related rates -

[eeelaynne]

We've been working on related rates in class and were assigned a set of
problems to do, but I feel like this one problem is missing some
information, but my teacher said no.
The question is:
At 3 PM a ship which is sailing due north at 12 knots is 5 miles west of a
westbound ship which is making 16 knots.
a) At what rate is the distance between teh ships changing at 3 PM?
b) At what time does the distance between the ships stop decreasing and
start increasing?


I know I have to differentiate the equation c^2 = a^2 + b^2 , and I've done
that, but I don't know how to draw the diagram for this problem. I know it
has to deal with a right triangle, but I can't seem to determine the values
for a, b, and c. I'm not sure if the two ships start level to each other,
meaning there would be no "c" or "b" value, just an
"a" value, being (I'm guessing here) the 5 miles. It doesn't make
sense though when I plug these values into the differentiation of the above
formaula though. Any help would be much appreciated

 

[daon]

We've been working on related rates in class and were assigned a set of
problems to do, but I feel like this one problem is missing some
information, but my teacher said no.
The question is:
At 3 PM a ship which is sailing due north at 12 knots is 5 miles west of a
westbound ship which is making 16 knots.
a) At what rate is the distance between teh ships changing at 3 PM?
b) At what time does the distance between the ships stop decreasing and
start increasing?


I know I have to differentiate the equation c^2 = a^2 + b^2 , and I've done
that, but I don't know how to draw the diagram for this problem. I know it
has to deal with a right triangle, but I can't seem to determine the values
for a, b, and c. I'm not sure if the two ships start level to each other,
meaning there would be no "c" or "b" value, just an
"a" value, being (I'm guessing here) the 5 miles. It doesn't make
sense though when I plug these values into the differentiation of the above
formaula though. Any help would be much appreciated

You know dh/dt is 12 and dx/dt = -16
you need to find d(d)/dt (should have used a different letter..).
note: 3PM is t=0, and i didn't fumble around with the units, so use caution

 

[skeeter]

At 3 PM a ship which is sailing due north at 12 knots is 5 miles west of a
westbound ship which is making 16 knots.
a) At what rate is the distance between teh ships changing at 3 PM?
b) At what time does the distance between the ships stop decreasing and
start increasing?

let x = horizontal distance between the ships
y = vertical distance between the ships
z = straight line distance between the ships

z^2 = x^2 + y^2

2z(dz/dt) = 2x(dx/dt) + 2y(dy/dt)

dz/dt = [x(dx/dt) + y(dy/dt)]/z

dz/dt = [x(dx/dt) + y(dy/dt)]/sqrt(x^2 + y^2)

(a) at t = 0, (3PM) ... x = 5, y = 0, dx/dt = -16 kts, dy/dt = 12 kts
dz/dt = [5(-16) + 0(12)]/sqrt(5^2 + 0^2) = -16 kts

(b) dz/dt = (-16x + 12y)/sqrt(x^2 + y^2)

since sqrt(x^2 + y^2) > 0, dz/dt > 0 when 12y > 16x or y > 4x/3 and
dz/dt < 0 when y < 4x/3.

dx/dt will start to be greater than 0 when y = 4x/3

as a function of time ...

x = 5 - 16t

y = 12t

12t = 4(5 - 16t)/3

36t = 20 - 64t

100t = 20

t = 1/5 hr

dz/dt will start to be positive at time t = 1/5 of an hour, or 3:12 PM

 

[eeelaynne]

as a function of time ...

x = 5 - 16t

y = 12t

12t = 4(5 - 16t)/3

36t = 20 - 64t

100t = 20

t = 1/5 hr

dz/dt will start to be positive at time t = 1/5 of an hour, or 3:12 PM

THank you for all of your help!! I was just wondering if I could ask you about the whole function of time section..I'm just not sure how you go to that point, with all of the substitutions