midterm2s

# midterm2s - 18.02 MIDTERM#2 SOLUTIONS BJORN POONEN 1:05pm...

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Unformatted text preview: 18.02 MIDTERM #2 SOLUTIONS BJORN POONEN October 21, 2010, 1:05pm to 1:55pm (50 minutes). 1) Suppose that the function T = x 2 + 2 xy + 2 y 2 gives the temperature at each point of the plane, and that one starts at the point (2 , 1). (a) (10 pts.) In which direction should one go to obtain the most rapid increase in T ? Express your answer as a unit vector. Solution: The direction of fastest increase is the direction of the gradient of T . We have ∇ T = h 2 x + 2 y, 2 x + 4 y i ( ∇ T )(2 , 1) = h 6 , 8 i , and its direction is the unit vector h 6 , 8 i √ 6 2 + 8 2 = h 6 , 8 i 10 = 3 5 , 4 5 . (b) (10 pts.) Approximately how far in that direction should one go to get an increase of . 1 in T ? Solution: The directional derivative in that direction is the length of the gradient, which is 10. Thus Δ T Δ s ≈ 10, where Δ s is distance travelled. To achieve Δ T = 0 . 1, we need Δ s ≈ Δ T 10 = . 1 10 = 0 . 01 . Remark: It is possible also to find the exact distance one should go, by finding the value of t such that T ( x,y ) evaluated at h 2 , 1 i + t 3 5 , 4 5 equals T (2 , 1) + 0 . 1, but this is much messier. 2) Let R be the circle of radius 2 centered at (0 , 0). Evaluate ZZ R ( x 2 + y 2 ) 2 dA . (Write your answer in simplest form.) Solution: Convert to polar coordinates, by substituting x 2 + y 2 = r 2 dA = r dr dθ. This leads to ZZ R ( x 2 + y 2 ) 2 dA = Z 2 π Z 2 r 4 r dr dθ = Z 2 π Z 2 r 5 dr dθ = Z 2 π r 6 6 2 dθ = Z 2 π 32 3 dθ = 32 3 (2 π ) = 64 3 π....
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## This note was uploaded on 09/14/2011 for the course MATH 18.02 taught by Professor Auroux during the Fall '08 term at MIT.

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midterm2s - 18.02 MIDTERM#2 SOLUTIONS BJORN POONEN 1:05pm...

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