Dr. Satish Shukla 1 of 50 Engg. Math. I Unit-I Differential Calculus Syllabus: Limits of functions, continuous functions, uniform continuity, monotone and inverse functions. Differentiable functions, Rolle s theorem, mean value theorems and Taylor s theorem, power series. Functions of several variables, partial derivatives, chain rule, Tangent planes and normals. Maxima, minima, saddle points, Lagrange multipliers, exact differentials. Rolle's Theorem. Let f be a function which is continuous everywhere on the interval [a, b] and has a derivative at each point of the open interval a, b. Also, assume that fa = fb. Then there is at least one point c in the interval a, b such that f c = 0. Proof. We prove the Rolle s theorem geometrically. x = a fa x = c fb x = b fa x = a x = c fb x = b fa c c 1 x = a fb x = b a b c Geometric Interpretation of Roll s Theorem Since fa = fb and function f is continuous in [a, b] we have the following three cases: Case a: Suppose that the function increases after point x = a. Since fa = fb and function f is continuous, there must exists a point c such that a < c < b and f has its maximum value at c. Therefore, we have f c = 0. Case b: Suppose that the function decreases after point x = a. Since fa = fb and function f is continuous, there must exists a point c such that a < c < b and f has its minimum value at c. Therefore, we have f c = 0. Case c: Suppose that the function increases after point x = a and then attains its maximum values and then decreases and attains its minimum value, i.e., function oscillates. Since fa = fb and function f is continuous, it finally returns to its initial value. Thus, we have more than one point c 1, c,... such that a < c 1, c,... < b and f has its maximum and minimum values at c 1, c,.... Therefore, we have f c 1 = f c = = 0. Thus, in each case we obtain the desired point c. Example 1. Verify the Rolle s theorem for fx = x in [ 1, 1].
Dr. Satish Shukla of 50 Sol: Here a = 1, b = 1. Given function fx is continuous in [ 1, 1] and fa = f1 = 1 = 1, fb = f 1 = 1 = 1, but we know that the function fx = x is not differentiable at point x = 0, and 0 [ 1, 1], therefore the Rolle s theorem cannot be verified. Example. Verify the Rolle s theorem for fx = e x sin x in [0, π]. Sol: Here a = 0, b = π. Given function fx is continuous in [0, π] and fa = f0 = e 0 sin 0 = 0, fb = fπ = e π sin π = 0, so fa = fb. Also, the function fx = e x sin x is differentiable at every point of the interval 0, π. Therefore, all the conditions of the Rolle s theorem are satisfied and by Rolle s theorem, there exists 0 < c < π such that f c = 0. Then f x = d dx ex sin x = e x sin x + e x cos x. Therefore, f c = 0 = e c sin c + e c cos c = 0 = e c [sin c + cos c] = 0 = sin c + cos c = 0 = sin c + cos c = 0 = sin c = 1 = c = π = c = π 4. Since c = π 4 0, π the Rolle s theorem is verified. [ Example. Verify the Rolle s theorem for fx = sin x in 0, π ]. Sol: Here a = 0, b = π [. Given function fx is continuous in 0, π ] and fa = f0 = π π sin 0 = 0, fb = f = sin = 0, so fa = fb. Also, the function fx = sin x is differentiable at every point of the interval 0, π. Therefore, all the conditions of the Rolle s theorem are satisfied and by Rolle s theorem, there exists c 0, π such that f c = 0. Then f x = d sin x = cos x. dx Therefore, f c = 0 = cos c = 0 = cos c = 0 = c = π = c = π 6. Since c = π 6 0, π the Rolle s theorem is verified. Example 4. Verify the Rolle s theorem for fx = cos x in [ π 4, π ]. 4
Dr. Satish Shukla of 50 Sol: Here a = π 4, b = π [ 4. Given function fx is continuous in π 4, π ] and fa = 4 f π = cos π π π = 0, fb = f = cos = 0, so fa = fb. Also, the 4 4 4 4 function fx = cos x is differentiable at every point of the interval π 4, π. Therefore, 4 all the conditions of the Rolle s theorem are satisfied and by Rolle s theorem, there exists c π 4, π such that f c = 0. Then 4 Therefore, f x = d cos x = sin x. dx f c = 0 = sin c = 0 = sin c = 0 = c = 0 = c = 0. Since c = 0 π 4, π the Rolle s theorem is verified. 4 Example 5. Verify the Rolle s theorem for fx = + x 1 / in [0, ]. Sol: Here a = 0, b =. Given function fx is continuous in [0, ] and fa = f 0 = + 0 1 / =, fb = f = + 1 / =, so fa = fb. Note that f is not differentiable in the interval 0,. Indeed: f x = d + x 1 / = dx x 1 1/. Therefore, f 1 does not exist and since 1 0,, therefore all the conditions of the Rolle s theorem are not satisfied, and so, it cannot be verified. Example 6. Verify the Rolle s theorem for fx = x 4x. Sol: Here the interval where the theorem is to be verified is not given. To find the interval put fx = 0, i.e., x 4x = xx 4 = 0 = x = 0, ±. So we obtain the intervals [, 0], [0, ] and [, ]. Given function fx is a polynomial in x, so, continuous and differentiable everywhere and f = f0 = f = 0. Therefore, all the conditions of the Rolle s theorem are satisfied and by Rolle s theorem, there exists c 0, such that f c = 0. Then Therefore, f x = d x 4x = x 4. dx f c = 0 = c 4 = 0 = c = ±. Since c =, 0 and c = 0, the Rolle s theorem is verified.
Dr. Satish Shukla 4 of 50 Mean Value Theorem OR Lagrange's Mean Value Theorem. Let f be a function which is continuous everywhere on the interval [a, b] and has a derivative at each point of the open interval a, b. Then there is at least one point c in the interval a, b such that f fb fa c =. b a Proof. We prove the this theorem with the help of Rolle,s theorem. A Tangent Cord B fa fb x = a x = c x = b Define a function F x by Geometric Interpretation of Mean Value Theorem F x = fx + αx 1 where α is an arbitrary constant. Then, we shall show that F satisfies all the conditions of Rolle s theorem. Then: I Since f is continuous in [a, b] and αx is a polynomial so it is continuous everywhere, and so, their sum F x = fx + αx is also continuous in [a, b]. II Since f is differentiable in a, b and αx is a polynomial so it is differentiable everywhere, and so, their sum F x = fx + αx is also differentiable in a, b. III Finally, since α was an arbitrary constant, choose α such that: F a = F b = fa + αa = fb + αb = fb fa α =. b a Thus, F satisfies all the conditions of Rolle s theorem. there exists c a, b such that Therefore, by Rolle s theorem Hence the proof is completes. F c = 0 = f c + α = 0 = f c = α = f fb fa c =. b a Example 7. Find the c of mean value theorem for the function fx = x 1x x in the interval [0, 4].
Dr. Satish Shukla 5 of 50 Sol: Here a = 0, b = 4 and the function f is a polynomial, and so, it is continuous and differentiable everywhere. Therefore, all the conditions of mean value theorem are satisfied. By mean value theorem there exists a point c 0, 4 such that f fb fa c =. b a Now Therefore, f x = x x + x 1x + x 1x = x 1x + 11. f c = fb fa b a = c f4 f0 1c + 11 = 4 0 = c 1c + 11 = 6 6 4 = c 1c + 11 = = c 1c + 8 = 0 = c = ±. Since c = 0, 4, hence the mean value theorem is verified. Example 8. Verify mean value theorem for the function fx = ln x in the interval 1 e x e. Sol: Here a = 1, b = e and the function f is logarithmic, and so, it is continuous in e the interval [ 1, e] and differentiable in the interval 1, e. Therefore, all the conditions of e e mean value theorem are satisfied. By mean value theorem there exists a point c 1, e e such that f fb fa c =. Now b a f x = 1 x. Therefore, f c = fb fa b a = 1 c = f e f 1 e e 1/e 1 = 1 ln e ln c = e e 1/e = 1 e1 1 = c e 1 = c = e 1. e Since c = e 1 e 1 e, e, hence the mean value theorem is verified.
