Multivariable Calculus - Partial Derivatives Assignment

[ConnorM]

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.

 

[Ray Vickson]

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.

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.

 

[ConnorM]

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.

 

[Ray Vickson]

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.

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.

 

[verty]

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.
...
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.

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?

 

[ConnorM]

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.

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?

 

[ConnorM]

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!

 

[HallsofIvy]

The gradient vector at each (x, y) point is, as you say -40x\vec{i}- 80y\vec{j}. 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.

 

[Ray Vickson]

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?

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 $k = e^{2C} = (\sqrt{500}/100)^2 = 500/100^2 = 1/20$.

 

[ConnorM]

OK thanks! You helped a lot! I actually understand this now thanks so much.