##\hat r = \frac {\vec r}{|\vec r|}## -- Does that help?Alexander350 said:I need to solve:
[tex]\dot{\mathbf{r}}=-kv\hat{r} - \dot{\mathbf{r}_s}[/tex]
However, I do not know how to deal with the fact that there is a unit vector. How can this be done? [itex]\dot{\mathbf{r}_s}[/itex] is a constant vector.
This is what it was written as initially, but the problem is that the absolute value of the vector also changes with time, so I did not know what to do.Mark44 said:##\hat r = \frac {\vec r}{|\vec r|}## -- Does that help?
v is just a constant number.Mark44 said:What is v in this problem? Is it ##|\vec v|##?
No, v is just a number. It would be the same question if you replaced kv with another constant c.Mark44 said:I asked the wrong question. What I meant was, could v be ##|\dot r|##?
How does this help when [itex]|r|[/itex] is not constant? And where does the [itex]\frac 1 {|r|}[/itex] come from on the constant?Mark44 said:Multiply ##\hat r## by 1, in the form of ##\frac{|r|}{|r|}##.
In other words, replace
##\dot{\mathbf{r}}=-kv\hat{r} - \dot{\mathbf{r}_s}##
with ##\dot{\mathbf{r}}=-kv\frac 1 {|r|} |r| \hat{r} - \dot{\mathbf{r}_s} = -\frac{kv}{|r|} \vec r - \dot{\mathbf{r}_s}##
Now you have a DE in terms of dr/dt and r, plus a constant.
BTW, you really should have posted this question in the Homework section.
Are you sure that it's not constant? You didn't provide the full problem, so there's the possibility that you're making an assumption that isn't necessary. For example, if a particle is moving in a circular path at constant speed, ##\dot r## is changing, but ##|\dot r|## is constant.Alexander350 said:How does this help when [itex]|r|[/itex] is not constant?
It shouldn't be there. When I started to write that, I was multiplying both sides by |r|, but I changed my mind, and multiplied one term of the equation by ##\frac{|r|}{|r|}##. I have edited my earlier response and the text that you quoted.Alexander350 said:And where does the [itex]\frac 1 {|r|}[/itex] come from on the constant?
It is definitely not constant. It is equal to the relative displacement between two things which are getting closer together over time. If it is not constant, does that make this approach impossible to solve?Mark44 said:Are you sure that it's not constant? You didn't provide the full problem, so there's the possibility that you're making an assumption that isn't necessary. For example, if a particle is moving in a circular path at constant speed, ##\dot r## is changing, but ##|\dot r|## is constant.
Smugglers set off in a ship in a direction perpendicular to a straight shore and move at a constant speed v. The coastguard's cutter is a distance a along the shore from the smugglers' ship, and leaves the shore at the same time. The cutter always moves at a constant speed directly towards the smugglers' ship, and catches up with it at a distance a from the shore. How many times greater is the speed of the coastguard's cutter than the smugglers' ship?Mark44 said:Can you post the actual problem statement? You've stated that v is "just a number," but I can't help thinking the v is somehow related to velocity or speed.
A unit vector is a vector with a magnitude of 1 and is used to indicate direction in a specific coordinate system. It is often denoted with a hat symbol on top (e.g. û ).
A unit vector can be used to represent a specific direction in a differential equation, such as the direction of a velocity or acceleration vector. It can also be used to simplify calculations and represent a normalized version of a vector.
Yes, a unit vector can be used in all types of differential equations, including ordinary differential equations, partial differential equations, and vector differential equations. It is a commonly used tool in mathematical modeling and problem-solving.
To incorporate a unit vector into a differential equation, you can substitute the unit vector for any direction or vector in the equation. You can also represent a vector as a combination of unit vectors, such as in polar or spherical coordinate systems.
There are no inherent limitations to using unit vectors in differential equations. However, it is important to accurately define the coordinate system and direction represented by the unit vector to avoid errors in calculations.