Dr. Satish Shukla 6 of 50 Example 9. Show that on the graph of any quadratic polynomial the chord joining the points for which x = a, x = b is parallel to the tangent line at the midpoint x = a + b. OR If fx = αx + βx + γ, where α, β, γ are constants and α 0, then find the value of c in Lagrange s mean value theorem in the interval [a, b]. Sol: The function f is polynomial, and so, it is continuous in the interval [a, b] and differentiable in the interval a, b. Therefore, all the conditions of mean value theorem are satisfied. By mean value theorem there exists a point c a, b such that f c = fb fa. Now b a f x = αx + β. Therefore, f c = fb fa b a = αc + β = f b f a b a = αc + β = αb + βb + γ αa + βa + γ b a = αc + β = αb a + βb a b a = αc + β = αb + a + β = c = b + a midpoint of a, b. Since f fb fa c =, hence the slope of tangent at midpoint c = b + a i.e., f c b a is equal to the slope of chord at the end points a, b. Therefore, the tangent and cord are parallel. Exercise Assignment Q.1 Discuss the conditions of Rolle s theorem for the function fx = tan x in the interval 0 x π. Ans. Since tan x is not continuous at x = π, the Rolle s theorem is not applicable. Q. Verify the Rolle s theorem for the function fx = x in the interval [ 1, 1]. Ans. c = 0. Q. Can Rolle s theorem be applied for the function fx = 1 x /. Hint. For the interval, put fx = 0, it gives the interval [, 4]. Then, since f is not differentiable at x =, 4, so, the Rolle s theorem cannot be verified. Q.4 Explain Rolle s theorem for the function fx = x a m x b n in the interval [a, b]. Ans. c = mb+na m+n a, b.
Dr. Satish Shukla 7 of 50 Q.5 Find the c of mean value theorem for the function fx = x in the interval [, ]. Ans. c = ±. Q.6 Verifiy mean value theorem for the function fx = x x 1 in the interval [0, 1]. Ans. c = 1. Taylor's Theorem. Suppose that the n 1th derivative f n 1 of f is continuous on the interval [a, b] and the nth derivative f n of f exists in the open interval a, b. Then for each x a in I there is a value c such that a < c < x and: fx = fa+ x a f a+ 1! The last term R n = form after n terms. x a f a+ +! x an f n a+ n! x an+1 f n+1 c. n + 1! x an+1 f n+1 c is called the remainder therm Lagrange s n + 1! Taylor s Series. Suppose R n 0 as n, then the expression for fx in the Taylor s theorem reduces into an infinite series and this series is called the Taylor s series or Taylor s series expansion of fx about the point x = a; and it is given by: fx = fa + x a f a + 1! Various forms of Taylor s series: x a f a + +! Maclaurin s series Put a = 0 in Taylor s series obtain: x an f n a +. n! fx = f0 + x 1! f 0 + x! f 0 + + xn n! f n 0 +. Expansion of fx + h in powers of x Replace x by x + h and a by h in Taylor s series obtain: fx + h = fh + x 1! f h + x! f h + + xn n! f n h +. Expansion of fx + h in powers of h Replace x, h by h, x respectively, in the previous series: Example 10. Expand ln fx + h = fx + h 1! f x + h! f x + + hn n! f n x +. 1 + x using Maclaurin s theorem. 1 x
Dr. Satish Shukla 8 of 50 Sol: Here fx = ln 1 + x. By Maclaurin s theorem we know that 1 x fx = f0 + x 1! f 0 + x! f 0 + + xn n! f n 0 +. Putting y = fx, y 0 = f0, y 1 0 = f 0, y 0 = f 0 etc., in the above we obtain: y = y 0 + x 1! y 1 0 + x! y 0 + + xn n! y n 0 +. Now differentiating successively and putting x = 0 we obtain: y = fx = ln 1 + x ln 1 x = y 0 = 0 y 1 = 1 1 + x + 1 1 x = y 1 0 = 1 y = 1 + x + 1 1 x = y 0 = 0 y = 1 + x + 1 x = y 0 = 4 6 y 4 = 1 + x + 6 4 1 x 4 = y 4 0 = 0 4 y 5 = 1 + x + 4 5 1 x 5 = y 5 0 = 48 and so on. Putting these values in we obtain: 1 + x ln = 0 + x x + 1 x 1!! It is the required series. 0 + x! = x + x + x5 5 +. 4 + x4 4! 0 + x5 5! 48 + Example 11. If ln sec x = 1 x + Ax 4 + Bx 6 +, then find the values of A and B. Sol: Since the given value of ln sec x is a series in powers of x, we will expand ln sec x by Maclaurin s series. Then, here fx = ln sec x and by Maclaurin s theorem we know that y = y 0 + x 1! y 1 0 + x! y 0 + + xn n! y n 0 +. Now differentiating successively and putting x = 0 we obtain: y 1 = y = fx = ln sec x = y 0 = 0 sec x tan x sec x = tan x = y 1 0 = 0 y = sec x = 1 + tan x = 1 + y 1 = y 0 = 1 y = y 1 y = y 0 = y 1 0 y 0 = 0 y 4 = y 1 y + y y = y 1 y + y = y 4 0 = y 1 0 y 0 + y 0 =
Dr. Satish Shukla 9 of 50 y 5 = y 1 y 4 + y y + 4y y = y 1 y 4 + 6y y = y 5 0 = y 1 0 y 4 0 + 6y 0 y 0 = 0 y 6 = y 1 y 5 + y y 4 + 6y y 4 + 6y y = y 1 y 5 + 8y y 4 + 6y = y 6 0 = y 1 0 y 5 0 + 8y 0 y 4 0 + 6y 0 = 16 and so on. Putting these values in we obtain: ln sec x = 0 + x 1! 0 + x! 1 + x! = 1 x + 1 1 x4 + 1 45 x6 +. 0 + x4 4! + x5 5! 0 + x6 6! 16 + On comparing the coefficients of various powers of x in the above and given series we obtain A = 1 1, B = 1 45. Sol: Example 1. Find the first five terms in the expansion of e sin x by Maclaurin s series. Here fx = e sin x. By Maclaurin s theorem we know that y = y 0 + x 1! y 1 0 + x! y 0 + + xn n! y n 0 +. 4 Now differentiating successively and putting x = 0 we obtain: y = fx = e sin x = y 0 = 1 y 1 = cos xe sin x = y cos x = y 1 0 = 1 y = y 1 cos x y sin x = y 0 = 1 y = y cos x y 1 sin x y cos x = y cos x y 1 sin x y 1 = y 0 = 0 y 4 = y cos x y sin x y 1 cos x y = y 4 0 = y 5 = y 4 cos x 4y sin x 5y cos x + y 1 sin x y = y 5 0 = 8 and so on. Putting these values in 4 we obtain: e sin x = 1 + x 1! 1 + x! 1 + x! 0 + x4 4! = 1 + x + x x4 8 x5 15 +. It is the required series. + x5 5! 8 + Example 1. Expand e ax cosbx by Maclaurin s theorem. Sol: Here fx = e ax cosbx. By Maclaurin s theorem we know that y = y 0 + x 1! y 1 0 + x! y 0 + + xn n! y n 0 +. 5 Now differentiating successively and putting x = 0 we obtain: y = fx = e ax cosbx = y 0 = 1 y 1 = ae ax cosbx be ax sinbx = ay be ax sinbx = y 1 0 = a y = ay 1 b e ax cosbx abe ax sinbx = ay 1 b y + ay 1 ay = ay 1 a + b y = y 0 = a b y = ay a + b y 1 = y 0 = aa b
Dr. Satish Shukla 10 of 50 and so on. Putting these values in 5 we obtain: It is the required series. e sin x = 1 + x 1! a + x! a b + x! aa b + = 1 + ax + a b x! + aa b x! +. Example 14. Expand e a sin 1 x by Maclaurin s theorem. Hence show that e θ = 1 + sin θ + 1! sin θ +! sin θ + Sol: where θ = sin 1 x. Here fx = e a sin 1 x. By Maclaurin s theorem we know that y = y 0 + x 1! y 1 0 + x! y 0 + + xn n! y n 0 +. 6 Since y = fx = e a sin 1 x we have y 0 = 1. Differentiating we get = y 1 = y 1 = e a sin 1 x ay 1 x a 1 x = 1 x y 1 = a y. 7 Therefore, y 1 0 = a. Again differentiating 7 we get: 1 x y xy 1 = a yy 1 = 1 x y xy 1 a y = 0. 8 Therefore, y 0 a y 0 = 0, i.e., y 0 = a. Again differentiating 8 we get: 1 x y xy 1 + a y 1 = 0. Therefore, y 0 1 + a y 1 0 = 0, i.e., y 0 = a1 + a values in 6 we obtain: e a sin 1 x = 1 + x 1! = 1 + ax + a x Putting a = 1 and sin 1 x = θ we get It is the required series. a + x! a + x! 1 + a +! e θ = 1 + sin θ + sin θ! + a1 + a x! + sin θ! +. +. and so on. Putting these Example 15. Find the first five terms in the expansion of ln1+sin x by Maclaurin s series.
