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Math Help - rate of change and tangent plane question

  1. #1
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    rate of change and tangent plane question

    Exam practice question to which I don't have an answer. Would be greatful for any help.


    Question.

    Consider the function f^2:R^3->R defined by

    f(x,y,z)=x^3+y^2-6xy+2z,

    and let P=(2,1,3).

    (i) Find the rate of change of f(x,y,z) at the point P in the direction v=(1,-1,1)^T.

    (ii) Write down the direction of maximum increase of f at P, and the rate of maximum increase. Show that P is on the surface f(x,y,z)=3 and write down the equation of the tangent plane to the surface at P.

    (iii) There is another point Q on the surface f(x,y,z)=3 where the tangent plane is parallel to the tangent plane at P. Find Q.


    Answer.

    (i) first, I find partial derivatives of f at P

    f_x=3x^2-6y=6, f_y=2y-6x=-10, f_z=2

    Let \varphi be the angle between the above vector (6,-10,2)^T and the unit vector v=(1,-1,1)^T. Then I can find alpha:

    |\nabla{f}||v|cos\varphi=<\nabla{f},v>

    cos\varphi=\frac{<\nabla{f},v>}{|\nabla{f}||v|}=\f  rac{6+10+2}{\sqrt{36+100+4}\sqrt{3}}=\frac{18}{2\s  qrt{105}}

    Then the rate of change of \nabla{f} in the direction of v=(1,-1,0)^T is just the length of the projection of this vector on the direction of v, which can be calculated as

    |\nabla{f}_u|=|\nabla{f}|cos\varphi=\sqrt{140}\fra  c{18}{2\sqrt{105}}=6\sqrt{3}

    (ii) the maximum rate of change of f at P is just the length of its gradient vector at this point, ie \sqrt{140} as per (i).

    Since coordinates of P satisfy the equation x^3+y^2-6xy+2x=3, it lies on the surface f(x,y,z)=3.

    The tangent plane to the surface f(x,y,z)=3 at P (2,1,3) is

    0=f_x(x-2)+f_y(y-1)+f_z(z-3)

    0=6(x-2)-10(y-1)+2(z-3)

    6x-10y+z=5 is the tangent plane to the surface f(x,y,z)=3 at P(2,1,3).


    For a parallel tangent plane, I understand that I need to find a point that satisfies the equation f(x,y,z)=3 AND an equation of a plane parallel to 6x-10y+z=5. However, I don't know yet how to find it.
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  2. #2
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    Quote Originally Posted by Volga View Post
    Exam practice question to which I don't have an answer. Would be greatful for any help.


    Question.

    Consider the function f^2:R^3->R defined by

    f(x,y,z)=x^3+y^2-6xy+2z,

    and let P=(2,1,3).

    (i) Find the rate of change of f(x,y,z) at the point P in the direction v=(1,-1,1)^T.

    (ii) Write down the direction of maximum increase of f at P, and the rate of maximum increase. Show that P is on the surface f(x,y,z)=3 and write down the equation of the tangent plane to the surface at P.

    (iii) There is another point Q on the surface f(x,y,z)=3 where the tangent plane is parallel to the tangent plane at P. Find Q.


    Answer.

    (i) first, I find partial derivatives of f at P

    f_x=3x^2-6y=6, f_y=2y-6x=-10, f_z=2

    Let \varphi be the angle between the above vector (6,-10,2)^T and the unit vector v=(1,-1,1)^T. Then I can find alpha:

    |\nabla{f}||v|cos\varphi=<\nabla{f},v>

    cos\varphi=\frac{<\nabla{f},v>}{|\nabla{f}||v|}=\f  rac{6+10+2}{\sqrt{36+100+4}\sqrt{3}}=\frac{18}{2\s  qrt{105}}

    Then the rate of change of \nabla{f} in the direction of v=(1,-1,0)^T is just the length of the projection of this vector on the direction of v, which can be calculated as

    |\nabla{f}_u|=|\nabla{f}|cos\varphi=\sqrt{140}\fra  c{18}{2\sqrt{105}}=6\sqrt{3}
    Not much point in using \nabla f three different times is there? Simpler to remember that the derivative of f in the direction of unit vector u is just \nabla f\cdot u. The unit vector in the direction of <1, -1, 1> is \left<1/\sqrt{3}, -1/\sqrt{3}, 1/\sqrt{3}\right> so the derivative in that direction is
    \left<6, -10, 2\right>\cdot<1/\sqrt{3}, -1/\sqrt{3}, 1/\sqrt{3}\right>= \frac{6+ 10+ 2}{\sqrt{3}}= \frac{18}{\sqrt{3}}= 6\sqrt{3} as you say.

    (ii) the maximum rate of change of f at P is just the length of its gradient vector at this point, ie \sqrt{140} as per (i).
    Yes that is correct.

    Since coordinates of P satisfy the equation x^3+y^2-6xy+2x=3, it lies on the surface f(x,y,z)=3.

    The tangent plane to the surface f(x,y,z)=3 at P (2,1,3) is

    0=f_x(x-2)+f_y(y-1)+f_z(z-3)

    0=6(x-2)-10(y-1)+2(z-3)

    6x-10y+z=5 is the tangent plane to the surface f(x,y,z)=3 at P(2,1,3).


    For a parallel tangent plane, I understand that I need to find a point that satisfies the equation f(x,y,z)=3 AND an equation of a plane parallel to 6x-10y+z=5. However, I don't know yet how to find it.
    You need to find another point on that surface so that f_x= 3x^2- 6y= 6, f_y= 2y- 6x= -10, and f_z= 2. The last is always true. Solve the first two for x and y. There will be two solutions, of course, one of which is x= 2, y= 1. Check that the other solution also satisfies f(x,y,z)= 3.
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  3. #3
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    Thank you!!

    So I solve 3x^2-6y=6, 2y-6x=-10 and I get

    x_1=2, x_2=4; y_1=7, y_2=1 ie the two points satisfying these equations are (2,1) and (4,7).

    Now what do I do for z?... Should i use the fact that z satisfies f_z=2?

    And, more importantly, how do I get closer to finding a tangent plane passing through this point that is parallel the one identified in (ii)?... Would that parallel plane has the same left side but different right side?
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  4. #4
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    I shouldn't have said "Check that the other solution also satisfies f(x,y,z)= 3". I should have said "Find z so that f(4, 7, z)= 3".
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  5. #5
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    Right. I think I got it now:

    Given x=4, y=7 I can find z that satisfied the equation f(x,y,z)=3: z=-29. Therefore, the point Q (4,7,-29) lies on that surface.

    Moreover, the normal vector to the surface at this point Q is (6,-10,2)^T, ie the same normal vector as normal vector to the same surface at the point P (2,1,3). Therefore, tangent planes at these two points (P and Q) are parallel.


    I don't need to find the equation of the tangent plane at Q, but suppose I do: would it be

    0=f_x(x-4)+f_y(y-7)+f_z(z+29)

    0=6(x-4)-10(y-7)+2(z+29), or

    6x-10y+2z=104

    Now, is there any connection between 5 (in the first tangent plane equation) and 104 here?...
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  6. #6
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    It is essentially saying that the distance between those two parallel planes is 104- 5= 99.
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