Hello, nivek516!

Here's the first one . . .

Two towers, $\displaystyle A\text{ and }B$. are 150m tall, 40m apart.

A horizontal tightrope (height of 60m) is attached to the two buildings.

A man walks across the tightrope at a constant $\displaystyle \tfrac{1}{3}$ m/sec.

A spotlight is on the top of $\displaystyle A$. How fast is the shadow of the man

moving up the wall of building $\displaystyle B$ when he is 8 m away from $\displaystyle B]$? First, make a sketch . . . Code:

A * * B
| * |
90 | * |
| * 40-x |
T + - - - - - * - - - - - + R
| x M * |
| * | y
60 | * |
| * S
| |
| |
C * - - - - - - - - - - - * D
40

The buildings are $\displaystyle AC\text{ and }BD\!:\;\;AC = BD = 150,\;CD = 40$

The tightrope is $\displaystyle TR\!:\;\;TR = 40,\;TC = 60.\;AT = 90$

The spotlight is at $\displaystyle A$, the man is at $\displaystyle M.$

. . Let: $\displaystyle TM \:=\: x \quad\Rightarrow\quad MR \:=\:40-x$

. . We are given: .$\displaystyle \tfrac{dx}{dt} = \tfrac{1}{3}$ m/sec.

The man's shadow is at $\displaystyle S\!:\;\;\text{let } RS = y.$

Since $\displaystyle \Delta ATM \sim \Delta SRM\!:\;\;\frac{x}{90} \:=\:\frac{40-x}{y} \quad\Rightarrow\quad y \:=\:90\,\frac{40-x}{x} \:=\:90\left(40x^{-1} - 1\right)$

Differentiate with respect to time: .$\displaystyle \frac{dy}{dt} \:=\:-\frac{3600}{x^2}\cdot\frac{dx}{dt}$

When $\displaystyle MR = 8\;(x = 32)\!:\;\;\frac{dy}{dt} \;=\;-\frac{3600}{32^2}\left(\frac{1}{3}\right) \;=\;-\frac{75}{64}$

The shadow is moving up building $\displaystyle B$ at $\displaystyle 1\tfrac{11}{64}\text{ m/sec}$