Dr. Satish Shukla 11 of 50 Sol: Here fx = ln1 + sin x. By Maclaurin s series we know that y = y 0 + x 1! y 1 0 + x! y 0 + + xn n! y n 0 +. 9 Since y = fx = ln1 + sin x we have y 0 = 0. Differentiating we get y 1 = cos x 1 + sin x = 1 + sin xy 1 = cos x. 10 Therefore, y 1 0 = 1. Again differentiating 10 we get: 1 + sin xy + y 1 cos x = sin x. 11 Therefore, 1 + 0y 0 + y 1 0 = 0, i.e., y 0 = 1. Again differentiating 11 we get: 1 + sin xy + y cos x y 1 sin x = cos x. 1 Therefore, 1 + 0y 0 + y 0 0 = 1, i.e., y 0 = 1. Again differentiating 1 we get: 1 + sin xy 4 + y cos x y sin x y 1 cos x = sin x. 1 Therefore, 1 + 0y 4 0 + y 0 0 y 1 0 = 0, i.e., y 4 0 =. Again differentiating 1 we get: 1 + sin xy 5 + 4y 4 cos x 6y sin x 4y cos x + y 1 sin x = cos x. 14 Therefore, 1 + 0y 5 0 + 4y 4 0 0 4y 0 + 0 = 1, i.e., y 5 0 = 5. Putting these values in 9 we obtain: ln1 + sin x = 0 + x 1! It is the required series. = x x! 1 + x! + x! 1 + x! x4 4! + 5x5 5! 1 + x4 4! +. + x4 5! 5 + Sol: Example 16. Expand tan 1 x in the ascending powers of x 1. By Taylor s series we know that fx = fa + x a f a + 1! Here fx = tan 1 x and a = 1, therefore: fx = f1 + x 1 f 1 + 1! x a f a +.! x 1 f 1 +. 15! Since fx = tan 1 x we have f1 = π 4. Differentiating we get: f x = 1 1 + x = f 1 = 1.
Dr. Satish Shukla 1 of 50 Rearranging the terms in the above we get: 1 + x f x = 1. Again differentiating we get: 1 + x f x + xf x = 0 = f 1 = 1. On putting these values in 15 we get tan 1 x = π 4 + x 1 1! x 1! +. Example 17. Expand sin x in powers of x π and hence evaluate sin 91 correct to four places of decimals. Sol: By Taylor s series we know that fx = fa + x a f a + 1! Here fx = sin x and a = π, therefore: x a f a +.! π fx = f + x π π x π f π + f +. 16 1!! π Since fx = sin x we have f = 1. Differentiating we get: Again differentiating we get: Again differentiating we get: Again differentiating we get: On putting these values in 16 we get Let x = 91, so that, π f x = cos x = f = 0. π f x = sin x = f = 1. π f x = cos x = f = 0. π f iv x = sin x = f iv = 1. sin x = 1 x π +! x π = 91 90 = 1 = π 180 x π 4. +. 4! radians = 0.0174 radians.
Dr. Satish Shukla 1 of 50 Putting the value of x π in the above series we obtain: sin 91 = 1 0.0174! = 0.9999 + 0.01744 4! correct up to four places of decimals. Example 18. Expand ln x in powers of x 1 and hence evaluate ln1.1 correct to four decimal places. Sol: By Taylor s series we know that fx = fa + x a f a + 1! Here fx = ln x and a = 1, therefore: fx = f 1+ x 1 f 1+ 1! x a f a +! x 1 f 1+! x a f a +! x 1 f 1+! Since fx = ln x we have f 1 = 0. Differentiating we get: x a4 f iv a +. 4! x 14 f iv 1+. 17 4! f x = 1 x = f 1 = 1. Again differentiating we get: f x = 1 x = f 1 = 1. Again differentiating we get: f x = x = f 1 =. Again differentiating we get: f iv x = 6 π x = f iv = 6. 4 On putting these values in 17 we get ln x = x 1 1 x 1 + 1 x 1 1 4 x 14 +. Putting x = 1.1 in the above we get: ln1.1 = 1.1 1 1 1.1 1 + 1 1.1 1 1 4 1.1 14 + = 0.1 0.005 + 0.000 0.0000 = 0.095 correct up to four places of decimals. Example 19. Expand x + 7x + x 1 in powers of x.
Dr. Satish Shukla 14 of 50 Sol: By Taylor s series we know that fx = fa + x a f a + 1! x a f a +! Here fx = x + 7x + x 1 and a =, therefore: fx = f + x f + 1! x f +! x a f a +! x f +! x a4 f iv a +. 4! x 14 f iv 1+. 18 4! Since fx = x + 7x + x 1 we have f = 45. Differentiating we get: Again differentiating we get: Again differentiating we get: f x = 6x + 14x + 1 = f = 5. f x = 1x + 14 = f = 8. f x = 1 = f = 1. All other higher order derivatives are zero. On putting these values in 18 we get x + 7x + x 1 = 45 + 5x + 19x + x. It is the required expansion. Example 0. Use Taylor s theorem to prove that tan 1 x + h = tan 1 x + h sin θ sin θ 1 where θ = cot 1 x. + 1 n 1 h sin θ n h sin θ sin nθ n + sin θ + h sin θ sin θ Sol: By Taylor s series we know that fx + h = fx + h 1! f x + h! f x + h! f x +. 19 Here fx + h = tan 1 x + h, and so, fx = tan 1 x therefore differentiating we get: Again differentiating w.r.t. x we get: f x = 1 1 + x = 1 1 + cot θ = sin θ. f x = sin θ cos θ dθ dx = sin θ d dx cot 1 x 1 = sin θ 1 + x = sin θ sin θ since x = cot θ.
