related rates

[cowgiljl]

a women is walking at a rate of 6 ft/sec across a bridge 40 ft above the water, a man in a boat is immediately below the woman on the bridge , at right angles to the bridge and at the rate of 4 ft /sec. how fast is the man and woman seprating at the end of 10 seconds

what i have figured out is the 2 deminationally not including the 40 ft
dx/dt = 6ft/sec dy/dt = 4 ft/sec dz/dt = ?
z^2 = x^2 + y^2 z = sq root 52 = 7.21

using implicate differenating dz/dt = 1/z (x(dx/dt) + y(dy/dt))

dz/dt = 1/7.21(4(10) +6(10)
= 100 / 7.21
= 13.87 ft/sec

I can drw it diagram but not sure what to do for the 40 ft

thanks joe

[arildno]

The rate of change of the distanceis found as follows:
The position of the woman relative to the man in the boat is:
\vec{r}(t)=6t\vec{i}+4t\vec{j}+40\vec{k}
The distance between them r(t), is given by the Pythagorean theorem:
r(t)=\sqrt{36t^{2}+16t^{2}+1600}
The separation velocity is now given as \frac{dr}{dt}