calculus
posted by Anonymous .
Could someone help me et up this problem?
Two people start from the same point. one walks east at 3mi/h and the other walks northeast at 2mi/h. How fast is the distance between the poeple changing after 15 minutes?
I think this is what I would do:::
First I would set up a triangle:
hypotenuse=h (northeast walking person)
height=y
leg=x (north walking person)
GIVEN: dx/dt=3mi/h
dh/dt=2mi/h
Need to find dy/dt when t=15.

East problem
East walker at 3 mi/hr
NE walker at 2 cos 45 mi/hr
East distance between = (3 1.41)t = 1.59 t
North distance between = 1.41 t
h = hypotenuse = t sqrt(1.59^2 + 1.41^2)
= 2.13 t
now find dh/dt
dh/dt = 2.13 mi/hr 
your work doesn't match the wording of the question.
Where does it mention a northwalking person? (your x)
There isn't even a right angled triangle
I drew a horizontal line to the east (to the right)
then from the same starting point a line at 45º to the other one (going northeast)
(make the horizontal a little longer, it represents 3 mph)
let the distance between their endpoints be x
let the time after they both leave be t hours.
then the 'goingeast' line has length 3t
the 'northeast' line is 2t
by Cosine Law
x^2 = (3t)^2 + (2t)^2  2(3t)(2t)cos 45º
= 13t^2  12t^2cos45
= 13t^2  6√2 t^2
dx/dt = 26t  12√2t
when t = 15 min = 1/4 hour
dx/dt = 26(1/4)  12√2(1/4)
= 2.26 mph 
thanx I thought I had to use the cosine law but I wasn't quite sure how to.
But how did you decide that the angle would be 45 degrees? 
where does a direction of NorthEast go in relation to East ?
What is the angle between North and East?