Euler Theorem Chain Rule PDF

8-Euler theorem chain rule VU Lecture No- 8 More About Euler Theorem Chain Rule In general, the order of differentiation in an nth order partial derivatice can be change witho...

8-Euler theorem chain rule VU Lecture No- 8 More About Euler Theorem Chain Rule In general, the order of differentiation in an nth order partial derivatice can be change without affecting the final result whenever the function and all its partial derivatives of order ≤ n are continuous. For example, if f and its partial order derivatives of the firs t, second, and third ord ers are continuous on an open set, then at each point o f the set, f xyy = f yxy = f yyx or in ano ther notation. ∂3 f ∂3 f ∂3 f = = ∂y 2∂x ∂y∂x∂y ∂x∂y 2 Order of differentiation For a function f (x,y) = y2x4ex + 2 5 ∂3 f 2 ∂y ∂x If we are interested to find , that is, differentiating in the order firstly w.r.t. x and then w.r.t. y, calculation will involve many steps making the job difficult. But if we differentiate this function with respect to y, firstly and then with respect to x secondly then the value of this fifth order derivative can be calculated in a few steps. ∂5 f = 0 ∂ x 2∂ y 3 EXAMPLE x+ y f ( x, y ) = x− y ∂ ∂ ( x − y) ( x + y) − ( x + y) ( x − y) f x ( x, y ) = ∂x ∂x ( x − y) 2 ( x − y )(1) − ( x + y )(1) = ( x − y)2 − 2y = ( x − y )2 ∂ ∂ ( x − y) ( x + y) − ( x + y) ( x − y) f y ( x, y ) = ∂y ∂y ( x − y) 2 ( x − y )(1) − ( x + y)(−1) = ( x − y )2 2x = ( x − y)2 42 © Copyright Virtual University of Pakistan 8-Euler theorem chain rule VU EXAMPLE f ( x, y ) = x 3e − y + y 3 sec x 1 f x ( x, y ) = 3 x 2e − y + y 3 sec x tan x 2 x f y ( x, y ) = − x 3e− y + 3 y 2 sec x EXAMPLE f ( x, y ) = x 2 ye xy f x ( x, y) = 2 xye xy + x 2 y 2e xy = xye xy ( 2 + xy ) f x (1,1) = (1)(1)e(1)(1) [ 2 + (1)(1)] = 3e f ( x, y ) = x 2 ye xy f y ( x, y) = x 2 e xy + x 3 ye xy = x 2e xy (1 + xy) f y (1,1) = (1)(1)e (1)(1) [1 + (1)(1)] = 2e Example f (x, y) = x2 cos( xy) f x (x, y) = 2x cos( xy) − x2 y sin( xy) f x ( 1 ,π ) = 2( 1 ) cos(π ) − ( 1 )2 (π ) sin(π ) 2 2 2 2 2 = −π 4 f y ( x, y ) = − x3 sin( xy ) f y ( 12 , π ) = −( 12 )3 sin( π 2 ) = − 18 43 © Copyright Virtual University of Pakistan 8-Euler theorem chain rule VU EXAMPLE w =∂w( 4 x − 3 y + 2 z ) 5 = 20( 4 x − 3 y + 2 z ) 4 ∂x 2 ∂w =− 24 ( 4 x − 3 y + 2z ) 3 ∂y∂x ∂ 3w =−1440(4x−3y+2z)2 ∂z∂y∂x ∂4 w = −576 ( 4 x − 3 y + 2 z ) ∂z 2∂y∂x Chain Rule in function of One variable Given that w= f(x) and x = g(t), we find dw as follows: dt From w = f(x), we get dw dx From x = g(t), we get dx dt Then dw dw dx = dt dx dt Example w = x + 4, x = Sint By Substitution w = Sint + 4 dw = Cost dt dw w=x+4 ⇒ =1 dx dx x = Sint ⇒ = Cost dt By Chain Rule dw dw dx = × = 1. × Cost = Cost dt dt dt 44 © Copyright Virtual University of Pakistan 8-Euler theorem chain rule VU Chain rule in function of one variable y is a function of u, u is a function of v v is a function of w, w is a function of z z is a function of x. Ultimately y is function of x dy so we can ta lk about dx and by cha in rule it is given by dy dy du dv dw dz = dx du dv dw dz dx w = f(x,y), x = g(t), y = f(t) Dependent variable w = f(x,y) ∂w ∂w ∂y ∂x Intermediate variables x y dx dy dt dt dw ∂w dx ∂w dy = + dt ∂x dt ∂y dt t Independent variables EXAMPLE BY SUBSTITUTION w = xy x = cost, y = sint w = cost sint 1 = 2 sint cot 2 1 = sin2t 2 dw 1 = cos2t.2 dt 2 = cos 2t 45 © Copyright Virtual University of Pakistan  8-Euler theorem chain rule VU EXAMPLE w = xy, x = cos t , and y = sin t ∂w ∂w =x =y ∂y ∂x dx dy = − sint, dt = cost, dt dw ∂w dx ∂w dy = + dt ∂x dt ∂y dt = (sin t )( − sin t ) + (cos t )(cos t ) 2 = - sint + cos2 t = cos 2t EXAMPLE z = 3x2 y3 x = t4 , y = t2 ∂z ∂z = 6xy3 , = 9 x2 y2 ∂x ∂y dx dy = 4t3 , = 2t dt dt dz ∂z dx ∂z dy = + dt ∂x dt ∂y dt = (6xy3) (4t3) + 9x2y2 (2t) = 6 (t4) (t 6) (4t3) + 9 (t8) (t4) (2t) = 24 t13 + 18t13 = 42t13 EXAMPLE z = 1 + x- 2xy4 x = ln y=t ∂z 1 = ∂x 2 1 + x- 2xy4 ∂z 1 3 - 4xy3 =.(- 8xy) = ∂y 2 1 + x- 2xy4 1 + x- 2xy4 dx 1 d = , = dt t dt 46 © Copyright Virtual University of Pakistan 8-Euler theorem chain rule VU dz ∂z dx ∂z dy = +. dt ∂x dt ∂y dt 1 – 2y4 1 4xy3 = -.1 2 1 + x - 2xy t t 4 1 + x - 2xy4 1 ⎡ 1– ⎤ = ⎢ 4 4 ⎣ 2t - 4xy3 ⎥ 1 + x - 2xy ⎦ 4 1 ⎡1-2t ⎤ = ⎢ - 4 (lnt) t3 ⎥ 4 1 + lint - 2 (lnt) t ⎣ 2t ⎦ 1 ⎡1 3 3 ⎤ = ⎢ - t - 4t lnt⎥ 4 1 + lnt - 2t lnt ⎣2t ⎦ EXAMPLE z = ln (2x2 + y) 2/3 x = t, y = t ∂z 1 4 = 2. 4x = 2 ∂x 2x + y 2x + y ∂z 1 dx 1 1 dy 2 -1/3 = 2 , = , = t ∂y 2x + y dt 2 t dt 3 w = f(x,y,z), x = g(t) ,y = f(t), z = h(t) ∂w w = f(x,y,z) Dependent variable ∂w ∂x ∂w ∂z ∂y y z dy dx dz dt dt dt Independent variables t Overview of Lecture#8 Chapter # 16 Topic # 16.4 Page # 799 Book Calculus By Haward Anton 47 © Copyright Virtual University of Pakistan 8-Euler theorem chain rule VU Lecture No - 9 Examples First of all we revise the example which we did in our 8th lecture. Consider w = f(x,y,z) Where x = g(t), y = f(t), z = h(t) Then dw ∂w dx ∂w dy ∂w dz = + + dt ∂x dt ∂y dt ∂z dt Example: w = x2 + y + z + 4 x = e t, y = cost, z= t+4 ∂w ∂w ∂w = 2x, = 1, =1 ∂x ∂y ∂z dx dy dz = et , = −Sint, =1 dt dt dt dw ∂w dx ∂w dy ∂w dz = + +. dt ∂x dt ∂y dt ∂z dt t = (2x) (e ) + (1). (− Sint) + (1) (1) = 2 (et ) (et ) − Sint + 1 = 2 e2t − Sint + 1 Consider w = f(x), where x = g(r, s). Now it is clear from the figure that “x” is intermediate variable and we can write. ∂w dw ∂x and ∂w dw ∂x Dependent variable = = ∂r dx ∂r ∂s dx ∂s w = f(x) Example: dw 2 w = Sin x + x , x = 3r + 4s dx dw = Cosx + 2x dx x Intermediate variables ∂x ∂x =3 =4 ∂r ∂s ∂x ∂x ∂w dw ∂x ∂r =. ∂s ∂r dx ∂r = (Cosx + 2x). 3 = 3 Cos (3r+4s) + 6 (3r + 4s) s = 3 Cos (3r + 4s) + 18r + 24s ∂w dw ∂x =. ∂s dx ∂s = (Cosx + 2x). 4 = 4 Cosx + 8x = 4 Cos (3r + 4s) + 8 (3r + 4s) = 4 Cos (3r + 4s) + 24r + 32s 48 © Copyright Virtual University of Pakistan 9-Examples VU Consider the function w = f(x,y), Where x = g(r, s), y = h(r, s) w = f(x,y) Dependent variable ∂w ∂w ∂x ∂y x y Intermediate variables ∂x ∂r ∂x ∂y ∂y ∂s ∂r ∂s r s r r ∂w ∂w ∂x ∂w ∂y = + ∂r ∂x ∂r ∂y ∂r Similarly if you differentiate the function “w” with respect to “s” we will get And we have ∂w ∂w ∂x ∂w ∂y = + ∂s ∂x ∂s ∂y ∂s 49 © Copyright Virtual University of Pakistan  9-Examples VU Consider the function w = f(x,y,z), Where x = g(r, s), y = h(r,s), z = k(r, s) w = f(x,y,z) Dependent variable x y z intermediate variables ∂x ∂y ∂y ∂x ∂z ∂z ∂r ∂r ∂s ∂s ∂r ∂s r p s r r s Independent variables Thus we have ∂w ∂w ∂x ∂w ∂y ∂w ∂z = + + ∂r ∂x ∂r ∂y ∂r ∂z ∂r Similarly if we differentiate with respect to “s” then we have, ∂w ∂w ∂x ∂w ∂y ∂w ∂z = + + ∂s ∂x ∂s ∂y ∂s ∂z ∂s Example: r Consider the function w = x + 2 y + z 2 , x = , y = r 2 + ln s, z = 2r First we will calculate s ∂w ∂w =1 = 2 ∂w = 2 z ∂x = 1 ∂y = 2r ∂z = 2 ∂x ∂y ∂z ∂r s ∂r ∂r ∂w ∂w ∂x ∂w ∂y ∂w ∂z Now as we know that = + + By putting the values from above we get ∂r ∂x ∂r ∂y ∂r ∂z ∂r δw ⎛1⎞ = (1) ⎜ ⎟ + (2)(2r ) + (2 z )(2) δr ⎝s⎠ 1 1 = + 4r + (4r )(2) = + 12r s s Now ∂x r ∂y 1 ∂z =− 2 = =0 ∂s s ∂s s ∂s So we can calculate ∂w ∂w ∂x ∂w ∂y ∂w ∂z = + + ∂s ∂x ∂r ∂y ∂r ∂z ∂r ⎛ r ⎞ ⎛1⎞ = (1) ⎜ − 2 ⎟ + (2) ⎜ ⎟ + (2 z )(0) ⎝ s ⎠ ⎝s⎠ 2 r = − 2 s s 50 © Copyright Virtual University of Pakistan 9-Examples VU Remembering the different Forms of the chain rule: The best thing to do is to draw appropriate tree diagram by placing the dependent variable on top, the intermediate variables in the middle, and the selected independent variable at the bottom. To find the derivative of dependent variable with respect to the selected independent variable, start at the dependent variable and read down each branch of the tree to the independent variable, calculating and multiplying the derivatives along the branch. Then add the products you found for the different branches. The Chain Rule for Functions of Many Variables S up p o se ω = f ( x, y, …., υ ) is a d iffe re nt iab le fu nc t io n o f t he var iab les x, y, ….., υ (a fin ite set) a nd t he x, y, … , υ are d iffe re nt iab le fu n ct io ns o f p , q , , t (ano t her fin it e set ). T he n ω is a d iffe re nt iab le fu nc t io n o f t he var iab les p t hro u g h t a nd t he p art ia l d er iv at iv es o f ω w it h resp ect to t he se var iab le s are g iv e n b y eq ua t io ns o f t he fo r m ∂ω ∂ω ∂x ∂ω ∂y ∂ω ∂υ = + + …… +. ∂p ∂x ∂p ∂y ∂p ∂υ ∂p The other equations are obtained by replacing p by q, …, t, one at a time. One way to remember last equation is to think of the right- hand side as the dot product of two vectors with components. ⎛∂ω ∂ω ∂ω⎞ ⎛∂x ∂y ∂υ⎞ ⎜ , …… ⎟ and ⎜ , …… ⎟ ⎝ ∂x ∂y ∂υ ⎠ ⎝∂p ∂p ∂p ⎠ Derivatives of ω with Derivatives of the intermedaite respect to the variables with respect to the intermedaite variables selected independent variable Example: w = ln(e r + e s + et + eu ) Taking “ln” of both sides of the given equation we get e w = e r + e s + et + eu Now Taking partial derivative with respect to “r, s , u , and t” we get u −w , e wu = e ⇒ wu = e w u e w wr = e r ⇒ wr = e r − w , e w ws = e s ⇒ ws = e s − w and e w wt = et ⇒ wt = et − w 51 © Copyright Virtual University of Pakistan 9-Examples VU Now since we have wr = e r − w Now Differentiate it partially w.r.t. “s” wrs = er − w (− ws ) (Here we use the value of ws ) = −e r − w e s − w wrs = −e r + s − 2 w Now differentiate it partially w.r.t. “t” and using the value of wt we get, wrst = −e r + s − 2 w (−2 wt ) = 2e r + s − 2 w e t − w wrst = 2e r + s +t −3 w Now differentiate it partially w.r.t. “u” we get, wrstu = 2e r + s −3w (−3wu ) and by putting the value of w , we u get, wrstu = −6e r + s +t −3 w (eu − w ) wrstu = −6e r + s +t + u − 4 w 52 © Copyright Virtual University of Pakistan 11-The triple scalar or Box product VU Lecture No -10 Introduction to vectors Some of things we measure are determined by their magnitude. But some times we need magnitude as well as direction to describe the quantities. For Example, To describe a force, We need direction in which that force is acting (Direction) as well as how large it is (Magnitude). Other Example is the body’s Velocity; we have to know where the body is headed as well as how fast it is. Quantities that have direction as well as magnitude are usually represented by arrows that point the direction of the action and whose lengths give magnitude of the action in term of a suitably chosen unit. A vector in the plane is a directed line segment. B v A v = AB Vectors are usually described by the single bold face roman letters or letter with an arrow. The vector defined by the directed line segment from point A to point B is written as AB. Magnitude or Length Of a Vector : Magnitude of the vector v is denoted by v = AB is the length of the line segment AB Unit vector Any Vector whose Magnitude or length is 1 is a unit vector. Unit vector in the direction of vector v is denoted by v. and is given by v v= v Addition Of Vectors B b r A O a This diagram shows three vectors, in two vectors one vector OA is connected with tail of vector AB. The tail of third vector OB is connected with the tail of OA and head is connected with the head of vector AB.This third vector is called Resultant vector. 53 © Copyright Virtual University of Pakistan 11-The triple scalar or Box product VU The resultant vector can be written as r=a+b Similarly r=a+b+c+d+e+f e D E d f C F c r b a A B O Equal vectors Two vectors are equal or same vectors if they have same magnitude and direction. a a Opposite vectors Two vectors are opposite vectors if they have same magnitude and opposite direction. a -a Parallel vectors Two vector are parallel if one vector is scalar multiple of the other. b = λa where λ is a non zero scalar. 54 © Copyright Virtual University of Pakistan  11-The triple scalar or Box product VU r = x i+ y j + z k Addition and subtraction of two vectors in rectangular component: Let a = a1i + a2 j + a3k and b = b1i + b2 j + b3k a + b = (a1i + a2j + a3k) + (b1i + b2 j + b3k) = (a1 + b1 )i + (a2 + b2 )j + (a3 + b3)k a - b = (a1i + a2j + a3k) - (b1i + b2j + b3k) = (a1 - b1 )i + (a2 - b2 )j + (a3 - b3)k I th component of first vector is added (subtracted) to the ith component of second vector, jth component of first vector is added (subtracted) to the jth component of second vector, similarly kth component of first vector is added (subtracted) to the kth component of second vector, Multiplication of a vector by a scalar a 2a 3a -2a Any vector a can be written as a = a aˆ Scalar product Scalar product (dot product) (“a dot b”) of vector a and b is the number a.b = |a| |b| cos θ. where θ is the angle between a and b. In word, a.b is the length of a times the length of b times the cosine of the angle between a and b. 55 © Copyright Virtual University of Pakistan 11-The triple scalar or Box product VU Remark:- a.b = b.a This is known as commutative law. Some Results of Scalar Product a.b = |a| |b| cos θ. ⊥ 1. a b If This means that if a is perpendicular to b. Then a.b=0 Also i.j= 0 = j.i j.k= 0 =k.j k.i= 0 =i.k 2. If a || b That means a is parallel to b. Then a.b = | a | | b | if we replace b by a then a.a = | a | | a | 2 a.a = | a | | a | = a.a so i.i=j.j=k.k=1 Example If a = 3k and b = 2i + 2 k , then a. = |a | |b | cos θ π = (3) (2) cos 4 2 =6. =3 2. 