calculus

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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.

• calculus -

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

• calculus -

your work doesn't match the wording of the question.
Where does it mention a north-walking 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 north-east)
(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 'going-east' line has length 3t
the 'north-east' 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

• calculus -

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?

• calculus -

where does a direction of North-East go in relation to East ?
What is the angle between North and East?

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