Trajectory using gradient and differential equations

In summary: \frac{1}{\text{time}}\left[x'(t)\right]_{t=0} = e^{-8\gamma\alpha t}\qquad [y'(t)\right]_{t=0} = e^{-2\gamma\alpha t}... sox'(0)=-8\gamma\alpha x(0)\qquad y'(0)=-2\gamma\alpha y(0)
  • #1
[SOLVED] Trajectory using gradient and differential equations

Homework Statement


A heat-seeking particle is located at the point P on a flat metal plate whose temperature
at a point (x, y) is T(x, y). Find parametric equations for the trajectory of the particle if
it moves continuously in the direction of maximum temperature increase.

T(x, y) = 5 − 4x[tex]^{2}[/tex] − y[tex]^{2}[/tex]; P(1, 4).


Homework Equations


velocity vector = [tex]\left \langle[/tex]x'(t), [tex]\right y'(t)\rangle[/tex]

[tex]\nabla T(x,y)[/tex] = [tex]\left\langle[/tex][tex]T_{x}[/tex], [tex]\right T_{y}\rangle[/tex]

The Attempt at a Solution


Let x = x(t), y = y(t) be parametric equations for the trajectory of the particle.
We want the particle to move continuously in the direction of maximum temperature
increase. Equivalently, we want the velocity vector [tex]\left \langle[/tex]x'(t), [tex]\right y'(t)\rangle[/tex] to point in the direction
of [tex]\nabla T(x(t),y(t))[/tex]. Since [tex]\nabla T(x,y)[/tex]= [tex]\left \langle[/tex] -8x, [tex]\right -2y\rangle[/tex], the previous condition will be satisfied if
[tex]\left \langle[/tex]x'(t), [tex]\right y'(t)\rangle[/tex] = [tex]\left \langle[/tex] -8x(t), [tex]\right -2y(t)\rangle[/tex] for all t. To find such functions x(t) and y(t), we must
solve the differential equations
x'(t) = -8x(t), y'(t) = -2y(t)

but this is where I get stuck...I know that to determine solution, we need the initial condition of x(0) = 1, y(0) = 4. But as to solve the differential equations, I have no idea. This is from my multivariable differential calculus class, and I think they assume that I know how to do this... sadly I do not. Any help would be greatly appreciated... thanks
 
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  • #2
issisoccer10 said:

The Attempt at a Solution


Let x = x(t), y = y(t) be parametric equations for the trajectory of the particle.
We want the particle to move continuously in the direction of maximum temperature
increase. Equivalently, we want the velocity vector [tex]\left \langle[/tex]x'(t), [tex]\right y'(t)\rangle[/tex] to point in the direction
of [tex]\nabla T(x(t),y(t))[/tex]. Since [tex]\nabla T(x,y)[/tex]= [tex]\left \langle[/tex] -8x, [tex]\right -2y\rangle[/tex], the previous condition will be satisfied if
[tex]\left \langle[/tex]x'(t), [tex]\right y'(t)\rangle[/tex] = [tex]\left \langle[/tex] -8x(t), [tex]\right -2y(t)\rangle[/tex] for all t.


You might want to introduce a constant factor here:

[tex]\left \langle x'(t),\right y'(t)\rangle =\alpha \nabla T(x(t),y(t))[/tex].

There are two reason for this. First, you only know that the particle moves in the direction of the tempertaure gradient, you do not know how strong it is coupled to that gradient. Second, in modelling physical situations you should be aware of the units you are using. Your velocity vector has units length/time while the temperature gradient is given in temperature/length; so in order for your equation to make sense [itex]\alpha[/itex] had better have units of (length/time) / (temperature/length) = length^2 / (time*temperature). You need not worry about this explicity but it is a point to consider.


issisoccer10 said:
To find such functions x(t) and y(t), we must
solve the differential equations
x'(t) = -8x(t), y'(t) = -2y(t)

Looks like x(t) and y(t) hardly change when you differentiate them, they just seem to pick up a factor -8 and -2 ... do you know any functions showing this strange behavior with respect to differentiation?
 
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  • #3
if x(t) = e[tex]^{-8t}[/tex] and y(t) = e[tex]^{-2t}[/tex] that would work right? so differential equations are just logic? at least at this level anyways..

but as for the constant [tex]\alpha[/tex] you were talking about.. that would change my parametrizations to x(t) = e[tex]^{-8t}[/tex] and y(t) = 4e[tex]^{-2t}[/tex] if it is to go through the point (1,4). But I'm not sure if that covers the [tex]\alpha[/tex].
 
