math103Exam2solutions

# math103Exam2solutions - Math 103 section 1 Exam 2...

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Math 103 section 1 - Exam 2 Instructions: You are not permitted to use a calculator or any other materials on this exam. Show all your work on the test. You may use the back of the pages if needed. Simplify all answers. Give exact answers (i.e. fractions, radicals, etc.) unless instructed otherwise. Sign the Honor Pledge after you ﬁnish the exam. I have neither given nor received any unauthorized help on this test, and I have conducted myself within the guidelines of the university honor code. Name and section: Pledge: Question: 1 2 3 4 5 6 Total Points: 7 11 14 7 10 18 67 Score:

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T at position ( x,y,z ) in space is given by the formula T = 50 + xyz . Suppose a bird is at position (2 , 3 , 10), if the bird ﬂies oﬀ in direction h 1 , - 1 , 2 i what is the rate of change of the temperature with respect to distance? In what direction should the bird ﬂy in order to get warmer the most quickly? Solution: The gradient of T is 5 T ( x,y,z ) = h yz,xz,xy i , and at the point (2 , 3 , 10), 5 T (2 , 3 , 10) = h 30 , 20 , 6 i and the length is 900 + 400 + 36 = 1336. The bird should ﬂy in the direction of h 30 , 20 , 6 i or 1 1336 h 30 , 20 , 6 i (which is just the unit vector in that direction) in order to get warm the fastest. So for the rate of change in the direction of the vector h 1 , - 1 , 2 i is u = 1 6 h 1 , - 1 , 2 i and the rate of change is D u T (2 , 3 , 10) = h 30 , 20 , 6 i ·
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## This note was uploaded on 04/19/2010 for the course MATH 103 taught by Professor Hain during the Fall '08 term at Duke.

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math103Exam2solutions - Math 103 section 1 Exam 2...

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