{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# quiz1sol - f x,y = x 2 2 y 3-6 y Solving f x = 2 x = 0 and...

This preview shows page 1. Sign up to view the full content.

Math 241 CD3/CD4 Quiz 1 Name: 1. (2 points) Let f ( x, y, z ) = (2 xz, 0 , - 3 y ). Find the curl and divergence of f . curl f = ∇ · f = ∂x , ∂y , ∂z · (2 xz, 0 , - 3 y ) = ∂x 2 xz + ∂y 0 + ∂z ( - 3 y ) = 2 z. div f = ∇ × f = i j k ∂x ∂y ∂z 2 xz 0 - 3 y = ( - 3 , 2 x, 0) . 2. Given a path x ( t ) = (cos(2 t ) , sin(2 t )) , 0 t π . (a) (2 points) Compute the arclength of x ( t ). (b) (6 points) Find unit tangent vector, principal normal vector, and curvature of x ( t ). (a) x 0 ( t ) = ( - 2 sin(2 t ) , 2 cos(2 t )) ⇒ k x 0 ( t ) k = p ( - 2 sin 2 t ) 2 + (2 cos 2 t ) 2 = 2. Length of x ( t ) = Z π 0 k x 0 ( t ) k dt = Z π 0 2 dt = Z π 0 2 dt = 2 t | π 0 = 2 π. (b) T = x 0 ( t ) k x 0 ( t ) k = ( - sin(2 t ) , cos(2 t )) d T dt = ( - 2 cos(2 t ) , - 2 sin(2 t )) ⇒ k d T dt k = 2. N = d T /dt k d T /dt k = ( - cos(2 t ) , - sin(2 t )). κ = d T ds = k d T /dt k ds/dt = 2 k x 0 ( t ) k = 1. ( or use κ = k v × a k k v k 3 .) 3. (5 points) Find and classify all the critical points for
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: f ( x,y ) = x 2 + 2 y 3-6 y . Solving f x = 2 x = 0 and f y = 6 y 2-6 y = 0, we have that the only critical points are (0,1) and (0,-1). Since f xx = 2 ,f xy = f yx = 0 and f yy = 12 y , Hf ( x,y ) = µ f xx ( x,y ) f xy ( x,y ) f yx ( x,y ) f yy ( x,y ) ¶ = µ 2 0 12 y ¶ . Hence Hf (0 , 1) = µ 2 0 12 ¶ . Since 2 > 0 and det Hf (0 , 1) > 0, (0,1) is a local minimum. And Hf (0 ,-1) = µ 2-12 ¶ . Since det Hf (0 ,-1) < 0, (0,-1) is a saddle point. 1...
View Full Document

{[ snackBarMessage ]}