  • #4
Yes, the exponential's are solutions to your equations. Well, of course there are actual methods to solve this kind of equation but I don't know how deeo you want to go into this. In easy cases it's often possible to ust guess an solution and that's perfectly acceptable - the only problem being that this method doesn't work in more complicated situations.

If you include the constant your differential equations become

[tex]
x'(t)=-8\alpha x(t)\qquad y'(t)=-2\alpha y(t)
[/tex]

I wrote above that alpha must have units of length^2 / (time*temperature) so the units in these equations are

[tex]
\frac{\tex{length}}{\text{time}}=[x'(t)]=-8[\alpha] [x(t)]=\frac{\text{length}^2}{\tex{time*temperature}}\text{length}=\frac{\text{length}^3}{\tex{time*temperature}}
[/tex]

So this seems not to work.. the reason for this is that the 8 in this equation is not dimensionless either (although in the previous calcuation I took it to be so).

If you consider your defintion of the temperature field

T(x, y) = 5 − 4x^2 − y^2,

then on the right hand side it should be temperatures, but the left hand side is not. Also note that being pedantic with units would make it impossible to actually have something like 5+x^2 .. so you could have yet another constant here to convert the left hand side to temperature

[tex]
T(x, y) = 5\beta-\gamma(4x^2-y^2)
[/tex]

with [tex][\beta]=\text{temperature},\qquad [\gamma]=\frac{\text{temperature}}{\text{length}^2}[/tex]

then the equation would allow you to consistently measure distance in units of length and temperature in -well - units of temperature.

Then including this constant as well we get:
[tex]
x'(t)=-8\gamma\alpha x(t)\qquad y'(t)=-2\gamma\alpha y(t)
[/tex]

note that [itex][\gamma\alpha]=\frac{1}{\text{time}}[/itex] so that's fine now.

So solving as you did gives

[tex]
x(t)=e^{-8\gamma\alpha t}\qquad y(t)=e^{-2\gamma\alpha t}
[/tex]

Here the argument of the exponential is now dimensionless which is very good :smile:

Including your initial condidion you multiply this solution for x(t) [y(t)] by 1 [4]. Considering that this 1 [4] has units of length, your final x(t) [y(t)] will also have units of length because the exponential is just a dimensionless factor.

That's it.

But actually nobody cares about that.. the most practical way is to just assume that these constants accounting for conversion of units are all set to one and hence don't actually appear, or you ignore units altogether and hope that it works out fine...however when it comes to interpreting your result in a physical way you must consider the issue of units in one way or another.
 
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  • #5
wow...units aren't cool...actually they make everything make sense and since they make sense, all is well... anyways thanks again
 

1. How can gradient and differential equations be used to calculate trajectory?

Gradient and differential equations can be used to calculate trajectory by determining the rate of change of an object's position over time. This is done by taking the derivative of the position function with respect to time, which gives the object's velocity. The integral of the velocity function then gives the object's position as a function of time, allowing for the calculation of the trajectory.

2. What is the relationship between gradient and trajectory?

The gradient of a function represents the direction and magnitude of its maximum rate of change. In the context of trajectory, this means that the gradient can be used to determine the direction and speed of an object's motion at any given point in time. This information is crucial for predicting and analyzing the path an object will take.

3. How does the initial position and velocity affect the trajectory of an object?

The initial position and velocity of an object both play important roles in determining its trajectory. The initial position sets the starting point for the object's motion, while the initial velocity determines the direction and speed at which the object moves from that starting point. These initial conditions, along with the forces acting on the object, are used in the gradient and differential equations to calculate the trajectory.

4. Can gradient and differential equations be used for all types of trajectories?

Yes, gradient and differential equations can be used for all types of trajectories, including linear, curved, and even complex trajectories. The equations can be adapted to account for different forces acting on the object, such as gravity or air resistance, allowing for the calculation of trajectories in various scenarios.

5. What are some real-world applications of using gradient and differential equations for trajectory?

There are many real-world applications of using gradient and differential equations for trajectory. Some examples include predicting the path of a rocket or satellite, determining the trajectory of a ball in sports, and analyzing the motion of objects in physics experiments. These equations are also used in fields such as engineering, astronomy, and aerodynamics to study and design various systems and structures.

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