2 56 © Copyright Virtual University of Pakistan 11-The triple scalar or Box product VU EXPRESSION FOR a.b IN COMPONENT FORM a = a 1 i + a 2 j + a 3 k a nd b = b1i + b2j + b3k a.b = ( a 1 i + a 2 j + a 3 k ). (b 1 i + b 2 j + b 3 k ) = a 1 i. (b 1 i + b 2 j + b 3 k ) + a 2 j. (b 1 i + b 2 j + b 3 k ) + a 3 k. (b 1 i + b 2 j + b 3 k ) = a 1 b 1 i. i + a 1 b 2 i. j + a 1 b 3 i. k + a 2 b j. i + a 2 b 2 j. j + a 2 b 3 j. k + a 3 b 1 k. i + + a 3 b2 k. j + a 3 b 3 k. k = a 1 b 1 ( 1 ) + a 1 b 2 (0 ) + a 1 b 3 ( 0 ) + a 2 b 1 (0 ) + a 2 b 2 (1 ) + a 2 b 3 (0 ) + a 3 b 1 (0 ) + + a 3 b 2 (0 ) + a 3 b 3 (1 ) = a 1 b 1 + a2 b 2 + a3 b 3 In dot product ith component of vector a will multiply with ith component of vector b , jth component of vector a will multiply with jth component of vector b and kth component of vector a will multiply with kth component of vector b A n g le B e t w e e n t w o v e c t o rs The angle between two nonzero vectors a and b is ⎛ a. b ⎞ θ = cos-1 ⎜ ⎝ |a ||b |⎠ Since the values of the arc lie in [0,i π ], above equation automatically gives the angle made by a and b. Example a = i − 2j − 2k and b = 6i + 3j + 2 k. a.b = (1)(6) + (− 2)(3) + (− 2) (2) =6−6−4=−4 |a | = (1)2 + (− 2)2 + (− 2)2 = 9 = 3 |b | = (6)2 + ( 3)2 + ( 2)2 = 49 = 7 ⎛ a.b ⎞ θ = cos-1 ⎜ ⎝ |a ||b |⎠ -1 ⎛ − 4 ⎞ -1 ⎛ 4⎞ = cos ⎜ = cos ⎜− ≈ 1.76 rad ⎝(3) (7) ⎠ ⎝ 21⎠ Perpendicular (Orthogonal) Vectors a and b are perpendicular if and only if a.b = 0. This has two parts If “a” and “b” are per perpendicular then a.b=0. And if a.b=0 “a” and “b” will be perpendicular. 57 © Copyright Virtual University of Pakistan 11-The triple scalar or Box product VU Vector Projectio Consider the Projection of a vector b on a vector a making an angle θ with each other B b θ O a C A From right angle triangle OCB COS θ =Base / hypotenuse → COS θ = | OC | |b| → |OC| = | b| Cosθ |b| |a| Cosθ = | a| b.a a = =b. | a| |a | a =b. | a| The number |b | cos θ is called the scalar component of B in the direction of a. a s ince |b | cos θ = B. , |a | we can find the scalar component by “dotting ” b with the direction of a Example Vector Projection of b = 6 i + 3 j + 2k onto a = i - 2 j − 2k is b.a a b= a.a a proj 6−6−4 = (i − 2 j − 2k) 1+4+4 4 = − (i − 2 j − 2k) 9 4 8 8 = i + j + k. 9 9 9 58 © Copyright Virtual University of Pakistan 11-The triple scalar or Box product VU of component he scalar b in T the direction of a a ⎛1 2 2 ⎞ |b |cosθ = b. = (6i + 3 j + 2k). ⎜ i − j − k |a | ⎝3 3 3 ⎠ 4 4 =2−2− =−. 3 3 T h e C ro s s P r o d u c t o f T w o V e c to r s in S p a c e Consider t wo nonzero vectors a and b in space. T he vec t or product a × b (“ a cross b ” ) to be the vector a × b = (| a | | b | sin θ ) n where n is a vector determined by right hand rule. Right-hand rule We start with two nonzero nonparallel vectors A and B.We select a unit vector n Perpendicular to the plane by the right handed rule. This means we choose n to be the unit vector that points the way your right thumb points when your fingers curl through the angle 0 from A to B. The vector A*B is orthogonal to both A and B. Some Results of Cross Product 59 © Copyright Virtual University of Pakistan 11-The triple scalar or Box product VU a × b = |a | |b | Sin θ ^ n If a ||b th e n a × b = 0 So a × a = 0 i× i = j× j= k × k = 0 If a ⊥ b th e n a × b = |a | |b | ^ n i × j = k, j× i= − k j × k = i, k × j = − i k × i= j i× k = − j N o te th a t th is p ro d u c t is n o t c o m m u ta tiv e. j i k The Area of a Parallelogram Because n is a unit the magnitude of a × b is |a × b | = |a | |b | |sin θ| |n| = |A| |b | sin θ This is the area of the ll l determined by a and b |a | being the base of the l θ | the and |b |ll|sin h i h 60 © Copyright Virtual University of Pakistan 11-The triple scalar or Box product VU a × b from the components of a and b Suppose a = a 1 i + a 2 j + a 3 k, b = b 1i + b2j + b 3k. a × b = (a 1 i + a2 j + a 3 k) × (b1 i + b 2 j + b 3 k ) = a1 b1 i × i + a 1 b2 i × j + a 1 b3 i × k + a 2 b 1 j × i + a 2 b2 j × j + a 2 b 3 j × k + a3 b1 k × i + a3 b2 k × j + a 3 b3 k × k a= =(a2ab13i −+ aa32bj2 )+i −a3(ak1,b3 − a3 b1 )j b =+ b(a11ib+2 −b2aj2 b+1)bk3.k. ⎪ i j k ⎪ ⎪ a2 a3 ⎪⎪ a × b = ⎪a1 ⎪b1 b2 b3⎪ Example: Let a = 2i + j + k and b = − 4i + 3j + k. Then i j k a×b = 2 1 1 −4 3 1 a × b = i (1 − 3) − j (2 + 4) + k (6 + 4) a × b = −2i − 6 j + 10k is the required cross product of a and b. Over view of Lecture # 10 Chapter# 14 Article # 14.3, 14.4 Page # 679 61 © Copyright Virtual University of Pakistan 11-The triple scalar or Box product VU Lecture No -11 The Triple Scalar or Box Product The product (a×b). c is called the triple scalar product of a, b, and c (in that order). As | (a×b).c | = |a×b| |c| |cos θ| the absolute value of the product is the volume of the parallelepiped (parallelogram-sided box) determin-ed by a,b and c By treating the planes of b and c and of c and a as he base planes of the parallelepiped determined by a, b and c we see that (a * b).c = (b×c).a=(c×a).b Since the dot product is commutative, ( a ×b).c =a.(b×c) a = a1i + a2j + a3k, b = b1i + b2j + b3k. c = c1i + c2j + c3k. ⎪i j k ⎪ a. (b × c) = a. ⎪ a1 a2 a3 ⎪ ⎪ b1 b2 b3 ⎪ =a. ⎡⎪b2 b3⎪ i - ⎪b1 b3 ⎪ j + ⎪b1 b2⎪k ⎤ ⎣⎪c2 c3 ⎪ ⎪c1 c3⎪ ⎪c1 c2 ⎪ ⎦ = a1 ⎪b2 b3⎪ − a ⎪b1 b3⎪ + a ⎪b1 b2⎪ ⎪c2 c3 ⎪ 2 ⎪c1 c3 ⎪ 3 ⎪c1 c2 ⎪ ⎪ a1 a2 a3 ⎪ = ⎪ b1 b2 b3⎪ Example ⎪ c1 c2 c3 ⎪ a = i + 2 j − k , b = − 2i + 3 k , c = 7 j − 4k. ⎪ 1 2 − 1⎪ ⎪ ⎪ a.(b ×c ) = ⎪− 2 0 3 ⎪ ⎪ 0 7 −4 ⎪ ⎪0 3 ⎪ ⎪− 2 3 ⎪ ⎪− 2 0 ⎪ =⎪ ⎪−2⎪ ⎪ −⎪ ⎪ ⎪7 − 4⎪ ⎪ 0 − 4⎪ ⎪ 0 7 ⎪ = − 21 − 16 + 14 = − 23 The volume is |a.(b × c )| = 23. 62 © Copyright Virtual University of Pakistan 11-The triple scalar or Box product VU When we solve a.(b×c) then answer is -23. if we get negative value then Absolute value make it positive and also volume is always positive. Gradient of a Scalar Function ∂ ∂ ∂ ∇≡i +j +k , ∂x ∂y ∂z ∇ is called “del operator” Gradient φ is a vector operator gradiantas φ is a vector operator defined defined as ⎡ ∂ ∂ ∂⎤ grad φ = ⎢i ⎛ ∂+ j ∂+ k ⎥∂φ⎞ grand φ =⎣ ∂⎜ix +∂jy +∂kz⎦ ⎟ φ = ∇⎝ φ∂x , ∂y ∂z⎠ = ∇ φ, ∂ ∂ ∂ ∇≡i +j +k , ∂x ∂y ∂z ∇ is called “del operator” ∇ “del operator” is a vector quantity. Grad means gradient. Gradient is also vector quantity. ∇ φ is vector and φ is scalar quantity, Every component of ∇ φ will operate with the. Directional Derivative If f(x,y) is differentiable at (x 0 ,y0 ), and if u = (u 1 , u 2 ) is a unit vector, then the directional derivative of f at (x0 , y 0 ) in the direction of u is defined by Du f(x0 ,y0 ) = fx (x0 ,y0 )u1 + fy (x0 ,y0 )u2 It should be kept in mind that there are infinitely many directional derivatives of z = f(x,y) at a point (x0 ,y0 ), one for each possible choice of the direction vector u Remarks (Geometrical interpretation) The directional derivative D u f(x0,y0) can be interpreted algebraically as the instantaneous rate of change in the direction of u at (x 0,y0 ) of z=f(x,y) with respect to the distance parameter s described above, or geometrically as the rise over the run of the tangent line to the curve C at the point Q 0 63 © Copyright Virtual University of Pakistan 11-The triple scalar or Box product VU Example T he directi onal derivative of f(x,y) = 3x 2y at the point (1, 2) in the direction of the vector a = 3i + 4j. f ( x ,y ) = 3 x 2 y fx ( x , y ) = 6 x y , fy ( x , y ) = 3 x 2 so that fx ( 1 , 2 ) = 1 2 , fy ( 1 , 2 ) = 3 a = 3i+4 j ^ a 1 Note: a= = (3i + 4 j) ||a|| 25 Formula for the directional 3 4 derivative