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B Solving a differential equation with a unit vector in it

  1. Nov 25, 2017 #1
    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.
     
  2. jcsd
  3. Nov 25, 2017 #2

    Mark44

    Staff: Mentor

    ##\hat r = \frac {\vec r}{|\vec r|}## -- Does that help?
     
  4. Nov 25, 2017 #3
    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.
     
  5. Nov 25, 2017 #4

    Mark44

    Staff: Mentor

    What is v in this problem? Is it ##|\vec v|##?
     
  6. Nov 25, 2017 #5
    v is just a constant number.
     
  7. Nov 25, 2017 #6

    Mark44

    Staff: Mentor

    I asked the wrong question. What I meant was, could v be ##|\dot r|##?
     
  8. Nov 26, 2017 #7
    No, v is just a number. It would be the same question if you replaced kv with another constant c.
     
  9. Nov 26, 2017 #8

    Mark44

    Staff: Mentor

    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.
     
    Last edited: Nov 26, 2017
  10. Nov 26, 2017 #9
    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?
     
    Last edited by a moderator: Nov 26, 2017
  11. Nov 26, 2017 #10

    Mark44

    Staff: Mentor

    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.
    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.
     
  12. Nov 26, 2017 #11
    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?
     
  13. Nov 26, 2017 #12

    Mark44

    Staff: Mentor

    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.
     
  14. Nov 26, 2017 #13
    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?
    I started with the equation [tex]\frac{d\mathbf{r_c}}{dt}=kv\frac{\mathbf{r_s}-\mathbf{r_c}}{|\mathbf{r_s}-\mathbf{r_c}|}[/tex] My logic behind this was that speed of the coastguard is some multiple of the speed of the smugglers and its velocity was in the direction of their relative displacement. Then someone recommended to set [itex]\mathbf{r}=\mathbf{r}_c-\mathbf{r}_s[/itex] and rewrite the equation as [itex]\dot{\mathbf{r}}=-kv\hat{r} - \dot{\mathbf{r}_s}[/itex]. I set [itex]\mathbf{r_s}=\left[ \begin{array}{ccc}0\\v\end{array}\right]t[/itex] and at [itex]t=0[/itex], [itex]\mathbf{r_c}=\left[\begin{array}{ccc}a\\0\end{array}\right][/itex]. I know there are other, probably better ways of solving this problem by finding components of velocity, but I wanted to try and solve it this way.
     
  15. Nov 26, 2017 #14

    Mark44

    Staff: Mentor

    I don't have time to look at this now, so maybe someone else will chime in. A possible way out is to get rid of the unit vector in your DE, since that's where the bottleneck is.
     
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