5) fy Cor. 3: If the given equation of the curve is in parametric form x = φ(t), y = ψ(t), dy ψ () dy ==dt 't dxdx φ '() t and whence the eqn (2) reduces to dt ψ't() Yt−ψ() =() Xt −φ() φ't() …(6) Example 1: Find the equation of the tangent at any point (x, y ) to the curve x2/3 + y2/3 = a2/3. Show that the portion of the tangent intercepted between the axes is of constant length. Solution: The given curve is x2/3 + y2/3 = a 2/3. 1/3 22−−dy dy=− y On differentiation with respect to x, we get xy1/3+= 1/3 0or . 33dx dx x Now equation of the tangent at (x, y) is Y – y = m(X – x), i.e. (Y – y) = –(y/x)1/3 (X – x) …(1) Further, intercept on x-axis, i.e. Y = 0 implies y12 2 2 1 2 1 Xx=− =+ xxy3333333 =() x + yx = ax dy …(2) dx Intercept on y-axis, i.e. X = 0 implies dy y 1/3 Yyx=− =+⋅ yx =+ yx2/3 y 1/3 dx x Geometrical Applications of Differentiation 99 = (x2/3 + y2/3)y1/3 = a2/3 y1/3 …(3) 2222 The square root of sum of square of the two intercepts =+()axy333() ==aa4/3 2/3 a Thus, the portion of the tangent intercepted between the axis is a which is a constant length. x y Example 2: Prove that +=1 touches the curve y = be–x/a at the ‘point’ where the curve ab crosses the axis of y. Solution: The curve y = be–x/a shall cross the y-axis at a point where x = 0, i.e. y = be–0/a = b Now at the point (0, b), dy − − ==−=−=−xa/11bbb be 0 dx()0,b a()0,b aa a a e ()0,b ∴ Tangent at (0, b) is b (Yb−=−−) () X0or aY + bX – ab = 0 a x y or +=1 (Because relation is true for all x, y). ab Example 3: Find the equation of the tangent to the curve y(x – 2)(x – 3) – x + 7 = 0 at the point where it cuts the x-axis. Solution: The curve cuts the x-axis at y = 0. But y = 0 implies –x + 7 = 0 or x = 7 Thereby means tangent is taken at the point (7, 0) The equation of the tangent at (7, 0), however, is given by ()−=−fx () − Yy Xx, where fx = y(2x – 5) – 1 and fy = (x – 2)(x – 3) fy (7,0) 1 or ()YX−=07() −, i.e. X – 20Y – 7 = 0 is the desired equation of tangent at (7, 0). 20 Example 4: Find the equation of the tangent at any point ‘θ’ on the curve whose equations are x = a(θ + sinθ); y = a(1 – cosθ). Solution: As the given equation is in parametric form whence the equation of tangent at any point ‘θ’ is given by y'()θ Yy−θ=() () Xx −θ() x'()θ 100 Engineering Mathematics through Applications dx dy Here =+θa()1cos , =θasin dθ dθ asinθ ∴Ya−−θ=()1cos{} Xa −θ+θ() sin a()1cos+θ θθ θ2sincosa Ya−=2sin2 22{} Xaa −θ−θ sin or θ 2 2cosa2 2 θθ θ or Ya−=2 sin2 tan() Xaa −θ−θ⋅ sin tan 22 2 θθ θ 22 Ya−=2sin tan ( Xa −θ− ) 2sin a 22 2 θθ or YXa=−θ()tanor()tan yxa =−θ 22 (on changing the current coordinates to general coordinates). m m x y Example 5: If the straight line p = x cosα + y sinα touch the curve +=1 , prove ab m m m that ()abpcosαm–1 +() sinαm–1 = m–1 . xm ym Solution: The given equation of the curve is +=1 …(1) abmm − xm−1 ydym 1 On differentiating with respect to x, we get mm+=0 abdxmm − m m1 dy bx or Slope of the tangent at ()xy,==− . dx a y ∴Equation of the tangent at (x, y) becomes − m m1 bx (Yy−=−) () Xx − ay − xxmm−1 yymm1 or XY+=+=1, (using (1)) …(2) ababmmmm Now if the line xcos α + y sin α = p touch the given curve then it should be identical to equation (2). cosαα sin p Whence == mm−−11 xy1 m abm Geometrical Applications of Differentiation 101 m−1 ααm−1 xa==cosy b sin ⇒ and ap bp mm Y ααmm−−11mm N abxcos+=+= sin y ∴ 1 N ppab o rm T´ whence the result. al P II Normal and its Equation Normal to a curve y = f (x) at any point P(x, y) is the straight Ψ line (NP, say) through P(x, y ) but it is perpendicular to the X O T 1 tangent at P. For m to be the slope of the tangent, − (say m’) Fig. 2.2 m will be the slope of the normal (since we know that for two perpendicular lines, slope of one is the negative reciprocal of the another). With this notion, it becomes easy to write the equation of the normal, if the equation of the curve is given, i.e. 1 dx (Yy−=−−) () Xx or ()Yy−=−() Xx − …(1) m dy dy f Cor. 1: If the equation of the curve is f(x, y) = 0, then =− x so that equation of the normal dx fy reduces to (Yy−) ()Xx− ( −=−) 1() − or = …(2) Yy Xx −f ffyx x fy Cor. 2: If the equation of the curve is in the parametric form x = φ(t), y = ψ(t) so that dψ dy ψ't() ==dt ,the equation of the normal becomes dxdφφ '() t dt 1 ()Yy−=−() Xx − or [X – φ(t)] φ'(t) + [Y – ψ(t)]ψ'(t) = 0 …(3) ψ't() φ't() III Lengths of Tangent, Normal, Subtangent, Subnormal, etc. Let y = f(x) be the equation of the curve with PTT' as tangent at point P(x, y) meeting the x-axis and y-axis at T and T' respectively. Let RPN be the normal at point P(x, y) meeting the x-axis at N and PM be the ordinate, OQ be the perpendicular from the origin to the normal (See dy Fig. 2.3). Let ∠MTP be Ψ so that tan Ψ= =y and ∠MPN = Ψ. Then we have the following dx 1 geometrical results: 102 Engineering Mathematics through Applications 1. Length of the tangent Y 2 =Ψ=+Ψ=+2 dx ()TP MPcosec y1 cot y 1 R dy Px(, y) (i.e. portion of the tangent between the curve Y Ψ and x-axis). 9 – 0 ° 2. Length of normal Ψ dy 2 y ()NP=Ψ=+Ψ=+ MPsec y 1 tan2 y 1 dx Ψ Y (i.e. portion of the normal between the curve X O T M N and x-axis) Q dx 3. Length of subtangent ()TM=Ψ= ycot y . T´ dy Fig. 2.3 dy 4.
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