can be written in the = i+ j following compact form using the 5 5 gradient notation ⎛3⎞ ⎛4⎞ Du f (1,2) = 12 ⎜ ⎟ + 3 ⎜ ⎟ ⎝5⎠ ⎝5⎠ Duf(x, y) = ∇ f(x, y). ^ u 48 The dot product of the gradient of f = 5 with a unit vector u^ produces the fx means that function f(x,y) is differentiating partially with respect to x and fy means that function f(x,y) is differentiating partially with respect to y. Example 2 2 f(x, y) = 2x + y , P 0 (− 1, 1) u = 3i − 4 j Another example, In this example we have to find directional derivative of the function |u| = 32 + (− 4)2 = 5 f ( x, y ) = 2 x 2 + y 2 at the point P0(-1,1) in ^u = 3 i − 4 j 5 5 the direction of u = 3i – 4j. To find the fx = 4x fx (− 1 , 1) = − 4 directional derivative we again use the fy = 2y fy (− 1, 1) = 2 above formula Du f (- 1,1) = fx(- 1,1)u1 + fy(- 1,1)u2 12 8 = − − =−4 5 5 Remarks If u = u1 i + u 2 jis a unit vector making an angle θ with the positive x - axis, then u 1 = cos θ and u 2 = sin θ D u f(x 0 ,y 0 )=f x (x 0 ,y 0 )u 1 + f y (x 0 ,y 0 )u 2 can be written in the form D u f(x 0 ,y 0 ) = f x (x 0 ,y 0 ) cos θ + f y (x 0 ,y 0 ) sin θ Example 64 © Copyright Virtual University of Pakistan 11-The triple scalar or Box product VU T h e d ir e c t io n a l d e r iv a t iv e o f e x y a t ( − 2 , 0 ) in t h e d ir e c t io n o f t h e u n it v e c to r u t h a t m a k e s a n a n g le o f π /3 w it h t h e p o s it iv e x - a x is. f(x,y) = exy fx(x.y) = yexy, fy (x, y) = x exy fx (− 2,0) = 0, fy (2,) = − 2 π π Duf(− 2, 0) = fx (− 2, 0) cos + fy(− 2, 0) sin 3 3 ⎛1⎞ ⎛ 3⎞ = 0 ⎜ ⎟ + (− 2) ⎜ ⎟ ⎝ ⎠ 2 ⎝ ⎠ 2 =− 3 Gradient of function If f is a function of x and then the gradient of is defined f b ∇f (x, y)= fx(x,y)i + f y(x,y)j Directional Derivative Formula for the directional derivative can be written in the following compact form using the gradient Du f(x, y) = ∇ f(x, y). ^ u The dot product of the gradient f with a unit ^ produces u directional derivative of f in direction u. EXAMPLE 2 ,In this example we have to find directional f(x,y) = 2xy− 3y , P0 (5,5) derivative of the function u = 4i + 3 j | u| = 4 + 3 = 5 2 2 f ( x, y ) = 2 x y − 3 y 2 at the point P0(5,5) in ^u = u = 4 i + 3 j the direction of u = 4i + 3j. To find the |u| 5 5 directional derivative we again use the fx = 2y, fy = 2x − by above formula fx (5, 5) = 10, fy (5,5) = − 20 ∇f = 10i − 20 j Du f(5,5) = ∇f. ^u ⎛4 ⎞ ⎛3⎞ = 10 ⎜ ⎟ − 20 ⎜ ⎟ ⎝5 ⎠ ⎝5⎠ =−4 65 © Copyright Virtual University of Pakistan  11-The triple scalar or Box product VU EXAMPLE Directional derivative of the function f(x, y) = xey+cos (xy) at the point (2,0) in the direction of a = 3i − 4 j. ^ a 3 4 u = = i − j. |a| 5 5 y fx ( x,y) = e − y sin (xy) y fy (x,y) = xe − xsin(xy) The partial derivatives of f at (2, 0) are 0 fx (2,0) = e − 0 = 1 0 fy (2,0) = 2e −2.0 = 2 The gradient of f at (2, 0) ∇f (2,0 = x (2, 0) i + y (2, 0) j =i+ j The derivative of f at 0) in the direction of a is ( u f) (2,0 = ∇f (2,0. ⎛3 4 ⎞ = j) ⎜ i − j ⎝5 5 ⎠ 3 8 − =− 5 5 Properties of Directional Derivatives Du f = ∇f. ^ u = |∇f| cos θ 1. The function f increases mostrapidly when cos θ = 1, or when u is the direction of ^ ∇f. That is, at each point P in its domain, f increases most rapidly in the direction of the gradient vector∇ f at P. The derivative in this direction is Du f = |∇ f| cos(0) = |∇ f|. 2. Similarly, f decreases most rapidly in the direction of − ∇f. The derivative in this direction is D u f=| ∇ f| cos ( π ) = − | ∇ f|. 66 © Copyright Virtual University of Pakistan 11-The triple scalar or Box product VU ^ orthogonal of the gradient is a direction of 3. Any direction u zero change in f because θ then equals π /2 and se Du f=|∇ f| cos (π /2)=|∇f|.0=0 x2 y2 f(x, y) = + 2 2 a) The direction of rapid change. The function increases most rapidly in the direction of ∇ f at (1,1). The gradient is (∇ f)(1,1) = (xi + y j)(1,1) = i + j. its direction is ^ i+ j u= |i + j| i+ j = (1)2 + (1)2 1 1 (1,1) are the directions orthogonal to ∇ f. = i+ j. 2 2 b) The directions of zero change The directions of zero change at ^=− n i+ j 2 2 1 1 1 1 and − ^ n= i− j. 2 2 rTT d apidly he he direction ecrease in directionsofof rapid −∇ f at(1,1), which is c). The function decreases most 1^ −= j.iu 2 67 © Copyright Virtual University of Pakistan 11-The triple scalar or Box product VU 68 © Copyright Virtual University of Pakistan  11-The triple scalar or Box product VU Lecture No -12 Tangent planes to the surfaces Normal line to the surfaces If C is a smooth parametric curve on three dimension, then tangent line to C at the point P0 is the line through P0 along the unit tangent vector to the C at the P0.The concept of a tangent plane builds on this definition. If P0(x0,y0,z0) is a point on the Surface S, and if the tangent lines at P0 to all the smooth curves that pass through P0 and lies on the surface S all lie in a common plane, then we shall regards that plane to be the tangent plane to the surface S at P0. Its normal (the straight line through P0 and perpendicular to the tangent) is called the surface normal of S at P0. Different forms of equation of straight line in two dimensional space 1. Slope intercept form of the Equation of a line. y = mx + c Where m is the slope and c is y intercept. 2.Point_Slope Form Let m be the slope and P0 ( x0 , y0 ) be the point of required line, then y – y = m (x – x ) 0 0 3. General Equation of straight line Ax + By + C = 0 Rise b Rise m = slope of line = = b Run a y − y0 = (x − x0 ) a Run 69 © Copyright Virtual University of Pakistan 12-Tangent planes to the surfaces VU Parametric equation of a line Parametric equation of a line in two dimensional space passing through the point (x0, y0) and parallel to the vector ai + bj is given by x = x0 + at, y = y0 + bt Eliminating t from both equation we get x − x0 y − y0 a = b b y − y0 = (x − x0 ) a Parametric vector form: r(t) = (x0 +at) i +(y0 +bt) j, Equation of line in three dimensional Parametric equation of a line in three dimensional space passing through the point (x0, y0, z0) and parallel to the vector ai + bj + ck is given by x = x0 + at, y = y0 + bt, z = z0 + ct Eliminating t from these equations we get x − x0 y − y0 z − z0 = = a b c EXAMPLE Parametric equations for the straight line through the point A (2, 4, 3) and parallel to the vector v = 4i + 0j – 7k. x = 2, y = 4, z = 3 0 0 0 and a = 4, b = 0 , c = - 7. The required parametric equations of the straight line are x = 2 + 4t, y = 4 + 0t, z = 3 – 7t Different form of equation of curve Curves in the plane are defined in different ways Explicit form: y = f(x) 70 © Copyright Virtual University of Pakistan 12-Tangent planes to the surfaces VU Example y= 9- x2 , -3 ≤ x ≤ 3. Implicit form: F(x, y) = 0 Example ]2 x + y2 = 9, - 3 ≤ x ≤ 3, 0≤y≤3 Parametric form: x= f(t) and y = g(t) Example x = 3cos θ, y = 3sinθ, 0≤θ≤π x = 3 cosθ, y = 3 sinθ 2 2 2 2 x + y = 9 cos θ + 9sin θ 2 2 = 9(Cos