We know the directional derivative of z=f(x,y) at (1,1) in the direction of (-1,-6) is the rate of change of z=f(x,y) at (1,1) in the direction of (-1,-6), which is
Question 1: Why is this directional derivative positive?
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Perhaps moving the black vector in the xy-plane so that it starts at the purple point might help.
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Expression 32: vector left parenthesis, left parenthesis, 1 , 1 , "f" left parenthesis, 1 , 1 , right parenthesis , right parenthesis , left parenthesis, 1 minus StartFraction, 1 Over StartRoot, 37 , EndRoot , EndFraction , 1 minus StartFraction, 6 Over StartRoot, 37 , EndRoot , EndFraction , "f" left parenthesis, 1 , 1 , right parenthesis , right parenthesis , right parenthesisvector1,1,f1,1,1−137,1−637,f1,1
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Answer:
Hide this folder from students.
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Homework: Compare this Desmos to Desmos 14.3 on Partial Derivatives.
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Question 2: From (1,1), can we head on a different direction to get a larger directional derivative?
