the problem is on the bottom. Please show how to work it out.?

Question

A balloon will pass directly under a street light suspended 18 feet above the ground. It is traveling toward
the street light at a rate of .75 feet per second and descending at a rate of .25 feet per second. When we
¯rst notice it, it is 20 feet from the light pole (i.e., measured horizontally) and 8 feet above ground

How fast is it's shadow approaching the point directly under the light when we first notice the ballon?

Answer 1

I drew a diagram and used trig as follows.

When we see the balloon it is 10' below the street light and 20' from the street light.

The angle it makes from the top of the street light,a, has
tan(a)=20/10

The shadow has the same angle with respect to the street light.
so tan(a)=D/18 where D is the distance of the shadow from the base of the pole.

D=36'

Now I looked at the rate of change of the balloon and related it as
Horizontal=20-3t/4
Vertical from the top of the pole is
=10+t/4

So the new a' has
tan(a')=(20-3t/4)/(10+t/4)
Which is related to the position of the shadow as
tan(a')=d'/18
so d'=
18*[(20-3t/4)/(10+t/4)]

now the average speed is
(D-d')/t
or (36-18*[(20-3t/4)/(10+t/4)])/t

as t->0
or dt this is the instantaneous velocity of the shadow, which I computed as 2.25 ft/sec using Excel.

I will try to simplify and re-post

OK, I'm back after scratching my head.

The equation
D-d'=
36-18*[(20-3t/4)/(10+t/4)]
is the position of the balloon with respect to the base of the pole

This is x(t)

velocity is dx/dt

so, I take the first derivative as
18*12.5/(100+5*t+.0625*t^2)
which is the instantaneous speed at any point until the balloon reaches the exact position under the light.

Since we only care about t=0
that is 18*12.5/100
or 2.25 ft/sec

If you can use calculus to solve the problem, this is a way to do it.

BTW: I used
http://www.webmath.com

to compute the derivative.

j

Answer 2

yes