Observing a submarine

[Randyyy]

a submarine is observed to be at the point (-1,1,-1/4) at t=0 and 45 minutes later it is observed to be at the point (2,4,-1/5). here the water surface is positioned in the xy-plane and z vertically up and with the unit in nautical miles.
Suppose now the submarine goes in a straight path with constant speed.

A: What´s the submarines velocity v and its speed [MATH]\mid V \mid [/MATH].
B: When and where does the submarine reach the surface?

For A my thinking was that [MATH]\Delta \vec{r}=r_1-r_0 \iff \Delta \vec{r} = (3,3,1/20)[/MATH], note that [MATH]r_1; (2,4,-1/5)[/MATH] and [MATH]r_0: (-1,1,-1/4)[/MATH]since Velocity=displacement/time I then have that [MATH]\vec{V}=\dfrac{\Delta r}{t}=\dfrac{\Delta r}{46 \cdot 60}=(1/900,1/900,1/5400)[/MATH]and so [MATH]\mid V \mid = \sqrt{\vec{V}.\vec{V}}=0,00157...[/MATH]
Could that be correct? It feels like the number is too small.

 

[Dr.Peterson]

Where does your 46*60 come from? There are two errors there! (One, I think, is just a typo; the other could conceivably be correct if you said what units you are using for speed!)

Never neglect units in your answers!

 

[Randyyy]

Ah yes, I meant to write 45*60, I thought that I should perhaps use SI-units and convert time to seconds, but I would think that my distance is in nautical miles so I have nautical miles/sec, I think.

 

[Dr.Peterson]

Ah yes, I meant to write 45*60, I thought that I should perhaps use SI-units and convert time to seconds, but I would think that my distance is in nautical miles so I have nautical miles/sec, I think.

Do you see that this is why the number you got was so small?

Speeds of boats are generally given in nautical miles per hour, because that gives more reasonable numbers.

You weren't told what units to use, so what you said is valid, but only if you state the units!

 

[Randyyy]

Yes, that makes perfect sense. If I instead use Nautical miles per hour which makes more sense I would then divide by 3/4 and so the dot product of V then gives me 5.65Naut/hour which seems a bit more what I expected to get initially.

For B: my thinking is that if it moves at a constant speed, and Z is the distance from the surface, then if we at t=0 started at z=-1/4 and then at t=45 we are at Z=-1/5 so we are getting closer to the surface by 0.05 units per 45min, hence the time it should take the submarine to reach the surface is 45*5=225 minutes. and If I follow the linearity of t=0 and t=45 the coordinates should be: (9,15,0).

 

[Dr.Peterson]

For B: my thinking is that if it moves at a constant speed, and Z is the distance from the surface, then if we at t=0 started at z=-1/4 and then at t=45 we are at Z=-1/5 so we are getting closer to the surface by 0.05 units per 45min, hence the time it should take the submarine to reach the surface is 45*5=225 minutes. and If I follow the linearity of t=0 and t=45 the coordinates should be: (9,15,0).

I think you're right, but your explanation wanders too much to be convincing, and you haven't shown your intermediate answers. How about stating the (vector) velocity in nm/h as requested, and using the vertical component to find the time in hours?

 

[Randyyy]

Okay, new attempt!

I know that [MATH]\Delta r[/MATH] really describes how much our position [MATH]\vec{r_0}[/MATH] changes per hour. We can also see this as [MATH]\vec{V}[/MATH] because if t=1 (one hour) we get [MATH]\vec{V}=\Delta r[/MATH]. this now allows us to express our new coordinates with a parameter t as follows: [MATH]\Delta r \cdot t + r_0[/MATH] = [MATH](-1+3t, 1+3t, - \dfrac{1}{4}+\dfrac{t}{20})[/MATH]. Because we want to know when z = 0 we solve the equation [MATH]- \dfrac{1}{4}+\dfrac{t}{20}=0 \iff t=5[/MATH]. Now we plug this into our equation above instead of t and this yields that the coordinate is (14,16,0). The time it takes is then t*45 = 5*45=225 min.

What I find weird is that if t=1 I get [MATH]\vec{V}=\Delta r[/MATH] which is actually (3,3,1/20) but if t=3/4, we had (4,4,1/15) which somehow is faster but somehow that doesn´t make sense to me. I feel like I have a few small errors in my calculations, I just can´t identify them at this very moment.

 

[Dr.Peterson]

Your velocity is in nmi/h, so you should do the work in hours and only convert to minutes, if you wish, at the end.

