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Math Help - directional derivatives!?

  1. #1
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    directional derivatives!?

    when f(x,y,z)= x^2+y^2-z and r(t)= [t/root2, t/root2, t^2]

    find the directional derivative of f along the curve r(t) at the point P where t= root2

    i have spent hours trying to work this one out but it is just beyond me! even just to know where to start would be a great help
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Ash_underpar View Post
    when f(x,y,z)= x^2+y^2-z and r(t)= [t/root2, t/root2, t^2]

    find the directional derivative of f along the curve r(t) at the point P where t= root2

    i have spent hours trying to work this one out but it is just beyond me! even just to know where to start would be a great help
    Recall that the directional derivative is defined as \nabla f(x_0,y_0,z_0)\cdot\bold u

    To find the point (x_0,y_0,z_0) find the component values of \bold r(t) at t=\sqrt{2} Thus, the point is (1,1,2).

    Now, lets find \nabla f(x,y,z)

    \nabla f(x,y,z)=\left<\frac{\partial}{\partial x}(x^2+y^2-z),\frac{\partial}{\partial y}(x^2+y^2-z), \frac{\partial}{\partial z}(x^2+y^2-z)\right> =\left<2x,2y,-1\right>

    Thus, \nabla f(1,1,2)=\left<2,2,-1\right>

    Now, let's find the unit vector \bold u.

    Since \bold r(\sqrt{2})=\left<1,1,2\right>, we need to normalize it.

    Its magnitude is \sqrt{1+1+4}=\sqrt{6}. As a result, \bold u=\left<\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\fra  c{2}{\sqrt{6}}\right>=\left<\frac{\sqrt{6}}{6},\fr  ac{\sqrt{6}}{6},\frac{\sqrt{6}}{3}\right>

    Now, find \nabla f(1,1,2)\cdot \bold u:

    \nabla f(1,1,2)\cdot \bold u=\left<2,2,-1\right>\cdot\left<\frac{\sqrt{6}}{6},\frac{\sqrt{  6}}{6},\frac{\sqrt{6}}{3}\right>=\frac{\sqrt{6}}{3  }+\frac{\sqrt{6}}{3}-\frac{\sqrt{6}}{3}=\color{red}\boxed{\frac{\sqrt{6  }}{3}}

    Does this make sense?

    --Chris
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  3. #3
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    cheers matey you make it sound so simple but this vector calculus section im doing at the minute is going right over my head!! thanks anyway
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