Chapter (2) Applications of Partial Differentiation 2.1 TANGENT PLANES AND NORMAL VECTORS Definition Assume that πΉ (π₯, π¦, π§) has continuous first-order partial derivatives and that π0 (π₯0 , π¦0 , π§0 ) is a point on the level surface π: πΉ(π₯, π¦, π§) = π. If ∇F(π₯0 , π¦0 , π§0 ) = 0, then π = ππ (ππ , ππ , ππ ) is a normal vector to π at π0 and the tangent plane to S at π0 is the plane with equation Fπ₯ (π₯0 , π¦0 , π§0 )(π₯ − π₯0 ) + Fπ¦ (π₯0 , π¦0 , π§0 )(π¦ − π¦0 ) + Fπ§ (π₯0 , π¦0 , π§0 )(π§ − π§0 ) = 0 Remark n = ∇F = Fπ₯ π + Fπ¦ π + Fπ§ π =< Fπ₯ , Fπ¦ , Fπ§ > where π normal vector at point (π₯0 , π¦0 , π§0 ) Remark The angle between two vectors π1 and π2 is 1 cos π = |π1 β π2 | βπ1 ββπ2 β Example 12 Consider the ellipsoid π₯ 2 + 4π¦ 2 + π§ 2 = 18. (a) Find an equation of the tangent plane to the ellipsoid at the point (1,2,1). (b) Find parametric equations of the line that is normal to the ellipsoid at the point (1,2,1). (c) Find the acute angle that the tangent plane at the point(1,2,1) makes with the xy-plane. Solution Solution (a). Since ∇F = Fπ₯ π + Fπ¦ π + Fπ§ π =< Fπ₯ , Fπ¦ , Fπ§ > 2 ∇F = 2xπ + 8yπ + 2zπ =< 2x, 8y, 2z > we have π = π΅π(1,2,1) =< 2, 16,2 > Hence, the equation of the tangent plane is Fπ₯ (π₯0 , π¦0 , π§0 )(π₯ − π₯0 ) + Fπ¦ (π₯0 , π¦0 , π§0 )(π¦ − π¦0 ) + Fπ§ (π₯0 , π¦0 , π§0 )(π§ − π§0 ) = 0 By substituting, then 2(π₯ − 1) + 16(π¦ − 2) + 2(π§ − 1) = 0 Or π₯ + 8π¦ + π§ = 18 Solution (b). Since π = π΅π(1,2,1) =< 2, 16,2 > at the point (1, 2, 1), it follows from parametric equations for the normal line to the ellipsoid at the point (π₯0 , π¦0 , π§0 ) = (1, 2, 1) are π₯ = 1 + 2π‘, π¦ = 2 + 16π‘, π§ = 1 + 2π‘ Solution (c). To find the acute angle θ between the tangent plane and the xy-plane (π§ = 0) , since π1 = π =< 2, 16,2 > and π2 =< 0, 0,1 >. This yields 3 cos π = |π1 β π2 | |2(0) + 16(0) + 2(1)| βπ1 ββπ2 β √(22 + 162 + 22 )√(02 + 02 + 12 ) cos π = 2 2√66 Thus, π = cos−1 ( 1 √66 ) ≈ 83° 2.2 Linear Approximation. Since the tangent plane to the surface π§ = π(π₯, π¦) at π is very close to the surface at least when it is near π, we may use the function defining the tangent plane as a linear approximation to π. Recall that the equation of the tangent plane to the graph of π(π₯, π¦) at π(π, π, π(π, π)) is π§ = π(π, π) + ππ₯ (π, π)(π₯ − π) + ππ¦ (π, π)(π¦ − π) Definition. 4 The linear function πΏ whose graph is this tangent plane is given by πΏ(π₯, π¦) = π(π, π) + ππ₯ (π, π)(π₯ − π) + ππ¦ (π, π)(π¦ − π) L is called the linearization of π at (π, π). The approximation π(π₯, π¦) ≈ πΏ(π₯, π¦) = π (π, π) + ππ₯ (π, π)(π₯ − π) + ππ¦ (π, π)(π¦ − π) is called the linear approximation or tangent plane approximation of π at (π, π). Example. 14 Let π(π₯, π¦) = π₯π π₯π¦ . Find the linearization of π at (1, 0). Use it to approximate π(1.1, −0.1). Solution ππ₯ = π π₯π¦ + π₯π¦π π₯π¦ , then ππ₯ (1,0) = 1 ππ¦ = π₯ 2 π π₯π¦ , then ππ¦ (1,0) = 1 and π(1, 0) = 1. Then πΏ(1,0) = π(1, 0) + ππ₯ (1, 0)(π₯ − 1) + ππ¦ (1, 0)(π¦ − 0) =π₯+π¦ The corresponding linear approximation is π(π₯, π¦) = π₯π π₯π¦ ≈ π₯ + π¦ Then 5 π(1.1, −0.1) = 1 The actual value of f(1.1, −0.1) is 0.98542 round up to 5 decimal places. 