Hello, Juggalomike!

1. A light is affixed to the top of a 12 foot tall lamppost.

A 6-foot tall man walks away from the lamp post at a rate of 5 ft/sec.

How fast is the length of his shadow increasing when he is 5 feet away? Code:

A
*
| * C
| *
12 | | *
| 6 | *
| | *
* - - - * - - - - - - - * - - -
B x D s E

The lampost is: $\displaystyle AB = 12$

The man is: $\displaystyle CD = 6$

His distance from the lampost is: $\displaystyle x = BD$, and $\displaystyle \frac{dx}{dt} = 5$

The length of his shadow is: $\displaystyle s = DE$

Since $\displaystyle \Delta CDE \sim \Delta ABE\!:\;\;\frac{s}{6} \:=\:\frac{x+s}{12} \quad\Rightarrow\quad s \:=\:x$

Differentiate with respect to time: .$\displaystyle \frac{ds}{dt} \:=\:\frac{dx}{dt}$

Therefore: .$\displaystyle \frac{ds}{dt} \:=\:5$

His shadow is lengthening at 5 ft/min.

2. A water tank has the shape of an inverted right circular cone

of altitude 12 feet and base radius 6 feet.

If water is being pumped into the tank at a rate of 10 gal/min,

at what rate is the water level rising when the water is 3 feet deep? Nasty! .Nasty! .They give the volume in __gallons__, then ask for answers in *feet*.

. . They could have given us a conversion formula at the very least!

I had to look it up: .$\displaystyle 1\text{ gal} \:\approx\:0.1337\text{ ft}^3$

So we have: .$\displaystyle \frac{dV}{dt} \:=\:10\text{ gal/min} \:=\:1.337\text{ ft}^3\text{/min}$

Look at the side view of the cone: Code:

A 6 D
- * - - - - * - - - - *
: \ | /
: \ | /
: \ B| r /
: * - - + - - * E
12 \ | /
: \ | /
: \ h| /
: \ | /
: \|/
- *
C

The height of the cone is: $\displaystyle AC = 12$

The radius of the cone is: $\displaystyle AD = 6$

The height of the water is: $\displaystyle h = BC$

The radius of the water is: $\displaystyle r = BE$

Since $\displaystyle \Delta EBC \sim \Delta DAC\!:\;\;\frac{r}{h} \:=\:\frac{6}{12}\quad\Rightarrow\quad r \:=\:\frac{h}{2}\;\;{\color{blue}[1]}$

The volume of the water is: .$\displaystyle V \:=\:\frac{\pi}{3}r^2h$

Substitute [1]: .$\displaystyle V \:=\:\frac{\pi}{3}\left(\frac{h}{2}\right)^2h \quad\Rightarrow\quad V \:=\:\frac{\pi}{12}h^3 $

Differentiate with respect to time: .$\displaystyle \frac{dV}{dt}\:=\:\frac{\pi}{4}h^2\,\frac{dh}{dt}$

Substitute our known values: .$\displaystyle 1.337 \:=\:\frac{\pi}{4}(3^2)\,\frac{dh}{dt}$

Therefore: .$\displaystyle \frac{dh}{dt} \;=\;\frac{4}{9\pi}(1.337) \;\approx\;0.05\text{ ft/min}$