shootforthestars
  • shootforthestars
"Minimizing the Total Time" Suk wants to go from point A to point B, as shown in the figure. Point B is 1 mile upstream and on the north side of the 0.4-mi-wide river. The speed of the current is 0.5 mph. Her boat will travel 4 mph in still water. When she gets to the north side she will travel by bicycle at 6 mph. Let (alpha) represent the bearing of her course and (beta) represent the drift angle, as shown in the figure. A) When (alpha)=12degrees, find (beta) and the actual speed of the boat. B) Find the total time for the trip when (alpha)=12degrees
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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shootforthestars
  • shootforthestars
C) Write the total time for the trip as a function of (alpha) and graph the function D) Find the approximate angle (alpha) that minimizes the total time for the trip and find the corresponding (beta)
amistre64
  • amistre64
do you have a best attempt at a picture?
shootforthestars
  • shootforthestars
|dw:1446400706262:dw|

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shootforthestars
  • shootforthestars
The answers are supposed to be: A) 7.02 and 3.87 B)15.5 C)T=( 1/((10cos[(alpha)+arcsin((cos(alpha))/8))) - ((1-.4tan(alpha))/6) D) 31.9 and 6.1
amistre64
  • amistre64
your drift is going to be on the other side of alpha, since your going upstream the current pushes you down stream
amistre64
  • amistre64
|dw:1446401437861:dw|
amistre64
  • amistre64
|dw:1446401524973:dw|
amistre64
  • amistre64
how would you spose we determine the drift angle now?
shootforthestars
  • shootforthestars
Would I have to use sin?
amistre64
  • amistre64
sin is a good start yes, sin(12) = x/4
shootforthestars
  • shootforthestars
so, that means that x would be 0.83?
amistre64
  • amistre64
good, and we know .5 of that is removed by the current, so whats left over? .33 right? but we might want to know the distance of the other leg .... x^2 + y^2 = 4^2 y = sqrt(4^2 - x^2) what is y now?
amistre64
  • amistre64
|dw:1446402127584:dw|
shootforthestars
  • shootforthestars
would it be 3.91?
amistre64
  • amistre64
another way to approach it is cos(12) = y/4 y = 4cos(12) = 3.91 yes now we can obtain the angle for theta, given that: theta+beta = 12 we can solve for beta
amistre64
  • amistre64
what trig can we inverse if we know the length of the legs?
shootforthestars
  • shootforthestars
cosine?
amistre64
  • amistre64
sin and cosine each use 1 leg ... and a hypotenuse. tangent uses both legs
shootforthestars
  • shootforthestars
ok
amistre64
  • amistre64
tan(theta) = (x-.5)/y and beta = 12-theta
shootforthestars
  • shootforthestars
beta would be 11.92?
amistre64
  • amistre64
so far x = 4sin(12) y = 4cos(12) tan(theta) = (4sin(12)-.5)/(4cos(12)) theta = atan((4sin(12)-.5)/(4cos(12))) ~= 4.85 degrees beta ~= 12 - 4.85
amistre64
  • amistre64
7.15 is close to your 'answers' all depending on when and how you approxiate
amistre64
  • amistre64
not sure why you have 2 answers provided for part A tho. any thoughts?
shootforthestars
  • shootforthestars
The other number is supposed to be the actual speed of the boat
amistre64
  • amistre64
oh, then we can determine that from theta, cos(theta) = y/s s = y/cos(theta)
amistre64
  • amistre64
with my approximation, that gives me a speed of 3.92
shootforthestars
  • shootforthestars
would the answer be 4?
shootforthestars
  • shootforthestars
oh ok
amistre64
  • amistre64
we could do pythag if we wanted to s^2 = (.33)^2 + (3.91)^2 s = 3.92 ...
amistre64
  • amistre64
let me know if this makes sense or not
shootforthestars
  • shootforthestars
ya, that makes more sense to me, thanks
amistre64
  • amistre64
so, the total time it takes to make the trip, distance/speed what are your thoughts for a solution process?
shootforthestars
  • shootforthestars
Well alpha is 12 and beta is 7.15, so the angle will be 19.15..?
shootforthestars
  • shootforthestars
I don't know where to begin
amistre64
  • amistre64
i dont know the definition of a 'drift angle' right off the top of my head. it is either the angle that displaces us away from 12 degrees, or it is the angle we would have taken to get to the same point .... but im guessing from your 'stated answers' that it is the displacement we get |dw:1446403844045:dw|
amistre64
  • amistre64
if we wanted to get to the same point in still water, that we arrived at going against the current; then the angle that an outside viewer would have seen us take is the relative angle (my own term). the displacement angle, is how far off course we move by the current, and is the error in our percieved approach
shootforthestars
  • shootforthestars
ok
amistre64
  • amistre64
if we perceived ourselves moving at an angle of 12 degrees, then we are getting displaced by 7.15 degrees off course. if someone was just up on a hill watching us, they would have seen us traveling (from their relative perspective) at an angle of 12-7.15 since 7.15 is close to your stated answers to start with, id assume 'beta' is our displacement angle
amistre64
  • amistre64
if t=1 time unit (an hour), then we would already have left the river, seeing that we would have come a distance of 3.91 miles across it, instead of .4 miles ... we need to adjust it by a certain scale (use similar triangles) to determine how far we have traveled |dw:1446404313441:dw|
amistre64
  • amistre64
can you think of a way to determine d1, and n ? we will need them both and then we will need to define them in terms of 12 degrees
amistre64
  • amistre64
forget that little 12 in the angle part ... its a slip of the brain at this point :)
shootforthestars
  • shootforthestars
d1 = .4 and n = 0.034 ?