Dr. Satish Shukla 15 of 50 Again differentiating we get: f x = cos θ sin θ sin θ sin θ cos θ dθ dx 1 = sin θ cos θ sin θ + sin θ cos θ 1 + x = sin θ sin θ since x = cot θ. On putting these values in 19 we get tan 1 x + h = tan 1 x + h 1! sin θ + h! sin θ sin θ + h! sin θ sin θ + = tan 1 x + h sin θ sin θ 1 + 1 n 1 h sin θ n h sin θ sin nθ n + sin θ + h sin θ sin θ Example 1. Expand tan x + π as far as the term x 4 and evaluate tan 46.5 to 4 four places of decimals. Sol: By Taylor s series we know that the expansion of fx + h in powers of x is: fx + h = fh + x 1! f h + x! f h + x! f h +. 0 Here fx + h = tan x + π, fx = tan x, h = π π 4 4, and so, f = 1. Differentiating 4 fx we get: π f x = sec x = 1 + tan x = 1 + [fx] = f =. 4 Again differentiating we get: π f x = fxf x = f = 4. 4 Again differentiating we get: f x = fxf x + f xf x = fxf x + [f x] π = f = 16. 4 Again differentiating we get: f iv x = fxf x + f xf x + 4f xf x = fxf x + 6f xf x π = f iv = 80. 4 On putting these values in 0 we get tan x + π 4 = 1 + x 1! + x! 4 + x! = 1 + x + x + 8x + 10x4 +. 16 + x4 4! 80 +
Dr. Satish Shukla 16 of 50 On putting x = 1.5 = 1.5 π radians = 0.06 approximately in the above equation 180 we get: tan 46.5 = 1 + 0.06 + 0.06 + 80.06 = 1.058. + 100.064 + Thus, tan 46.5 = 1.058 correct to four places of decimals. Example. Find the value of 10. Sol: Let fx + h = x + h. By Taylor s series we know that the expansion of fx + h in powers of h is: fx + h = fx + h 1! f x + h! f x + h! f x +. 1 Here fx + h = x + h, and so, f x = x. Differentiating fx we get: Again differentiating we get: Again differentiating we get: On putting these values in 1 we get f x = 1 x. f x = 1 4x /. f x = 8x 5/. h x + h = x + x h 8x + h / 16x +. 5/ On putting x = 9, h = 1 in the above equation we get: 1 10 = 9 + 9 1 8 9 + 1 / 16 9 + 5/ = + 0.16667 0.0046 + 0.0005 =.169. Thus, 10 =.16 correct to four places of decimals. Exercise Assignment Q.1 Expand Ans. e x 1 + e x in Maclaurin s series as far as the terms x. e x 1+e x = 1 + x 4 8x! +. Q. Expand e x cos x in Maclaurin s series. Ans. e x cos x = 1 + x + x x.
Dr. Satish Shukla 17 of 50 Q. Prove that: sin 1 x =! x + x 4 +. 4! Hint. Use Maclaurin s series for y = sin 1 x. Q.4 Prove that: ln 1 + e x = ln + 1 x + 1 8 x 1 19 x4 +. Hint. Use Maclaurin s series for y = ln 1 + e x. Q.5 Prove that: e x sin x = x + x +! x 5! x5 +. Hint. Use Maclaurin s series for y = e x sin x. Q.6 Find the Maclaurin s series for y = sin m sin 1 x. Ans. y = mx + mm 1 x +.! Q.7 Expand tan x in powers of x π 4. Ans. tan x = 1 + x π 4 + x π 4 +. Q.8 Expand 7x 6 x 5 + x + in powers of x 1. Ans. 7x 6 x 5 + x + = 7 + 9x 1 + 76x 1 + 110x 1 + 90x 1 4 + 9x 1 5 + 7x 1 6. Q.9 Find the Taylor s series expansion of lncos x about the point π. Ans. lncos x = ln 1 x π 4! x π. Q.10 Prove that lnx + h = ln x + h x h x + h x. Hint. Use Taylor s series and expand fx + h in powers of h. Q.11 Calculate the value of 5 correct to four places of decimals by taking first four terms in Taylor s series. Hint. Use Taylor s series and expand fx + h = x + h in powers of h, and put x = 4, h = 1. 0 L Hospital s Rule 0 form. Let f t and g t exist and g t 0 for all t a, b. If fc = gc = 0 for some c a, b, then fx lim x c gx = lim f x x c g x. Proof. Suppose a < c < x < b or a < x < c < b. Define a new function h: [c, x] R by: ht = ft fx gx gt.
Dr. Satish Shukla 18 of 50 Then hc = hx = 0 and the function h is continuous in the interval [c, x] and differentiable in the interval c, x. Therefore, by the Rolle s theorem there exists ζ such that c < ζ < x and h ζ = 0, i.e., f ζ fx gx g ζ = 0 = fx gx = f ζ g ζ. Since c < ζ < x or x < ζ < c, as x c we have ζ c, and so, the above inequality yields: fx lim x c gx = lim f x x c g x which proves the required result. Remark 1. If fc = gc =, then the L Hospital s rule remains true form. Also, the forms 0, 0 0, 0, 1 etc., can be converted into either or 0 0 form, and then can be solved by L Hospital s rule. Example. Evaluate the following limits: i lim x 0 iv lim x x tan x x + x + 1 ii lim1 + x 1/x x π iii lim x 0 x π/ cos x x a ftdt x+ v lim [tan x + π/4]1/x vi x 0 lim x a x a. Sol: x i Given limits is: L = lim x 0 tan x = lim sec x = 1. x 0 1 0 0 form ii Given limits is: Thus, lnl = 1, and so, L = e. iii Given limits is: L = lim x 0 1 + x 1/x ln1 + x = lnl = lim x 0 x 1 = lim x 0 1 + x = 1. L = lim x π/ = lim x π/ =. x π cos x sin x 1 form 0 0 form 0 0 form
Dr. Satish Shukla 19 of 50 iv Given limits is: x+ x x + 1 + /x 1 + /x L = lim = lim lim x x + 1 x 1 + 1/x x 1 + 1/x x 1 + /x = lim 1 x 1 + 1/x x 1 + /x = lim 1 form x 1 + 1/x Therefore: 1 + /x ln 1 + 1/x lnl = lim x 1/x 1 + 1/xx = lim [ 1 + ] 1/x/x + 1 + /x1/x x 1 + /x 1 + 1/x [ ] 1 + 1/x + 1 + /x = lim x 1 + /x1 + 1/x [ = lim + 1 ] x 1 1 = 1. 0 0 form Thus, lnl = 1, and so, L = e. v Given limits is: L = lim x 0 [tan x + π/4] 1/x ln [tanx + π/4] = lnl = lim x 0 x = lim x 0 =. Thus, lnl =, and so, L = e. sec x + π/4 1 tanx + π/4 vi Given limits is: x a L = lim ftdt x a x a d [ x a = lim dx ftdt] x a d x a dx fx = lim x a 1 = fa. 1 form 0 0 form 0 0 form Here we assumed that the function is continuous so that the integral x a fx = fa. lim x a ftdt exists and a x b x Example 4. Evaluate: lim. x 0 x
Dr. Satish Shukla 0 of 50 Sol: Given limits is: a x b x 0 L = lim x 0 x 0 form a x ln a b x ln b = lim x a 1 = ln a ln b. Functions of Several Variables Suppose, a particle is moving parallel to the earth surface, then at any instant its energy depends only upon its velocity surely, we neglect the effect of other celestial and terrestrial bodies on the energy of particle. Precisely, the energy of particle Ev = 1 mv + K 0 where v is the velocity of particle and K 0 is its potential energy which is constant. Thus, the Ev depends only on the velocity v. We say that the energy E of particle is an output, while its velocity is the input for this output function, and for various values of input we obtain the different outputs. Now consider same particle but with a different situation. Suppose, the particle is moving in such a way that its hight from the earth surface changes continuously. Then, at any instant its energy depends upon its velocity v, as well as, its hight h from the earth surface. Precisely, the energy of particle Ev, h = 1 mv + mgh. What we see? We now see that the output function E depends on the two inputs, namely, the velocity v and the hight h of the particle. We say that the energy function E is a function of a single variable v, while the energy function E is a function of two variables v, h. In general, we say that a quantity y is a function f of n variables if its value depends on n variables x 1, x,..., x n. Mathematically, we represent this fact by: y = fx 1, x,..., x n. Partial Derivatives. Suppose, y = fx is a function of single variable. If we draw a graph of this function by taking the values of x on X-axis and of y on Y -axis, then we get a two-dimensional curve. Clearly, the input x can change only along the X-axis either towards left or towards right, and so, we can find the rate of change of y only along X-axis. This rate is called the derivative total derivative of y with respect to x and denoted by dy dx. Graph of a function of single variable Y Graph of a function of two variable Z 0. X 1 0.5 0.5 1 0. 5 Y 5 5 5 5 5 X
