Application of calculus rates on trucks and cars. Willing to help?

Question

At noon on a certain day, a truck is 250 miles due east of a car. The truck is travelling west at a constant speed of 25 mi/hr while the car is travelling north at 50 mi/hr.

a. What rate is the distance between them changing after four hours?
b. At what time is the distance between the car and the truck neither increasing nor decreasing?

Answer 1

Draw a coordinate system with x-axis from east to west and y-axis from south to north. Let the car start at the origin. It travels along the y-axis with
y = 50*t
The truck travels along the x-axis, starting 250 miles left the origin, hence:
x = -250+25*t
The points of car and truck and the origin form a right triangle, with the hypotenuse representing the distance between car and truck. The momentary distance is:
s = sqrt[x²+y²]
= sqrt[(25t-250)²+(50t)²]
= 25*sqrt[5t²-20t+100]
with s in miles, t in hours

a.
Calculate the time derivative of the distance:
s' = ds/dt = 25/2 * (10t-20) / sqrt[5t²-20t+100]
for t = 4 hr
s'= 2 mi/hr²

b.
That means the rate of distance change is zero
s' = 0
<=>
25/2 * (10t-20) / sqrt[5t²-20t+100] = 0
=>
(10t-20) = 0
=>
t = 2hr

Answer 2

Let
x = east/west distance between car and truck
y = north/south distance between car and truck
z = distance between car and truck
t = time in hours

Given

dx/dt = -25 miles/hour
dy/dt = 50 miles/hour
x = 250 at t = 0
y = 0 at t = 0

Find
dz/dt when t = 4
t when dz/dt = 0

By the Pythagorean Theorem

At t = 4
x = 250 - 25t = 250 - 25*4 = 150
y = 50t = 50*4 = 200
z² = x² + y² = 150² + 200² = 62500
z = 250

dz/dt = (dz/dx)(dx/dt) + (dz/dy)(dy/dt)
2z(dz/dt) = 2x(-25) + 2y(50)
dz/dt = (-25x + 50y)/z = (-25*150 + 50*200)/250
= (-25*3 + 50*4)/5 = 25 miles/hour
__________________________

dz/dt = (-25x + 50y)/z = 0
-25x + 50y = 0
25x = 50y
x = 2y

x = 250 - 25t
y = 50t

250 - 25t = 2(50t) = 100t
250 = 125t
t = 2 hours

Answer 3

c^2 = a^2 + b^2

c= s = distance between the vehicles

a = 250mi + 25mph*T
b = 0mi + 50mph*T

square both a&b... add them together then find the derivative of the distance equation

2c' = 2a'+2b'
and solve for the c'(T)

plug-in T=4 for rate at 4 hours
plug-in 0= a' +b' to calculate the T at which the rate is 0.

Answer 4

a)
4*25=100
4*50=200
d=sqrt(40000+22500)
d=250 miles