Mat 241 Homework Set 7key Due Professor David Schultz

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1 Mat 1 Homework Set 7ke Due Proessor David Schultz Directions: Show all algebraic steps neatl and concisel using proper mathematical smbolism. When graphs and technolog are to be implemented, do so appropriatel. Mechanics:, #1. Consider the unction deined b A. Determine all critical/stationar points. (, ) = = ( + + ) ( + + 1) ( + + 1) ( + + 1) = = = ( + + 1) ( + + 1) ( + + 1) ( + + 1) ( + + 1) = = + + = = = = 18 = 0 18 = 0 = 0, = 0 using these in note;i both & = 0, then = 1 does not work. I = 0, then cannot. critical points ( 1, 0),( 1, 0) = 0, we get the ollowing = 1 ields non-real solutions or. Thus, can be zero but B. Classi each point as a relative maimum, relative minimum, or a saddle point. Justi our responses.

2 = ; = Test or = = = ( + + 1) ( + + 1) ( + + ) ( ) ( + + ) ( + + ) ( + + 1) ( + + 1) ( 18) ( 18) ( + + 1) ( + + 1) ( + + 1) ( 18 ) ( ) ( + + 1) ( + + 1) (, ) ( 1, 0) ( 18) ( 0) 1 0 = =. 5 < 0 (, ) ( 18) ( 0) ( 0) 1, 0 = =. 5 ( 0) ( 9 + 9) ( ) ( 0) 1 0 = = 0 ( 1, 0) ( 1, 0) ( 1, 0) ( 1, 0) (. 5)(. 5) D = = = Since < 0 and D > 0, ( 1, 0) is a local maimu m

3 Test or (, ) ( 1, 0) ( 18) ( 0) ( ) 1, 0 = =. 5 < 0 ( 18) ( 0) ( 0) 1, 0 = =. 5 ( 0) ( 9 + 9) ( ) ( 0) 1 0 = = 0 ( 1, 0) ( 1, 0) ( 1, 0) ( 1, 0) ( 5) D = = = Since > 0 and D > 0, ( 1, 0) is a local minimum C. Find the equation o the tangent plane or each relative maimum/minimum , 0 = ; P1 = 1, 0, 9 9 ( 1, 0) = ; P1 = 1, 0, 9 9 Plane : z = Plane : z = 9 9 1, 0, 1, 0, D. Plot the unction and an horizontal tangent planes using a suitable 3-D grapher.

4 #. Let R be the triangular region in the -plane with vertices (-1, -), (-1, ) and (3, ). A metal plate in the shape o R is heated so that the temperature at (, ) is given b: T(, ) = in degrees Celsius. A. Sketch the region R in the -plane. B. Determine all critical points within the region R.

5 T(, ) = T = ; T = T = + ; T = T = 1 8 From T = 0 & T = 0, we have critical point, T, = C. Determine all critical points on the boundar. The boundar is made up o three lines: We the unction or each line: = -1, =, and = 1. T(, ) = = T ( ) = 1; T ( ) = 0 = and T ( ) = 1 1, is a local minimum. Furthermore, T( 1, ) = 1; T( 1, ) = 5; T(, ) = T ( ) = ; T ( ) = 0 = and T ( ) = 1, is a local minimum. Furthermore, T( 1, ) = 1; T( 3, ) = 13;

6 T(, ) = + + = T ( ) = 3 T ( ) = 0 = and T ( ) = 3 1, is a local minimum. Furthermore, T( 1, ) = 5; T( 3, ) = 13; D. At what point in R or on its boundar is the temperature maimized? At what point is the temperature minimized? What are the etreme temperatures? The suraces has its maimum temperature on the boundar point (3, ) and its minimum temperature occurring within the region at absolute maimum and minimum are 13 and -1/7 respectivel. 8 7, 7. The (, ) = ln( + ). #3. Consider the unction A. Sketch the unction s domain in R.

7 B. Determine all critical point(s) over its domain. = + (, ) ln 1 1 = 1 ; = = 0 1 = 0 + = 1 + = = 0 1 = =, = C. Classi the critical point(s) ound in part B. (, ) = ln( + ) = = 1; 1 1 = =, = + = ; D = = = Since > 0 and D < 0,, is a saddle point. Conceptual Development

8 #. Two suraces F(,,z) = 0 & G(,,z) = 0 are said to be orthogonal at a point P i F & G are nonzero at P and the normal lines to the suraces are perpendicular at P. From this it can be shown that: Two suraces are orthogonal at a point P i F G + F G + F G = 0 Use this result to show that the sphere z = + are orthogonal at the point (0,, ). z z + + z = 8 and the cone F ( 0,, ) 0,, G ( 0,, ) 0,, F,, z = + + z 8; F,, z =,, z = G,, z = + z ; G,, z =,, z = F 0,, G 0,, = FG + FG + FzGz = = 0 #5. In Calculus I i is a continuous unction o one variable, sa or eample, ( ), with two relative maima on an interval, then there must be a relative minimum between the relative maima. (Convince oursel b drawing some pictures in R ). The purpose o this problem is to show ou that this does not etend to unctions o two variables. Show that (, ) e e = has two relative maima but no other critical points! (Based on the article, Two Mountains Without a Valle, the Mathematics Magazine, Vol.60 No/1 Februar 1987, p.8)

