MIDTERM 2 REVIEW: ADDITIONAL PROBLEMS
I MPLICIT D IFFERENTIATION
1) If
√
x+
√
y = 1, find y 00 .
Solution. Taking the derivative (with respect to x) of both sides of the given equation, we find that
1
1
√ + √ · y 0 = 0,
2 x 2 y
which gives
√
y
0
y = −√ .
x
√
√
Notice that the original equation implies y = 1 − x, so we can actually write y 0
entirely in terms of x, as
√
1− x
1
0
y =− √
= − √ + 1 = −x−1/2 + 1.
x
x
Taking derivatives again, we then have
1
1
y 00 = x−3/2 = √ .
2
2x x
If you hadn’t noticed that y 0 could be written entirely in terms of x, then you would
have computed
√
√
x · 2√1 y · y 0 − y 2√1 x
√
y 00 = −
( x)2
√
y
Plugging y 0 = − √x into this equation would have then given
√
√
√
√
− y
√
y
√
1
√
x · 2√1 y · √x − y 2√1 x
−
−
x+ y
1
2
2 x
00
√
y =−
=−
=
= √ .
x
x
2x x
2x x
2)* Find all points on the curve x2 y 2 + xy = 2 where the slope of the tangent line is −1.
Solution. First we use implicit differentiation to find y 0 at any point. Taking derivatives (with respect to x) on both sides of the given equation, we find
2xy 2 + x2 · 2y · y 0 + 1 · y + x · 1 · y 0 = 0,
which can be rearranged to give
y0 =
−2xy 2 − y
.
2x2 y − x
Problems marked with a * are extra challenging.
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We now want to find those points (x, y) that for which y 0 = −1 and which lie on
the given curve (i.e., satisfy the equation x2 y 2 + xy = 2). If we first set y 0 = −1 in
the above equation, we find that
−2xy 2 − y
= −1.
2x2 y − x
At this point the algebra gets a bit tricky. First, notice that the above equation can
be simplified to
2x2 y − 2xy 2 + x − y = 0,
which happens to factor as
(2xy + 1)(x − y) = 0.
So, we must have either xy = − 12 or x = y. However, if xy = − 12 , then notice that
the equation for our curve becomes (−1/2)2 + (−1/2) = 2, which is not true! So,
we must actually have x = y. If we plug this into the equation for our curve, we
find that x4 + x2 = 2, or equivalently x4 + x2 − 2 = 0. This equation factors as
(x2 − 1)(x2 + 2) = 0, and so we must either have x2 = 1 or x2 = −2. The second
case is impossible, so it must be that x2 = 1, and hence x = ±1. Since x = y, this
means that the two solutions are (1, 1) and (−1, −1). Those are the only two points
on our curve where the tangent line has slope −1.
R ELATED R ATES
3) Gold dust is being poured into a conical pile at a rate of 30 cubic feet per minute.
The diameter of the base of the pile is always equal in length to the height of the
pile. How quickly is the height of the pile increasing when the height is 10 feet?
(Hint: Recall that the volume of a cone with base radius r and height h is V =
1
πr2 h.)
3
Solution. The base of the conical pile of gold dust has radius r = h/2, where h is
the height of the pile. It follows that the volume of gold dust is
2
1 2
1
h
1
V = πr h = π
h = πh3 .
3
3
2
12
Differentiating this equation with respect to time t gives
dV
1
dh
= πh2
dt
4
dt
We are given that
dV
dt
= 30 cubic feet per minute. So, when h = 10 feet we have
1
dh
30 = π(10)2 ,
4
dt
and so
dh
dt
=
30
25π
≈ 0.38 feet per minute.
2
4)* Suppose the minute hand on a clock is 8 inches long, and the hour hand is 4” long.
Find the rate at which the distance between the tips of the hands is increasing
when the clock reads 1:00.
(Hint: Recall the Law of Cosines, which says that if θ is an angle in a triangle with
adjacent edges a, b and opposite edge c, then c2 = a2 + b2 − 2ab cos(θ).)
Solution. Let c be the distance between the tips of the hands of the clock, and let θ
be the angle between those hands (measured clockwise from the minute hand to
the hour hand). By the Law of Cosines, we have
c2 = 42 + 82 − 2 · 4 · 8 cos(θ) = 80 − 64 cos(θ).
Differentiating with respect to time t, we then have
dθ
dc
2c = 64 sin(θ) .
dt
dt
dθ
What is dt ? The minute hand travels once around the circular clock face every
hour, and so rotates at 2π radians per hour, or 2π/60 = π/30 radians per minute.
