Homework 4 - Math 1451-008 (Howle)
Due Monday 3/26/2012 in class
Name:
R Number:
This cover sheet must be attached as the top page of your homework.
See homework requirements in the syllabus.
1. Let f (x) = cos x. Use linear approximation of f (x) to estimate the value of cos
π
2
+ 0.01 .
2. Let g(x) = x2 − 2. Do two steps of the Newton-Raphson method starting with initial guess
x0 = 1.
3. Suppose that the edge lengths x, y, z of a closed rectangular box are changing at the following
rates:
dx
dy
dz
= 1 m/sec,
= −2 m/sec,
= 1 m/sec.
dt
dt
dt
(a) Find the rate at which the box's volume is changing at the instant when x = 4, y = 3,
and z = 2.
(b) Find the rate at which the box's surface area is changing at the instant when x = 4,
y = 3, and z = 2.
4. Given the function f (x) = 4x3 − x4 .
(a)
(b)
(c)
(d)
Find all critical points of f (x).
Identify all intervals on which f (x) is increasing and on which it is decreasing.
Find the relative extreme values of f (x).
Identify all intervals on which f (x) is concave up and on which it is concave down. Find
any inection points of f (x).
(e) Sketch a graph of f (x) clearly indicating and labeling the information found in parts (a)
through (d).
1
1.
Let
f (x) = cos x.
cos + 0.01 .
Solution: Let x0 =
π
2
Use linear approximation of
π
2
f (x)
to estimate the value of
and nd the formula for the tangent line to f (x) at the point (x0 , y0 ).
y(x) = f 0 (x0 )(x − x0 ) + f (x0 )
= − sin(x0 )(x − x0 ) + cos(x0 )
π
π
π
x−
+ cos
= − sin
2
2
2
π
= −1 x −
+0
2
π
= − x−
2
We can then approximate f
π
2 + 0.01.
π
2
f
+ 0.01 with the value of the tangent line evaluated at x =
f (x) ≈ y(x)
π
π
+ 0.01
≈ y
+ 0.01
2
2
π
π
= −
+ 0.01 −
2
2
= −0.01
Note that this is a fairly good approximation. If I calculate cos( π2 + 0.01) on a calculator, I
get approximately -0.0099998.
2.
Let g(x) = x2 −2. Do two steps of the Newton-Raphson method starting with initial
guess x0 = 1.
Solution: Newton-Raphson iterates to nd roots of the given function g(x). If we start with
an initial guess x0 , the next guess x1 is given by x1 = x0 − g(x0 )/g 0 (x0 ). Given this new x
value, the next guess x2 is similarly given by x2 = x1 − g(x1 )/g 0 (x1 ). So for this function with
the given initial guess we have:
g(x0 )
g 0 (x0 )
12 − 2
1
3
g(1)
= 1− 0
= 1−
= 1+ =
g (1)
2(1)
2
2
g(x1 )
= x1 − 0
g (x1 )
x1 = x0 −
x2
=
=
g 32
3
3
− = −
2 g0 3
2
2
2
3
2
2
−2
3
2
3 9/4 − 2
3 1/4
3
1
17
−
= −
= −
=
2
3
2
3
2 12
12
2
3.
Suppose that the edge lengths
the following rates:
x, y, z
of a closed rectangular box are changing at
dx
dy
dz
= 1 m/sec,
= −2 m/sec,
= 1 m/sec.
dt
dt
dt
(a)
Find the rate at which the box's
y = 3, and z = 2.
volume
is changing at the instant when x = 4,
Solution:
V
dV
dt
dV
dt
= xyz
dx
d
+ x (yz)
dt
dt
dx
dz
dy
= yz
+ xy
+ xz
dt
dt
dt
= (3)(2)(1) + (4)(3)(1) + (4)(2)(−2)
= (yz)
= 6 + 12 − 16
= 2
(b)
Find the rate at which the box's
x = 4, y = 3, and z = 2.
surface area
is changing at the instant when
Solution:
A = 2xy + 2xz + 2yz
dA
dy
dx
dz
dx
dz
dy
=
2x + 2y
+ 2x + 2z
+ 2y
+ 2z
dt
dt
dt
dt
dt
dt
dt
= 2(4)(−2) + 2(3)(1) + 2(4)(1) + 2(2)(1) + 2(3)(1) + 2(2)(−2)
= −16 + 6 + 8 + 4 + 6 − 8
= 0
3
4.
Given the function f (x) = 4x3 − x4 .
(a)
Find all critical points of f (x).
Solution:
f 0 (x) = 12x2 − 4x3 = 0
0 = 4x2 (3 − x)
x = 0, 3
(b)
Identify all intervals on which f (x) is increasing and on which it is decreasing.
Using the solution to (a), since x2 ≥ 0 for all x, we see that f 0 (x) > 0 when
x < 3 (but not equal to 0), and f 0 (x) < 0 when x > 3. So f (x) is increasing on
(−∞, 0) ∪ (0, 3) and decreasing on (3, ∞).
Solution:
(c)
Find the relative extreme values of f (x).
by the 1st derivative test, we have a relative max at x = 3. (x = 0 is neither
a max nor a min.)
Solution:
(d)
Identify all intervals on which f (x) is concave up and on which it is concave
down. Find any inection points of f (x).
Solution:
f 00 (x) = 24x − 12x2 = 0
0 = 12x(2 − x)
x = 0, 2
So f 00 (x) > 0 when x > 0 and x < 2, i.e., on (0, 2)
or when x < 0 and x > 0 (which can't happen).
f 00 (x) < 0 when x < 0 and x < 2, i.e., when x < 0,
or when x > 0 and x > 0, i.e., when x > 2.
Therefore, f (x) is concave up on (0, 2) and concave down on (−∞, 0) ∪ (2, ∞). There
are inection points at x = 0 and x = 2.
(e)
Sketch a graph of f (x) clearly indicating and labeling the information found
in parts (a) through (d).
4