Time Required to Go from an Island to a Town.

[robthebear]

Time Required to Go from an Island to a Town. An island is 3 miles
from the nearest point P on a straight shoreline. A town is located
20 miles down the shore from P.

(a) A person has a boat that averages 12 miles per hour and the same
person can run 5 miles per hour. Express the time T that is takes to
go from the island to town, as a function of x, were x is the
distance from P to where the person lands the boat. Give the domain.

(b) How long will it take to travel from the island to town if you
land the boat 8 miles from P?

(c) How long will it take if you land the boat 12 miles from P?

(d) Graph the function T=T (x).

(e) Create A TABLE with TBLstart = 0 and (increment sign,
Triangle symbol) Tbl = 1. To the nearest mile, determine which value
of x results in the least time.

(f) Using MINIMUM, what value of x results in the least time?

(g) The least time occurs by heading directly to town from the
island. Explain why this solution makes sense.
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Answers
a) T (x) = (20-x)/5 + sqrt(9+x^2)/12 Domain: {x! 0(< or =)x(< or =)20}
b) 3.1 hours
c) 2.6 hours
d) Graph of T (x) = (20-x)/5 + (sprt 9+x^2)/12
e) x = 20 miles
f) x = 20 miles
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part A: I have been pulling my hair out trying to get this part. I
just do not see it...

I tried to do the e and f part but I do not understand it eather.

part E: I put the T(x) in my ti-83 for the Y= and then i set my
TBLset and then use the table. I do not understand the answer. On my
Table when x=20 my Y1 = 1.6853 not the one mile it is asking.
Because when x = 25 my table says Y1 = 1.0983 and that seems like it
would be closer to one then the answer my books gives.

part F: The book says that x = 20 miles for this one. Ok so if you
take the derivative and set equal to zero to find the min.

I would

(20-x)/5 + (sqrt 9+x^2)/12 = 0

(20-x)/5 = -(sqrt 9+x^2)/12

12(20-x) = 5(-3-x)

240 - 12x = -15 + -5x

255 = 7x

36.4286 = x

I am still not getting the 20 miles the books is getting... am i
right or is the book?

But when you put it in the table on my TI-83 you get x = 36.4286 and
your Y1 = -.2397

 

[ting]

For, a.

'An island is 3 miles from the nearest point P on a straight shoreline. '

This tell you that the perpendicular distance from the island to point P on the shore line is 3 miles.

Let the place where the boat lands be L.

The distance from P to L is x.

The distance from L to the town is 20-x.

You have a right-angled triangle so using pythagoras' theorem, the distance the boat travels from
the island to L is, sqrt(3^2 + x^2).

Time = distance/ rate.

T (x) = (20-x)/5 + sqrt(9+x^2)/12

X is the distance from be the shotest distance is 0 and the greatest is 20.

For b and c, substitute the values in and solve or put the function into your calculator and find the values of y when x = 8 and when x = 12.

Can you do the rest?

 

[ting]

'I do not understand the answer. On my
Table when x=20 my Y1 = 1.6853'


This is correct, the y means it takes 1.6.. hours. This is the shortest time. When x = 20 you are travelling directly from the island to the town.

 

[Denis]

"part E: I put the T(x) in my ti-83 for the Y= and then i set my
TBLset and then use the table. I do not understand the answer. On my
Table when x=20 my Y1 = 1.6853 not the one mile it is asking.
Because when x = 25 my table says Y1 = 1.0983 and that seems like it
would be closer to one then the answer my books gives."

If x=25, then your table should say 3.0983;
takes 2.0983 to travel to point 25 by boat;
then you're 5 miles past the Town: takes 1 hour to walk back!
You need to use the absolute function with 20-x.


"part F: The book says that x = 20 miles for this one. Ok so if you
take the derivative and set equal to zero to find the min.
I would
(20-x)/5 + (sqrt 9+x^2)/12 = 0
(20-x)/5 = -(sqrt 9+x^2)/12
12(20-x) = 5(-3-x)"

how in heck did -sqrt(9+x^2) become (-3-x)?!
(-3-x)^2 = 9 + 6x + x^2