Mat 267 Engineering Calculus III Updated on 9/19/2010

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1 Chapter 11 Partial Derivatives Section 11.1 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 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 a unique real number denoted b (,, z ) Eamples: 1. Find the domain o a) (, ) 1 and b) (, ) 1 Solution: a) The domain o is D {(, ) } b) The domain o is D {(, ) 1 0, }. Find the domain and range o (, ) 6 Solution: The domain o is the circle o radius 6. And the range o the unction o is D {(, ) 6} that is all points inside and on { 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 traces are parabolas. The domain is all the ordered pairs (, ) in, 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 represents spheres as level suraces. See eample 15, page # 897 at our tet. Section 11. Limits and Continuit Deinition: Let (, ) be a unction o two variables whose domain D includes points arbitraril close to ( ab, ). Then the limit o (, ) as (, ) approaches ( ab, ) is L and is written as s and continuit lim (, ) (, ) ( a, b) i or ever number 0 there is a corresponding number 0 such that (, ) L whenever (, ) D and 0 ( a) ( b ) L Deinition: Continuous unction The unction (, ) is continuous on D i is continuous at ever point ( ab, ) in D. Eamples: 1. Given (, ). Find the limits when (, ) (0,0) a) the ais b) the ais c) the line = d) the line = - e) the parabola (0) Solution: a) Along ais = 0: lim (,0) 0 (, ) (0,0) 0 (0) b) Along ais = 0: lim (0, ) 0 (, ) (0,0) 0 c) Along the line = : d) Along the line =- : lim (, ) 1/ lim (, ) 1/ (, ) (0,0) (, ) (0,0)

3 e) Along the parabola lim (, ) 0 (, ) (0,0) 4. Given (, ). Find the limit i eists. In Eample 1, we have seen dierent values along dierent lines/curves, thus the limit does not eist.. Evaluate lim (, ) (1,) 4. Evaluate lim ( )ln( ) Solution: Where 5. Evaluate (, ) (0,0) ln r 1/ r 1/ r lim ( ) ln( ) lim r ln r lim lim 0 (, ) (0,0) r 0 r 0 r 0 r lim (, ) (0,0) sin Solution: DNE. sin 0 lim 1/ 0 (, ) (0,0) along ais and 0 sin lim 1. Limit (0) (, ) (0,0) 6. Evaluate 7. Evaluate lim (, ) (0,0) lim (, ) (0,0) lim (, ) (0,0) lim (, ) (0,0) Answer: lim (,, z) (0,0,0) lim (, ) (0,0) z z 6

4 1. (, ), (, ) (0,0) 0, (, ) (0,0) The limit lim (, ) (0,0) Section 11. Partial Derivatives Eamples; does not eist, thereore it is discontinuous. 1. Find (1,), (1,), (1,), (1,), (1,), (1,) or (, ) 4 (1,) Solution: (1,) , ind the rest (, ), ind, (, ) e, ind, 4. (, t) arctan( t ), ind, 5. (, ) ln( 5 ), ind,,,, t 6. Find the partial derivatives o the unction (1,) (1,) You can Find,,,, Solution: From undamental theorem o calculus I, we know that cos( 6), cos( 6). Now ou can ind the rest. Section 11.4 Tangent Planes and Linear Approimations Suppose the unction (, ) has continuous partial derivatives. An equation o the tangent plane to the surace z (, ) at the point P( 0, 0, z0) is Eamples: 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)

