and ( x, y) in a domain D R a unique real number denoted x y and b) = x y = {(, ) + 36} that is all points inside and on

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1 Mat 7 Calculus III Updated on 10/4/07 Dr. Firoz Chapter 14 Partial Derivatives Section 14.1 Functions o Several Variables Deinition: A unction o two variables is a rule that assigns to each ordered pair o real numbers (, ) in a set D a unique real number denoted b (, ). The set D is the domain o and its range is the set o values that takes on, that is { (, ) (, ) D}. Deinition: I is a unction o two variables with domain D, then the graph o is the set o all points (,, z) in R such that z = (, ) and (, ) D. Deinition: The Level Curves (contour curve) o a unction o two variables are the curves with equation (, ) = k where k is a constant (in the range o ) Functions o three or more variables: A unction o three variables, is a rule that assigns to each ordered triple (,, z) in a domain D R a unique real number denoted b (,, z ) Eamples: 1. Find the domain o a) (, ) = and b) (, ) = Solution: a) The domain o is D = {(, ) } b) The domain o is D = {(, ) , }. Find the domain and range o (, ) = 6 Solution: The domain o is D the circle o radius 6. = {(, ) + 6} that is all points inside and on And the range o the unction o is { z z = 6,(, ) D}. Find the domain and range o is z = h(, ) = 4 + Solution: We have seen in chapter 1 that the unction h(, ) is an elliptic paraboloid with verte at (0, 0, 0), and opens upward. Horizontal traces are ellipses and vertical

2 Mat 7 Calculus III Updated on 10/4/07 Dr. Firoz traces are parabolas. The domain is all the ordered pairs (, ) in R, that is the plane. The range is the set [0, ) o all nonnegative real numbers. 4. Sketch all the level curves o the unction 5. Find the level suraces o the unction (, ) 6 = or k = 0,1,, (,, z) = + + z Solution: Choose dierent numerical values o (,, z) and observe that k z our tet. = + + represents spheres as level suraces. See eample 15, page # 897 at Section 14. Limits and Continuit Deinition: Let (, ) be a unction o two variables whose domain D includes points arbitraril close to ( a, b ). Then the it o (, ) as (, ) approaches ( a, b) is L and is written as s and continuit i or ever number 0 (, ) = L (, ) ( a, b) ε > there is a corresponding number δ > 0 such that (, ) L ε whenever (, ) D and 0 ( a) ( b) < < + < δ Deinition: Continuous unction The unction (, ) is continuous on D i is continuous at ever point ( a, b) in D. Eamples: 1. Given (, ) =. Find the its when (, ) (0,0) along + a) the ais b) the ais c) the line = d) the line = - e) the parabola = (0) Solution: a) Along ais = 0: (,0) = = 0 (, ) (0,0) + 0 (0) b) Along ais = 0: (0, ) = = 0 (, ) (0,0) 0 + c) Along the line = : d) Along the line =- : (, ) = = 1/ + (, ) = = 1/ + (, ) (0,0) (, ) (0,0)

3 Mat 7 Calculus III Updated on 10/4/07 Dr. Firoz e) Along the parabola = + (, ) = = 0 (, ) (0,0) 4. Given (, ) =. Find the it i eists. + In Eample 1, we have seen dierent values along dierent lines/curves, thus the it does not eist.. Evaluate (, ) (1,) + 4. Evaluate Solution: Where 5. Evaluate ( + ) ln( + ) (, ) (0,0) r r ln r 1/ r 1/ r ( + ) ln( + ) = ln = = = 0 (, ) (0,0) r 0 r 0 r 0 + = r + sin + (, ) (0,0) Solution: DNE. + sin 0 = 1/ + 0 (, ) (0,0) along ais and 0 + sin = 1. Limit (0) + (, ) (0,0) 6. Evaluate 7. Evaluate (, ) (0,0) (, ) (0,0) + Answer: 0 8. Homework problems: (, ) (0,0) (, ) (0,0) 8 + (,, z) (0,0,0) (, ) (0,0) z + + z