9 (, ) = 3 (, ) : (, ) : (, ) (, ) = : (, ) = 16 e e = 8e 8 = 8e = 8e From (1), = implies that e e e e e = e = e = = e e = e e = /* (1) hence, e = 0 ( e = ). Substituting in ( /* () must be zero which is not possible, 8 = 0 1 = 1 1+ = 0 = ± 1, = 0 ( 1 0) ( 1 0) So, we have the critical points, &,. Now, we classi them: ): (, ) : (, ) : (, ) = ( 16)( 1) 8 = 18, ( 1, 0) < 0 & D( 1, 0) 0, ( 1, 0) is a relati (, ) : (, ) : (, ) 1 0) ( 16)( 1) 8 = 18, ( 1, 0) 0 & D( 1, 0) > 0 ( 1 0) 1 0 = 8 = = = 16 = 1 D( 1, 0) Since > 1 0 = 8 = = = 16 = 1 D(, = Since < ve maimum.,, is a relative maimum.

10 #6. Use Lagrange multipliers to determine the dimensions o a rectangular bo, having an open-top with volume o 3 t 3, and requiring the least amount o material or its construction. SA(,, z) minimize = + z + z 3 = z constraint G,, z = z 3 SA = + z, + z, + : λ G = zλ, zλ, λ SA = λ G + z = zλ; + z = zλ; + = λ 1 1 λ = + & λ = + = z z 1 λ = + & λ = + z = = z = z = = = = =, =, z = 3 3 3, 6 The bo should be. * So how do we know that this indeed a minimum? #7.The igure below shows the intersection o the elliptical paraboloid z = + and the right circular clinder + = 1.

11 A. Use Lagrange multipliers to ind the highest and lowest points on the curve o the intersection. That is, ind the maima and minima or z = + subject to the constraint F(, ) = +, G, = = 1. F(, ) =, 8 ; λ G(, ) = λ, λ F(, ) = λ G(, ) = λ and 8 = λ i then λ = and = + = we get 0, 1 0. Using = 1 ( 1, 0),( 1, 0) i, then λ = and =. Using + = we get = 1 ( 0, 1),( 0, 1) F( 1, 0) = = 1; F( 1, 0) = = 1 F( 0, 1) = = ; F( 0, 1) = = The maumum height is and the minimum height is1. B. Determine parametric equations or the curve o intersection. + = 1 and z = + = cos, = sin, = cos + sin let t t t t then z t t t C. What is the domain or t in our equations ound in part B? D :( t t ) D. Using z(t) rom part B, what is the maimum height and minimum height it can take on and how does this compare to part A s answer? z t t t = cos + sin z t = cos t sint + 8sint cos t = 6sint cost π z ( t ) = 0 : 6sint cos t = 0 t= kπ, k = 0, ± 1, ±,...; t = + kπ, k = 0, ± 1, ±,... When t = π, ( π ) = 1, π = 0 z = = 1 π π π When t =, 0, 1 z ( 0) 1 = = = + = z has a maumum height o and a minimum height o 1. As epected.

12 A Challenge or ou I was sitting in m oice and started plotting several paraboloids. I plotted: z = + 1 z = z = + The outcome is shown: I realized that there could ver well be a plane that is simultaneousl tangent to all three paraboloids. Indeed, there is and we shall ind this together. Step 1. Name the suraces: S1,, z : z 3 (,, ) : + = 0 S z z = 0 (,, ) : S z + 1 z = 0 Suppose the plane we seek touches the suraces at the points (,, ), (,, ), and (,, ) P z P z P z respectivel Determine the gradient vectors or the three suraces at the points P, P, and P. 1 3

13 Step. The corresponding components rom the three gradient vectors are equivalent. Using that act and the surace equations S & S 3, show that: = + 1, =, z = z = 1, 3 = 1 + 1, z3 = z1 P & P in terms o 1, 1,& z 1. Could ou see how this same Now write 3 result could be deduced just rom looking at the equations o S, S, and S? 1 3 Step 3. Since we now have three points on the plane all in terms,,& z, create the two vectors in the plane P 1 P and P 1 P 3. o Step. Determine the cross-product o the vectors ound in Step 3. This vector is a normal vector to the plane. It is also collinear with S,, z =, = 1, and z = Use that act as well as S 1 to show that Step 5. Using the inormation rom step show that the equation o the tangent plane which is tangent to all three paraboloids is given b: z + 5 = 0