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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)
leg=x (north walking person)

GIVEN: dx/dt=3mi/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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