The hour hand travels once around the circular clock face twelve hours, and so rotates at 2π/12 = π/6 radians per hour, or π/360 radians per minute. The difference
between their rates of rotation is therefore π/360 − π/30 = −11π/360 ≈ −0.096
≈ −0.096 radians per minute.
radians per minute. So, dθ
dt
Now we just need to determine θ and c when the clock reads 1:00. At that time,
the minute hand is pointing straight up (at 12 on the clock face), and the hour hand
is pointing at 1 on the clock face. The angle θ between the two is 2π/12 = π/6. So,
θ = π/6. If we substitute this into our original formula relating c and θ, we find
that c2 = 80 − 64 cos(π/6) ≈ 24.57, and so c ≈ 4.96 inches. Plugging these values of
c, θ, and dθ
into our above equation gives
dt
dc
= 64 sin(π/6)(−0.096),
dt
≈ −0.310 inches per minute.
2(4.96)
which gives
dc
dt
L INEAR A PPROXIMATIONS AND D IFFERENTIALS
5) Estimate (1.01)11 using a linear approximation.
Solution. There are many possible solutions to this problem, but the following is
probably the most straightforward. If we let f (x) = x11 , then the quantity we are
interested in estimating is f (1.01). We can easily compute f (1) = 111 = 1, so it
seems reasonable to use a linearization of f at a = 1 to make this estimate, i.e., to
use the approximation
f (x) ≈ f 0 (1)(x − 1) + f (1)
for x near 1, to estimate f (1.01). We compute f 0 (x) = 11x10 , so f 0 (1) = 11. We
therefore have
f (x) ≈ 11(x − 1) + 1
for x near 1. In particular,
(1.01)11 = f (1.01) ≈ 11(1.01 − 1) + 1 = 11(0.01) + 1 = 1.11.
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6)* Use differentials to estimate the volume of a thin cylindrical shell with height h,
inner radius r, and thickness ∆r. What is the error in this estimate?
Solution. The exact volume of the shell would be the difference between the volume of the outer cylinder (with radius r +∆r) and the volume of the inner cylinder
(with radius r), namely
volume of shell = π(r + ∆r)2 h − πr2 h.
If V (r) denotes the volume of a cylinder with radius r and (fixed) height h, then
we can rewrite this as
volume of shell = V (r + ∆r) − V (r) = ∆V.
In particular, we can estimate this volume using the approximation dV ≈ ∆V . A
quick computation gives
dV = 2πrh dr
(recall that the height h is some fixed constant). So, our estimate is
volume of shell ≈ 2πrh dr = 2πrh ∆r.
What is the error in this estimate? Well, the exact volume of the shell is ∆V =
π(r + ∆r)2 h − πr2 h, so the error in our estimate is
error = ∆V − dV
= π(r + ∆r)2 h − πr2 h − 2πrh∆r
= π(r2 + 2r∆r + (∆r)2 )h − πr2 h − 2πrh∆r
= 2πrh∆r + π(∆r)2 h − 2πrh∆r
= π(∆r)2 h.
A BSOLUTE M AXIMA AND M INIMA
7) Find the absolute maximum and minimum of the function f (x) = x5 − x3 + 2 on
the interval [−1, 1].
Solution. We first find the critical points of f on the given interval. We compute
f 0 (x) = 5x4 − 3x2 = x2 (5x2 − 3).
4
p
So, f 0 (x) = 0 exactly when either x = 0 or x = ± 3/5 ≈ ±0.77. We now compare
the values of the function f at these critical points and the end points:
f (−1) = (−1)5 − (−1)3 + 2 = −1 − (−1) + 2 = 2
r
r
r
p
p
p
3
3
3
6
3
9
+
+2=
+ 2 ≈ 2.19
f (− 3/5) = (− 3/5)5 − (− 3/5)3 + 2 = −
25 5 5 5
25 5
f (0) = 05 − 03 + 2 = 2
r
r
r
p
p
p
9
3
3
3
6
3
f ( 3/5) = ( 3/5)5 − ( 3/5)3 + 2 =
−
+2=−
+ 2 ≈ 1.81
25 5 5 5
25 5
f (1) = 15 − 13 + 2 = 2.
q
6
3
So, the maximum value of f on the interval is 25
+ 2 ≈ 2.19, and the minimum
5
q
6
3
value is − 25
+ 2 ≈ 1.81.
5
8)* Find the absolute maximum and minimum of the function f (x) = x4 −3x3 +3x2 −x
on the interval [0, 2].
Solution. We first find the critical points of f on the given interval. We compute
f 0 (x) = 4x3 − 9x2 + 6x − 1.
We need to determine when f 0 (x) = 0, which means we need to find the roots of
the above polynomial. For a general cubic polynomial, this can be quite difficult.