2.3. Maximum and Minimum Definition A point (a, b) is said to be relative maximum of a function f(x, y), if (π, π) in the domain of π such that π(π, π) ≥ π(π₯, π¦), for all (x, y) in a neighborhood of (a, b). and π is said to have an absolute maximum at (π, π) if π (π, π) ≥ π(π₯, π¦)for all points (π₯, π¦) in the domain of π. Definition A point (a, b) is said to be relative Minimum of a function f(x, y), if (π, π) in the domain of π such that π(π, π) ≤ π(π₯, π¦), for all (π₯, π¦) in a neighborhood of (π, π). and π is said to have an absolute Minimum at (π, π) if π(π, π) ≤ π(π₯, π¦), for all points (π₯, π¦) in the domain of π. 6 If f has a relative maximum or a relative minimum at (π, π),, then we say that π has a relative extremum at (π, π),, and if π has an absolute maximum or absolute minimum at (π, π), then we say that π has an absolute extremum at (π, π). Necessary Condition A necessary condition that π(π₯, π¦) have a relative extremum at (π, π), an interior of the domain of π, is ππ₯ (π, π) = ππ¦ (π, π) = 0 Or one (or both) of partials does not exist Definition (Critical point) The point (π, π) is called critical point of π if ππ₯ (π, π) = ππ¦ (π, π) = 0 or if either partials does not exist at (π, π). Definition (saddle point) 7 the function π has neither a relative maximum nor a relative minimum at critical point (a, b),the point (a, b) is called a saddle point of π. Example (14) Let π(π₯, π¦) = 4 − π₯ 2 −π¦ 2 , find the extreme values of π Solution The partial exist every where, so the extreme values occur only where ππ₯ (π₯, π¦) = −2π₯ = 0, ππ¦ (π₯, π¦) = −2π¦ = 0 Thus the only critical point is (0,0). The graph π clearly indicates that π(0,0) = 4 ≥ π(π₯, π¦), is the absolute maximum. Second Partial Test Suppose that π has continuous second partials on a disc about critical point (π, π) and that ππ₯ (π, π) = ππ¦ (π, π) = 0 Define a new function πΉ (π₯, π¦) = ππ₯π₯ (π₯, π¦)ππ¦π¦ (π₯, π¦) − [ππ₯π¦ (π₯, π¦)] 8 2 (1) If πΉ (π, π) > 0, and ππ₯π₯ (π, π) < 0, then π(π, π) is a relative Maximum (2) If πΉ (π, π) > 0, and ππ₯π₯ (π, π) > 0, then π(π, π) is a relative Minimum (3) If πΉ (π, π) < 0, then π has a saddle point at (π, π). (4) If πΉ (π, π) = 0, then the test fails. Remarks (a) Since π has continuous second partials, ππ₯π¦ = ππ¦π₯ so either may be used for defining If πΉ (π, π) > 0, then ππ₯π₯ and ππ¦π¦ must have the (b) same sign. Example (15) Find the a relative maximum and the a relative minimum of π(π₯, π¦) = π₯ 3 + π¦ 3 − 3π₯ − 12π¦ + 20 Solution ππ₯ (π₯, π¦) = 3π₯ 2 − 3, ππ¦ (π₯, π¦) = 3π¦ 2 − 12 Therefore the only critical points are where, 3π₯ 2 − 3 = 0 and 3π¦ 2 − 12 = 0 Thus π₯ = ±1, π¦ = ±2 9 Thus the critical points are (1,2), (1, −2), (−1,2) and (−1, −2) ππ₯π₯ = 6π₯, ππ¦π¦ = 6π¦, ππ₯π¦ = 0 πΉ (π₯, π¦) = ππ₯π₯ (π₯, π¦)ππ¦π¦ (π₯, π¦) − [ππ₯π¦ (π₯, π¦)] 2 πΉ (π₯, π¦) = (6π₯ )(6π¦) − [0]2 = 36π₯π¦ Thus πΉ (1,2) = 72 > 0, ππ₯π₯ (1,2) = 6 > 0 Thus (1,2) is relative Minimum Thus πΉ (−1, −2) = 72 > 0, ππ₯π₯ (−1, −2) = −6 < 0 Thus (−1, −2) relative maximum πΉ (−1,2) = −72 < 0, thus (−1,2) ππ π π πππππ πππππ‘ πΉ (1, −2) = −72 < 0, thus (1, −2) ππ π π πππππ ππππ FINDING ABSOLUTE EXTREMA ON CLOSED AND BOUNDED SETS How to Find the Absolute Extrema of