An equation for the path that the shark will swim on

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Homework Help Overview

The problem involves determining the path a shark will swim based on the concentration of blood in water, modeled by a given concentration function. The original poster attempts to set up a differential equation using the gradient of the concentration function to find the direction of greatest increase.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster discusses using the gradient to establish a differential equation and attempts to solve it through separation of variables. They express difficulty with integration and mention the use of computational tools. Other participants question the correctness of the derivatives used, suggesting that errors may complicate the integration process.

Discussion Status

Participants are actively exploring the problem, with some providing feedback on potential errors in the original poster's approach. There is an ongoing examination of the derivatives and their impact on the integral, indicating a productive dialogue without a clear consensus on the solution.

Contextual Notes

The original poster expresses uncertainty about the integration process and the implications of using computational tools, which may not align with textbook expectations. There is a focus on verifying the correctness of the derivatives before proceeding further.

kostoglotov
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Homework Statement


[/B]
A shark will in the direction of the most rapidly increasing concentration of blood in water.

Suppose a shark is at a point x_0,y_0 when it first detects blood in the water. Find an equation for the path that the shark will follow by setting up and solving a differential equation.

Homework Equations



C(x,y) = e^{-(x^2+2y^2)/10^4}: blood conc. in ppm at (x,y)

The Attempt at a Solution



\nabla C(x,y) = \left<-xe^{-(x^2+2y^2)/5\times 10^7},-ye^{-(x^2+2y^2)/2.5\times 10^7}\right> will give the vector in the direction of greatest increase of [blood] at any given (x,y).

Thus \frac{dy}{dx} = \frac{\frac{\partial C}{\partial y}}{\frac{\partial C}{\partial x}} will be the tangent to the curve of the path of the shark at any given point (x,y), which we can set up as our differential equation and solve by separation of variables.

\frac{dy}{dx} = \frac{y}{x}e^{-(x^2+2y^2)/5\times 10^7}

\frac{dy}{dx} = \frac{y}{x}e^{-x^2/5\times 10^7}e^{-2y^2/5\times 10^7}

\frac{1}{y}e^{2y^2/5\times 10^7}dy = \frac{1}{x}e^{-x^2/5\times 10^7}dx

\int \frac{1}{y}e^{2y^2/5\times 10^7}dy = \int \frac{1}{x}e^{-x^2/5\times 10^7}dx

I can't solve this last equation. I can't figure out how to do the integration, and CASs (like Matlab and Mathematica) show me something called ei() and series expansions that I don't understand, and that I'm fairly sure are not the answer the textbook intended.

I've tried parametrizing the dy/dx function as per elliptic parametrization which gives

m(t) = \frac{1}{2} \tan{t} e^{-(2\cos^2{t}+\sin^2{t})/10^8}

But integrating that wrt t is just as problematic.

Is there something with taking the natural logs of both sides that I can do to eliminate that exponential function?

How do I solve this?
 
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Something went wrong with your derivatives. That makes the integral harder than it should be.
 
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mfb said:
Something went wrong with your derivatives. That makes the integral harder than it should be.

I found the derivatives using Matlab's gradient() function.

I'll take them manually. See how I go...thanks :)
 
mfb said:
Something went wrong with your derivatives. That makes the integral harder than it should be.
Such a stupid oversight...thanks again!
 

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