In this case, however, we are going to be a bit luck. First, notice that f 0 (1) =
4 − 9 + 6 − 1 = 0, so we know x − 1 is one of the factors of the f 0 (x). If we perform
polynomial long division, we find that
f 0 (x) = (x − 1)(4x2 − 5x + 1) = (x − 1)(4x − 1)(x − 1) = (x − 1)2 (4x − 1).
So, we can now see that f 0 (x) = 0 exactly when x = 14 , 1. We now compare the
values of f at the endpoints and these critical points:
f (0) = 0
f (1/4) = (1/4)4 − 3(1/4)3 + 3(1/4)2 − (1/4) = −27/256 ≈ −0.11
f (1) = 1 − 3 + 3 − 1 = 0
f (2) = 24 − 3(2)3 + 3(2)2 − 2 = 2.
So, the maximum value of f on the interval is 2, and the minimum value is −27/256 ≈
−0.11.
M EAN VALUE T HEOREM
9) Two runners start a race at the same time and end in a tie. Prove that at some point
during the race they were running at exactly the same speed.
Solution. Let f (t) and g(t) be the positions of the first and second runners at time t,
respectively. Let h(t) = f (t) − g(t), and suppose the runners finish the race at time
T . We are given that h(0) = 0 (since the runners start the race at the same position)
and h(T ) = 0 (since the runners end in a tie, and so are at the same position at
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the end time T ). Assuming the positions of the runners define continuous and
differentiable functions (which seems reasonable), it follows from Rolle’s Theorem
that h0 (c) = 0 at some time c between time 0 and T . This implies f 0 (c) = g 0 (c), i.e.,
the two runners were running at exactly the same speed at time c.
G RAPHING
10) Follow our usual checklist of properties to sketch y = (4 − x2 )5 .
Solution. We first note that the function f (x) = (4 − x2 )5 is defined for all real
numbers. When x = 0, we have y = 45 = 1024, so the y intercept is (0, 1024). When
y = 0, we have 0 = (4 − x2 )5 , whose only solutions are x = ±2. So, the x-intercepts
are (−2, 0) and (2, 0). Next note that f (−x) = (4 − (−x)2 )5 = (4 − x2 )5 = f (x),
so the function f is even. This means that the graph will be symmetric about the
y-axis. As for asymptotes, there can be no vertical asymptotes (since f is defined
everywhere). There are also no horizontal asymptotes, since lim f (x) = ∞.
x→±∞
We now move on to checking where f is increasing and decreasing. We first
compute
f 0 (x) = 5(4 − x2 )4 · (−2x) = −10x(4 − x2 )4 .
It follows that f 0 (x) = 0 exactly when x = 0, ±2. These points divide up our
domain as (−∞, −2)∪(−2, 0)∪(0, 2)∪(2, ∞). Checking test points in each interval,
we find f 0 (−3) > 0, f 0 (−1) > 0, f 0 (1) < 0, and f 0 (3) < 0. It follows that f is
increasing on (−∞, −2) and (−2, 0), and decreasing on (0, 2) and (2, ∞). There is a
local maximum at x = 0 (where f (0) = 45 = 1024).
Lastly, we check the concavity of the graph. Computing the second derivative,
we find
f 00 (x) = −10(4 − x2 )4 − 40x(4 − x2 )3 · (−2x) = −10(4 − x2 )4 + 80x2 (4 − x2 )3
= 10(4 − x2 )3 (−(4 − x2 ) + 8x2 ) = 10(4 − x2 )3 (9x2 − 4).
In particular, we can see that f 00 (x) = 0 exactly when x = ±2, ± 32 . These points
divide up our domain as (−∞, −2) ∪ (−2, − 32 ) ∪ (− 23 , 23 ) ∪ ( 23 , 2) ∪ (2, ∞). Checking
test points, we find that f 00 (−3) < 0, f 00 (−1) > 0, f 00 (0) < 0, f 00 (1) > 0, and f 00 (3) <
0. So, f is concave down on (−∞, −2), (− 32 , 23 ), and (2, ∞), and concave up on
(−2, − 32 ) and ( 32 , 2). There are inflection points at x = ±2, ± 32 .