5 Where (1,1) (1,1) 8/ 9. z w e, ind dw wd wd wzdz. Find the linearization o the unction at the point (, ). Use this linearization to estimate (1.9,.1) Solution: (, ) 1.5 (, ). We have the linearization L(, ) z 4 (, )( ) (, )( ) 4 1.5( ) 1.5( ) Now L(1.9,.1) z 4 1.5(1.9 ) 1.5(.1 ) 4 4. The dimensions o a closed rectangular bo are measured as 90 centimeters, 50 centimeters, and 60 centimeters, respectivel, with the error in each measurement at most. centimeters. Use dierentials to estimate the maimum error in calculating the surace area o the bo. Solution: Suppose that the bo has dimension V z, the surace area is S z z. We have now ds S d S d S dz ( z) d ( z) d ( ) dz 160 z Section 11.5 The Chain Rule d d d 1. Given ( ), g( t ), then dt d dt dz d d. Given z (, ), g( t), h( t ) then dt dt dt. Given z (, ), g( s, t), h( s, t ) then z s s s and z t t t d F 4. Implicit dierentiation:, F(, ) d F 0, ( ) Eamples: 1. Let Compute w ( 4, ) s Solution: w w w w z s s s z s ( z) t ( z) te For s 4, t we have 8 8, e, z 4 st

6 Now plug all those values to ind w ( 4,) s Given z 4, cos t, sin t, ind dz dt. Given z 1 z tan ( ), s t, sln t, ind,and s Solution: z z z 4st ln t s s s 1 ( ) 1 ( ) z z z s s / t t t t 1 ( ) 1 ( ) z t and 4. 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. When dz d d t =, (,7) g () (,7) h () 6 dt dt dt z M M 5. Given M e, uv, u v, z u v ind, when u, v 1 u v M M M M z Solution: use u u u z u and M M M M z v v v z v 6. Consider the curve Find the equation o the tangent line to the curve at the point (1,1) in the orm o m b. Solution: The equation o the tangent line 9 8 is 1 ( 1) The radius o a right circular cone is increasing at a rate o inches per second and its height is decreasing at a rate o inches per second. At what rate is the volume o the cone changing when the radius is 10 inches and the height is 0 inches? Solution: 1 dv dr 1 dh V r h rh r (10)(0)() (10 )( ) dt dt dt

7 Section 11.6 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 uj, 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 cos where is the angle between the gradient vector and the given unit vector. For maimum value o directional derivative cos must be equal to 1, which occurs when 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, z 0). 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 derivative o at the point (4, 4, -) in the direction o the origin. 4 4 Solution: Du (, ) (4,4, ),,

8 . Find the directional derivatives: 1 a) (, ), u i j at the point (1, ) b) (, ) e, u cos i sin j, / at the point (-, 0) c) (,, z) z z, a i j k at the point (1, -, 0) d) (,, z), at the point P(, 1, -1) in the direction rom P to Q(-1,, 0) z 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 e D (, ),ind ma u (,0) (, ),ind min u (,0) e D 6. Find the equation o the tangent plane to 4 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 n n cos cos (1/ 66) 1 Section 11.7 Maimum and Minimum Values Deinition: A unction o two variables has local ma at ( ab, ) i (, ) ( a, b) where (, ) is near ( ab, ). The number ( a, b) is called the local maimum value o On the other hand i (, ) ( a, b) where (, ) is near ( ab, ). The number ( a, b) is called the local minimum value o Theorem: I has a local maimum or minimum at ( ab, ) and the irst partial derivatives o eist then ( a, b) 0, ( a, b ) 0. The point ( ab, ) 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 ( ab, ), 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

9 ) 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 ( ab, ) is called a SADDLE POINT. And graph o crosses its tangent plane at ( ab, ) 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, then z (, ) 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) 5. 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 (, ) 6 7 on the closed triangular region R with vertices P(0, 0), Q(, 0) and M(0, 5) Section 11.8 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,, zand such that (,, z) 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:

10 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. 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 g z. Now g s( s )( s z), s( s )( s z) (,, z) A s( s )( s )( s z), z p, where (, ) and s( s )( s z), s( s )( s ) 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 ( 1) ( ) ( z ) with z 4. We consider (,, z) ( 1) ( ) ( z ), g(,, z) z Now set g ( 1),( ),( z ), z 4. Solving we get 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 V, z 5/ 4V