Expert: Scotto - 11/19/2008

Question
A man is walking on a bridge that is 20 ft above a river, as a boat passes directly under the center of the bridge(at a rt angle to the bridge) at 10 feet per Second. At that moment the man is 50 feet from the center and approaching it at a rate of 5 feet per second.

a) At what rate is the distance Between the boat and the man changing at that moment?

b) Is the rate at which they are approaching or separating increasing or is it decreasing?

THANK YOU SO MUCH SERIOUSLY you are amazing if you can step by step me through this

Answer
Boat: directly below at 10ft/s, so x=0 and dx/dt = 10 (x is how far up the river I can find the boat).

Man on bridge: 50 ft from center, speed equals 5 ft/s, so y=50 and dy/dt = -5 (y is how far from the center of the river).

Bridge: 20 feet over water, so I will let z=20 (z is the height).  Note that dz/dt is 0 because the height is not changing.

D = √(x²+y²+z²).

dD/dt = (2x dx/dt + 2y dy/dt + 2z dz/dt)/[0.5√(x²+y²+z²)].

The first term 2x dx/dt is 0 since x is 0.
The second term, 2y dy/dt is 2(-5)(50).
The third term is zero since dz/dt is 0.

a) The change of distance, dD/dt, is then 2(-5)(50)/[0.5(√(50²+20²)].

b) At some given time t, x(t) is 10t, y(t) is 50 - 5t, and z(t) is 20 with no change.

The total distance was given in the first problem.

The equations for x(t), y(t), and z(t) can be put to get an equation for D(t).  The derivative was also found in A and this is the speed at which the difference between the two is changing.

Various values of t can be put into the total distance equation to find the distance between them.  Various values of t can be put into the rate of change equation to find whether the distance between them is increasing or decreasing.