Jack and Jill are out walking around in a field. At noon, Jack is located 800 feet west of Jill. Jack walks east at the rate of 5 feet/sec. while Jill walks north at 4 feet/sec. How fast is the distance between them changing at 12:02 p.m.?
At a time of t seconds after noon,
Jack will have walked 5t ft, while Jill will have walked 4t ft
let the distance between them be d ft
I see a right angled triangle with a base of
800-5t, a height of 4 t and a hypotenuse of d
d^2 = (800-5t)^2 + (4t)^2
2d dd/dt = 2(800-5t)(-5) + 2(4t)(4)
at 12:02 , t = 2
d^2 = 624164
d = appr 790.04
dd/dt = (2(790)(-5) + 64)/(2(790.04))
= -4.959..
the distance is DECREASING at appr 4.96 ft/second at that moment
=
To find the rate at which the distance between Jack and Jill is changing, we can use the concept of the derivative.
Let's denote the distance between Jack and Jill at any given time as "d". We can break down this distance into two components: the northward distance "y" and the eastward distance "x".
At any time t, we have the following relationships:
x^2 + y^2 = d^2 -- (1) (applying the Pythagorean theorem)
x = 800 - 5t -- (2) (Jack's position as a function of time)
y = 4t -- (3) (Jill's position as a function of time)
We want to find dd/dt, the rate at which d is changing with respect to time t, at 12:02 p.m. (which is 2 minutes after noon).
First, let's find x and y at 12:00 p.m. (noon):
x = 800 - 5t
= 800 - 5(0)
= 800 feet
y = 4t
= 4(0)
= 0 feet
Using equations (1), (2), and (3), we can find d at 12:00 p.m.:
800^2 + 0^2 = d^2
640,000 = d^2
d ≈ 800 feet
Now, let's find x and y at 12:02 p.m.:
x = 800 - 5t
= 800 - 5(2/60)
= 799.6667 feet
y = 4t
= 4(2/60)
= 0.1333 feet
Using equations (1), (2), and (3), we can find d at 12:02 p.m.:
(799.6667)^2 + (0.1333)^2 = d^2
639,466.7222 = d^2
d ≈ 799.6667 feet
Now, let's differentiate equation (1) with respect to time t:
2*x*dx/dt + 2*y*dy/dt = 2*d*dd/dt
Since dx/dt = -5 (Jack is moving eastward), dy/dt = 4 (Jill is moving northward), and we want to find dd/dt, we can rearrange the equation:
2*-5*799.6667 + 2*4*0.1333 = 2*799.6667*dd/dt
Simplifying this equation:
-7996.667 + 0.5332 = 1599.3334*dd/dt
-7996.1338 = 1599.3334*dd/dt
Finally, let's solve for dd/dt:
dd/dt = -7996.1338 / 1599.3334
dd/dt ≈ -5 feet/sec
Therefore, at 12:02 p.m., the distance between Jack and Jill is changing at a rate of approximately -5 feet/sec. The negative sign indicates that the distance is decreasing.