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Multivariable Calculus - Partial Derivatives Assignment

  1. Jul 11, 2014 #1
    1. Marine biologists have determined that when a shark detects the presence of blood in the water, it will swim in the direction in which the concentration of the blood increases most rapidly. Suppose that in a certain case, the concentration of blood at a point P(x; y) on the surface of the seawater is given by,

    f(x; y) = 10^8 - 20x^2 - 40y^2

    where x and y are measured in meters in a rectangular coordinate system with the blood source at
    the origin. Suppose a shark is at the point (100;500) on the surface of the water when it first detects the presence of blood. Find the equation of the shark's path towards the blood source. [Hint: You can use the fact that if y=g(x) satisfies g'(x) =(a/x)*g(x), then g(x) =Cx^a for some constant C.

    2. So far in this class we have learned about limits and continuity of multi-variable functions, tangents planes, linear approximations, gradient vectors, directional vectors, and partial derivatives.[/b]


    3. I know that the gradient vector points toward where the function is increasing most rapidly, to obtain the gradient vector I differentiated the equation f(x; y) for x to obtain fx = -40x and for y to obtain fy = -80y. Next I subbed in the sharks initial position (100;500) and found the gradient vector to be ∇f (100;500) = (-4000; -40000). The thing is my question asks for the equation for the sharks path to the blood source. I was thinking and wouldn't that mean its path is a straight line? If anyone has some guidance or knows what the next step is that would be much appreciated!
    Thanks.
     
    Last edited by a moderator: Jul 11, 2014
  2. jcsd
  3. Jul 11, 2014 #2

    Ray Vickson

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    In general the path would not be a straight line; from certain, special starting points it would be a straight line, but not if you start from more-or-less random positions. Why? Well, the direction of the shark's instantaneous velocity is given by the gradient, so the direction changes as x and/or y change.
     
  4. Jul 11, 2014 #3
    So if the sharks path is going to be changing due to x and y changing, how am I supposed to create an equation to model that? Using the gradient vector ∇f(100;500) I know the direction of travel, also the hint at the bottom shows that y is a function of x called g(x). Would I have to use some 1st order differentiation to solve this problem? With the skills we have learned so far in class I don't know how to find the equation that the shark will follow.
     
  5. Jul 11, 2014 #4

    Ray Vickson

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    Starting from ##(x,y)##, let the shark go along a short segment ##(\Delta x, \Delta y)##. What is the relationship between ##\Delta x## and ##\Delta y##? So, if the shark follows a path ##y = f(x)##, what does all this say about the function ##f##? Use the hint supplied.
     
  6. Jul 11, 2014 #5

    verty

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    This question had me stumped for a good while. You have calculated the gradient ##\nabla f(x,y)##. Now any path of steepest ascent ##y = g(x)## has a tangent that is related to ##\nabla f## how?
     
  7. Jul 11, 2014 #6
    OK! So what if I use my gradient vector ∇f=( -40x, -80y ) and say that from y = g(x) being a path of steepest ascent its derivative will be equal to the slope of the gradient vector, so g(x)'s derivative will be g'(x) = -80y/-40x = 2y/x. From here I have a differential equation I can solve.

    g'(x) = 2y/x

    (1/2y)*g'(x) = 1/x

    ...solving this I got...

    (1/2)*ln(y) = ln(x) + C , from here would I sub in my initial value (100 , 500) to find C and the resulting equation would be the sharks path?

    (1/2)*ln(500) = ln(100) + C

    C = 1.497 ≈ 1.5

    then subbing in C and simplifying I got,

    (1/2)*ln(y) = ln(x) + 1.5

    e^((1/2)*ln(y)) = e^(ln(x) + 1.5)

    (y^1/2 )^2= (e^(ln(x) + 1.5))^2

    y=(e^(ln(x) + 1.5))^2

    Is this correct or at least on the right track?
     
    Last edited: Jul 12, 2014
  8. Jul 12, 2014 #7
    Oops! I messed up some of the simplification, my new answer is

    y= Cx^2

    subbing in (100;500)

    500=C(100)^2

    C=1/20

    y=x^2 / 20

    This is also similar to what the hint says!
     
  9. Jul 12, 2014 #8

    HallsofIvy

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    The gradient vector at each (x, y) point is, as you say [itex]-40x\vec{i}- 80y\vec{j}[/itex]. At each (x, y) that points in a direction such that the tangent of the angle with the x-axis is (-80y)/(-40x)= 2y/x. Since the tangent of the angle is the derivative, that is the same as dy/dx= 2y/x. Solve that differential equation (it is "separable" so it is just a matter of integrating) to find the shark's path.
     
  10. Jul 12, 2014 #9

    Ray Vickson

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    It is OK, but unnecessarily complicated. Just use the fact that ##\left(e^{\ln x + C}\right)^2 = k x^2##, where ##k = e^{2C}##. Of course, ##C = (1/2) \ln 500 - \ln 100 = \ln(\sqrt{500}/100)##, so ##k = e^{2C} = (\sqrt{500}/100)^2 = 500/100^2 = 1/20##.
     
  11. Jul 12, 2014 #10
    OK thanks! You helped a lot! I actually understand this now thanks so much.
     
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