Question

Considere a função f(x,y)=ln(x2+y2). O gradiente de f em P(1,2) é igual a:

a) 0,8i+0,4j
b) 2i+4j
c) 0,4i+0,8j
d) 2i-4j
e) 0,4i-0,8j

Answer 1

$\boxed{\boxed{\nabla f(x,y) = \frac{\partial f}{\partial x}\;i + \frac{\partial f}{\partial y}\;j }}$

encontrando as derivadas parciais
$\bmatrix {\\ f(x)=ln(x^2+y^2)\\\\ \frac{\partial f}{\partial x}= \frac{1}{x^2+y^2}*(2x+0) = \frac{2x}{x^2+y^2} \\\\\ \frac{\partial f}{\partial x}= \frac{1}{x^2+y^2}*(0+2y) = \frac{2y}{x^2+y^2} \\ \end$

temos o gradiente da função
$\nabla f(x,y)= \frac{2x}{x^2+y^2} \;i + \frac{2y}{x^2+y^2} \; j\\\\ \text{no ponto p(1,2)}\\\\ \nabla f(1,2)= \frac{2*1}{1^2+2^2} \;i + \frac{2*2}{1^2+2^2} \; j \\\\ \boxed{\boxed{\nabla f(1,2)= \frac{2}{5} \; i + \frac{4}{5} \; j =0,4\;i + 0,8\;j}}$