Please help me with this rate of change problem!!! It's very interesting! :P?

Question

Two cards are travelling along perpendicular roads, car A at 40mph, car B at 60mph. At Noon, when car A reaches the intersection, car B is 90 miles away, and moving toward it. At 1 PM the distance between the cars is changing, in mph, at the rate of what?

Please show your work.

Answer 1

At noon
A = (0, 0), B = (-90, 0)

At time t hours
A = (0, 40t) B = (-90 + 60t, 0)

So |AB| = √[(40t)² + (-90 + 60t)²]
= 10√[16t² + 81 - 108t + 36t²]
= 10√[52t² - 108t + 81]
d|AB|/dt = 5(104t - 108)/√[52t² - 108t + 81]

When t = 1 d|AB|/dt = 5 * (-4)/√[52 - 108 + 81]
= -20/√25
= -4 mph ie they are getting closer at a speed of 4 mph

Answer 2

To figure the rate, you need to get an equation of the distance between the cars as a function of time. The derivative of this is the rate fo change of the distance.

Call noon time zero (t=0) and express t in hours. If car A is moving in the plus x direction from the intersection (origin) its position is:

Ax = 40t, Ay = 0

At time zero, car B is 90 miles away perpendicular to the x-axis. At t = 0, y = 90. Since it approaches the origin at 60 mph, its y distance decreases 60 miles per hour. So the location over time is:

Bx = 0, By = 90 - 60t

The distance between the two, by the Pythagorean theorem, is:

D = sqrt((Ax - Bx)^2 + (Ay - By)^2) = sqrt((40 t)^2 + (90 - 60 t)^2)

Expanding:

D = sqrt(5200*t^2 - 10800*t + 8100)

Taking the derivative with respect to time:

D' = (520*t - 540)/sqrt(52*t^2 - 108*t + 81)

At 1 PM t = 1 so:

D' = (520 - 540)/sqrt(52 - 108 + 81) = -4 mph.

Answer 3

Make a right triangle

One leg has a length of 90 - 60t, the other leg has a length of 40t

The distance squared between the cars at any particular time t is
s^2 = (90-60t)^2 + (40t)^2

Now take the derivative

2s ds/dt = 2 (90-60t) (-60) + 3200 t
2s ds/dt = 2 (-5400 + 3600 t ) + 3200 t

When t = 1 s = 50

100 ds/dt = -10800 + 7200 + 3200 = -400

ds/dt = -4 mph

Answer 4

Let card A be at point [0,0] at t=0, let card B be at point [0,-90] at t=0,
Now d=sqrt((Va*t)^2 + (-90+Vb*t)^2), where Va=40, Vb=60, t = any time >0;
d’=(Va^2*t+(-90+Vb*t)*Vb)/d = {for t=1}= (40*40+(-90+60)*60)/sqrt(40*40+30*30) =
= -4mph; yours is card A – 6 spades, mine card B – ace of spades.