Gradient of function

**taurus** · April 28th 2008, 10:23 PM

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

**Danshader** · April 28th 2008, 11:44 PM

when you say gradient do you mean this gradient: :
$\nabla f(x,y,z)\text{?}$

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)$

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