4 Mat 7 Calculus III Updated on 10/4/07 Dr. Firoz 6. (, ), (, ) = + + (0,0) 0, (, ) = (0,0) The it does not eist, thereore it is discontinuous. (, ) (0,0) + + Section 14. Partial Derivatives Eamples; 1. Find (1,), (1,), (1,), (1,), (1,), (1,) or (, ) = (1,) Solution: (1,) = = = 58, ind the rest = + +, ind, (, ) (, ) = e, ind, 4. (, t) = arctan( t ), ind, 5. (, ) = ln( + 5 ), ind,,,, t (1,) (1,) = = 4 + = 14. You can Section 14.4 Tangent Planes and Linear Approimations Suppose the unction (, ) has continuous partial derivatives. An equation o the tangent plane to the surace z (, ) P(,, z ) is Eamples: = at the point z z = (, )( ) + (, )( ) Given (, ) =, ind equation o the tangent plane at (1,1,1) Solution: The equation is z 1 = (1,1)( 1) + (1,1)( 1) Where (1,1) = (1,1) = 8 / 9. w z = e, ind = + + z dw w d w d w dz

5 Mat 7 Calculus III Updated on 10/4/07 Dr. Firoz Section 14.5 The Chain Rule Forms: d d d 1. Given = ( ), = g( t), then = dt d dt dz d d. Given z = (, ), = g( t), = h( t) then = + dt dt dt z. Given z = (, ), = g( s, t), = h( s, t) then = + s s s and z = + t t t d F 4. Implicit dierentiation: =, F(, ) = 0, = ( ) d F Eamples: 1. Given. Given z = + 4, = cos t, = sin t, ind dz dt 1 z z z = tan ( + ), = s t, = sln t, ind,and s t z z z 4st ln t Solution: = + = + s s s 1 + ( + ) 1 + ( + ) z z z s s / t = + = + t t t 1 + ( + ) 1 + ( + ) and. Given z = (, ), = g( t), = h( t), g() =, h() = 7, g () = 5, h () = 4 dz (, 7) = 6, (, 7) = 8, ind, t = dt dz d d Solution: and are unctions o one variable onl. We have = +. dt dt dt dz d d When t =, = + = (,7) g () + (,7) h () = 6 dt dt dt z M M 4. Given M = e, = uv, = u v, z = u + v ind, when u =, v = 1 u v M M M M z M M M M z Solution: use = + + and = + + u u u z u v v v z v

6 Mat 7 Calculus III Updated on 10/4/07 Dr. Firoz Section 14.6 Directional Derivatives In this section we etend the concept o a partial derivative to the more general notion o a directional derivative. The partial derivatives o a unction give the instantaneous rates o change o that unction in directions parallel to the coordinate aes. Directional derivatives allow us to compute the rates o change o a unction with respect to distance in an direction. The partial derivatives (, ) and (, ) represent the rates o change o (, ) in the directions parallel to the -ais and -ais. In this section we investigate the rates o change o (, ) in other directions given. Deinition: I (, ) is a unction o and and a given unit vector u = u1i + u j, then the directional derivative o (, ) in the direction o u at ( 0, 0) is given b D (, ) = (, ) u + (, ) u = u u where =<, > called the gradient o (, ). The gradient o a scalar unction is a vector which is orthogonal to the level curves. Maimum directional derivative: We have Du ( 0, 0) = u = u cosθ where θ is the angle between the gradient vector and the given unit vector. For maimum value o directional derivative cosθ must be equal to 1, which occurs when θ = 0 and then the unit vector has the same direction as the gradient vector. Theorem: Suppose (, ) is a dierentiable unction o two variables. The maimum value o the directional derivative Du (, ) is direction as the gradient vector =<, > and it occurs when u has the same Tangent Plane and Normal Vector: Assume that F(,, z) continuous irst order partial derivatives and let c = F( 0, 0, z0). I F( 0, 0, z0) O then F( 0, 0, z0) is a normal vector to the surace c = F(,, z) at the point P0 ( 0, 0, z0) and the tangent plane to the surace at P0 ( 0, 0, z0) is ( ) (,, z ) + ( ) (,, z ) + ( z z ) (,, z ) = z Eamples 1. Find the directional derivatives: 1 a) (, ) =, u = i + j at the point (1, ) b) (, ) = e, u = cosθi + sin θ j, θ = π / at the point (-, 0) c) (,, z) z z, a i j k = + = + at the point (1, -, 0)

