# Homework Help: An equation for the path that the shark will swim on

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1. Apr 3, 2015

### kostoglotov

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. 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?

2. Apr 3, 2015

### Staff: Mentor

Something went wrong with your derivatives. That makes the integral harder than it should be.

3. Apr 3, 2015

### kostoglotov

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

I'll take them manually. See how I go...thanks :)

4. Apr 3, 2015

### kostoglotov

Such a stupid oversight...thanks again!