ps2B-2009

# ps2B-2009 - x y =(1 1 to(0 95 95 by assuming that the hiker...

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18.02A Problem Set 2B – Fall 2009 due (along with Part 2A) Thursday Nov.19/06, 12:45 in 2-106 Part I (5 points) Recitation. Mon. Nov. 13 Functions of several variables; graphs, level curves. Read: 19.1 Work: 2A - 1abe Lecture 9. Tues. Nov. 14 Gradient; directional derivative. Read: 19.5 to p. 683 Work: 2D - 1 acd, 2a, 4, 7, 9bcde Lecture 10. Thurs. Nov. 16 Gradients in 3D; tangent planes. Read: rest of 19.5, Notes P Part II (15 points) Directions. Try each problem alone for 15 minutes. If you collaborate subsequently, your solutions must be written up independently. It is illegal to consult problem sets from previous semesters. Problem 1. (Tues. 7pts: 2+2+3) A hiker climbs a mountain whose height is given by z = 1000 - 2 x 2 - 3 y 2 . a) Find the tangent plane to the mountain at the point (1 , 1 , 995). Deduce an approximation for the vertical distance climbed in moving from (
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Unformatted text preview: x, y ) = (1 , 1) to (0 . 95 , . 95) by assuming that the hiker moves along this tangent plane. What is the error resulting from this approximation? b) When the hiker is at the point (1 , 1 , 995), in what direction should she move in order to ascend as rapidly as possible? c) If she continues to move on a path of steepest ascent, show that the projection of this path on the xy-plane is y = x 3 / 2 . Problem 2. (Tues. 8pts: 4 + 4) a) Find the tangent vector at the point (1 , 1 , 2) to the curve of intersection of the surfaces z = x 2 + y 2 and z = x + y . b) Generalize this result to ﬁnd an expression for the tangent vector of the curve of intersection of any two surfaces z = f ( x, y ) and z = g ( x, y ) at a point of intersection P = ( x , y , z )....
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## This note was uploaded on 05/14/2010 for the course 18.02A 18.02A taught by Professor Johnbush during the Winter '10 term at MIT.

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