This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Prelim 3 November 27, 2001 Show all your work. Your reasoning is important. No calculators are allowed. You are allowed to use a 5 by 8 index card. The number of points each problem is worth is indicated in parantheses. Keep in mind there may be simpler methods than direct applications of calculus. Good luck. 1. (16 points) Find the directional derivative of f ( x, y ) = 2 x 2 + 3 xy y 2 at the point (2 , 3) in the same direction as v = i + 3 j . SOLUTION: Step 1 Find the direction. The direction of v is 1 10 i + 3 10 j . Step 2 Find the gradient of f . f ( x, y ) = h 4 x + 3 y, 3 x 2 y i f (2 , 3) = h 17 , i Step 3 Find the directional derivative. D v f (2 , 3) = f (2 , 3) 1 10 , 3 10 = 17 10 2. (16 points) Find the centroid (ie. center of mass with uniform density) of the region bounded by y = 1 x 2 and the x axis. SOLUTION: Step 1 Find the area of the region. The formula for the area of a parabola is Area = 2 3 base height. Our parabola has a base of 2 and a height of 1 so its area is 4 3 . Step 2 Find the centroid. Since our region is symmetric about the yaxis we must have that the centroid lies on the yaxis. So x = 0. Now we need to find y . M x = Z 1 1 Z 1 x 2 y dydx = Z 1 1 (1 x 2 ) 2 2 dx = Z 1 (1 x 2 ) 2 dx = Z 1 1 2 x 2 + x 4 dx = 1 2 3 + 1 5 = 8 15 So y = M x M = 8 15 3 4 = 2 5 . So the centroid is ( x, y ) = , 2 5 . 3. Let g ( x, y ) = x 2 + 2 xy + 4 y 2 , where x = 2 cos and y = sin ....
View
Full
Document
This note was uploaded on 09/29/2009 for the course 1920 1920 at Cornell University (Engineering School).
 '10
 CONNELLY

Click to edit the document details