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How did he get the value inside the circle f(x, y)=x² + sin(xy) Vf=x*+sin(xy)) i+ d(x²+sin(xy)), dx ду =(2x + y cos(xy))i + (0 +x cos(xy))j (Vf)u.0)=(2x + y cos(xy))i + (0 + x cos(xy))j :(2 ·1 + 0 cos(0))i + (0 + 1 cos(0))j =2i +j Now let (xi + yj) be the unit vector( in the direction of the direction vector A. Therefore the directional derivative of the function in the direction of the vector A is given by; Direct Deriv=Vf AL =(2i +j) (xi + yj) =2x + y Now it is given that the directional derivativ at this point. Therefore; 2x + y = 1

How did he get the value inside the circle f(x, y)=x² + sin(xy) Vf=x*+sin(xy)) i+ d(x²+sin(xy)), dx ду =(2x + y cos(xy))i + (0 +x cos(xy))j (Vf)u.0)=(2x + y cos(xy))i + (0 + x cos(xy))j :(2 ·1 + 0 cos(0))i + (0 + 1 cos(0))j =2i +j Now let (xi + yj) be the unit vector( in the direction of the direction vector A. Therefore the directional derivative of the function in the direction of the vector A is given by; Direct Deriv=Vf AL =(2i +j) (xi + yj) =2x + y Now it is given that the directional derivativ at this point. Therefore; 2x + y = 1

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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How did he get the value inside the circle f(x, y)=x² + sin(xy) Vf=- a(x²+sin(xy)) ; + a(x*+sin(-v)) ; dx ду =(2x + y cos(xy))i + (0 +x cos(xy))j (Vf)(1,0)=(2x + y cos(xy))i + (0 + x cos(xy))j =(2 ·1 + 0 cos(0))i + (0 + 1 cos(0))j =2i +j Now let (xi + yj) be the unit vector( in the direction of the direction vector A. Therefore the directional derivative of the function in the direction of the vector A is given by; Direct Deriv=Vf A =(2i + j) (xi + yj) =2x + y Now it is given that the directional derivativ at this point. Therefore; 2x + y = 1
Transcribed Image Text:How did he get the value inside the circle f(x, y)=x² + sin(xy) Vf=- a(x²+sin(xy)) ; + a(x*+sin(-v)) ; dx ду =(2x + y cos(xy))i + (0 +x cos(xy))j (Vf)(1,0)=(2x + y cos(xy))i + (0 + x cos(xy))j =(2 ·1 + 0 cos(0))i + (0 + 1 cos(0))j =2i +j Now let (xi + yj) be the unit vector( in the direction of the direction vector A. Therefore the directional derivative of the function in the direction of the vector A is given by; Direct Deriv=Vf A =(2i + j) (xi + yj) =2x + y Now it is given that the directional derivativ at this point. Therefore; 2x + y = 1
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