2214.17 Related Lamppost And Shadow

[karush]

View attachment 7331
ok just seeing if this is ok
where does the 35 come in ?
I tried to follow an example but??
is not x=35

 

[MarkFL]

Let's generalize a bit and let $P$ be the height of the post, and $M$ be the height of the man, where $M<P$. Using the provided diagram as a guide, we see from similarity that:

\(\displaystyle \frac{x+y}{P}=\frac{y}{M}\)

Arrange this as:

\(\displaystyle y=\frac{M}{P-M}x\)

Differentiate w.r.t time $t$:

\(\displaystyle \d{y}{t}=\frac{M}{P-M}\d{x}{t}\)

Do you see now that \(\displaystyle \d{y}{t}\) is independent of $x$ (the distance between the man and the pole)? It depends only on the ratio of the man's height to the difference between the height of the pole and the man, and on \(\displaystyle \d{x}{t}\). That's why you don't see 35 in the solution.

Suppose two men are walking at the same constant speed along a line connecting them and the pole...and one man is taller than the other...whose shadow changes length more quickly?

 

[karush]

ok, see your point guess they just put it in there to brain twist some

so I presume that $3 ft/sec$ is correct

I looked at 5 examples they all did something different?

 

[MarkFL]

Suppose two men are walking at the same constant speed along a line connecting them and the pole...and one man is taller than the other...whose shadow changes length more quickly?

Let's look at our formula:

\(\displaystyle y=\frac{M}{P-M}x\)

Now suppose we define:

\(\displaystyle f(M)=\frac{M}{P-M}\) where $0<M<P$

Hence:

\(\displaystyle f'(M)=\frac{P}{(P-M)^2}>0\)

So, we see that as $M$ increases, so must $f$...and so the taller man's shadow will change length at a quicker rate than that of the shorter man. :)

What happens to the rate of change of the shadow's length as the man's height approaches that of the pole?