Page 294 - Calculus Demystified

Chapter 3

Fig. S3.2 281

The roots of this equation are h = 2, 6. Of course height 6 gives a trivial
cylinder, as does height 0. We find that the solution of our problem is height
2, radius 2.
3. We know that
V = · w · h
hence
dV d dw dh
= · w · h + · · h + · w ·
dt dt dt dt
= 1 · 60 · 15 + 100 · (−0.5) · 15 + 100 · 60 · 0.3
= 900 − 750 + 1800 = 1950 in/min.
4. We know that v 0 =−15. Therefore the position of the body is given by

2
p(t) =−16t − 15t + h 0 .
Since
2
0 = p(5) =−16 · 5 − 15 · 5 + h 0 ,
we find that h 0 = 475. The body has initial height 475 feet.
5. We know that
1
2
V = · πr · h.
3
Therefore
d 1 dh 1 dr
2
0 = V = · π · r · + · π · 2r · · h.
dt 3 dt 3 dt
At the moment of the problem, dh/dt = 3, r = 5, h = 12/(5π). Hence

dr 12
2 2
0 = π · 5 · 3 + π · (2 · 5) · ·
dt 5π