Putting all of the information we’ve discovered together, we should have a
graph that looks approximately like this:
6
1000
500
-3
-2
1
-1
2
3
-500
-1000
11) Follow our usual checklist of properties to sketch y =
x
.
x2 −9
Solution. We first note that the domain of the function f (x) = x2x−9 is all real numbers except ±3, i.e., the domain is (−∞, −3) ∪ (−3, 3) ∪ (3, ∞). When x = 0, we
have y = 0, and so the y-intercept is (0, 0). When y = 0, we have x = 0, so the
point (0, 0) is also the only x-intercept. This function is neither even nor odd, so
the graph doesn’t have any special symmetry. As for asymptotes, there are vertical
asymptotes at x = ±3, where one can check
lim f (x) = −∞
lim f (x) = ∞
x→−3−
x→3−
lim f (x) = ∞
lim f (x) = −∞.
x→3+
x→−3+
We also have a horizontal asymptote at y = 0, since lim f (x) = 0.
x→±∞
We now move on to checking where f is increasing and decreasing. We first
compute
f 0 (x) =
(x2 − 9)(1) − (x)(2x)
−(2x2 + 9)
=
.
(x2 − 9)2
(x2 − 9)2
Since the numerator is never zero, there are no points where f 0 (x) = 0. In fact, we
always have f 0 (x) < 0. So, the function f is always decreasing.
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Lastly, we check the concavity of f . We first compute
(x2 − 9)2 (−4x) − (−2x2 − 9)(2(x2 − 9)(2x))
(x2 − 9)4
−4x(x2 − 9)2 + 4x(2x2 + 9)(x2 − 9)
=
(x2 − 9)4
4x(x2 − 9)(−(x2 − 9) + (2x2 + 9))
=
(x2 − 9)4
4x(x2 + 18)
=
.
(x2 − 9)3
f 00 (x) =
It follows that f 00 (x) = 0 exactly when x = 0. This divides up our domain into
(−∞, −3) ∪ (−3, 0) ∪ (0, 3) ∪ (3, ∞). Cecking test points in each interval, we find
f 00 (−4) < 0, f 00 (−1) > 0, f 00 (1) < 0, and f 00 (4) > 0, so f is concave down on
(−∞, −3) and (0, 3), and concave up on (−3, 0) and (3, ∞). There is an inflection
point at x = 0 (where f (0) = 0).
Putting all of this information together, we should sketch a graph that looks
approximately like the following:
4
2
-4
2
-2
4
-2
-4
12) Follow our usual checklist of properties to sketch y = 1 +
1
x
+
1
.
x2
2
Solution. We first note that the domain of the function f (x) = 1 + x1 + x12 = x +x+1
x2
is all real numbers except 0, i.e., the domain is (−∞, 0) ∪ (0, ∞). There is no yintercept, since the function is not defined for x = 0. When y = 0, we have x2 +
x + 1 = 0 (using the second form of f ), which has no roots. So, there is also no
x-intercept. This function is neither even nor odd, so the graph doesn’t have any
special symmetry. As for asymptotes, the graph has a vertical asymptote at x = 0,
where one can check lim f (x) = ∞. The graph also has a horizontal asymptote at
x→0
y = 1, since lim f (x) = 1.
x→±∞
8
We now move on to finding where f is increasing and decreasing. We first compute
x2 (2x + 1) − (x2 + x + 1)(2x)
−x2 − 2x
−x − 2
=
=
.
2
2
4
(x )
x
x3
So, there is a single critical point at x = −2. Recalling that our domain is (−∞, 0) ∪
(0, ∞), this divides up our domain into (−∞, −2) ∪ (−2, 0) ∪ (0, ∞). Checking
1
3−2
= −27
< 0, f 0 (−1) = 1−2
= 1 > 0, and
some test points, observe that f 0 (−3) = −27
−1
−1−2
0
f (1) = 1 = −3 < 0. So, f is decreasing on (−∞, −2) < 0, increasing on (−2, 0),
and decreasing on (0, ∞). Notice that f has a local minimum at x = −2 (where the
value is f (−2) = 34 ).
Lastly, we check the concavity of f . We compute
f 0 (x) =
x3 (−1) − (−x − 2)(3x2 )
2x3 + 6x2
2(x + 3)
=
=
.
3
2
6
(x )
x
x4
So, the only time f 00 (x) = 0 is when x = −3. This divides up our domain as
(−∞, −3) ∪ (−3, 0) ∪ (0, ∞). Checking some test points, observe that f 00 (−4) =
2(−4+3)
< 0, f 00 (−1) = 2(−1+3)
> 0, and f 00 (1) = 2(1+3)
> 0. So, f is concave down on
(−4)4
(−1)4
(1)4
(−∞, −3), concave up on (−3, 0), and concave up on (0, ∞). Notice that f has an
inflection point at x = −3 (where the value is f (−3) = 97 ).
Putting all of this information together, we should sketch a graph that looks
approximately like the following:
f 00 (x) =
6
5
4
3
2
1
-4
-3
-2
1
-1
2
-1
It is a bit difficult to see in this graph, but there is an inflection point at x = −3 and
a local minimum at x = −2.
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