θ + Sin θ) 2 2 x +y =9 Parametric vector form: r(t) = f(t) i +g(t) j, a ≤ t ≤ b. : r ( t) = 3 cosθi + 3 sinθ j, 0 < θ < π. Equation of a plane A plane can be completely determined if we know its one point and direction of perpendicular (normal) to it. Let a plane passing through the point P0 (x0, y0, z0) and the direction of normal to it is along the vector n = ai + bj + ck Let P (x, y, z) be any point on the plane then the line lies on it so that n ⊥ P0 P (⊥ means perpendicular to ) 71 © Copyright Virtual University of Pakistan 12-Tangent planes to the surfaces VU → P0 P = (x − x0 ) i + (y − y0 ) j + (z − z0 ) k → n. P0 P = 0 a (x − x0 ) + b (y − y0 ) + c (z − z0 ) = 0 is the required equation of the plane Here we use the theorem ,let a and b be two vectors, if a and b are perpendicular then a.b=0 so n and P0 P are perpendicular vector so n.P0P=0 REMARKS Point normal form of equation of plane is a (x – x ) + b (y – y ) + c (z – z ) = 0 0 0 0 We can write this equation as ax + by + cz – ax – by – cz = 0 0 0 0 ax + by + cz + d = 0 where d = – ax – by – cz 0 0 0 Which is the equation of plane EXAMPLE An equation of the plane passing through the point (3, - 1, 7) and perpendicular to the vector n = 4i + 2j - 5k. A point-normal form of the equation is 4(x – 3) + 2 (y + 1) – 5 (z – 7) = 0 4x + 2y – 5z + 25 = 0 Which is the same form of the equation of plane ax + by + cz + d = 0 The general equation of straight line is ax + by + c = 0 Let (x1, y1) and (x2, y2) be two points on this line then ax1 + by1 + c = 0 ax2 + by2 + c = 0 Subtracting above equation a(x2 − x1) + b (y2 − y1) = 0 72 © Copyright Virtual University of Pakistan  12-Tangent planes to the surfaces VU v = (x2 − x1 )i + (y2 − y1 )j is a vector in the direction of line φ (x, y) = ax + by φ x = a, φy = b ∇φ = ai + bj = n ∇φ. r = 0 Then n and v are perpendicular The ge neral equat ion of plane is ax + by + cz + d = 0 For any two points (x1 , y1 , z1 ) and (x2 , y2 , z2 ) lying on this plane we have ax1 + by1 + cz1 + d = 0 (1) a x2 + by2 + cz2 + d = 0 (2) Subtracting equation (1) from (2) have a (x2− x1)+b (y2−y1) + c (z2 − z1) = 0 [ (ai + b j + ck ). (X2-X1) i+(Y2-Y1) j +(Z2-Z1) k ] Here we use the definition of dot product of two vectors. φ = ax + by + cz φ x = a, φ y = b, φz = c ∇ φ = ai + bj + ck Where v = (x2 − x1 )i+(y2 −y1 ) j + (z 2 − z 1 )k ∇ φ is always normal to the plane. Gradients and Tangents to Surfaces f (x, y) = c z = f(x,y), z = c If a differ entiable function f(x,y) has a constant value c along a smooth curve having parametric equation x = g(t), y = h(t), r = g(t)i+ h (t)j differentiating both sides of this equation with respect to t leads to the equation d d f (g(t), h(t)) = ( c ) dt dt 73 © Copyright Virtual University of Pakistan 12-Tangent planes to the surfaces VU ∂f dg ∂f dh + =0 Chain Rule ∂x dt ∂y dt ⎛ ∂f ∂f ⎞ ⎛dg dh ⎞ ⎜ i + j.⎜ i + j = 0 ⎝ ∂x ∂y ⎠ ⎝ dt dt ⎠ dr ∇f. = 0 dt ∇f is normal to the tangent vector d r/ dt, so it is normal to the curve through (x0 , y 0 ). Tangent Plane and Normal Line Consider all the curves through the point P0(x0, y0, z0) on a surface f(x, y, z) = 0. The plane containing all the tangents to these curves at the point P0(x0, y0, z0) is called the tangent plane to the surface at the point P0. The straight lines perpendicular to all these tangent lines at P0 is called the normal line to the surface at P0 if fx, fy, fz are all continuous at P0 and not all of them are zero, then gradient f (i.e fxi + fyj + fzk) at P0 gives the direction of this normal vector to the surface at P0. Tangent plane Let P0 (x 0 , y0 , z 0 ) be any point on the S urface f(x,y,z) = 0. If f(x,y,z) is differentiable at p o (x0 ,y 0 ,z0 ) then the tangents plane at the point P 0 (x0 ,y 0 ,z0 ) has the equation EXAMPLE 2 2 2 9x + 4y − z = 36 P (2,3,6). 2 2 2 f(x,y,z) = 9x + 4y − z − 36 fx = 18x, fy = 8y, fz = − 2z Equationsf of(P) Tangent = 36, Plane f (P)to=the 24,surface f (P) = -12 x y z 74 © Copyright Virtual University of Pakistan  12-Tangent planes to the surfaces VU through P is 36(x – 2) + 24(y – 3) –12(z – 6) = 0 3x + 2y – z – 6 = 0 EXAMPLE x z = x cos y − ye (0,0,0). x cos y − ye –z = 0 f (x,y,z) = cos y − yex –z fx (0,0, 0) = (cos y − yex )(0,0) = 1 − 0.1 = 1 fy (0,0, 0) = (− x sin y − ex )(0,0) = 0 − 1 = − 1. fz (0,0, 0) = - 1 The tangent plane is fxh(0,0,0)(x f − 0)+fy (0,0,0)(y − 0) + fz(0,0,0)(z−0)=0 1(x−0)−1 (y−0)−1(z−0) = 0, x − y − z = 0. 75 © Copyright Virtual University of Pakistan 13-Orthogonal surfaces VU Lecture No -13 Orthogonal Surface In this Lecture we will study the following topics Normal line Orthogonal Surface Total differential for function of one variable Total differential for function of two variables Normal line Let P0 (x0,y0,z0) be any point on the surface f(x,y,z)=0 If f(x,y,z) is differentiable at P o(x 0,y0,z 0) then the normal line at the point P (x 0,y ,z ) has the equation o 0 0 x = x0 +fx (P0 )t, y = y0 +fy (P0 )t, z = z0 +fz(P0 )t Here fx means that the function f(x,y,z) is partially differentiate with respect to x And fx(P0) means that the function f(x,y,z) is partially differentiate with respect to x at the point P0(x0,y0,z0) fy means that the function f(x,y,z) is partially differentiate with respect to y And fy(P0) means that the function f(x,y,z) is partially differentiate with respect to y at the point P0(x0,y0,z0) Similarly fz means that the function f(x,y,z) is partially differentiate with respect to z And fz(P0) means that the function f(x,y,z) is partially differentiate with respect to z at the point P0(x0,y0,z0) EXAMPLE Find the Equation of the tangent plane and normal of the surface f(x,y,z)= x2+y2+z2-4 at the point P(1,-2,3) 2 2 2 f(x,y,z) = x +y +z − 14 P (1, − 2, 3). fx = 2x, fy = 2y, fz = 2z fx (p0 ) = 2, fy (P0 ) = − 4, fz (P 0 ) = 6 Equation of the tangent p lane to the surface at P is 76 © Copyright Virtual University of Pakistan 13-Orthogonal surfaces VU 2(x − 1) − 4 (y + 2) + 6 (z − 3) = 0 x − 2y + 3z− 14 = 0 Equations of the normal line of the surface through P are x− 1 y+2 z− 3 = = 2 −4 6 x− 1 y+2 z− 3 = = EXAMPLE 1 −2 3 Find the equation of the tangent plane and normal plane 2 2 2 4x − y + 3z = 10 P (2,− 3,1) 2 2 2 f(x,y,z) = 4x − y + 3z − 10 fx = 8x, fy = − 2y, fz = 6z fx (P) = 16, fy (P) = 6, fz (P) = 6 Equations of Tangent Plane to the surface through P is 16(x − 2) + 6 (y+3) + 6 (z − 1) = 0 8x + 3y + 3z = 10 Equations of the normal line to the surface through P are x− 2 y + 3 z− 1 = = 16 6 6 x− 2 y + 3 z− 1 = = 8 3 3 Example 1 z= x7 y-2 2 1 f(x,y,z) = x7 y-2 – z 2 7 6 -2 fx = x.y , fy = - x7.y-3, fz = -1 2 7 fx (2, 4, 4) = (2)6 (4) = 896 2 fy (2, 4, 4) =− (2)7 (4)-3 = − 2 fz (2, 4, 4) =-1 Equation of Tangent at (2, 4, 4) is given by fx(2,4,4)(x−2)+ fy (2,4,4)(y−4)+ fz (2,4,4)(z−4) = 0 896 (x− 2) + (− 2) (y− 4) − (z − 4) = 0 896x− 2y − z − 1788 = 0 The normal line has equation s x = 2+fx(2,4,4)t, y = 4+fy(2,4,4) t, z = 4+fz(2,4,4)t x = 2 + 896t, y = 4− 2t, z = 4− t 77 © Copyright Virtual University of Pakistan  13-Orthogonal surfaces VU ORTHOGONAL SURFACES Two surfaces are said to be orthogonal at a point of their intersection if their normals l are orthogonal at that point are orthogonal. They are Said to intersect orthogonally if they at every point common to them. CONDITION FOR ORTHOGONAL SURFACES Let (x, y, z) be any point of intersection of f (x, y, z) = 0---- (1) and g (x, y, z) = 0 -----(2) Direction ratios of a line normal to (1) are f x, fy , fz Similarly, direction rations of a line normal to (2) are g x, gy, gz The two normal lines are orthogonal if and only if fx gx + fy gy + fz gz = 0 EXAMPLE Show that given two surfaces are orthogonal or not f(x,y,z) = x2 + y2 + z − 16 (1) g(x,y,z) = x2 + y2 − 638 (2) Adding (1) and (2) 63 1 x2 + y2 = , z= (3) 4 4 fx = 2x, fy = 2y fz = 1 gx = 2x, gy = 2y gz = − 63 fx gx + fy gy + fz gz = 4 (x2 + y2 ) − 63 using (3) fx gx + fy gy + fz gz ⎛63⎞ = 4 ⎜ ⎟ − 63 = 0 ⎝4⎠ Since they satisfied the condition of orthogonality so they are orthogonal. Differentials of a functions For a function / y = f(x) dy = f (x) d x is called the differential of functions f(x) dx the differential of x is the same as the actual change in x i.e. dx = ∇ x where as dy the differential of y is the approximate change in the value of the functions which is different from the actual change ∇y in the value of the functions. 78 © Copyright Virtual University of Pakistan 13-Orthogonal surfaces VU Distinction between the increment n ∆y and the differential dy Approximation to the curve If f is differentiable at x , then the tangent line to the curve y = f(x) at x0 is a reasonably good approximation 0 to the curve y = f(x) for value of x near x0. Since the tangent line passes / through the point (x 0 , f(x0 )) and has slope f (x0 ), the point-slope form of its equation is y − f (x0 ) = f(x0 )(x − x0 ) or / y = f(x0 ) + f (x0 ) (x − x0 ) / EXAMPLE f(x) = x x = 4 and dx∆=x = 3y = 3 ∆y = x +∆x − x = 7 − 4 ≈.65 If y = x , then dy 1 1 = so dy = dx dx 2 x 2 x 1 3 = (3) = =.75 2 x 4 79 © Copyright Virtual University of Pakistan 13-Orthogonal surfaces VU EXAMPLE Using differentials approximation for the value of cos 61 °. Let y = cosx and x = 6 0° then dx = 61° − 60° = 1° ∆y ≈ dy = − sinx dx = − sin60° (1°) 3 ⎛ 1 ⎞ = ⎜ π 2 ⎝180 ⎠ Now y = cos x y+∆ y = cos (x+ ∆x) = cos (x+dx) = cos (60° +1° )=cos61° cos61° = y+∆ y = cosx + ∆ y 3 ⎛ 1 ⎞ ≈ cos60° − ⎜ π⎟ 2 ⎝180 ⎠ 1 3 ⎛ 1 ⎞ cos61° ≈ − ⎜ π⎟ 2 2 ⎝180 ⎠ = 0.5 − 0.01511 = 0.48489 cos61° ≈ 0.48489 EXAMPLE A box with a square base has its height twice is width. If the width of the box is 8.5 inches with a possible error of ± 0.3 inches. Let x and h be the width and the height of the box respectively, then its volume V is given by 2 V=x h Since h = 2x, so (1) take the form 3 V = 2x 2 dV = 6x dx Since x = 8.5, dx = ±0.3, so putting these values in (2), we have 2 dV = 6 (8.5) (±0.3) = ± 130.05 This shows that the possible error in the volume of the box is ±130.05. TOTAL DIFFERENTIAL If we move from (x0 , y0 ) to a point (x0 + dx, y0 + dy) nearby, the resulting differential in f is df = fx (x0 , y0 ) dx + fy (x0 , y0 ) dy This change in the linearization of 80 © Copyright Virtual University of Pakistan  13-Orthogonal surfaces VU f is called the total differential of f. EXACT CHANGE Area = xy x = 10, y = 8 Area = 80 x = 10.03 y = 8.02 Area = 80.4406 Exact Change in area = 80.4406 − 80 = 0.4406 EXAMPLE A rectangular plate expands in such a way that its length changes from 10 to 10.03 and its breadth changes from 8 to 8.02. Let x and y the length and breadth of the rectangle respectively, then its area is A = xy dA = Ax dx + Ay dy = ydx + xdy By the given conditions x = 10, dx = 0.03, y = 8, dy = 0.02. dA = 8(0.03) + 10(0.02) = 0.44 Which is exact Change EXAMPLE The volume of a rectangular parallelepiped is given by the formula V = xyz. If this solid is compressed from above so that z is decreased by 2% while x and y each is increased by 0.75% approximately V = xyz. dV = Vx d x + Vy dy + Vzdx dV = yzdx + xzdy + zydz (1) 0.75 0.75 2 dx= x, dy = y, dz= − z 100 100 100 Putting these values in (1), we have 0.75 0.75 2 dV = xyz+ xyz − xyz 100 100 100 0.5 0.5 =− xyz = − V 100 100 This shows that there is 0.5 % decrease in the volume. 81 © Copyright Virtual University of Pakistan 13-Orthogonal surfaces VU EXAMPLE A formula for the area ∆ of a triangle is 1 ∆ = ab sin C. Approximately what error is 2 made in computing ∆ if a is taken to be 9.1 instead of 9, b is taken to be 4.08 instead of 4 and C is taken to be 30°3/ instead of 30°. By the given conditions a = 9, b = 4, C = 30°, da = 9.1 − 9 = 0.1, db = 4.08 − 4 = 0.08 ⎛ 3 ⎞° dC = 30°3′ − 3′ = ⎜ ⎟ ⎝60⎠ 3 π = × radians 60 180 Putting these values in (1), we have 1 ∆ = ab sin C 2 ∂ ⎛1 ⎞ ∂ ⎛1 ⎞ d∆ = ⎜ ab sin C⎟ da + ⎜ ab sin C⎟ db ∂a ⎝2 ⎠ ∂b ⎝2 ⎠ ∂ ⎛1 ⎞ + ⎜ ab sin C⎟ dC ∂C ⎝2 ⎠ 1 1 d∆ = b sin Cda + a sin C db 2 2 1 + ab cos CdC 2 1 1 d∆= 4 sin 30°(0.1)+ 9 sin 30°(0.08) 2 2 1 ⎛ π ⎞ + 36cos 30° ⎜ ⎟ 2 ⎝3600⎠ ⎛1⎞ 1 ⎛1⎞ d∆=2⎜ ⎟(0.1)+ ⎜ ⎟ (0.08) ⎝2⎠ 2 ⎝2⎠ ⎛ 3⎞⎛ 3.14 ⎞ +18 ⎜ ⎟⎜ ⎟ = 0.293 ⎝ 2 ⎠⎝3600⎠ 0.293 % change in area = × 100 ∆ 0.293 = × 10 = 3.25% 9 82 © Copyright Virtual University of Pakistan 14-Extrema of functions of two variables VU Lecture No -14 Extrema of Functions of Two Variables In this lecture we shall fined the techniques for finding the highest and lowest points on the graph of a function or , equivalently, the largest and smallest values of the function. The graph of many functions form hills and valleys. The tops of the hills are relative maxima and the bottom of the valleys are called relative minima. Just as the top of a hill on the earth’s terrain need not be the highest point on the earth , so a relative maximum need not be the highest point on the entire graph. Absolute maximum 2 A function f of two variables on a subset of R is said to have an D absolute (global) maximum value on if there is some point x y ) of D such that value of on D D ( 0, 0 f f (x0 , y0 ) > f (x, y) for all (x, y) ∈ D In such a case f ( x0, y 0) is the absolute maximum Relative extremum and absolute extremum , y ), then we say that f has a If f has a relative maximum or a relative minimum at (x0 0 relative extremum at (x 0 ,y 0), and if f has an absolute maximum or absolute minimum at (x0,y 0 ), then we say that f has an absolute extremum at (x 0,y0 ). Absolute minimum 2 absolute (global) A function f of two variables on a subset D of R is said to have an minimum value on D if there is some point (x , y ) of D such that 0 0 f ( x0, y0 ) ≤ f ( x, y) for all (x, y ) ∈ D. In such a case f (x0, y0 ) is the absolute minimum value of f on D. R elative (local) maximum The function f is said to have a relative (local ) maximum at some point (x0,y0) of its domain D if there exists an open disc K centered