Math 165 – worksheet for ch. 3/4 – solutions
Problems on this worksheet cover topics
• Linearization (3.11)
• Related Rates (3.10)
• Increasing / decreasing
• Concave up / concave down
• L’Hopital’s rule (4.5)
Section 4.6 was covered by an earlier worksheet in more detail.
Problem 1 Find the linearization of the function f (x) = x3 − π4 sin(2πx) + 2
at a = 1. Then use the linearization to approximate f (0.9997).
Solution a) The linearization at a is a function L(x) given by
L(x) = f 0 (a)(x − a) + f (a)
so the graph of L(x) is the tangent line at x = a. Here, a = 1 and
f 0 (x) = 3x2 − 8 cos(2πx),
which means f 0 (1) = −5. Also, f (1) = 3. Therefore,
L(x) = −5(x − 1) + 3.
b) f (0.9997) ≈ L(0.9997) = 3.0015.
Problem 2 A man who is 6 ft tall walks with a speed of 8 ft/s away from
a street light that is atop an 18-ft pole. How fast is the tip of his shadow
moving along the ground when he is 100 ft from the light pole?
Solution We make a sketch and indicate the distance x of the man from
the light pole, distance z of the tip of his shadow from the light pole. Then
we use similar triangles to get
z
z−x
=
.
18
6
you can set up the ratio equation in many ways, but in this way it is easiest
to differentiate both sides (so if you need the quotient rule for differentiating,
it might be worthwhile first simplifying your equation). Differentiating gives
z0
z 0 − x0
=
.
18
6
Multiply by 18, substitute x = 8 and solve for z 0 to get z 0 = 12 (in feet per
second) .
Not part of this problem. It is remarkable that the clue x = 100 is never
needed. This means that the tip of the shadow has a constant velocity, just
like the man - only the shadow’s tip advances faster than the man, which is
what makes the shadow grow ever longer and longer faster.
Problem 3 Consider the function
f (x) = x4 − 4x2
Find all intevals where f (x) is increasing, decreasing, concave up/ concave
down, and its points of inflection.
Solution First,
f 0 (x) = 4(x3 − 2x)
and second,
f 00 (x)12x2 − 8.
p
and the sign pattern is + − +. Therefore f (x)
So f 00 (x) = 0 for x = ±p 2/3,p
is concave down in [− 2/3, 2/3], and concave up on the outside of this
interval. At the endpoints of this interval, f 00 (x) changes sign and therefore
we get points of inflection
p
P = (± 2/3, −20/9).
√
√
Next, f 0 (x) = 0 for x = 0, 2, − 2. The sign pattern is − + −+ with the
aforementioned
x-values√showing where the sign changes.
increas√ So f (x) is √
√
ing in [− 2, 0] and in [ 2, ∞), decreasing in (−∞, − 2] and in [0, 2].
Problem 4 Find the limit
(x − 1) sin(πx)
x→1
tan2 (πx)
L = lim
Solution This is a limit of the form 0/0, so we may apply l’Hopital’s Rule.
We get
d
(x − 1) sin(πx) = sin(πx) + π(x − 1) cos(πx)
dx
d
tan2 (πx) = 2 tan(πx) sec2 (πx)
dx
And the ratio of these two is still of the form 0/0. So we apply l’Hopital’s
Rule again. Differentiating gives
d
sin(πx) + π(x − 1) cos(πx) = 2π cos(πx) − π 2 (x − 1) sin(πx)
dx
d
d 2π sin(πx)
2π tan(πx) sec2 (πx) =
dx
dx cos3 (πx)
2π 2 (cos4 πx + 3 cos2 πx sin πx)
.
=
cos6 (πx)
Now we can take the limit of both these functions, getting 2π and 2π 2 ,
respectively. The ratio of these limits is 1/π, and that is then also the answer
to our original problem.