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Math Help - directional derivative

  1. #1
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    directional derivative

    Find the directional derivative of f x, y = 3 x^2 y at the point (1,2) in the direction of v= 3 i+4 j .

    after i find the unit vector i am lost...can anyone help???
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by chris25 View Post
    Find the directional derivative of f x, y = 3 x^2 y at the point (1,2) in the direction of v= 3 i+4 j .

    after i find the unit vector i am lost...can anyone help???
    The directional derivative can be defined in terms of the gradient:

    \bold{D}_{\bold u} f(x_0,y_0)=\nabla f(x_0,y_0)\cdot \bold{u}

    Well, once we get the unit vector \bold{u}=\frac{3}{5}\bold{i} +\frac{4}{5}\bold{j}, we need to find \nabla f(1,2)

    Note that the \nabla operator [in this case] refers to \frac{\partial}{\partial x}\bold i +\frac{\partial}{\partial y}\bold j

    Thus \nabla f(x,y)=\frac{\partial}{\partial x}(3x^2y)\bold i+\frac{\partial}{\partial y}(3x^2y)\bold j=6xy\bold i +3x^2 \bold j

    Now we can see that:

    \nabla f(1,2)=6(1)(2)\bold i +3(1)^2 \bold j=12\bold i +6 \bold j

    Thus,

    \bold{D}_{\bold u} f(1,2)=\nabla f(1,2)\cdot \bold{u}=\underbrace{\bigg(12\bold i +6\bold j\bigg)}_{\nabla f(1,2)}\cdot \underbrace{\bigg(\frac{3}{5}\bold i +\frac{4}{5}\bold j\bigg)}_{\bold u}=\frac{36}{5}+\frac{24}{5}=\frac{60}{5}=\color{b  lue}\boxed{12}

    Does this make sense?

    --Chris
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    The directional derivative can be defined in terms of the gradient:

    \bold{D}_{\bold u} f(x_0,y_0)=\nabla f(x_0,y_0)\cdot \bold{u}

    Well, once we get the unit vector \bold{u}=\frac{3}{5}\bold{i} +\frac{4}{5}\bold{j}, we need to find \nabla f(1,2)

    Note that the \nabla operator [in this case] refers to \frac{\partial}{\partial x}\bold i +\frac{\partial}{\partial y}\bold j

    Thus \nabla f(x,y)=\frac{\partial}{\partial x}(3x^2y)\bold i+\frac{\partial}{\partial y}(3x^2y)\bold j=6xy\bold i +3x^2 \bold j

    Now we can see that:

    \nabla f(1,2)=6(1)(2)\bold i +3(1)^2 \bold j=12\bold i +6 \bold j

    Thus,

    \bold{D}_{\bold u} f(1,2)=\nabla f(1,2)\cdot \bold{u}=\underbrace{\bigg(12\bold i +6\bold j\bigg)}_{\nabla f(1,2)}\cdot \underbrace{\bigg(\frac{3}{5}\bold i +\frac{4}{5}\bold j\bigg)}_{\bold u}=\frac{36}{5}+\frac{24}{5}=\frac{60}{5}=\color{b  lue}\boxed{12}

    Does this make sense?

    --Chris
    f(x,y)=3x^2y

    \nabla{f}=\left\langle{6xy,3x^2}\right\rangle

    \nabla{f}(1,2)=\left\langle{6(1)(2),3(1)^2}\right\  rangle=\left\langle{12,{\color{red}3}}\right\rangl  e

    \nabla{f}(1,2)\cdot\bold{\hat{u}}=12\cdot\frac{3}{  5}+3\cdot\frac{4}{5}=\frac{48}{5}
    \therefore{\bold{D}_{\bold{u}}}=\frac{48}{5}\quad\  because\nabla{f}(1,2)\cdot\bold{\hat{u}}=\frac{48}  {5}
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