at (x0,y0) and of radius r 2 2 2 2 K ={( x, y ) ∈ R : ( x − x0 ) + ( y − y0 ) < r } With K ∈ D such that f(x0, y0) ≥ f (x, y) for all (x, y ) Relative (local) minimum 83 © Copyright Virtual University of Pakistan 14-Extrema of functions of two variables VU The function f is said to have a relative(local) minimum at some point (x 0, y 0) of D if there exists an open discKcentred at (x0, y0 ) and of radius r with K ⊂ D such that f (x 0, y0) ≤ f (x , y) for all (x , y) ∈ K. Extreme Value Theorem If f (x, y) is continuous on a closed and bounded set R, then f has both an absolute maximum and on absolute minimum on R. Remarks If any of the conditions the Extreme Value Theorem fail to hold, then there is no guarantee that an absolute maximum or absolute minimum exists on the region R. Thus, a discontinuous function on a closed and bounded set need not have any absolute extrema, and a continuous function on a set that is not closed and bounded also need not have any absolute extrema. Extreme values or extrema of f The maximum and minimum values of f are referred to as extreme values of extrema of f.Let a function f of two variables be defined on an open disc 2 2 2 K = {(x, y ): (x − x0) + (y − y0) < r }. Suppose f ( x0 , y ) and f ( x 0 , y0 ) both exist on K x 0 y If f has relative extrema at (x0,y0),then f x (x0, y0) = 0 = f y (x0, y0 ). 84 © Copyright Virtual University of Pakistan 14-Extrema of functions of two variables VU Saddle Point A differentiable function f(x, y) has a saddle has a saddle point (a, b) if in every open disk centered at (a, b) there are domain points (x, y) where f (x, y) > f (a, b) and domain points (x, y) where f (x, y) < f (a, b). The corresponding point (a, b, f (a, b)) on the surface z = f (x, y) is called a saddle point of the surface Remarks Thus, the only points where a function f(x,y) can assume extreme values are critical points and boundary points. As with differentiable functions if a single variable, not every critical point gives rise to o a local extremum. A differentiable function of a single variable might have a point of inflection. A differentiable function of two variable might have a saddle point. EXAMPLE Fine the critical points of the given function f (x, y) = x 3 + y3 − 3axy, a > 0. f x , f y exist at all points of the domain of f. f x = 3x2 − 3ay, f y = 3y2 − 3ax For critical points f x = f y = 0. Therefore, x2 − ay = 0 (1) and ax − y2 = 0 (2) Substituting the value of x from (2) into (1), we have y4 − ay = 0 a2 y(y3 − a3) = 0 y= 0, y= a and so x = 0, x= a. The critical points are (0, 0) and (a, a). 85 © Copyright Virtual University of Pakistan 14-Extrema of functions of two variables VU Overview of lecture # 14 Topic Article # page # Extrema of Functions of Two Variables 16.9 833 Absolute maximum 16.9.1 833 Absolute manimum 16.9.2 833 Extreme Value Theorem 16.9.3 834 Exercise set Q#1,3,5,7,9,11,13,15,17 841 Book CALCULUS by HOWARD ANTON 86 © Copyright Virtual University of Pakistan 15-Examples VU Lecture No - 15 Example EXAMPLE f ( x , y) = x 2 + y 2 x f x ( x, y) = 2 x + y2 y f y ( x, y) = 2 x + y2 The partial derivatives exist at all points of the domain of f except at the origin which is in the domain of f. Thus (0, 0) is a critical point of f Now fx(x, y) = 0 only if x = 0 and fy(x, y) = 0 only if y = 0 The only critical point is (0,0) and f(0,0)=0 Since f (x, y) ≥ 0 for all (x, y), f (0, 0) = 0 is the absolute minimum value of f. Example 2 2 z = f(x, y) = x + y (Paraboloid) fx (x, y) = 2x, f y (x, y) = 2y when f x (x, y) = 0, fy (x, y) = 0 we have (0, 0) as critical point. 87 © Copyright Virtual University of Pakistan 15-Examples VU EXAMPLE z = g(x, y) = 1− x2 − y2 (Paraboloid) gx (x, y) = − 2x, gy (x, y) =− 2y whengx (x, y) =0, gy (x, y) =0 we have (0, 0) as critical point. EXAMPLE 2 2 z = h(x,y)=y −x (Hyperbolic paraboloid) hx (x, y) = − 2x, hy (x, y) = 2y when hx (x, y) = 0, hy (x, y) = 0 we have (0, 0) as critical point. EXAMPLE f(x, y) = x2 + y2 x y fx = fy = 2 2 x +y x + y2 2 The point (0,0) is critical point of f because the partial derivatives do not both exist. It is evident geometrically that fx(0,.0) does not exist because the trace of the cone in the plane y=0 has a corner at the origin. The fact that fx(0,0) does not exist canalso be seen algebraically by noting that fx(0,0) canbe interpreted as the derivative with respect to x of the function f (x, 0) = x2 + 0 = |x| at x = 0. 88 © Copyright Virtual University of Pakistan 15-Examples VU But |x| is not differentiable at x = 0, so f x(0,0) does not exist. Similarly, fy(0,0) does not exist. The function f has a relative minimum at the critical point (0,0). The Second Partial Derivative Test Let f be a function of two variables with continuous second order partial derivatives in some circle centered at a critical point (x0, y 0), and let D = f xx (x0, y0) f yy (x0, y0) − f 2 xy ( x0, y0) (a) If D > 0 and fxx(x 0,y0) > 0 , then f has a relative minimum at (x0,y0). (b) If D > 0 and fxx(x0,y0) < 0 , then f has a relative maximumat (x0,y0). (c) If D < 0 , then f has a saddle point at (x0,y0). (d) If D = 0 , then no conclusion can be drawn. REMARKS If a function f of two variables has an absolute extremum (either an absolute maximum or an absolute minimum) at an interior point of its domain, then this extremum occurs at a critical point. EXAMPLE f(x,y) = 2x2 − 4x + xy2 − 1 2 fx (x, y) = 4x − 4 + y , fxx (x, y) = 4 fy (x, y) = 2xy, fyy (x, y) = 2x fxy (x, y) = fyx (x, y) = 2y For critical points, we set the first partial derivatives equal to zero. Then 2 4x − 4 + y = 0 (1) and 2xy = 0 (2) we have x = 0 or y = 0 x = 0, then from (1), y = ± 2. y = 0, then from (1), x = 1. Thus the critical points are (1,0), (0, 2), (0, − 2). 89 © Copyright Virtual University of Pakistan 15-Examples VU We check the nature of each point. fxx(1,0) = 4, fyy (1, 0) = 2, fxy (1, 0) = 0 D= fxx(1, 0).fyy (1, 0) - [fxy (1, 0)]2 =8>0 and fxx (1, 0) is positive. Thus f has a relative minimum at (1, 0). fxx(0,−2) = 4, fyy (0,−2) = 0, fxy(0,−2) = −4 D= fxx(0, − 2).fyy (0, − 2) - [fxy (0, − 2)]2 = − 16 < 0. fxx (0, 2) = 4, fyy (0, 2) = 0, fxy (0, 2) = 4 D= fxx(0, 2).fyy (0, 2) - [fxy (0, 2)]2 = − 16 < 0. Therefore, f has a saddle point at (0,2). Therefore, f has a saddle point at (0, − 2). EXAMPLE 2+y2 +2x) f(x,y) = e-(x -(x 2+y 2+2x) fx(x, y)=−2 (x+1)e , -(x 2+y 2+2x) fy(x, y) = − 2ye For critical points fx (x,y) = 0, x + 1 = 0, x =− 1 and fy (x, y) = 0, y = 0 Hence critical point is −( 1,0). 2 -(x2+y 2+2x) fxx(x,y) = [(− 2x − 2) − 2]e fxx( −1, 0) = - , 2 -(x2+y 2+2x) fyy (x,y) = [4y − 2]e fyy (− 1, 0) = - -(x2+y 2+2x) fxy(x,y) = − 2y (− 2x − 2)e fxy (− 1, 0) = 0 2 D = fxx(−1,0) fyy(−1, 0) − f xy (− 1, 0) = (-2e ) (-2e ) > 0 This shows that f is maximum at (−1, 0). 