"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

- shootforthestars

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- 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

do you have a best attempt at a picture?

- shootforthestars

|dw:1446400706262:dw|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- 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

your drift is going to be on the other side of alpha, since your going upstream the current pushes you down stream

- amistre64

|dw:1446401437861:dw|

- amistre64

|dw:1446401524973:dw|

- amistre64

how would you spose we determine the drift angle now?

- shootforthestars

Would I have to use sin?

- amistre64

sin is a good start yes, sin(12) = x/4

- shootforthestars

so, that means that x would be 0.83?

- 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

|dw:1446402127584:dw|

- shootforthestars

would it be 3.91?

- 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

what trig can we inverse if we know the length of the legs?

- shootforthestars

cosine?

- amistre64

sin and cosine each use 1 leg ... and a hypotenuse.
tangent uses both legs

- shootforthestars

ok

- amistre64

tan(theta) = (x-.5)/y
and beta = 12-theta

- shootforthestars

beta would be 11.92?

- 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

7.15 is close to your 'answers' all depending on when and how you approxiate

- amistre64

not sure why you have 2 answers provided for part A tho. any thoughts?

- shootforthestars

The other number is supposed to be the actual speed of the boat

- amistre64

oh, then we can determine that from theta,
cos(theta) = y/s
s = y/cos(theta)

- amistre64

with my approximation, that gives me a speed of 3.92

- shootforthestars

would the answer be 4?

- shootforthestars

oh ok

- amistre64

we could do pythag if we wanted to
s^2 = (.33)^2 + (3.91)^2
s = 3.92 ...

- amistre64

let me know if this makes sense or not

- shootforthestars

ya, that makes more sense to me, thanks

- amistre64

so, the total time it takes to make the trip, distance/speed
what are your thoughts for a solution process?

- shootforthestars

Well alpha is 12 and beta is 7.15, so the angle will be 19.15..?

- shootforthestars

I don't know where to begin

- 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

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

ok

- 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

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

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

forget that little 12 in the angle part ... its a slip of the brain at this point :)

- shootforthestars

d1 = .4 and n = 0.034 ?

- 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

|dw:1446404869504:dw|
time = d1/s + d2/6

- amistre64

hunh, yeah, or speed for d1 simplifies to k :)

- amistre64

our time for d1 that is ...

- shootforthestars

well, d1 is .401, right?

- amistre64

no, d1 = 3.92*k = 3.92*.4/3.91
d1/3.92 = k = .4/3.91

- 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

Ok, I got .161 for that

- amistre64

total time is the sum of the times of each part :)
and then convert hours to minutes if wanted

- amistre64

.263 of an hour is; .263 of 60 minutes ... for about 16 minutes

- amistre64

so, the tricky part now, is constructing a suitable equation from what we have worked thru using 12 degrees

- amistre64

\[time_{sec}=60*\left(\frac{d_1}{s_1}+\frac{d_2}{s_2}\right)\]
etc ...

- shootforthestars

wait, for part b, how did you get .263?

- amistre64

well, we covered how to find the times ... then add up the times

- amistre64

d1/3.92 = .102
d2/6 = .161
-----------------
adds to: .263

- shootforthestars

ok, so how would I get 15.5 from there?

- 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

it isnt going to be 15.5, but its pretty close assuming differences in approximations

- shootforthestars

oh! ok, so 15.78, right?

- amistre64

.26*60 = 15.6 so yeah, we are in the same ball park id say

- shootforthestars

ok, thanks!

- amistre64

youre welcome

- shootforthestars

so for part c, you said:
timesec=60âˆ—(d1s1+d2s2)

- shootforthestars

how would I go about the rest?

- 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

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

|dw:1446409102346:dw|

- shootforthestars

I will try

- shootforthestars

##### 1 Attachment

- amistre64

press the [attach file] button, and then pic the file from the dialog .. you got it

- amistre64

|dw:1446409318281:dw|
so this one looks like it conforms best; 12 degrees from straight across, as oppose to from shore

- shootforthestars

ok

- 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

ok, so it is just what we have been doing?

- amistre64

x = 4sin(12)
y = 4cos(12)
|dw:1446409634388:dw|
yeah, its been what we did to come to the values we used

- 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

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

simplfy to your hearts content and replace 12 by alpha, then take the derivative to find minimalized times

- shootforthestars

ok, that will work i think

- amistre64

with any luck :)

- shootforthestars

Ok, so now all I have left is problem D

Looking for something else?

Not the answer you are looking for? Search for more explanations.