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Gradient Vectors

  1. Mar 25, 2012 #1
    1. The problem statement, all variables and given/known data
    An artificial hill has altitude given by the function A(x,y)=300e^-(x^2 +y^2)/100 where the positive y-axis points north and the positive x-axis points east.

    a.)What would be the instantaneous rate of change of her altitude if she walks precisely northwest, starting from the point (8,6)?

    b.) Which way should she start walking to go downhill most rapidly from the point (8,6) and what would the rate of change of her altitude be if she walked in that direction?

    c.) Which ways could she start walking to stay at a constant altitude, starting from (8,6)?
    (Note that of course she is in a location with 3 coordinates, so I'm specifying only her x and y coordinates in this problem.)

    Could you review all my answers?
    2. Relevant equations
    Directional derivatives.

    3. The attempt at a solution
    Problem 3a., the "instantaneous rate of change of her altitude if she walks precisely northwest, starting from the point (8,6)?" can be interpreted as find the directional derivative of A(x,y) at (8,6) in the direction of u= -cos(pi/4)i +sin(pi/4)j, since Northwest would be 135 degrees.

    Ax (8,6)=-6xe^(-x^2 -y^2)/100 =-48e^-1
    Ay(8,6)=-6ye^(-x^2 -y^2)/100=-36e^-1

    D_u A(x,y)=(-48e^-1)(-cos(pi/4)i)+(-36e^-1)(sint(pi/4)j ), D as in directional derivative.

    I found that D= 3.121560451

    Problem 3b.)
    Which way should she start walking to go downhill most rapidly from the point (8,6)?
    Downhill most rapidly from (8,6) means the -|gradA|
    gradA(x,y)=Ax(x,y)i+Ay(x,y)j
    gradA(x,y)=-48e^(-1)i -36e^(-1)j, she should start walking in this direction to go uphill most rapidly and -gradA(x,y) to go downhill most rapidly.
    Namely,
    48e^(-1)i +36e^(-1)j

    what would the rate of change of her altitude be if she walked in that direction?
    Rate of change of her altitude would be |gradA(x,y)|=22.07276647

    Problem c.)
    Which ways could she start walking to stay at a constant altitude, starting from (8,6)?
    (Note that of course she is in a location with 3 coordinates, so I'm specifying only her x and y coordinates in this problem.)
    Domain of x and y are all reals, but z=A(x,y) can only be [0,300]
    So there are only upwards<110.363824, 300]
    and downwards from <110.36824, 0] ways.


    Thank you
     
  2. jcsd
  3. Mar 25, 2012 #2
    I think you're fine with parts A and B. For part C, you want to find a vector [itex]\vec{r}[/itex] such that [itex]\vec{\nabla}A(x,y)\cdot\vec{r}=0[/itex]. When you find that you have one equation with two unknowns, recall that you only need the direction, so [itex]\vec{r}\cdot\vec{r}=1[/itex]. Now you have two equations with two unknowns. You should expect to find one direction and its opposite to satisfy these relations.
     
  4. Mar 25, 2012 #3
    ## \nabla A\left( 8,6\right) = < \left( -48e^{-1}\right) ,\left( -36e^{-1}\right) > ##

    I looked up the dot product of the same vector and there are some problems with it. So, since I only need the direction r.r=1 ? I don't understand, how can you disregard the vectors magnitudes?
     
  5. Mar 25, 2012 #4
    Now that you have [itex]\vec{\nabla}A(8,6)[/itex], dot it with [itex]\vec{r}=\{r_x,r_y\}[/itex]. In solving for [itex]r_x[/itex] and [itex]r_y[/itex], you can use [itex]r_x^2+r_y^2=1[/itex].
     
  6. Mar 25, 2012 #5
    Why do you say that the dot product is 1?
     
  7. Mar 25, 2012 #6
    The length of [itex]\vec{r}[/itex] is of no concern, only its direction matters. In fact, here's what you really need to solve: [itex]\vec{\nabla}A(8,6)\cdot\hat{r}=0[/itex]. Is this less ambiguous?
     
  8. Mar 25, 2012 #7
    See if this contour plot of [itex]A(x,y)[/itex] helps. The blue vector represents the unit vector pointed in the direction of [itex]\vec{\nabla}A(x,y)[/itex]. The black vectors represent unit vectors in the direction of no ascent - they correspond to each [itex]\hat{r}[/itex] you need to find.

    attachment.php?attachmentid=45510&stc=1&d=1332721994.png
     

    Attached Files:

  9. Mar 25, 2012 #8
    Yes. It is less ambiguous, but still I am confused.

    So you want:
    ## < -48e^{-1},-36e^{-1}>\dfrac {.\overline {r}} {||\overline {r}\parallel }= ## 0

    Then,
    ## -48e^{-1}r_{1}=36e^{-1}r_{2} ##

    I don't know, how to proceed.
     
  10. Mar 25, 2012 #9
    So that's one equation in two unknowns. Now use that [itex]\hat{r}[/itex] is a unit vector: [itex]r_x^2+r_y^2=1[/itex]. Then you have two equations and two unknowns.
     
  11. Mar 25, 2012 #10
    How do you derive rx^2+ry^2 =1 from the unit vector?
     
  12. Mar 25, 2012 #11
    Your vector's length is [itex]|\hat{r}|=\sqrt{r_x^2+r_y^2}=1[/itex]. Square both sides and [itex]1^2=1[/itex].
     
  13. Mar 25, 2012 #12
    Ok, I agree that the unit vector's length is 1.
    Solving both equations yield:
    r1=(-3/4) SQRT(4/3)
    r2=SQRT(4/3)
    The directions are:
    r=r1+r2, and it's oppposite
    -r=-r1+-r2
    ?
     
  14. Mar 25, 2012 #13
    Check your math.
     
  15. Mar 25, 2012 #14
    Sorry about that,
    r1=(-3/4)SQRT(16/7)
    r2=SQRT(16/7)

    r=r1+r2 and opposite
    -r=-r1-r2
     
  16. Mar 25, 2012 #15
    Check your math.
     
  17. Mar 25, 2012 #16
    Do you just suggest me to simplify the expression?
    ## r_{1}=\dfrac {36e^{-1}r_{2}} {-48e^{-1}}=\dfrac {-3} {4}r_{2} ##
    ## r_{1}^{2}+r_{2}^{2}=1,\left( -\dfrac {3} {4}r_{2}\right) ^{2}+r_{2}^{2}=1 ##
    ## r_{2}=\left( \sqrt {\dfrac {1} {7}}\right) 4 ##
    and r1=-3SQRT(1/7)
     
  18. Mar 25, 2012 #17
    No, I suggest that your answer is wrong. So far, I agree with [itex]r_1=-(3/4)r_2[/itex], but you somehow go awry when plugging this into [itex]r_1^2+r_2^2=1[/itex].
     
  19. Mar 25, 2012 #18
    Wow, I'm sorry.
    r2=4/5, r1=-3/5

    r=r1+r2 and opposite direction
    -r=-r1-r2
     
  20. Mar 25, 2012 #19
    Great, so you can either state these in vector format: [itex]\hat{r}=-(3/5)\hat{x}+(4/5)\hat{y}[/itex] (and similarly for [itex]-\hat{r}[/itex]), or in angular format as measured from the positive x-axis.
     
  21. Mar 25, 2012 #20
    Thank you. According, to a classmate part A from this problem can be answered as having the vector direction (-1,1). He gets a different result than mine.

    Do you agree with my result?
     
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