At time t = 1, a particle is located at position (x, y) = (5, 2). If it moves in the velocity field F(x, y) = xy − 1, y2 − 11 find its appro

Question

At time t = 1, a particle is located at position (x, y) = (5, 2). If it moves in the velocity field F(x, y) = xy − 1, y2 − 11 find its approximate location at time t = 1.02.

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Thiên Thanh 3 years 2021-08-14T04:27:41+00:00 1 Answers 40 views 0

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  1. Helga
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    2021-08-14T04:29:07+00:00
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    Answer:

    Its approx location is (5.18,1.9)

    Explanation:

    Using F( 5,2) = ( xy-1, y²-11)

    = ( 5*2-¹, 2²-11)

    = (9,-5)

    = so at point t=1.02

    (5,2)+(1.02-1)*(9,-5)

    (5,2)+( 0.02)*(9,-5)

    (5+0.18, 2-0.1)

    = ( 5.18, 1.9)

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )