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Math Help - function of two variables, directional derivatives

  1. #1
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    function of two variables, directional derivatives

    Let f(x,y) be a function that is differtiable everywhere. At a certain point P in the xy plane, the directional derivative of F in the direction of i - j is sqrt(2) and the directional derivative of f in the direction of i + j is 3sqrt(2). What is the maximum directional derivative at P?

    A) 3sqrt(2) B) 2sqrt(5) C) 43sqrt(5) D) 6 E) 8

    I know the directional derivative is the del of f(x,y) = <fx, fy>. I also know that the directional derivative also points in the direction of maximum increase.

    So should i have <fx, fy> = i - j meaning fx = 1 and fy = -1 evaluated at P?

    Same for <fx, fy> = i + j meaning fx = 1 and fy = 1 at P?

    I am not sure what to do can someone help me out
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by ldaniels241 View Post
    Let f(x,y) be a function that is differtiable everywhere. At a certain point P in the xy plane, the directional derivative of F in the direction of i - j is sqrt(2) and the directional derivative of f in the direction of i + j is 3sqrt(2). What is the maximum directional derivative at P?

    A) 3sqrt(2) B) 2sqrt(5) C) 43sqrt(5) D) 6 E) 8

    I know the directional derivative is the del of f(x,y) = <fx, fy>. I also know that the directional derivative also points in the direction of maximum increase.

    So should i have <fx, fy> = i - j meaning fx = 1 and fy = -1 evaluated at P?

    Same for <fx, fy> = i + j meaning fx = 1 and fy = 1 at P?

    I am not sure what to do can someone help me out
    Since they are using an arbitary point P we might as well make it the center of our coordinate system.

    This diagram may help

    Attachment 16398

    The two vectors in black can be used as a basis for all points on the unit circle in the picture. i.e any green line on the unit circle can be written as a linear combination of these two vectos

    x\vec{i}+y\vec{j}=\frac{c_1}{\sqrt{2}}(\vec{i}-\vec{j})+\frac{c_2}{\sqrt{2}}(\vec{i}+\vec{j})

    Provided that c_1^2+c_2^2=1

    or in shorter notation

    (x,y)=c_1v_1+c_2v_2

    We know that

    \nabla f(0,0)\cdot \underbrace{\frac{1}{\sqrt{2}}(\vec{i}-\vec{j})}_{v_1}=\sqrt{2}
    and
    \nabla f(0,0)\cdot \underbrace{\frac{1}{\sqrt{2}}(\vec{i}+\vec{j})}_{  v_2}=3\sqrt{2}

    Now the function we want to maximize is

    M(c_1,c_2)=\nabla f(0,0)\cdot (x\vec{i}+y\vec{j})=\nabla f(0,0)\cdot (c_1v_1+c_2v_2)=
    c_1\nabla f(0,0)\cdot v_1+c_2 \nabla f(0,0)\cdot c_2=\sqrt{2}c_1+3\sqrt{2}c_2

    Now just maximize this function with the constraint that c_1^2+c_2^2=1


    Now when I solve for c_1,c_2 I get

    c_1=\frac{1}{\sqrt{10}},c_2=\frac{3}{\sqrt{10}}

    This gives the maximum of the function M=2\sqrt{5}
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  3. #3
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    Thank you so much. I just wish I could view the picture though.
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