90 © Copyright Virtual University of Pakistan 15-Examples VU EXAMPLE f(x,y) =3 2x4 + y2 − x2 − 2y fx(x, y) = 8x − 2x, fy (x, y) = 2y − 2 2 fxx (x, y) = 24x − 2, f yy (x,y) = 2, fxy (x, y) = 0 For critical points fx(x, y) = 0, 2 2x (4x − 1) = 0, x = 0,1/2,-1/2 fy (x, y) = 0, − 2y 2 = 0, y=1 Solving above equation we have the critical ⎛ 1 ⎞ ⎛1 ⎞ points (0,1), ⎜− , 1⎟ ⎜ , 1⎟. ⎝ 2 ⎠ ⎝2 ⎠ fxx (0,1) = − 2, fyy (0, 1) = 2, fxy (0, 1) = 0 2 D = fx(0, 1) fyy (0, 1) − f xy (0, 1) = (− 2)(2) − 0 = −4 < 0 This shows that (0, 1) is a saddle point. ⎛1 ⎞ ⎛1 ⎞ fxx ⎜ , 1 = 4, fyy = ⎜ ,1 =2 ⎝2 ⎠ ⎝2 ⎠ ⎛1 ⎞ fxy⎜ , 1 = 0 ⎝2 ⎠ ⎛1 ⎞ ⎛1 ⎞ 2 ⎛ 1 ⎞ D = fxx⎜ ,1 fyy⎜ ,1 − f xy⎜− ,1 ⎝2 ⎠ ⎝2 ⎠ ⎝ 2 ⎠ = (4) (2) − 0 = 8 > 0 ⎛1 ⎞ ⎛1 ⎞ fxx⎜ ,1 = 4 > 0, so f is minimum at ⎜ ,1. ⎝2 ⎠ ⎝2 ⎠ 91 © Copyright Virtual University of Pakistan 15-Examples VU Example Locate all relative extrema and saddle points of f (x, y) = 4xy − x4 − y4. fx(x, y) = 4y − 4x3, fy (x, y) = 4x − 4y3 For critical points fx (x, y) = 0 4y − 4x3 = 0 (1) y = x3 fy (x, y) = 0 4x − 4y3 = 0 (2) x = y3 Solving (1) and (2), we have the critical points (0,0), (1, 1),(−1, −1). 2 Now fxx (x, y) = − 12x , fxx (0, 0) = 0 2 fyy (x, y) = − 12y , fyy (0,0) = 0 fxy (x, y) = 4, fxy (0, 0) = 4 2 D = fxx (0,0) fy (0,0) − f xy (0,0) 2 = (0) (0) − (4) = - 16 < 0 This shows that (0,0) is the saddle point. 2 fxx (x, y) = − 12x , fxx (1,1) = − 12 < 0 2 fyy (x,y) = − 12y , fy (1,1) = − 12 fxy (x, y) = 4, fxy (1,1) = 4 2 D = fxx (1,1) fyy (1,1) − f xy (1, 1) 2 = (− 12) (− 12) − (4) = 128 > 0 This shows that f has relative maximum at (1,1). 2 fxx (x,y) = −12x , fxx (−1, −1) = − 12 < 0 2 fy (x, y) = − 2y , fy (−1, −1) = − 12 fxy (x, y) = 4, fxy (− 1, − 1) = 4 2 D=fxx (−1,−1) fyy (−1,−1)−f xy(−1,−1) 2 = (− 12) (− 12) − (4) = 128 > 0 This shows that f has relative maximum (−1, − 1). Over view of lecture # 15 Book Calculus by HOWARD ANTON Topic # Article # Page # Example 3 836 Graph of f(x,y) 16.9.4 836 The Second Partial Derivative Test 16.9.5 836 Example 5 837 92 © Copyright Virtual University of Pakistan 16-Extreme valued theorem VU Lecture No -16 Extreme Valued Theorem EXTREME VALUED THEOREM If the function f is continuous on the closed interval [a, b], then f has an absolute maximum value and an absolute minimum value on [a, b] Remarks An absolute extremum of a function on a closed interval must be either a relative extremum or a function value at an end point of the interval. Since a necessary condition for a function to have a relative extremum at a point C is that C be a critical point, we may determine the absolute maximum value and the absolute minimum value of a continuous function f on a closed interval [a, b] by the following procedure. 1. Find the critical points of f on [a, b] and the function values at these critical. 2. Find the values of f (a) and f (b). 3. The largest and the smallest of the above calculated values are the absolute maximum value and the absolute minimum value respectively Example Find the absolute extrema of f(x)= x3+ x2-x+1 on [-2,1/2] Since f is continuous on [-2,1/2], the extreme value theorem is applicable. For this f /(x) =3 x2+2x-1 This shows that f(x) exists for all real numbers, and so the only critical numbers of f will be the values of x for which f (x)=0.. / Setting f (x) = 0, we have (3x − 1) (x + 1) = 0 from which we obtain 1 x=−1 and x =3 1 The critical points of f are −1 and 3 , and each of these points is in the given closed interval (-2, 1 ) We find the function values at the critical points and at the end 2 points of the interval, which are given below. f(− 2) = − 1, f (− 1) = 2, ⎛1⎞ 22 ⎛1⎞ 7 f = , f⎜ = ⎝3⎠ 27 ⎝ 2⎠ 8 1 The absolute maximum value of f on(-2, 2 ) is therefore 2, which occurs at− 1, and the absolute min. value of f on 1 ) − which occurs at the left end point− 2. 2 is 1, (-2, Find the absolute extrema of − 93 © Copyright Virtual University of Pakistan 16-Extreme valued theorem VU f (x) = (x 2) 2/3 on [1, 5]. Since f is continuous on [1. 5], the extreme-value ttheorem is applicable. Differentiating f with respect to x, we get / 2 f (x) = 3 (x − 2)1/3 / / There is no value of x for which f (x) = 0. However, since f (x) does not exist at 2, we conclude that 2 is a critical point of f, so that the absolute extrema occur either at 2 or at one of the end points of the interval. The function values at these points are given below. 3 f(1) = 1 , f(2) = 0, f(5)= 9 From these values we conclude that the absolute minimum value of f on [1,5] is 0, occurring at 2, and the absolute maximum value of f on [1, 5] is 3 9 ,occurring at 5. Find the absolute extrema of h(x) = x2/3 on [− 2, 3]. / 2 -1/3 2 h (x) = x = 1/3 3 3x / h (x) has no zeros but is undefined at x = 0. The values of h at this one critical point and at the endpoints x = − 2 and x = 3 are h(0) = 0 2/3 1/3 h (− 2) = (− 2) = 4 2/3 1/3 h(3) = (3) = 9. 1/3 The absolute maximum value is 9 assumed at x = 3; the absolute minimum is 0, assumed at x = 0. How to Find the Absolute Extrema of a Continuous Function f of Two Variables on a Closed and Bounded Region R. Step 1. Find the critical points of f that lie in the interior of R. Setp 2. Find all boundary points at which the absolute extrema can occur, Step 3. Evaluate f(x,y) at the points obtained in the previous steps. The largest of these values is the absolute maximum and the smallest the absolute minimum. Find the absolute maximum and minimum value of 94 © Copyright Virtual University of Pakistan 16-Extreme valued theorem VU f(x,y) = 2 + 2x +2y-x2-y2 On the triangular plate in the first quadrant bounded by the lines x=0,y=0,y=9-x Since f is a differentiable, the only places where f can assume these values are points inside the triangle having vertices at O(0,0), A(9,0)and B(0,9) where fx = fy=0 and points of boundary. B ( 0, 9).(9/2,9/2) x=0 y=9-x.(1,1) O y=0 A( 9,0) For interior points: We have fx=2-2x=0 and fy=2-2y=0 yielding the single point (1,1) For boundary points we take take the triangle one side at time : 1. On the segment OA, y=0 U(x) = f(x, 0)=2+2x-x2 may be regarded as function of x defined on the closed interval 0≤x≤9 Its extreme values may occur at the endpoints x=0 and x=9 which corresponds to points (0, 0) and (9, 0) and U(x) has critical point where U/(x) = 2-2x=0 Then x=1 On the segment OB, x=0 and V(y)=f(0,y)= 2+2y-y2 Using symmetry of function f, possible points are (0,0 ),(0,9) and (0,1) 3. The interior points of AB. With y = 9 - x, we have f(x, y) = 2+2x+2(9-x)–x2–(9- x)2 W(x) = f(x, 9-x) = - 61 +18x – 2x2 Setting w′(x)= 18 -4x = 0, x = 9/2. At this value of x, y = 9 – 9/2 9 9 Therefore we have ( , ) as a critical point. 2 2 (x, y) ⎛9 9⎞ (0,0) (9,0) (1, 0) ⎜ , ⎟ ⎝2 2⎠ 41 f(x,y) 2 3 − − 61 2 95 (x, y) (0, © 9) Copyright (0,1) (1,1) Virtual University of Pakistan f(x,y) − 61 3 4 16-Extreme valued theorem VU The absolute maximum is 4 which f assumes at the point (1,1) The absolute minimum is -61 which f assumes at the points (0, 9) and (9,0) EXAMPLE Find the absolute maximum and the a