Question

Calcule a derivada parcial abaixo e substitua no ponto

w = ln √ x2 + y² + z² , ( 3, 0, 4 )

Escolha uma:
a. wx = 3/25; wy = -1; wz = 2/25
b. wx = 4/25; wy = 0; wz = 3/25
c. wx = 3/25; wy = 0; wz = 4/25
d. wx = 2/25; wy = -1; wz = 4/25

Answer 1

Derivada parcial no ponto (3,0,4) de

$\displaystyle \text W = \text {ln}(\sqrt{\text x^2 + \text y^2 + \text z^2})}$

Derivando parcialmente :

$\displaystyle \frac{\partial \text w}{\partial \text x}[\text{\ ln}(\sqrt{\text x^2 + \text y^2 + \text z^2}) ] = \frac{1}{\sqrt{\text x^2 + \text y^2 + \text z^2 }}.[(\text x^2+\text y^2+z^2)^{\frac{1}{2}}]'$  

$\displaystyle \frac{.[(\text x^2+\text y^2+z^2)^{\frac{1}{2}}]'}{\sqrt{\text x^2 + \text y^2 + \text z^2 }} = \frac{1}{2}.\frac{2x}{\sqrt{\text x^2 + \text y^2 + \text z^2 }.\sqrt{\text x^2 + \text y^2 + \text z^2 }}$

a raiz no denominador vai sumir porque vai ficar ao quadrado e o 2 simplifica com o 2.

Portanto :  

• $\displaystyle \frac{\partial \text w}{\partial \text x} = \frac{\text x}{\text x^2+\text y^2 + \text z^2 }$

analogamente os outros serão assim :

• $\displaystyle \frac{\partial \text w}{\partial \text y} = \frac{\text y}{\text x^2+\text y^2 + \text z^2 }$

• $\displaystyle \frac{\partial \text w}{\partial \text z} = \frac{\text z}{\text x^2+\text y^2 + \text z^2 }$  

Substituindo o ponto (3,0,4) :

$\displaystyle \frac{\partial \text w}{\partial \text x} = \frac{\text x}{\text x^2+\text y^2 + \text z^2 } = \frac{3}{3^2+0^2+4^2 } = \boxed{\frac{3}{25}}$

$\displaystyle \frac{\partial \text w}{\partial \text y} = \frac{\text 0}{\text 3^2+\text 0^2 + \text 4^2 } = \boxed{0}$  

$\displaystyle \frac{\partial \text w}{\partial \text z} = \frac{\text z}{\text x^2+\text y^2 + \text z^2 } = \frac{4}{3^2+0^2+4^2} = \boxed{\frac{4}{25}}$

Portanto :

$\displaystyle \huge\boxed{\frac{\partial \text w}{\partial \text x} = \frac{3}{25}} \ ; \ \boxed{ \frac{\partial \text w}{\partial \text y} = 0 }} \ ; \huge\boxed{\frac{\partial \text w}{\partial \text z} = \frac{4}{25} }$

Letra C