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Answers for “方向 (5,2) 的特征线的斜率是多少?”

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The slope of the characteristic lines is 2/52/5. This is derived from the direction vector (5,2), where the change in y is 2 and the change in x is 5, so dy/dx=2/5dy/dx = 2/5.

Conditions: The direction vector is (5,2).; The characteristic lines are straight lines in the xy-plane.

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The equation (5,2)·∇u=0u=0 implies that the directional derivative of u in the direction of the vector (5,2) is zero. This means that the rate of change of u is zero when moving in that specific direction.

Conditions: The direction vector is (5,2).; ∇u is the gradient of u.