a Continuous Function f of two Variables on a Closed and Bounded Set π 10 Step 1. Find the critical points of π that lie in the interior of R. Step 2. Find all boundary points at which the absolute extrema can occur Step 3. Evaluate π(π₯, π¦) at the points obtained in the preceding steps. The largest of these values is the absolute maximum and the smallest the absolute minimum Example Find the absolute maximum and minimum values of π (π₯, π¦) = 3π₯π¦ − 6π₯ − 3π¦ + 7 (1) on the closed triangular region π with vertices (0, 0), (3, 0), and (0, 5). Solution. The region R is shown in Figure 11 We have ππ₯ = 3π¦ − 6 = 0 πππ ππ¦ = 3π₯ − 3 = 0 Solving these equations yields π₯ = 1 and π¦ = 2, so (1, 2) is the only critical point. As shown in Figure, this critical point is in the interior of R. Next we want to determine the locations of the points on the boundary of R at which the absolute extrema might occur. 1. The line segment between (0, 0) and (3, 0): On this line segment we have π¦ = 0, so (1) simplifies to a function of the single variable π₯, π’(π₯) = π(π₯, 0) = −6π₯ + 7, 0 ≤ π₯ ≤ 3 This function has no critical points because π’′ (π₯) = −6 is nonzero for all π₯. Thus the extreme values of π’(π₯) occur at the endpoints π₯ = 0 and π₯ = 3, which correspond to the points (0, 0) and (3, 0) of R. 2. The line segment between (0, 0) and (0, 5): On this line segment we have π₯ = 0, so (1) simplifies to a function of the single variable π¦, π£(π¦) = π(0, π¦) = −3π¦ + 7, 0 ≤ π¦ ≤ 5. This function has no critical points because π£ ′ (π₯) = −3 is nonzero for all π¦. Thus, the extreme values of π£(π¦) occur at the 12 endpoints π¦ = 0 and π¦ = 5, which correspond to the points (0, 0) and (0, 5) of π . 3. The line segment between (3, 0) πππ (0, 5): In the π₯π¦-plane, an equation for this line segment is π¦−5 0−5 = π₯−0 3−0 or −5 π = π + 5, 0 ≤ π ≤ 3 (2) 3 so (*) simplifies to a function of the single variable x, −5 π€ (π₯ ) = π (π, π + 5) 3 −5 = 3π ( π + 5) − 6π 3 −5 − 3( π + 5) + 7 3 = −5ππ + 14π − 8, 0 ≤ π ≤ 3 Since π€ ′ (π₯) = −10π₯ + 14, the equation π€ ′ (π₯ ) = 0 yields π₯ = 7 5 as the only critical point of π€. Thus, the extreme values of w occur either at the critical point π₯= 7 5 or at the endpoints x = 0 and x = 3. The endpoints correspond to the points (0, 5) and (3, 0) of R, and from (2) the critical point corresponds to( 7 5 8 , ) . 3 Finally, Table lists the values of π(π₯, π¦) at the interior critical point and at the points on the boundary where an absolute extremum can occur. From the table we 13 conclude that the absolute maximum value of f is π(0, 0) = 7 and the absolute minimum value is π(3, 0) = −11. (x, y) (0, 0) (3, 0) (0, 5) ( f(x, y) 7 -11 -8 7 8 , ) 5 3 9 5 (1, 2) 1 3.4 Total Differential. Let π§ = π(π₯; π¦). As in the case of functions of one variable, we take the differentials ππ₯ and ππ¦ to be independent variables. Definition. The differential ππ§, or the total differential, is defined to be ππ§ = ππ₯ (π₯; π¦)ππ₯ + ππ¦ (π₯; π¦)ππ¦ Consider the differential of π at the point (π; π). The tangent plane approximation of π at (π; π) implies that for a small change ππ₯ of π and a small change ππ¦ of π, the actual change βπ§ of π§ is approximately equal to ππ§. In other words, βπ§ ≈ ππ§ = ππ₯ (π; π)ππ₯ + ππ¦ (π; π)ππ¦ Similarly, for a function w = f(x, y, z) of three variables we have the total differential of w at (π; π; π) 14 βπ€ ≈ ππ€ = ππ₯ (π; π, π)ππ₯ + ππ¦ (π; π; π)ππ¦ + ππ§ (π; π; π)ππ§ which is also referred to as the total differential of π at (π, π, π). Example Let π(π₯; π¦) = π₯ 2 + 3π₯π¦ − π¦ 2 . ππ§. Find If π₯ changes from 2 to 2.05 and π¦ changes from 3 to 2.96, compare the values of βπ§ and ππ§. Solution ππ§ = ππ§ ππ§ ππ₯ + ππ¦ = (2π₯ + 3π¦)ππ₯ + (3π₯ − 2π¦)ππ¦ ππ₯ ππ¦ At the point (2; 3), ππ§ = ((2)(2) + 3(3))ππ₯ + ((3)(2) − (2)(3))ππ¦ That is ππ§ = 13ππ₯. Now we take ππ₯ = 2.05 − 2 = 0.05 and ππ¦ = 2.96 − 3 = −0.04. Thus, ππ§ = 13(0.05) = 0.65. For the actual change in z, we have βπ§ = π(2.05; 2.96) − π(2; 3) = 0.6449: Remark if βπ₯ and .βπ¦ are close to 0, then the magnitude of the error |βπ§ − ππ§| will be much smaller than the distance √βπ₯ 2 + βπ¦ 2 between (π, π) and (π + βπ₯, π + βπ¦) Example For the function π§ = π₯π¦ 2 15 1. Find the total differential for π§ = π₯π¦ 2 2. Use βπ§ ≈ ππ§ to approximate the change in π§ from its value at (0.5, 1.0) to its value at (0.503, 1.004). 3. Compute the error in approximating βπ§ by ππ§ 4. Compare the magnitude of the error in this approximation with the distance between the points (0.5, 1.0) and (0.503, 1.004). Solution 1. For π§ = π₯π¦ 2 , we have ππ§ = π¦ 2 ππ₯ + 2π¦π₯ππ¦. 2. Evaluating this differential at (π₯, π¦) = (0.5,1.0), ππ₯ = βπ₯ = 0.503 − 0.5 = 0.003, and ππ¦ = π¦ = 1.004 − 1.0 = 0.004 yields ππ§ = 1.02 (0.003) + 2(0.5)(1.0)(0.004)=0.007 π§ = 0.5 at Since (π₯, π¦) = (0.5, 1.0) and π§ = 0.507032048 at (π₯, π¦) = (0.503, 1.004), we have βπ§ = 0.507032048 − 0.5 = 0.007032048 3. The error in approximating βπ§ by ππ§ has magnitude |ππ§ − βπ§| = |0.007 − 0.007032048| = 0.000032048 4. Since the distance between (0.5,1.0) and (0.503,1.004) = (0.5 + βπ₯, 1.0 + βπ¦) is 16 √(βπ₯)2 + (βπ¦)2 = √(0.003)2 + (0.004)2 = √0.000025 = 0.005 We have |ππ§ − βπ§| = 0.000032048 < √(βπ₯)2 + (βπ¦)2 == 0.005 Lagrange Multipliers In this section we consider the problem of maximizing or minimizing a function π(π₯; π¦) subject to a constraint π(π₯; π¦) = 0. 1. Find all π₯; π¦; π§ and π such that ππ₯ = πππ₯ ; ππ¦ = πππ¦ ; ππ§ = πππ§ The number π is called a Lagrange Multiplier. 2. Evaluate π at all points (π₯; π¦; π§) obtained in (1). The largest of these values is the absolute maximum of π; the smallest is the absolute minimum of π. Example. Find the extreme values of π(π₯; π¦) = π₯ 2 + 2π¦ 2 on the circle π₯ 2 + π¦ 2 = 1. See figure 17 Solution Let π (π₯; π¦) = π₯ 2 + 2π¦ 2 πππ π(π₯, π¦) = π₯ 2 + π¦ 2 − 1 ππ₯ = πππ₯ ; ππ¦ = πππ¦ ; equivalent to 2π₯ = 2π π₯; 4π¦ = 2ππ¦ we need to solve the following system of equations in π₯; π¦ and π 2π₯ (π − 1) = 0 2π¦(π − 2) = 0 π₯ 2 + π¦ 2 = 1 (constraint equation) The first equation gives either π₯ = 0 ππ π = 1. 