Dr. Satish Shukla 1 of 50 We call the inputs as independent variable and the output as dependent variable. Now consider a different case, when the dependent variable z is a function of two independent variables x, y. We write z = fx, y. Now if we draw the graph of this function by taking the values of x, y and z on three mutually perpendicular axes, we obtain a three dimensional surface. Then, apart from the previous case the independent variables x, y the inputs now can change in the XY -plane in any direction right or left, up or down; or in any direction different from these two, and so, we can find the rate of change of z along any such direction. Such rate of change is called the directional derivative of f. In particular, we are interested in finding the rate of change directional derivative of f in two directions i along X-axis; and ii along Y -axis, and so, we get two directional derivatives along these two axes. The rate of change of f or z along X-axis is called the partial derivative of f or z with respect to x and it is denoted by f. Similarly, the rate of change of f or z along Y -axis is called the partial derivative of f or z with respect to y and it is denoted by f y. Because, in moving along X-axis y remains constant, and f is the rate of change of f along X axis, we have: f = lim h 0 fx + h, y fx, y. h Similarly, in moving along Y -axis x remains constant, and f is the rate of change of f y along Y axis, we have: f y = lim fx, y + k fx, y. k 0 k In similar way, we can define the partial derivatives of higher orders. To find the partial derivative of z = fx, y with respect to x we differentiate z by usual rules of differentiation with respect to x, but treat the variable y as constant. Similarly, when we find the partial derivative of z = fx, y with respect to y we differentiate z by usual rules of differentiation with respect to y, but treat the variable x as constant. If u = fx, y, z is a function of three variables, then find the partial derivative of u = fx, y, z with respect to x we differentiate u by usual rules of differentiation with respect to x, but treat all other variables y and z as constant, and so on.. Sol: Example 5. Find the first and second partial derivatives of the function z = x + y axy. Given function is Differentiating partially with respect to x we get: Differentiating partially with respect to y we get: z = x + y axy. z = x ay. z y = y ax. 4
Dr. Satish Shukla of 50 Differentiating partially with respect to x and y we get: z = 6x; z y = z y = a. Differentiating 4 partially with respect to y we get: z y = 6y. Sol: Example 6. If zx + y = x + y, then show that Given function is z z = 4 1 z y z. y Differentiating 5 partially with respect to x we get: z = x + y x + y. 5 z = x + yx x + y = x + xy y. 6 x + y x + y Differentiating 5 partially with respect to y we get: From 6 and 7 we obtain: z = x + yy x + y = y + xy x. 7 x + y x + y z z = y = = = [ x + xy y y + xy x x + y x + y [ ] x y x + y [ x yx + y x + y 4x y x + y ] ] and 4 1 z z y ] = 4 [1 x + xy y y + xy x x + y x + y [ ] x + y x + xy y y + xy x = 4 x + y [ ] x + y + xy x + xy y y + xy x = 4 x + y [ ] x + y xy = 4 x + y 4x y = x + y.
Dr. Satish Shukla of 50 Therefore: y Example 7. If u = x tan 1 x and u y = u y. z z = 4 1 z y z. y x y tan 1, then show that u y y = x y x + y Sol: Differentiating given function partially with respect to y we get: = x 1 y 1 + y/x 1 [ x x y tan 1 + y 1 y 1 + x/y = x x + y y tan 1. = x y tan 1 x y x y + xy x + y xy ] Differentiating the above equation partially with respect to x we get: u = [ ] x x y tan 1 y y 1 = 1 y 1 + x/y 1 y = x y x + y. Similarly, y = x tan 1 y. Differentiating with respect to y we get: x u y = x 1 1 + y/x 1 x 1 = x y x + y. Therefore: u y = u y = x y x + y. Example 8. If v = x + y + z 1/, then prove that v + v y + v z = 0. Sol: Differentiating given function partially with respect to x we get: v = 1 x + y + z / x = x x + y + z /. Again differentiating with respect to x we obtain: v = x + y + z / x x + y + z 5/ x = x + y + z 5/ [ x x + y + z ] = x + y + z 5/ x y z.
Dr. Satish Shukla 4 of 50 Using symmetry of v in x, y and z we obtain: v y = x + y + z 5/ y x z v and z = x + y + z 5/ z x z. Adding the above three we get: v + v y + v = x + y + z 5/ x y z + x + y + z 5/ y x z z + x + y + z 5/ z x z = x + y + z 5/ x y z + y x z + z x z = x + y + z 5/ 0 = 0. Example 9. If u = ln x + y + z xyz, then show that + y + 9 u = z x + y + z. Sol: Differentiating given function partially with respect to x we get: = x yz x + y + z xyz. Using symmetry of u in x, y and z we obtain: and Adding the above three we get: y = y xz x + y + z xyz z = z xy x + y + z xyz. + y + z Thus: = x yz x + y + z xyz + y xz x + y + z xyz + = x yz + y xz + z xy x + y + z xyz = x + y + z yz xz xy x + y + z xyz x + y + z yz xz xy = x + y + zx + y + z yz xz xy = x + y + z. + y + z = x + y + z. z xy x + y + z xyz
Dr. Satish Shukla 5 of 50 Therefore: + y + u = z + y + z + y + z = + y + z x + y + z = + + x + y + z y x + y + z z x + y + z = x + y + z x + y + z x + y + z = 9 x + y + z. Sol: Example 0. If u = e xyz, then show that u y z = 1 + xyz + x y z e xyz. Differentiating given function partially with respect to x we get: z Again differentiating with respect to y we get: u y z = e xyz xy. = e xyz xz xy + e xyz x = e xyz x + x yz. Again differentiating with respect to x we get: u y z = e xyz yz x + x yz + e xyz 1 + xyz. = 1 + xyz + x y z e xyz. Example 1. If x x y y z z = c, then show that z y = x ln ex 1 at point x = y = z. Sol: Given that: x x y y z z = c. Taking logarithm we obtain: x ln x + y ln y + z ln z = ln c. Differentiating given function partially with respect to x note that, z is a function of x and y both, so, it will not be treated as constant we get: x 1 x + ln x + z 1 z z + ln z z = 0 = 1 + ln x + 1 + ln z z = 0 = z = 1 + ln x 1 + ln z. By symmetry of the function z in the variables x and y we obtain: z y = 1 + ln y 1 + ln z.
