Solving Distance and Direction-Based Homework Problems

[munkachunka]

Ok

like ...sqrt(1.49t-30.61)^2+(0-73.9-5.6t)^2

 

[learningphysics]

Ok

like ...sqrt(1.49t-30.61)^2+(0-73.9-5.6t)^2

exactly... be careful of the second parentheses... you made a mistake with signs...

that's the formula for distance... we need to find the time when this distance is minimum...

The square root makes this a little messy... so what we can do is find the time when distance^2 is minimum... that's the same as when distance is minimized... this just allows us to get rid of the square root...

Multiply out and simplify the expression for distance^2...

 

[munkachunka]

d^2=1.49^2*t^2-30.61^2+-73.9^2-5.6^2*t^2 =

d^2=2.22*t^2-936.9+-5461.21-31.36*t^2 =

d^2 = -6427.25*t^4

not sure I have done that right beacuse the answer should be positive,

I can't see where the sign is wrong

 

[learningphysics]

distance^2 = (1.49t-30.61)^2+(0-(73.9-5.6t))^2

distance^2 = (1.49t-30.61)^2+(0-73.9+5.6t)^2

distance^2 = (1.49t-30.61)^2 + (-73.9 + 5.6t)^2

distance^2 = (1.49t-30.61)^2 + (5.6t -73.9)^2

You didn't multiply it out correctly... remember:

(a+b)^2 = a^2 + 2ab + b^2
(a-b)^2 = a^2 - 2ab + b^2

 

[munkachunka]

ah yes, my silly mistake, it is getting late here now

distance^2 = (1.49t-30.61)^2 + (5.6t -73.9)^2

==d^2 = 1.49t^2-2*1.49t*30.61+-30.61^2 + 5.6t^2-2*5.6t*-73.9+-73.9^2

= d^2 = 1.49t^2-91.2t+936.9 +5.6t^2+827.7t+5461.21

=d^2 = 7.09t^2+736.5t+6398.11

is this right so far?

I have to get some sleep now, it is 2300 this side of the pond, I look forward to finishing this tomorrow with a fresh mind!

thankyou so much for your help today, it has been great

 

[learningphysics]

Still a few careless mistakes there.

You're welcome. Great for me too. get some sleep. See you tomorrow...

 

[munkachunka]

good morning

I have woken up now and am fresh minded and raring to go.

distance^2 = (1.49t-30.61)^2 + (5.6t -73.9)^2

D^2= 33.56t^2-918.88t+6397.57

where to now?

do I sqrt the right hand side?

 

[learningphysics]

Cool. First thing I want you to notice is the shape of distance^2... it's a parabola... more specifically it's a parabola with a minimum...

Now there are 2 ways to go about finding when distance^2 is a minimum... you can either take the derivative and set it to 0, then solve for t... or you can find the vertex... this would be done by completing the square...

It's up to you which method is preferable.

 

[munkachunka]

hi,

I attempted to solve by completing the square but it all went a bit crazy.

got it to

33.56t^2-918.88t+211085.1=217482.7

I do not know how to convert the left hand side to square form, the times I have done this before I could just see the answer.

I am not sure how to solve using the derivative, do you mean differentiate?

i.e D^2= 33.56t^2-918.88t+6397.57 = D^2=67.12t-918.88
(or would it be 2D=67.12t-918.88)?

 

[learningphysics]

hi,

I attempted to solve by completing the square but it all went a bit crazy.

got it to

33.56t^2-918.88t+211085.1=217482.7

I do not know how to convert the left hand side to square form, the times I have done this before I could just see the answer.

I am not sure how to solve using the derivative, do you mean differentiate?

i.e D^2= 33.56t^2-918.88t+6397.57 = D^2=67.12t-918.88
(or would it be 2D=67.12t-918.88)?

Yeah, let's go with the derivative method... it's much simpler here...

I'd just use a new variable for distance^2... let's say R. so

R = 33.56t^2-918.88t+6397.57

As you calculated:
\frac{dR}{dt} = 67.12t-918.88

The minimum or maximum of something occurs when it's derivative equals 0. So the minimum of R (we know from the shape of the parabola that R doesn't have a maximum... just one minimum) occurs when \frac{dR}{dt} = 0 in other words when 67.12t-918.88 = 0.

If you solve for t... then get R at the t value you get... then you can take the square root to get distance...

