Worksheet #5
Math 221
1. Use the definition to determine the derivative of the given function at a general point
a.
(a) f (x) = 3x2 + x + 1
Solution:
3(a + h)2 + a + h + 1 − (3a2 + a + 1)
f (a + h) − f (a)
= lim
f (a) = lim
h→0
h→0
h
h
2
2
3a + 3h + 6ah + a + h + 1 − 3a2 − a − 1
3h2 + 6ah + h
= lim
= lim
h→0
h→0
h
h
h(3h + 6a + 1)
= lim (3h + 6a + 1) = 6a + 1.
= lim
h→0
h→0
h
0
(b) f (x) = 2x+1
3x+4
Solution:
1 2(a + h) + 1 2a + 1
f (a + h) − f (a)
= lim
−
f (a) = lim
h→0 h
h→0
h
3(a + h) + 4 3a + 4
1
5h
5
= lim
= lim
h→0 h
h→0 (3a + 4)(3a + 3h + 4)
(3a + 4)(3a + 3h + 4)
5
=
.
(3a + 1)2
√
(c) f (x) = x2 + 1
Solution:
p
√
(a + h)2 + 1 − a2 + 1
0
f (a) = lim
h→0
h
p
p
√
√
2
(a + h) + 1 − a2 + 1 (a + h)2 + 1 + a2 + 1
p
√
= lim
h→0
h
(a + h)2 + 1 + a2 + 1
0
1 (a + h)2 + 1 − (a2 + 1)
1
h2 + 2ah
p
√
√
= lim p
h→0 h
(a + h)2 + 1 + a2 + 1 h→0 h (a + h)2 + 1 + a2 + 1
a
=√
2
a +1
= lim
2. A vase is made from an inverted right circular cone of height 20 inches and radius 5
1
inches. In general, the volume of a right cone of height h and radius r is 31 πr2 h.
(a) Find a formula for V (h), the volume of liquid in the vase if it is h inches deep.
Solution: Note that using properties of similar triangles,
r
5
=
⇒ r = h/4.
h
20
1
1
V (h) = πr2 h = πh3 .
3
48
(b) Plot V (h) and
Solution:
dV
.
dh
1
dV
= V 0 (h) = πh2 .
dh
16
(c) If the vase is being filled at a constant rate, plot V (t) and dV
as well as h(t) and
dt
dh
.
dt
Solution: Filling at a constant rate meant V 0 (t) = c for some constant c. Therefore V (t) = ct.
1/3
1 3
48c
ct = πh ⇒ h(t) =
t
48
π
−2/3
1/3
1 48c
48c
48c
1
0
h (t) =
t
·
=
3
π
π
π
3t2/3
2
3. Now consider the vase pictured (also of height 20 inches and obtained by rotating a
curve):
(a) Sketch the plots of V (h) and
Solution:
dV
.
dh
(b) If the vase is being filled at a constant rate, plot V (t) and
dh
.
dt
Solution:
3
dV
dt
as well as h(t) and