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Math Help - Gradient of function

  1. #1
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    Question Gradient of function

    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
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  2. #2
    Member Danshader's Avatar
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    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)<br />
    <br />
=(\frac{\partial}{\partial x}\overrightarrow{i} + \frac{\partial}{\partial y}\overrightarrow{j} + \frac{\partial}{\partial z}\overrightarrow{k})(f(x,y,z))<br />
    <br />
=\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)<br />

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

    substituting into (1) gives:
    <br />
=\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}<br />
    <br />
=2xyz^4\overrightarrow{i} +x^2z^4\overrightarrow{j} +4x^2yz^3\overrightarrow{k} \longrightarrow(2)<br />

    entering the value of x,y,and z into equation (2) gives:
    <br />
 \nabla f(x,y,z)<br />
=2(-1)(3)(-1)^4\overrightarrow{i} +(-1)^2(-1)^4\overrightarrow{j}+4(-1)^2(3)(-1)^3\overrightarrow{k}<br />
    <br />
=6\overrightarrow{i}+\overrightarrow{j}-12\overrightarrow{k}<br />
    <br />
=\left(\begin{array} {ccc} 6 & 1 & -12\\ \end{array} \right)\left( \begin{array}{c} \overrightarrow{i} \\ \overrightarrow{j}\\ \overrightarrow{k} \end{array} \right)
    Last edited by Danshader; April 29th 2008 at 12:58 AM.
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