7 Mat 7 Calculus III Updated on 10/4/07 Dr. Firoz d) (,, z) =, + z at the point P(, 1, -1) in the direction rom P to Q(-1,, 0) 4 4. Suppose that Du (1,) = 5, Dv (1,) = 10, u = i j, v = i + j. ind (1,), (1,),and Du (, ) in the direction o origin.. 4. e D (, ) =,ind ma u (,0) (, ) =,ind min u (,0) e D 5. Find the equation o the tangent plane to z = 18 at P(1,, 1) and determine the acute angle that the plane makes with the plane. Solution: F(,, z) =<,8,z >. Now F(1,,1) =<,16, > The plane thru P(1,, 1) has the equation ( 1) + 16( ) + ( z 1) = 0 The angle between two planes is the angle between the normal to the planes. Let us call the normals F(1,,1) =<,16, >= n1 and on plane < 0, 0,1 >= n. The angle n n θ = = n1 n cos cos (1/ 66) Section 14.7 Maimum and Minimum Values Deinition: A unction o two variables has local ma at ( a, b) i (, ) ( a, b) where (, ) is near ( a, b ). The number ( a, b) is called the local maimum value o On the other hand i (, ) ( a, b) where (, ) is near ( a, b ). The number ( a, b) is called the local minimum value o Theorem: I has a local maimum or minimum at ( a, b) and the irst partial derivatives o eist then ( a, b) = 0, ( a, b) = 0. The point ( a, b) is called a stationar point or a critical point. Second derivative test: Suppose the second partial derivatives o are continuous on a disk with center at ( a, b ), and suppose that ( a,) b = 0, ( a, b) = 0. Let us deine that D = D( a, b) = ( a, b) ( a, b) ( a, b) = 1) I D > 0, and ( a, b) > 0, then ( a, b) is a local minimum ) I D > 0, and ( a, b) < 0, then ( a, b) is a local maimum ) I D < 0, then, then ( a, b) is neither a local minimum nor a local maimum. In this case the point ( a, b) is called a SADDLE POINT. And graph o crosses its tangent plane at ( a, b )

8 Mat 7 Calculus III Updated on 10/4/07 Dr. Firoz Finding Absolute etrema on a closed and bounded set R Step 1 Find the critical points o that lie in the interior o R Step Find all boundar points at which the absolute etrema can occur Step Evaluate (, ) at the points obtained in the preceding steps. The largest o these values is the absolute maimum and the smallest o these values is the absolute minimum. Etreme Value Theorem: I is continuous on a closed bounded set D in attains both an absolute ma and an absolute min on D. Eamples: 1. The surace. The surace. The surace 4. The surace z R, then = (, ) = + has relative min (absolute min) at (0, 0) z = (, ) = 1 ( + ) has relative ma (absolute ma) at (0, 0) z = (, ) = + has relative min (absolute min) at (0, 0) z (, ) 8 = = + has relative min at (, 6) The surace z = (, ) = has relative min at (1, 1) and at (-1, -1) and a saddle point at (0, 0) 6. Find all absolute ma and min o z = (, ) = on the closed triangular region R with vertices P(0, 0), Q(, 0) and M(0, 5) Section 14.8 The Lagrange Multiplier Method To ind maimum and minimum values o (,, z) subject to the constraint g(,, z) = k, where g O on the surace g(,, z) = k 1. Find all values o,, z and λ such that (,, z) = λ g(,, z) and g(,, z) = k. Evaluate at all the points (,, z) that result rom step 1. The largest o these values is the absolute maimum and the smallest o these values is the absolute minimum. Eamples: 1. At what point or points on the circle + = 1 does (, ) = have an absolute maimum, and what is that maimum? Answer: (1/,1/ ), ( 1/, 1/ ), ½. Use Lagrange multiplier method to prove that the triangle with maimum area that has a given perimeter P is equilateral.

9 Mat 7 Calculus III Updated on 10/4/07 Dr. Firoz Solution: p = + + z, s = p /, where s is the hal o perimeter = constant. We need to maimize area A = s( s )( s )( s z). Let us consider (,, z) = A = s( s )( s )( s z), + + z = p, where g(, ) = + + z. Now = λ g s( s )( s z) = λ, s( s )( s z) = λ = and s( s )( s z) = λ, s( s )( s ) = λ = z. Use Lagrange multiplier method to ind the point on the plane + z = 4 that is closest to the point (1,,) Solution: Let us consider a point (,, z) on the given plane. To minimize d z = ( 1) + ( ) + ( ) with z 4 + =. We consider (,, z) = ( 1) + ( ) + ( z ), g(,, z) = + z Now set = λ g ( 1) = λ, ( ) = λ,( z ) = λ, + z = 4. Solving we get λ = 4 /, = 5/, = 4 /, z = 11/, which is the point on the plane that has minimum distance rom the given point. 4. The base o an aquarium with given volume V is made o slate and the sides are made o glass. I slate costs ive times as much (per unit area) as glass. Use Lagrange multiplier method to ind the dimensions o the aquarium that minimize the cost o the materials. Solution: Hint volume V = z, C(,, z) = 5 + ( z + z) = cost to minimize. Answer: 5. = = / 5 V, z = 5/ 4V