Hello, Super Mallow!

I use a slightly different set-up . . .

2) A balloon is rising vertically above a level, straight road at a constant rate of 1 ft/sec.

Just when the balloon is 65 feet about the ground,

a bicycle moving at a constant rate of 17 ft/sec passes under it.

How fast is the distance s(t) between the bicycle and the balloon increasing 3 sec later? Code:

B *
↑\
↑ \
t ↑ \
↑ \
↑ \ s(t)
P * \
| \
65 | \
| \
* → → → → *
A 17t C

The balloon is at $\displaystyle P$ when the cycle passes under it (at $\displaystyle A$): .$\displaystyle AP = 65$

In the next $\displaystyle t$ seconds, the balloon rises $\displaystyle t$ feet to point $\displaystyle B$: .$\displaystyle BP = t$

In the same $\displaystyle t$ seconds, the bicycle moves $\displaystyle 17t$ feet to point $\displaystyle C$: .$\displaystyle AC = 17t$

Using Pythagorus: .$\displaystyle s(t) \;=\;\sqrt{(17t)^2 + (t+65)^2} \;=\;\left(290t^2 + 130t + 4225\right)^{\frac{1}{2}}$

Differentiate with respect to time:

. . $\displaystyle \frac{ds}{dt} \;=\;\frac{1}{2}\left(290t^2+130t+4225\right)^{-\frac{1}{2}}(580t + 130) \;=\;\frac{290t+65}{\sqrt{290t^2+130t+4225}} $

And now let $\displaystyle t = 3.$