Given the function f(x,y)= arctan(xy) (A) Find the gradient vector f(x,y) (B) Compute the directional derivative of f at the point (1,2) in the direction of the vector
v = 5i + 10j
"\\displaystyle \nf(x,y) = arc\\tan(xy)\\\\\n\\frac{\\partial f(x,y)}{\\partial x}=\\frac{y}{1+(xy)^2}, \\qquad \\frac{\\partial f(x,y)}{\\partial x}=\\frac{2}{5}\\\\\n\\frac{\\partial f(x,y)}{\\partial y}=\\frac{x}{1+(xy)^2}, \\qquad \\frac{\\partial f(x,y)}{\\partial x}=\\frac{1}{5}\\\\\n\\text{Next, we find the unit vector of the given direction}\\\\\nu=\\frac{5i+10j}{\\sqrt{125}}= \\frac{5i+10j}{5\\sqrt{5}}\\\\\n\\text{the directional derivative is given by $\\nabla f(x)\\cdot u$}\\\\\n= \\frac{3\\sqrt{5}}{25}"
Comments
Leave a comment