1. Volume = pr2h Part (a) dV/dt = -5p(h)1/2 V = 25ph dV/dt = 25p(dh/dt)
(Remember, the radius is a constant.)
dV/dt = -5p(h)1/2
V = 25ph
dV/dt = 25p(dh/dt)
-5p(h)1/2 = 25p (dh/dt)
-5p(h)1/2 /25p = (dh/dt)
-(h)1/2 /5 = (dh/dt)

2. Part (b)
-5 dh = h1/2 dt
Cross multiply
Separate the variables
h-1/2 dh = -1/5 dt

3. Make the substitutions now
h-1/2 dh = -1/5 dt
Integrate both sides...
2h1/2 = -1/5t + C
Make the substitutions now
2(17)1/2 = -1/5(0) + C
C = 2(17)1/2
2h1/2 = -1/5t + 2(17)1/2

4. h1/2 = -1/10t + 171/2
Divide all terms by 2
Square both sides
h = (-1/10t + 171/2)2
2h1/2 = -1/5t + 2(17)1/2

5. The coffee pot would be empty when h= 0.
Part (c)
The coffee pot would be empty when h= 0.
h = (-1/10t + 171/2)2
Answer from Part (b)
0 = (-1/10t + 171/2)2
1/10t = 171/2
t = 10(171/2)
The coffee pot will be empty in sec.