• Apr 28th 2008, 10:23 PM
taurus
I got a question that i dont know how to answer could someone help:
Code:

Evaluate the gradient of f at the point P, where f(x, y, z) = x2yz4,    P(-1, 3, -1) Please enter your answer as a row vector, using the list syntax.
thanks
• Apr 28th 2008, 11:44 PM
$\nabla f(x,y,z)\text{?}$

if that is so then:
$\nabla f(x,y,z)
$

$
=(\frac{\partial}{\partial x}\overrightarrow{i} + \frac{\partial}{\partial y}\overrightarrow{j} + \frac{\partial}{\partial z}\overrightarrow{k})(f(x,y,z))
$

$
=\frac{\partial}{\partial x}f(x,y,z)\overrightarrow{i} +\frac{\partial}{\partial y}f(x,y,z)\overrightarrow{j} +\frac{\partial}{\partial z}f(x,y,z)\overrightarrow{k} \longrightarrow(1)
$

since f(x,y,z) is given as:
$f(x,y,z)= x^2yz^4$

substituting into (1) gives:
$
=\frac{\partial}{\partial x}(x^2yz^4)\overrightarrow{i} +\frac{\partial}{\partial y}(x^2yz^4)\overrightarrow{j} +\frac{\partial}{\partial z}(x^2yz^4)\overrightarrow{k}
$

$
=2xyz^4\overrightarrow{i} +x^2z^4\overrightarrow{j} +4x^2yz^3\overrightarrow{k} \longrightarrow(2)
$

entering the value of x,y,and z into equation (2) gives:
$
\nabla f(x,y,z)
=2(-1)(3)(-1)^4\overrightarrow{i} +(-1)^2(-1)^4\overrightarrow{j}+4(-1)^2(3)(-1)^3\overrightarrow{k}
$

$
=6\overrightarrow{i}+\overrightarrow{j}-12\overrightarrow{k}
$

$
=\left(\begin{array} {ccc} 6 & 1 & -12\\ \end{array} \right)\left( \begin{array}{c} \overrightarrow{i} \\ \overrightarrow{j}\\ \overrightarrow{k} \end{array} \right)$