Dr. Satish Shukla 6 of 50 Differentiating the above equation partially with respect to x we obtain: z = [ 1 + ln y ] y 1 + ln z 1 = 1 + ln y 1 + ln z 1 z z 1 = 1 + ln y 1 + ln z 1 [ z 1 + ln x ] 1 + ln z 1 + ln x1 + ln y =. z1 + ln z Putting x = y = z in the above equation we get: z y 1 + ln x1 + ln x = z1 + ln x 1 = z1 + ln x 1 = zln e + ln x = x ln ex 1. Sol: Example. If v = r n, where r = x + y + z, then show that v xx + v yy + v zz = nn + 1r n. Given that: r = x + y + z. Differentiating partially with respect to x we get: r r = x = r = x r. By symmetry of the function r in the variables x and y we obtain: r y = y r, r z = z r. Now, given that v = r n. Differentiating partially with respect to x we obtain: v Again differentiating with respect to x we get: Again by symmetry we obtain: r = nrn 1 = nrn 1 x r = nxr n. v = nxrn = nr n n r + nn xr = nr n + nn xr n x r = nr [ n 4 r + n x ] v y = nrn 4 [ r + n y ]
Dr. Satish Shukla 7 of 50 and v z = [ nrn 4 r + n z ]. Adding the above three equalities we obtain: v xx + v yy + v zz = v + v y + v z = nr n 4 [ r + n x ] + nr n 4 [ r + n y ] +nr n 4 [ r + n z ] = nr n 4 [ r + n x + y + z ] = nr n 4 [ r + n r ] = nr n 4 n + 1 r = nn + 1r n. Example. If u = fr and x = r cos θ, y = r sin θ, then prove that u + u y = f r + 1 r f r. Sol: Given that x = r cos θ and y = r sin θ. On squaring and adding these two we obtain r = x + y. Differentiating the above equation with respect to x partially we obtain: r r Similarly we obtain: r = x r. r y = y r. Given that u = fr. Differentiating u with respect to x partially we get: = f r r = = x r f r. = x, i.e. Again differentiating the above equation with respect to x partially and using the formula d dx f 1f f = f 1f f + f 1 f f + f 1 f f we get: u = [ x r] r f = u = x r f r r + 1 r f r x r r f r = u = x r f r + 1 r f r x r f r. 8 Since the given functions are symmetric in x and y we obtain: u y = y r f r + 1 r f r y r f r. 9
Dr. Satish Shukla 8 of 50 Adding equations 8 and 9 we obtain u + u y = 1 r f r x + y + r f r 1 r f r x + y = 1 r f r r + r f r 1 r f r r = f r + r f r 1 r f r = f r + 1 r f r. Example 4. If x = r cos θ, y = r sin θ, then prove that r = r and 1 r θ = r θ. Sol: Given that x = r cos θ 0 y = r sin θ. 1 Squaring and adding these two we obtain: r = x + y. So, as we found in the previous r example: = x. Again dividing 1 by 0 we get r y x = sin θ cos θ = tan θ y = θ = tan 1. x Differentiating the above equation partially with respect to θ we obtain: θ = 1 1 + y y x x = y sin θ = r x + y r = sin θ r. On differentiating equation 0 partially with respect to r we get: r = r = cos θ = x r = r. Again differentiating equation 0 partially with respect to θ we get: = r sin θ = r r θ θ = 1 = r θ r θ. Chain Rule for Partial Differentiation. Suppose z = fx, y be a function of two variable, where x = xt, y = yt are functions of another variable t. Suppose there is a small change δt in the variable t, due to which there are small changes δx and δy in the variables x and y respectively. Because of these changes in x and y, suppose there is a small change δz in the function z = fx, y. Then, the rate of change of z in X direction will be z,
Dr. Satish Shukla 9 of 50 and so the change in z along the X direction will be z δx. Similarly, the change in z in Y direction will be z δy. Since the changes are very small the total approximate change y in z will be: δz z z δx + y δy. Therefore, the rate of change of z with respect to t: δz δt z δx δt + z δy y δt. For instantaneous rate of change, letting δ 0, and so, δx, δy 0 in the above inequality we obtain: dz dt = z dx dt + z dy y dt. In a general case, if z = fx, y, x = xr, s and y = yr, s, then we have: z r z s = z = z r + z y y r s + z y y s. The above results can be generalized for a function of n variables. Sol: and Example 5. If u = fx, y and x = r cos θ, y = r sin θ, then prove that + = y Since x = r cos θ, y = r sin θ we have: Now by chain rule we have: and Therefore: + 1 r r = θ θ r = cos θ, θ y r = sin θ, y θ r = + 1 r r = r sin θ = r cos θ. r + y y r = cos θ + sin θ y = θ + y y θ = r sin θ + r cos θ y. cos θ + sin θ + 1 y r = cos θ + sin θ = +. y. θ r sin θ + sin θ + cos θ y + r cos θ y
Dr. Satish Shukla 0 of 50 Example 6. If u = fx y, y z, z x, then prove that + y + z = 0. Sol: Let X = x y Y = y z Z = z x. 4 Then we have u = fx, Y, Z, i.e., u is a function of X, Y, Z. Differentiating, and 4 with respect to x, y, z we get: X = 1, X y = 1, X z = 0 Y = 0, Y y = 1, Y z = 1 Z = 1, Z y = 0, Z z = 1. Now by chain rule of partial differentiation we get: and = X y z X + Y Y + Z Z = X 1 + Y 0 + Z 1 = X Z, = X X y + Y Y y + Z Z y = 1 + X Y 1 + Z 0 = X + Y = X X z + Y Y z + Z Z z = X 0 + 1 + Y Z 1 = Y + Z. Adding the above three equalities we get + y + z = X Z X + Y Y + Z = 0. Example 7. Transform the Laplace equation u + u = 0 into the polar coordinates. y
Dr. Satish Shukla 1 of 50 Sol: We know that the relation between cartesian coordinates x, y and the polar coordinates r, θ are given by x = r cos θ, y = r sin θ, i.e., r = x + y and θ = tan 1. y x Therefore: r = x r = cos θ, r y = y r = sin θ and θ = sin θ r, θ y = cos θ r. Now by chain rule we have: = r = r + θ θ cos θ r sin θ r The above relation is true for all functions u, and so: Similarly, we have y = r = cos θ r sin θ r r y + θ θ y sin θ r + cos θ r The above relation is true for all functions u, and so: Therefore: u = = Similarly, cos θ r sin θ r = cos θ r y sin θ r + cos θ r θ cos θ r sin θ = cos θ u sin θ cos θ r + sin θ r u = y y = cos θ θ + sin θ u θ y sin θ r + cos θ r = cos θ r sin θ r u. θ θ. = sin θ r + cos θ r u. θ θ. θ θ cos θ r sin θ r θ sin θ cos θ r θ r θ r sin θ r 1 r θ + 1 u sin θ r r θ r θ = sin θ u + sin θ cos θ r + cos θ r sin θ r + cos θ r 1 r sin θ θ + cos θ u θ θ + 1 r. u r θ θ + cos θ r θ sin θ r + cos θ u r θ cos θ r + sin θ u r θ
Dr. Satish Shukla of 50 Therefore the Laplace equation will be: u + u y = 0 = cos θ + sin θ u r + sin θ + cos θ r = u r + 1 r r + 1 r u θ = 0. r + sin θ + cos θ r u θ = 0
