# Related Rates Problems (two)

• Nov 2nd 2008, 08:51 AM
bakeit1
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.
http://www.webassign.net/larson/2_06-26.gif
(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 http://www.webassign.net/www28/symIm...2b82d56e22.gif 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.
• Nov 2nd 2008, 09:37 AM
skeeter
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.

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

base equals height.

the volume of water in the trough is ...

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

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.

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

$V = x^3$

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

btw, where did "r" come from?