amistre64
  • amistre64
we have a ratio of 3.91 to .4 3.91 * k = .4 ; when k= .4/3.91 is our scalar so d1 = 3.92*.4/3.91 = .401 or so yes n = .33*.4/3.91 = .034 yes so time taken is distance, divided by speed; and our speed is 3.92 (relatively speaking)
amistre64
  • amistre64
|dw:1446404869504:dw| time = d1/s + d2/6
amistre64
  • amistre64
hunh, yeah, or speed for d1 simplifies to k :)
amistre64
  • amistre64
our time for d1 that is ...
shootforthestars
  • shootforthestars
well, d1 is .401, right?
amistre64
  • amistre64
no, d1 = 3.92*k = 3.92*.4/3.91 d1/3.92 = k = .4/3.91
amistre64
  • amistre64
.102 hours is the time needed to cross d1, given we started at an angle of 12 degree 1-n is the distance needed (d2) traveling at 6mph so the time needed to finish that part is going to be (1-.034)/6
shootforthestars
  • shootforthestars
Ok, I got .161 for that
amistre64
  • amistre64
total time is the sum of the times of each part :) and then convert hours to minutes if wanted
amistre64
  • amistre64
.263 of an hour is; .263 of 60 minutes ... for about 16 minutes
amistre64
  • amistre64
so, the tricky part now, is constructing a suitable equation from what we have worked thru using 12 degrees
amistre64
  • amistre64
\[time_{sec}=60*\left(\frac{d_1}{s_1}+\frac{d_2}{s_2}\right)\] etc ...
shootforthestars
  • shootforthestars
wait, for part b, how did you get .263?
amistre64
  • amistre64
well, we covered how to find the times ... then add up the times
amistre64
  • amistre64
d1/3.92 = .102 d2/6 = .161 ----------------- adds to: .263
shootforthestars
  • shootforthestars
ok, so how would I get 15.5 from there?
amistre64
  • amistre64
.263 of an hour, well an hour is equal to 60 minutes so its the same ratio .... .263 of 60 minutes. what is .263*(60 minutes) ?
amistre64
  • amistre64
it isnt going to be 15.5, but its pretty close assuming differences in approximations
shootforthestars
  • shootforthestars
oh! ok, so 15.78, right?
amistre64
  • amistre64
.26*60 = 15.6 so yeah, we are in the same ball park id say
shootforthestars
  • shootforthestars
ok, thanks!
amistre64
  • amistre64
youre welcome
shootforthestars
  • shootforthestars
so for part c, you said: timesec=60∗(d1s1+d2s2)
shootforthestars
  • shootforthestars
how would I go about the rest?
amistre64
  • amistre64
you simply define the terms with reference to the 12 degrees that we started out from, and then youd generalize it by letting rewriting 12 as alpha
amistre64
  • amistre64
but, i cant really say that i have a clear picture in my head. can you screenshot the figure? or attach a picture file of it?
amistre64
  • amistre64
|dw:1446409102346:dw|
shootforthestars
  • shootforthestars
I will try
shootforthestars
  • shootforthestars
amistre64
  • amistre64
press the [attach file] button, and then pic the file from the dialog .. you got it
amistre64
  • amistre64
|dw:1446409318281:dw| so this one looks like it conforms best; 12 degrees from straight across, as oppose to from shore
shootforthestars
  • shootforthestars
ok
amistre64
  • amistre64
the formula is just writing out what we did along the way really determine a distance triangle of at a 1 hour time interval scale it to match the distances required of the river |dw:1446409493031:dw|
shootforthestars
  • shootforthestars
ok, so it is just what we have been doing?
amistre64
  • amistre64
x = 4sin(12) y = 4cos(12) |dw:1446409634388:dw| yeah, its been what we did to come to the values we used
amistre64
  • amistre64
\[T=t_1+t_2\] \[T=\frac{d_1}{s_1}+{d_2}{s_2}\] \[\frac{s_1}{4cos(12)}=\frac{d_1}{.4}\] \[\frac{.4}{4cos(12)}=\frac{d_1}{s_1}=t_1\] \[\frac{4sin(12)-.5}{4cos(12)}=\frac{n_1}{.4}\] \[\frac{.4(4sin(12)-.5)}{4cos(12)}=n_1\] \[d_2=1-n_1=1-\frac{.4(4sin(12)-.5)}{4cos(12)}\] \[\frac{d_2}{s_2}=\frac{1-\frac{.4(4sin(12)-.5)}{4cos(12)}}{6}\]
amistre64
  • amistre64
lets see if that works 60*((.4)/(4cos(12)) + (1-(.4(4sin(12)-.5))/(4cos(12)))/(6)) http://www.wolframalpha.com/input/?i=60*%28%28.4%29%2F%284cos%2812%29%29+%2B+%281-%28.4%284sin%2812%29-.5%29%29%2F%284cos%2812%29%29%29%2F%286%29%29
amistre64
  • amistre64
simplfy to your hearts content and replace 12 by alpha, then take the derivative to find minimalized times
shootforthestars
  • shootforthestars
ok, that will work i think
amistre64
  • amistre64
with any luck :)
shootforthestars
  • shootforthestars
Ok, so now all I have left is problem D

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