# Math Help - Related Rates #2

1. ## Related Rates #2

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?

11

2. We are looking for ds/dt.

Find the relation of s to the givens so that we can have an equation involving
the s.

The kite is rising vertically, the bicycle is running horizontally, the distance, s, between the kite and the bicycle is moving "diagonally".
So the figure is that of a right triangle whose parts after time t are:
---vertical leg = (65 +1*t) = (65 +t)
---horizontal leg = 17*t = 17t
---hypotenuse = s

By Pythagorean theorem,
s^2 = (65 +t)^2 +(17t)^2 ------------------------(i)
Differentiate both sides with respect to time t,
2s(ds/dt) = 2(65 +t)(1 dt/dt) +2(17t)(17 dt/dt) <----I purposely showed dt/dt---just for comparison.
s(ds/dt) = (65 +t) +(289t)
s(ds/dt) = 290t +65 -----------------------(ii)

We were given t = 3 sec

We can compute s when t = 3 sec.
s^2 = (65 +t)^2 +(17t)^2 -------------------------(i)
s^2 = (65 +3)^2 +(17*3)^2
s^2 = 7225
s = sqrt(7225) = 85 ft

So, substituting those into (ii),
s(ds/dt) = 290t +65
85(ds/dt) = 290*3 +65
85(ds/dt) = 935
ds/dt = 935/85 = 11 ft/sec

Therefore, at that instant, s is increasing at the rate of 11 ft/sec. ----answer.

3. 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 $P$ when the cycle passes under it (at $A$): . $AP = 65$

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

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

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

Differentiate with respect to time:
. . $\frac{ds}{dt} \;=\;\frac{1}{2}\left(290t^2+130t+4225\right)^{-\frac{1}{2}}(580t + 130) \;=\;\frac{290t+65}{\sqrt{290t^2+130t+4225}}$

And now let $t = 3.$