Answers to Worksheet 5, Math 272

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1 Answers to Worksheet 5, Math 7 1. Calculate the directional derivative of the function f(, y, z) = cos y sin z at the point a = (1, π/, 5π/6) in the direction u = (3, 0, 1). The gradient of f at a is ( ) 6 f( a) = (cos y sin z, sin y sin z, cos y cos z) (1,π/,5π/6) =,, ( ) u 6 D u f( a) = f( a) u =,, (3, 0, 1) 10 = Find the equation for the line tangent to the curve 3 +y 3 = 7 y at the point a = (, 1). Let f(, y) = 3 + y y The gradient vector of f at a is f( a) = (3 + 7 y, 3y + 7 ) (,1) = ( 17, 10) Hence the equation for the line tangent to the level set f(, y) = 0 at the point a = (, 1) is ((, y) (, 1)) ( 17, 10) = 0 That is 17 ( ) + 10(y + 1) = 0 3. Find the point on the paraboloid z 3y = 0 at which the tangent plane is parallel to the plane 5 + y + z = 1. Let f(, y, z) = z 3y. The gradient vector of f is f = (, 6y, z) Hence, there eists a constant k such that (, 6y, z) = k(5, 1, 1). This yields k = 5, y = 1 15 z = 1 5 Finally, we have ( 1 5 ) 3( 1 15 ) = 0 This gives us = 75. So the point is ( 75, 1 15, 1 5 ) 1

2 . Calculate the divergence curl of the function ( F (, y, z) = y, y z, z ) div F (, y, z) = 1 y + 1 z + 1 curlf (, y, z) = i j k y y y z z z = ( y z, z, ) y 5. Find the second-degree Taylor polynomial of the function f(, y, z) = cos(yz) at the point a = (, π/, 1) We have f(, π/, 1) =. Further f(, y, z) = (cos(yz), z sin(yz), y sin(yz)) 0 z sin(yz) y sin(yz) Hf(, y, z) = z sin(yz) z cos(yz) sin(yz) yz cos(yz) y sin(yz) sin(yz) yz cos(yz) y cos(yz) Hence, f(, π/, 1) = ( Hf(, π/, 1) = π,, ) 0 π ( + π) π ( + π) π The second-degree Taylor polynomial at a = (, π/, 1) is p(, y, z) = π + (,, ) ( +, y π, z 1) + 1 [ ] 0 + y π z 1 ( + π) π ( + π) π π + y π z 1 6. Categorize the following quadratic forms as positive definite, negative definite, indefinite, or none of these

3 (a) f(, y, z) = + y z + 5y 3yz The symmetric matri is 1 5/ 0 A = 5/ 1 3/. 3/ 0 1 Because a 11 = 1 > 0, 1 5/ 0 det 5/ 1 3/ < 0, det(a) > 0, 3/ 0 1 By Sylvester Theorem, this quadratic form is indefinite. (b) f(, y) = 3 y + 6y The symmetric matri is 3 A =. 6 Because a 11 = 3 > 0 det(a) > 0, By Sylvester Theorem, this quadratic form is positive definite. (c) f(, y, z) = y + z + 3y + 6yz + z The symmetric matri is 1 5 A = Because a 11 = 1 > 0, () 1 5 det > 0 det(a) > By Sylvester Theorem, this quadratic form is positive definite. 7. Show that if a, b, c are all nonzero 0 a b A = a 0 c, b c 0 then p( ) = T A is indefinite. It is easy to know p( ) = (ay + bz + cyz) 3

4 Let s take two nonzero vectors = (,, 0) y = (,, 0) for nonzero, we have p( ) = a p( y) = a. Since a is nonzero, we know that p( ) p( y) have opposite signs, by the definition, p( ) = T A is indefinite.. Sketch the level curves of the function f(, y) = cos y for the indicated values of c = 0, 1/, 1/. We skip the plot of the level curves. We only give the solutions of f(, y) = cos y = c. If f(, y) = cos y = 0, then y = ± π + kπ. If f(, y) = cos y = 1/, then y = ± π 3 + kπ. If f(, y) = cos y = 1/, then y = ± π + kπ. 9. For the following functions, find all the critical points use the second derivative tests to identify the local etrema. f(, y) = + 3y + 1y Solving equation 0 = f(, y) = (, 6y + 1) we get one critical point (1, ). The Hessian of f at the critical point, 0 Hf(1, ) =. 0 6 is positive definite. Thus, (1, ) is a local minimum. f(, y) = + y ln(y). Solving equation 0 = f(, y) = (1 1, 1 1 y ) we get one critical point (1, 1). The Hessian of f at the critical point, 1 0 Hf(1, 1) =. 0 1 is positive definite. Thus, (1, 1) is a local minimum.

5 10. Find a point on the plane + y + z = 1 that is closest to the point (1,1,1). We need to find point (, y, z) on the given plane that minimizes the distance d = ( 1) + (y 1) + (z 1). Equivalently we can minimize d. Thus, we can formulate the problem as minimize f(, y, z) = ( 1) + (y 1) + (z 1) subject to g(, y, z) = + y + z 1 = 0. Using the Lagrange multiplier rule we know a potential minimizing point must satisfy f(, ȳ, z) = λ g(, ȳ, z), that is, ( 1, y 1, z 1) = λ(1, 1, 1). Clearly, = ȳ = z. Since g(, ȳ, z) = + ȳ + z 1 = 0 we obtain = ȳ = z = 1/3. That is (1/3, 1/3, 1/3) is the only cidate, therefore, (1/3, 1/3, 1/3) is the closest point to (1, 1, 1) on the plane + y + z = 1. 5