Two cars start from the same point at the same time. One travels north at 25 mph, and the other travels east at 60 mph. How fast is the distance between them increasing at the end of 1 hr? Find D after 1 Hr, solve D^2=25^2+ 60^2

simple way:

Let the time passed since they left be t hours
So distance traveled by northbound car is 25t miles
distance traveled by eastbound car is 60 t miles

Now make a sketch calling the distance between them D
D^2 = (25t)^2 + (60)t^2 = 4224t^2
D = √4224 t
dD/dt = √4224 = 65 mph

or (your way)

D^2 = x^2 + y^2
when t = 1 , x=60, y=25
D^2 = 3600 + 625 = 4224
D = √4224

2D dD/dt = 2x dx/dt + 2y dy/dt
dD/dt = (x dx/dt + y dy/dt)/D

when t=1 , D = √4224 , dx/dt = 60 , dy/dt = 25 , x=60 , y = 25

dD/dt = (60(60) + 25(25))/√4225
= 4225/√4225
= 65

Often the choice of variables will simplify or complicate your solution. Notice that my way not only gave us the answer more quickly, but it also pointed out an interesting fact.
The rate does not even depend on the time, the two cars would be separating at 65 mph at any given time of t.

Nice to see that we both do these in that simple way.

Nice catch on the constant speed of separation. I hadn't noticed that, and was vaguely surprised that the 65 popped up again at the end.

you had nothing about the angle of separation. can you please answer this question: two vehicles leave point c travelling at 60km/h the angle between the two points is 60 degrees how far apart will A and B be after 5 hours.

To find the distance between the two cars after 1 hour, we need to consider their velocities and the Pythagorean theorem.

Let's assume the starting point is the origin (0,0) on a coordinate plane. The car traveling north has a constant velocity of 25 mph, so after 1 hour, it will be 25 miles north of the origin, which can be represented by the point (0,25).

Similarly, the car traveling east has a constant velocity of 60 mph, so after 1 hour, it will be 60 miles east of the origin, which can be represented by the point (60,0).

Using the Pythagorean theorem, we can find the distance between these two points (0,25) and (60,0):

d^2 = (change in x)^2 + (change in y)^2

d^2 = (60 - 0)^2 + (0 - 25)^2
d^2 = 3600 + 625
d^2 = 4225

Taking the square root of both sides, we find:

d = sqrt(4225)
d = 65

So, after 1 hour, the distance between the two cars is 65 miles.

To find the rate at which the distance is increasing, we can differentiate the equation with respect to time (t):

d^2 = (25t)^2 + (60t)^2

2d * dd/dt = 2(25t)(25) + 2(60t)(60)

dd/dt = (25t)(25) + (60t)(60) / d

Plugging in t = 1 (since we're looking for the rate after 1 hour) and d = 65 (from the previous calculation), we get:

dd/dt = (25)(25) + (60)(60) / 65
dd/dt = 625 + 3600 / 65
dd/dt = 4225 / 65
dd/dt = 65 mph

Therefore, the distance between the two cars is increasing at a rate of 65 mph after 1 hour.

if the cars start at (0,0), then their positions at time t are

one: (0,25t)
two: (60t,0)

distance d:

d^2 = (60t)^2 + (25t)^2

when t=1,

d^2 = 3600+625 = 4225
d = 65

2d dd/dt = 2(60t)(60) + 2(25t)(25)
at t=1,

2(65) dd/dt = 7200+1250
dd/dt = 65 mph