Math 210, Exam 2, Spring 2010 Problem 1 Solution

Transcription

1 Math, Exam, Spring Problem Solution. Find and classify the critical points of the function f(x,y) x 3 +3xy y 3. Solution: By definition, an interior point (a,b) in the domain of f is a critical point of f if either () f x (a,b) f y (a,b), or () one (or both) of f x or f y does not exist at (a,b). The partial derivatives of f(x,y) x 3 + 3xy y 3 are f x 3x + 3y and f y 3x 3y. These derivatives exist for all (x,y) in R. Thus, the critical points of f are the solutions to the system of equations: f x 3x +3y () f y 3x 3y () Solving Equation () for y we get: y x (3) Substituting this into Equation () and solving for x we get: 3x 3y 3x 3 ( x ) 3x 3x 4 3x( x 3 ) We observe that the above equation is satisfied if either x or x 3 x. We find the corresponding y-values using Equation (3): y x. If x, then y. If x, then y (). Thus, the critical points are (,) and (, ). We now use the Second Derivative Test to classify the critical points. The second derivatives of f are: f xx 6x, f yy 6y, f xy 3

2 The discriminant function D(x, y) is then: D(x,y) f xx f yy f xy D(x,y) (6x)( 6y) (3) D(x,y) 36xy 9 The values of D(x,y) at the critical points and the conclusions of the Second Derivative Test are shown in the table below. (a,b) D(a,b) f xx (a,b) Conclusion (, ) 9 Saddle Point (, ) 7 6 Local Minimum Recall that (a,b) is a saddle point if D(a,b) < and that (a,b) corresponds to a local minimum of f if D(a,b) > and f xx (a,b) >..5 y Figure : Pictured above are level curves of f(x,y). Darker colors correspond to smaller values of f(x,y). It is apparent that (,) is a saddle point and (, ) corresponds to a local minimum. x

3 Math, Exam, Spring Problem Solution. Sketch the region of integration and compute Solution: y y e y dxdy. x R x y The integral is evaluated as follows: x y e y dxdy [xe y] y dy ye y dy [ ] e y [ ] e [ ] e e

4 Math, Exam, Spring Problem 3 Solution 3. Compute z dv A where A is the region inside the sphere x + y + z, inside the cylinder x + y, and above the xy-plane. Solution: The region A is plotted below. z y x We use Cylindrical Coordinates to evaluate the triple integral. The equations for the sphere and cylinder are then: Sphere : r + z z r Cylinder : r r The surface that bounds A from below is z (the xy-plane) and the surface that bounds A from above is z r (the sphere). The projection of the region A onto the xy-plane is the disk r, θ π. Using the fact that dv r dz dr dθ in Cylindrical Coordinates, the value of the triple integral is:

5 A z dv 3 8 π r π r π π π [ θ ] π [ z zr dz dr dθ ] r dr dθ r ( r ) dr dθ [ r ] 4 r4 dθ 3 4 dθ 3π 4

6 Math, Exam, Spring Problem 4 Solution 4. Compute the integral of the field F(x,y) (x + y)î + ĵ along the curve c (θ) cosθî+sinθĵ. Solution: By definition, the line integral of a vector field F along a curve C with parameterization c (θ) x(θ),y(θ), a θ b is given by the formula: b F T ds F c (θ)dθ C a From the given parameterization c (θ) cosθî+sinθĵ we have: c (θ) sinθî+cosθĵ and, using the fact that x(θ) cosθ and y(θ) sinθ, the function F can be rewritten as: F (x+y)î+ĵ F (cosθ+sinθ)î+ĵ Assuming an interval θ π, the value of the line integral is then: b F T ds F c (θ)dθ C a π π π ((cosθ +sinθ) î+ĵ) ( sinθî+cosθĵ) dθ (cosθ+sinθ)( sinθ)dθ ( sinθcosθ sin θ)dθ [ cos θ θ+ ] π 4 sin(θ) [ cos (π) (π)+ ] 4 sin(4π) π [ cos ()+ ] 4 sin()

7 Math, Exam, Spring Problem 5 Solution 5. Find the minimum and maximum of the function f(x,y) x+y subject to the condition x +y. Solution: We find the minimum and maximum using the method of Lagrange Multipliers. First, we recognize that x + y is compact which guarantees the existence of absolute extrema of f. Then, let g(x,y) x + y. We look for solutions to the following system of equations: f x λg x, f y λg y, g(x,y) which, when applied to our functions f and g, give us: We begin by noting that Equation () gives us: λ(4x) () y λ(y) () x +y (3) y λ(y) y λ(y) y( λ) From this equation we either have y or λ. Let s consider each case separately. Case : Let y. We find the corresponding x-values using Equation (3). x +y x + x x ± Thus, the points of interest are (,) and (,). Case : Let λ. Plugging this into Equation () we get: λ(4x) (4x) x 4 We find the corresponding y-values using Equation (3). x +y ( 4) +y 8 +y y 7 8 y ± 7 8

8 7 Thus, the points of interest are (, ) and 4 8 (, 7 ). 4 8 We now evaluate f(x,y) x+y at each point of interest obtained by Cases and. f(,) f(,) f( 4, 7 8 ) 9 8 f( 4, 7 8 ) 9 8 From the values above we observe that f attains an absolute maximum of 9 8 minimum of. and an absolute.5..5 y Figure : Shown in the figure are the level curves of f(x,y) x + y and the ellipse x + y (thick, black curve). Darker colors correspond to smaller values of f(x,y). Notice that () the parabola f(x,y) x + y 9 is tangent to the ellipse at the points 8 7 (, ) and 4 8 (, 7) which correspond to the absolute maximum and () the parabola 4 8 f(x,y) x+y is tangent to the ellipse at the point (,) which corresponds to the absolute minimum. x

9 Math, Exam, Spring Problem 6 Solution 6. For the vector field F(x,y) (x+y)î+(x y)ĵ, find a function ϕ(x,y) with gradϕ F or use the partial derivative test to show that such a function does not exist. Solution: In order for the vector field F f(x,y),g(x,y) to have a potential function ϕ(x,y) such that gradϕ F, it must be the case that: f y g x Using f(x,y) x+y and g(x,y) x y we get: f y, g x which verifies the existence of a potential function for the given vector field. If gradϕ F, then it must be the case that: f(x,y) x () g(x,y) y () Using f(x,y) x+y and integrating both sides of Equation () with respect to x we get: x f(x,y) x x+y x dx (x+y) dx ϕ(x,y) x +xy +h(y) (3) We obtain the function h(y) using Equation (). Using g(x,y) x y we get the equation: y g(x,y) y x y We now use Equation (3) to obtain the left hand side of the above equation. Simplifying we get: ( ) y x +xy +h(y) x y x+h (y) x y h (y) y

10 Now integrate both sides with respect to y to get: h (y)dy ydy h(y) y +C Letting C, we find that a potential function for F is: ϕ(x,y) x +xy y