Question

Give both answers
Give both answers

Answer 1

`\hat u = < -0.9439,-0.3303 >`

Explanation:

To find the direction in which the maximum rate of change occurs for the function f(x,y)=2xsin(xy) at the point (1, 3), we need to find the gradient vector and normalize it to obtain a unit vector.

The gradient vector of a function f(x,y) is given as:

`\\abla f(x,y)=\Big < (\partial f (x,y))/(\partial x),(\partial f (x,y))/(\partial y) \Big >`

Taking the partial derivatives.

`(\partial f)/(\partial x) = 2\sin(xy)+2xy\cos(xy)\\\\\\\\(\partial f)/(\partial y) = 2x^2\cos(xy)`

Evaluate these at the point (1,3).

`(\partial f)/(\partial x)(1,3) = 2\sin((1)(3))+2(1)(3)\cos((1)(3))\\\\\\\ \Longrightarrow \boxed{(\partial f)/(\partial x)(1,3) =2\sin(3)+6\cos(3)}\\\\\\\\(\partial f)/(\partial y)(1,3) = 2(1)^2\cos((1)(3)) \\ \\ \\ \Longrightarrow \boxed{(\partial f)/(\partial y)(1,3) = 2\cos(3)}`

Thus, we have:

`\\abla f(1,3)=\Big < 2\sin(3)+6\cos(3), \ 2\cos(3) \Big >`

To obtain the unit vector in the direction of maximum rate of change, we need to normalize the gradient vector:

`\hat u =( \\abla f(1,3))/(||\\abla f(1,3)||)\\\\\\\\\Longrightarrow \hat u = (\Big < 2\sin(3)+6\cos(3), \ 2\cos(3) \Big > )/(√((2\sin(3)+6\cos(3))^2+(2\cos(3))^2) ) \\\\\\\\\Longrightarrow \hat u =( < -5.6577,-1.9800 > )/(5.9942) \\\\ \\\\\therefore \boxed{\boxed{\hat u = < -0.9439,-0.3303 > }}`

Thus, the problem is solved.