Dr. Satish Shukla of 50 Exercise Assignment Q.1 If z = fx, y, x = e u + e v, y = e u + e v, then prove that: z z v = x z y z y. Hint: Use the chain rule of partial differentiation. Tangent Plane and Normal Normal Line Definition 1. Given a point P on a surface S, the tangent plane of S at point P, is the plane passing through P which contains the tangent lines of all the curves on S passing through P. Formula for tangent plane. Let fx, y, z = 0 be the equation of surface S, and P x 0, y 0, z 0 be a point on S. Then the equation of any plane passing through P is given by: ˆn r r 0 = 0 5 where r = xî + yĵ + zˆk and r 0 = x 0 î + y 0 ĵ + z 0ˆk and ˆn is the unit normal vector to the surface S at point P. Since P x 0, y 0, z 0 is given, the vector r 0 is known and we have to find only ˆn. Then, it is sufficient to obtain the direction cosines/ratios of ˆn. For this, we know that the normal vector to the surface S at point P is given by: gradf = f = î + ĵ y + ˆk z f = î f + ĵ f y + ˆk f z, where all the partial derivatives are calculated at point P. Therefore, it follows from 5 that the equation of tangent plane: î f + ĵ f f + ˆk [îx x0 + ĵy y 0 + y z ˆkz ] z 0 = 0 or f x x 0 + f y y y 0 + f z z z 0 = 0. 6 Definition. Given a point P on a surface S, the normal line or normal to S at point P, is the line passing through P which is perpendicular to the tangent plane at point P. Formula for Normal. Let fx, y, z = 0 be the equation of surface S, and P x 0, y 0, z 0 be a point on S. Then the equation normal to the surface S passing through P is given by: where f x = f, f y = f y, f z = f z x x 0 f x = y y 0 f y = z z 0 f z 7 are calculated at point P. In parametric form: x = x 0 + f x t y = y 0 + f y t z = z 0 + f z t.
Dr. Satish Shukla 4 of 50 Example 8. Find the equation of the tangent plane and the normal to the surface z = 41 + x + y at point,, 6. Sol. The equation of given surface can be written as fx, y, z = z 41 + x + y = 0. Therefore, f f f f f = 8x, = 8y, = z. At point,, 6 we have = 16, y z y = 16, f z = 1. Also, here x 0 =, y 0 =, z 0 = 6, therefore, by 6 the equation of tangent plane of the surface S will be: f x x 0 + f y y y 0 + f z z z 0 = 0 = 16x 16y + 1z 6 = 0 = 4x + 4y z + = 0. By 7 the equation of normal at point P will be x x 0 = y y 0 = z z 0 f x f y f z = x 4 = y 4 = z 6. Example 9. Find the equation of tangent plane and the normal to each of the following surfaces at the given points: A x + y = z at, 1, ; B x + y + xyz = at 1, 1,. Sol. A. The equation of given surface is fx, y, z = x + y + z = 0, and the point is x 0 =, y 0 = 1, z 0 =. Therefore, f f f = 4x, = y, = and at point, 1,, y z f f f = 8, =, =. Therefore, the equation of tangent plane at point, 1, is y z f x x 0 + f y y y 0 + f z z z 0 = 0 = 8x + y 1 + z + = 0 = 4x + y + z 6 = 0. By 7 the equation of normal at point P will be = x 4 x x 0 f x = y y 0 f y = z z 0 f z = y 1 = z +. Sol. B. The equation of given surface is fx, y, z = x + y + xyz = 0, and the point is x 0 = 1, y 0 =, z 0 = 1. Therefore, f = x + yz, f y = y + xz, f z = xy and at point 1,, 1, f =, f y = 9, f z = 6. Therefore, the equation of tangent
Dr. Satish Shukla 5 of 50 plane at point 1,, 1 is f x x 0 + f y y y 0 + f z z z 0 = 0 = x 1 + 9y + 6z + 1 = 0 = x y z + = 0. By 7 the equation of normal at point P will be x x 0 f x = y y 0 f y = z z 0 f z = x 1 = y 9 = z + 1 6. Example 40. Show that the plane ax + by + cz + d = 0 touches the surface px + qy a + z = 0, if p + b + cd = 0. q Sol. Given equations of surface and plane are: fx, y, z = px + qy + z = 0 8 ax + by + cz + d = 0. 9 Suppose, plane 9 is a tangent plane of surface 8 at point P x 0, y 0, z 0. Then, we have f f f = px, = qy, y z = and at point P x 0, y 0, z 0, f = px 0, f y = qy 0, f z =. Therefore, the equation of tangent plane to the surface 8 at point P will be: f x x 0 + f y y y 0 + f z z z 0 = 0 = px 0 x x 0 + qy 0 y y 0 + z z 0 = 0 = px 0 x + qy 0 y + z px 0 + qy 0 + z 0 = 0. 40 But, by assumption, the tangent plane of surface 8 at point P is the plane 9, so, comparing the equations 9 and 40 we obtain: a = px 0 b = qy 0 41 c = d = px 0 + qy0 + z 0. Since point P x 0, y 0, z 0 is situated on the surface 8 it will satisfy 8, therefore we have px 0 + qy 0 + z 0 = 0. 4 Now, L.H.S. = a p + b q + cd = px 0 + qy 0 px 0 + qy0 + z 0 using 41 p q = 4px 0 + 4qy0 4px 0 + qy0 + z 0 = 4px 0 + qy 0 + z 0 = 0 using 4 = R.H.S.
Dr. Satish Shukla 6 of 50 Exercise Assignment Q.1 Find the equation of the normal to the surface x + y + z = a at any point P x 0, y 0, z 0. Q. Show that the plane x+1y 6z 17 = 0 touches the conicoid x 6y +9z +17 = 0. Find also the point of contact. Q. Find the tangent plane and normal to the surface xz xy 4x = 7 at point P 1, 1,. Q.4 Find the tangent plane and normal to the surface xyz = a at point P x 1, y 1, z 1. Q.5 Derive the formula for a tangent plane and normal to a given surface. Maxima and Minima of function of two variables Necessary condition for maxima or minima of a function of two variables: Suppose z = fx, y is a function of two variables x and y. We say that there is a maxima of function f at point a, b if fa + h, b + k fa, b < 0 for all h, k positive or negative. Similarly, say that there is a minima of function f at point a, b if fa + h, b + k fa, b > 0 for all h, k positive or negative. We discuss the necessary condition for maxima or minima of f analytically and geometrically. The Taylor s series for the function f about the point a, b is given by: fa + h, b + k = fa, b + [ h + k y ] fa, b + 1! [ h + k y ] fa, b + Neglecting the higher order terms we get: fa+h, b+k = fa, b+hf x a, b+kf y a, b+ 1 [ h f xx a, b + hkf xy a, b + k f yy a, b ]! or fa + h, b + k fa, b = hf x a, b + kf y a, b + 1 [ h f xx a, b + hkf xy a, b + k f yy a, b ]. 4! If there is a maxima or minima at a, b the LHS of the above equation is negative or positive. Therefore the RHS must be negative or positive for all values of h and k. Note that, the first two terms of the RHS changes their sign with change in the signs of h and k as h and k becomes positive and negative, and so, LHS will be negative or positive for all h and k if the first two terms becomes zero, i.e., f x a, b = f y a, b = 0. Geometrically, since at maxima or minima, the tangent plane to the surface z = fx, y becomes parallel to the XY -plane, its normal at point a, b must be in Z-direction. Since the direction ratios of normal are f x a, b, f y a, b and f z a, b, at maxima or minima we must have f x a, b = f y a, b = 0. Second Derivative Test. Therefore putting f x a, b = f y a, b = 0 in equation 4 we obtain: fa + h, b + k fa, b = 1 [ h f xx a, b + hkf xy a, b + k f yy a, b ]. 44!
Dr. Satish Shukla 7 of 50 Let r = f xx a, b, s = f xy a, b, t = f yy a, b, then: h f xx a, b + hkf xy a, b + k f yy a, b = h r + hks + k t = 1 r hr + ks + k t s. r On putting this value in 44 we get: fa + h, b + k fa, b = 1! [ 1 r hr + ks + k ] t s. r For maxima the LHS, and so the RHS should be negative and it is possible if r < 0 and t s r < 0, i.e., rt s > 0. For minima the LHS, and so the RHS should be positive and it is possible if r > 0 and t s r > 0, i.e., rt s > 0. For saddle point the LHS, and so the RHS should be positive as well as negative should change the sign and it is possible in the following two ways: i if r > 0 and t s r < 0, i.e., rt s < 0. ii if r < 0 and t s r > 0, i.e., rt s < 0. Thus, for saddle point we must have rt s < 0. Finally, if rt s = 0, then the neglected terms in the series becomes effective and we need the further investigation. Working Rules for finding the Maxima and Minima. Let We follow the following steps: Dx, y = f xx x, yf yy x, y [ f xy x, y ]. 1 Find the first derivatives f x x, y and f y x, y and solve the equations: f x x, y = 0 f y x, y = 0. Solutions of the above system is are the critical points. Suppose, a critical point is a, b; if Da, b > 0 and f xx a, b > 0, then fx, y has a local minimum at a, b; if Da, b > 0 and f xx a, b < 0, then fx, y has a local maximum at a, b; 4 if Da, b < 0, then fx, y has a saddle point at a, b; 5 if Da, b = 0, then we cannot draw any conclusions and further investigations are required. Example 41. Discuss the maxima and minima of fx, y = x + y axy. Sol: Given function is fx, y = x + y axy. Differentiating partially with respect to x and y we get: f x x, y = x ay, f y x, y = y ax.
Dr. Satish Shukla 8 of 50 First we find the critical point. Then: f x x, y = 0, f y x, y = 0 = x ay = 0, y ax = 0. Since f is symmetric in x and y, a solution of the above system is x = y. Putting x = y in the above equation we get: x ax = 0 = xx a = 0 = x = 0, a. Since x = y, we get two critical points 0, 0 and a, a. Now, by differentiating f x x, y and f y x, y again with respect to x and y we get: f xx x, y = 6x, f xy x, y = a, f yy x, y = 6y. Now we find D at each critical point. Then: i. D0, 0 = f xx 0, 0f yy 0, 0 [ f xy 0, 0 ] = 0 0 [ a ] = 9a < 0. Since D0, 0 < 0, the critical point 0, 0 is a saddle point. ii. Da, a = f xx a, af yy a, a [ f xy a, a ] = 6a 6a [ a ] = 6a 9a = 7a > 0. Since D0, 0 > 0, there is a maxima or minima at the critical point a, a. We consider two cases: If a < 0, then f xx a, a = 6a < 0 and so there is a maxima of function f and its maximum value is f max = fa, a = a + a a a a = a. If a > 0, then f xx a, a = 6a > 0 and so there is a minima of function f and its minimum value is f min = fa, a = a + a a a a = a. Example 4. Discuss the maxima and minima of u = xy + a x + a y. Sol: Given function is u = xy + a x + a. Differentiating partially with respect to x and y y we get: First we find the critical point. Then: u x x, y = y a x, u yx, y = x a y. u x x, y = 0, u y x, y = 0 = y a a = 0, x x y = 0.
Dr. Satish Shukla 9 of 50 Since f is symmetric in x and y, a solution of the above system is x = y. Putting x = y in the above equation we get: x a x = 0 = x a = 0 = x = a. Since x = y, we get the critical point a, a. Now, by differentiating u x x, y and u y x, y again with respect to x and y we get: u xx x, y = a x, u xyx, y = 1, u yy x, y = a y. Now we find D at each critical point. Then: Da, a = u xx a, au yy a, a [ u xy a, a ] = [ 1 ] = > 0. Since Da, a > 0, there is a maxima or minima at critical point a, a. Then u x,x a, a = > 0, and so, there is a minima of function u and its minimum value is u min = ua, a = a + a + a = a. Example 4. Discuss the maxima and minima of fx, y = xya x y. Sol: Given function is fx, y = xya x y = axy x y xy. Differentiating partially with respect to x and y we get: f x x, y = ay xy y, f y x, y = ax x xy. First we find the critical point. Then: f x x, y = 0, f y x, y = 0 = ay xy y = 0, ax x xy = 0. Since f is symmetric in x and y, a solution of the above system is x = y. Putting x = y in the above equation we get: ax x x x = 0 = xa x = 0 = x = 0, a. a Since x = y, we get two critical points, a and a, a. Now, by differentiating f x x, y and f y x, y again with respect to x and y we get: f xx x, y = y, f xy x, y = a x y, f yy x, y = x. Now we find D at each critical point. Then: i. D0, 0 = f xx 0, 0f yy 0, 0 [ f xy 0, 0 ] = 0 0 [ a ] = a < 0.
Dr. Satish Shukla 40 of 50 Since D0, 0 < 0, the critical point 0, 0 is a saddle point. ii. a D, a a = f xx, a a f yy, a [ a f xy, a = a a [ a ] a Since D, a = 4a 9 a 9 > 0. ] > 0, there is a maxima or minima at the critical point consider two cases: a If a > 0, then f xx, a = a maximum value is a f max = f, a = a 9 a If a < 0, then f xx, a = a minimum value is a f min = f, a = a 9 a, a. We < 0 and so there is a maxima of function f and its a a a = a 7. > 0 and so there is a minima of function f and its a a a = a 7. Example 44. Discuss the maxima and minima of fx, y = x y 1 x y. Sol: Given function is fx, y = x y 1 x y = x y x 4 y x y. Differentiating partially with respect to x and y we get: f x x, y = x y 4x y x y, f y x, y = x y x 4 y x y. First we find the critical point. Then: f x x, y = 0, f y x, y = 0 implies x y 4x y x y = 0; x y x 4 y x y = 0 = x y 4x y = 0; x y x y = 0 = 4x + y = ; x + y =. On solving the above equations we get x = 1, y = 1. Therefore, the critical point is 1, 1. Now, by differentiating f x x, y and f y x, y again with respect to x and y we get: f xx x, y = 6xy 1x y 6xy, f xy x, y = 6x y 8x y 9x y, f yy x, y = x x 4 6x y. 1 Now at critical point, 1 we have 1 D, 1 1 = f xx, 1 1 f yy, 1 [ 1 f xy, 1 ] = 1 1 1 = 1 9 8 1 7 1 144 > 0.