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
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 North-East go in relation to East ?
What is the angle between North and East?
To solve this problem, we need to use the concepts of related rates. We will differentiate the equation that relates the three sides of the triangle, which is the Pythagorean theorem:
h^2 = x^2 + y^2
First, let's differentiate both sides of the equation with respect to time (t):
2h * dh/dt = 2x * dx/dt + 2y * dy/dt
Given that dx/dt = 3 mi/h and dh/dt = 2 mi/h, we can substitute these values into the equation:
2h * 2 = 2x * 3 + 2y * dy/dt
Simplifying the equation, we have:
4h = 6x + 2y * dy/dt
We also know that the initial conditions are h = 0 (since both people start from the same point), x = 0 (since the person walking north does not move horizontally), and h = 15 minutes. We can substitute these values into the equation:
4(15) = 6(0) + 2y * dy/dt
60 = 2y * dy/dt
To find dy/dt, the rate at which the distance between the people is changing after 15 minutes, we need to solve for it. Rearranging the equation:
dy/dt = 60 / (2y)
Now, we only need to calculate the value of y at t = 15 minutes to find the rate of change of distance between the two people. This involves finding the values of x and y using the given information.
Since dx/dt = 3 mi/h and the time t = 15 minutes, we can calculate that x = dx/dt * t = 3 mi/h * (15 min/60 min) = 0.75 miles.
Now, using the Pythagorean theorem, we can find the value of y:
h^2 = x^2 + y^2
(0.75)^2 + y^2 = 15^2
0.5625 + y^2 = 225
y^2 = 224.4375
y ≈ 14.98 miles
Substituting this value of y into the equation for dy/dt:
dy/dt = 60 / (2 * 14.98)
dy/dt ≈ 2 mi/h
Thus, the rate at which the distance between the two people is changing after 15 minutes is approximately 2 miles per hour.
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