The trick we're using here is that the time when distance becomes a minimum is the same time that distance^2 is a minimum...

You could do the whole thing with taking the derivative of D, instead of R = D^2... but the derivative is just messier because of the square root...

 

[munkachunka]

67.12t-918.88 = 0.

therefore t=918.88/67.12

t=13.69

not sure what you mean by "then get R at the t value you get" so close but so far

 

[learningphysics]

67.12t-918.88 = 0.

therefore t=918.88/67.12

t=13.69

not sure what you mean by "then get R at the t value you get" so close but so far

You have the formula for R (ie: D^2) for any time t... plug in t=13.69 and get R... then you can take the square root to get distance...

basically all you need to do now is get the distance at t = 13.69..

 

[munkachunka]

ok think I got it

pluging t=13.67 back into

d^2=22.56t^2-918.88t+6397.57

=d^2=6271.32-12561+6397.57

=d^2=107.89

d=sqrt107.89

d=10.38m

 

[learningphysics]

That's right... Only one small thing... you might be off a little due to not carrying enough decimal places... I get around 10.6...

more accurately this:
33.56t^2-918.88t+6397.57

I think should actually be:
33.58t^2 - 918.90t + 6398.18

so using this I get... t = 13.68, and R = 111.86, and d = 10.58

I just didn't want you to lose points on something small like that after working so hard on this.

 

[munkachunka]

thats great! I can now see what the method is for this type of question, would I be able to get just a few more mins of your time, could you explain why I had to differentiate and set it to 0?, I know the distance formula was used to get the closest point of the 2 vectors and the result of this was a quadratic which we then differentiated and solved for D.
I think my main issue is that I don't understand how the 2 vectors get turned into a parabola which gives the 2 points. does each parabola's side represent 1 vector?

thanks a million

 

[learningphysics]

thats great! I can now see what the method is for this type of question, would I be able to get just a few more mins of your time, could you explain why I had to differentiate and set it to 0?, I know the distance formula was used to get the closest point of the 2 vectors and the result of this was a quadratic which we then differentiated and solved for D.
I think my main issue is that I don't understand how the 2 vectors get turned into a parabola which gives the 2 points. does each parabola's side represent 1 vector?

thanks a million

No prob. Remember the parabola of R vs. t, is nothing to do with the coordinates of the boats... it is distance^2 vs. time... so you should not think that the boats follow a parabolic path or anything like that... it just gives the distance^2 between the two boats at any time t... it just happened to turn out to be a parabola... we didn't actual turn it into a parabola...

Don't think of them as two vectors... think of them as two points... each point representing the position...

So A is at position (1.49t, 0) B is at position (30.61,73.9-5.6t)

does it make sense to you that

distance = sqrt[(1.49t-30.61)^2 + (5.6t -73.9)^2]

gives the distance at any time t

So at this point, we are no longer dealing with (x, y) coordinates of the boats... we are dealing with distance and time...

So if you plotted this, the graph would give the distance between boats A and B at any time t...

The reason I mentioned R vs. t was a parabola was to point out that there indeed as a minimum... A parabola is just a function where the maximum power is 2... So say y = x^2 +3... that's a parabola... parabola's can go upward (when they have a minimum) or downward (when they have a maximum). The way to recognize if it's a parabola with a min or a max is to look at the coefficient of x^2. If it's positive, that means it has a minimum, if it is negative, that means it has a maximum... so y = -2x^2 + 4x + 3 has maximum, not a minimum...

When you have a function say y that varies with x... and you want to find the point when y is a maximum or a minimum... you need to find the point where the derivative of y with respect to x is 0... If this doesn't sound familiar, I highly recommend going through your calculus text and reviewing this concept...

Usually once when we find the points (x,y) where the first derivative is 0... we still need to calculate the second derivative at that point, to find out if it's a maximum or a minimum... but with simple functions like a parabola we don't need to worry about this, because we can recognize instantly from the equation whether it has a maximum or a minimum...

If you've got a calculus text handy, definitely check out this section... they'll probably explain it much better than I could...

 

[learningphysics]

Here's an example of this idea:

http://cs.gmu.edu/cne/modules/dau/calculus/derivatives/deriv_max_frm.html

 

[munkachunka]

i think you explained that very well yourself, thanks again for your time and definatly your patience!

 

[learningphysics]

i think you explained that very well yourself, thanks again for your time and definatly your patience!

No prob. My pleasure.