1. Related Rates Problems (two)

Alright, I will need some assistance for two problems that I face. Thanks in advance.

A trough is 12 feet long and 3 feet across the top (see figure). Its ends are isosceles triangles with altitudes of 3 feet.

(a) If water is being pumped into the trough at 2 cubic feet per minute, how fast is the water level rising when h is 1.9 foot deep?

(b) If the water is rising at a rate of inch per minute when h = 2.5, determine the rate at which water is being pumped into the trough.

Honestly I'm not sure how to do this one at all, I need an explanation.

AND

All edges of a cube are expanding at a rate of 6 centimeters per second. (a) How fast is the volume changing when each edge is 3 centimeter(s)?

-I set dV/dt as 6 cm/s
-We must find dx/dt when x=3cm

The eq. is V=x^3

So dV/dt = 3x^2 (dr/dt)

so I solved for dr/dt, dr/dt = dV/dt*1/3x^2. Then I plugged in 3 for x, but this wasn't correct, where did I go wrong? Thanks a lot guys, much appreciated.

2. look at the small triangle of water at the end of the trough ... it is similar to the triangle that forms the end base of the entire trough.

$\displaystyle \frac{b}{h} = \frac{3}{3} = \frac{1}{1}$

base equals height.

the volume of water in the trough is ...

$\displaystyle V = \frac{1}{2}bh \cdot 12 = 6h^2$

now you have a water volume formula strictly in terms of h ... take the time derivative, substitute your given info, then solve for what you need.

for the second problem, the edge is increasing at 6 cm/sec, not the volume.

$\displaystyle \frac{dx}{dt} = 6$

$\displaystyle V = x^3$

take the time derivative, sub in your given values, calculate $\displaystyle \frac{dV}{dt}$

btw, where did "r" come from?