18 If π₯ = 0, then the constraint equation gives π¦ = ±1. Thus we have the solutions (0; −1); (0; 1). If π = 1, then the second equation gives π¦ = 0. Thus, by the constraint equation, we have the solutions (−1; 0); (1; 0). Consequently, we have four solutions (0; 1); (0; −1); (−1; 0); (1; 0) Now π (0; 1) = 2; π (0; −1) = 2; π(−1; 0) = 1; π(1; 0) = 1. Therefore, the absolute maximum value is 2 at (0; 1), (0; −1) and the absolute minimum value is 1 at (−1; 0), (1; 0) Example Find the point on the sphere π₯ 2 + π¦2 + π§2 = 4 that are closest to and farthest from the point (3; 1; −1). Solution 19 we will find points on the sphere that minimize and maximize the square of the distance to (3, 1, −1). Thus, we want to find the relative extrema of π (π₯, π¦, π§) = (π₯ − 3)2 + (π¦ − 1)2 + (π§ + 1)2 subject to the constraint π(π₯, π¦, π§) = π₯ 2 + π¦ 2 + π§ 2 = π ππ₯ = πππ₯ ; ππ¦ = πππ¦ ππ§ = πππ§ ; which leads to the equations 2(π₯ − 3) = 2π₯π, 2(π¦ − 1) = 2π¦π, 2(π§ + 1) = 2π§π Or (π₯ − 3) (π¦ − 1) (π§ + 1) = π; = π; =π π₯ π¦ π§ The first two equations imply that (π₯ − 3) (π¦ − 1) = π₯ π¦ 20 from which it follows that 3π¦ = π₯ Similarly, the first and third equations imply that (π¦ − 1) (π§ + 1) = π¦ π§ from which it follows that π¦ = −π§ Substituting 3π¦ = π₯ and π¦ = −π§ in the constraint equation, we obtain 9π¦ 2 + π¦ 2 + π¦ 2 = 4 11π¦ 2 = 4 ππ π¦ = ± 2 √11 Substituting these values in π₯ = 3π¦and π§ = −π¦ yields two points 6 2 −2 −6 −2 2 ( , , ),( , , ) √11 √11 √11 √11 √11 √11 Since π( 6 , 2 , −2 √11 √11 √11 ) = √11 – 2,π ( it follows that ( 6 , 2 , −2 −6 , −2 , 2 √11 √11 √11 ) = √11 + 2 ) is the point on the √11 √11 √11 sphere closest to (3, 1, -1), and ( −6 , −2 , 2 √11 √11 √11 that is farthest Exercises 21 ) is the point 1. Consider the surface π₯π§ − π¦π§ 3 + π¦π§ 2 = 2. (a) Find an equation of the tangent plane to the surface at the point (2, −1, 1). (b) Find parametric equations of the line that is normal to the surface at the point (2, −1,1). (c) Find the acute angle that the tangent plane at the point (2, −1,1) makes with the π₯π¦ −plane. 2. Locate all relative maxima, relative minima, and saddle points, if any (i) π (x, y) = y 2 + xy + 3y + 2x + 3 (ii) π(π₯, π¦) = π₯ 2 + π₯π¦ − 2π¦ − 2π₯ + 1 (iii) π(π₯, π¦) = π¦ sin π¦ (iv) π(π₯, π¦) = π π₯ sin π¦ (v) π(π₯, π¦) = π₯π π¦ (vi) π(π₯, π¦) = π₯ 2 + π¦ − π π¦ 3. Find the absolute extrema of the given function on the indicated closed and bounded set R. i. f(x, y) = xy − x − 3y; R is the triangular region with vertices (0, 0), (0, 4), and (5, 0). ii. π(π₯, π¦) = π₯π¦ − 2π₯; π is the triangular region with vertices (0, 0), (0, 4), and (4, 0) iii. f(x, y) = x 2 − 3y 2 − 2x + 6y; R is the region bounded by the square with vertices (0, 0), (0, 2), (2, 2), and (2, 0). 22 iv. π (π₯, π¦) = π₯ 2 + 2π¦ 2 − π₯; R is the disk π₯ 2 + π¦ 2 ≤ 4 v. π (π₯, π¦) = π₯π¦ 2 ; π is the region that satisfies the inequalities π₯ ≥ 0, π¦ ≥ 0, and π₯ 2 + π¦ 2 ≤ 1 4. Use a total differential to approximate the change in the values of π from π to π. Compare your estimate with the actual change in π . i. π (π₯, π¦) = π₯ 2 + 2π₯π¦ − 4π₯; π (1,2), π(1.01,2.04) ii. π (π₯, π¦, π§) = 2π₯π¦ 2 π§ 3 ; π(1, −1, 2), π(0.99, −1.02,2.02) 5. At what point or points on the circle π₯ 2 + π¦ 2 = 1 does π(π₯, π¦) = π₯π¦ have an absolute maximum, and what is that maximum 6. Find the points on the sphere π₯ 2 + π¦ 2 + π§ 2 = 36 that are closest to and farthest from the point (1, 2, 2). 23