Part of what I was asking for was your velocity vector, which you haven't yet actually shown, but it's implied by what you do show: (3, 3, 1/20). But that is not the velocity; it's the [MATH]\Delta r[/MATH]. And this is the movement not over an hour, but over 3/4 hour. Try that again.

If you found t = 5, doesn't that mean it surfaces after 5 hours? On what grounds do you multiply by 45 minutes? You're still not using units properly.

Again, I think you got the right answer before, but you weren't using vectors properly to do it, and that is what you are supposed to be learning.

 

[Randyyy]

Okay, let's try this again. [MATH]\vec{V}=\dfrac{\Delta r}{t}[/MATH]. [MATH]\Delta r = r_1-r_0[/MATH] where [MATH]r_0: (-1,1,-\dfrac{1}{4})[/MATH] and [MATH]r_2: (2,4,-\dfrac{1}{5})[/MATH] and the time is 3/4th an hour because this is t=0 and t=45 min. Let´s now find our velocity vector and express it in hours. [MATH]\vec{V}=\dfrac{\Delta r}{3/4} = (4,4,\dfrac{1}{5})[/MATH] and this is nmi/h. Now we now that [MATH]d=r_0+vt[/MATH]. d is the coordinate when we hit the surface. Rewriting using our vectors we get that [MATH]d=r_0+\vec{V} \cdot t \iff d=(-1+4t, 1+4t, -\dfrac{1}{4}+ \dfrac{t}{15})[/MATH]. Because we want to find the time where we hit the surface we now solve the equation [MATH]-\dfrac{1}{4}+ \dfrac{t}{15}=0 \iff t=\dfrac{15}{4}[/MATH] and t is in hours because everything else is in nmi/h. So it takes us 15/4hours to reach the surface. We plug this in now to find our coordinate d.
[MATH]d=r_0+\vec{V} \cdot t \iff d=(14,16,0)[/MATH]
I am not sure how I got my first answer anymore because this to me seems to be correct unless I once again messed up the units but this time I am confident that my units are all correct and that I utilized my vectors properly. Also, I agree, I should not solve this using anything other than vectors because that is what the task was created for. If I take "shortcuts" and solve them any other way I won´t learn how to use vectors properly and will likely struggle a lot moving forward so I appreciate your time and help greatly as always!

 

[Dr.Peterson]

Okay, let's try this again. [MATH]\vec{V}=\dfrac{\Delta r}{t}[/MATH]. [MATH]\Delta r = r_1-r_0[/MATH] where [MATH]r_0: (-1,1,-\dfrac{1}{4})[/MATH] and [MATH]r_2: (2,4,-\dfrac{1}{5})[/MATH] and the time is 3/4th an hour because this is t=0 and t=45 min. Let´s now find our velocity vector and express it in hours. [MATH]\vec{V}=\dfrac{\Delta r}{3/4} = (4,4,\dfrac{1}{5})[/MATH] and this is nmi/h. Now we now that [MATH]d=r_0+vt[/MATH]. d is the coordinate when we hit the surface. Rewriting using our vectors we get that [MATH]d=r_0+\vec{V} \cdot t \iff d=(-1+4t, 1+4t, -\dfrac{1}{4}+ \dfrac{t}{15})[/MATH]. Because we want to find the time where we hit the surface we now solve the equation [MATH]-\dfrac{1}{4}+ \dfrac{t}{15}=0 \iff t=\dfrac{15}{4}[/MATH] and t is in hours because everything else is in nmi/h. So it takes us 15/4hours to reach the surface. We plug this in now to find our coordinate d.
[MATH]d=r_0+\vec{V} \cdot t \iff d=(14,16,0)[/MATH]
I am not sure how I got my first answer anymore because this to me seems to be correct unless I once again messed up the units but this time I am confident that my units are all correct and that I utilized my vectors properly. Also, I agree, I should not solve this using anything other than vectors because that is what the task was created for. If I take "shortcuts" and solve them any other way I won´t learn how to use vectors properly and will likely struggle a lot moving forward so I appreciate your time and help greatly as always!

You've got it.

Your answer of 15/4 hours, of course, is 15/4 * 60 = 225 minutes, which is what you said before.

What you were doing before, in effect, is taking 45 minutes as your unit of time, which would have been valid if you had said what you meant, but is quite awkward.

 

[Randyyy]

You've got it.

Your answer of 15/4 hours, of course, is 15/4 * 60 = 225 minutes, which is what you said before.

What you were doing before, in effect, is taking 45 minutes as your unit of time, which would have been valid if you had said what you meant, but is quite awkward.

Yes, this was much easier and made more sense, thank you once again Dr.Peterson!