Math 2250 Written HW #12 Solutions
1. A large international jet (like a Boeing 747 or an Airbus A340) typically has a takeoff speed at
sea level of about 80 m/s (which is approximately 180 mph), and needs to get up to that speed
within the 3000 meters of a typical international runway. Assuming the engines provide a
constant acceleration and the plane starts from rest, what’s the minimum acceleration needed
to ensure the plane will take off?
Answer: Let s(t) be the position of the plane on the runway t seconds after it starts accelerating down the runway. Let v(t) = s0 (t) be the plane’s velocity at time t, and let
a(t) = v 0 (t) = s00 (t) be the acceleration. Then we know that
a(t) = A
for some constant A which we’re trying to figure out.
Moreover, we know that the plane starts at one end of the runway, so
s(0) = 0,
and that it starts from rest, so
v(0) = 0.
Now, since v 0 (t) = a(t) = A, we just have to find some function with derivative A and we’ll
d
know that v(t) differs from that function by a constant. Clearly, dt
(At) = A, so we know
that
v(t) = At + C
for some constant C. Now, since v(0) = 0, we know that
0 = A(0) + C = C,
so in fact v(t) = At. In turn, since s0 (t) = v(t) and
s(t) =
d
dt
A 2
2t
= At, this means that
A 2
t +D
2
for some constant D. Since s(0) = 0, we see that
0=
so D = 0 and we have that s(t) =
A 2
(0) + D = D,
2
A 2
2t .
Of course, we still need to figure out what A is. To do that, we need to use the fact that the
plane should be going 80 m/s when it is 3000 meters down the runway. In other words, if we
call the time when the plane reaches the end of the runway t0 , we should have that
s(t0 ) = 3000
and
v(t0 ) = 80.
In other words,
A 2
t
2 0
80 = At0 ,
3000 =
1
so we see that t0 =
so A =
3200
3000
=
80
A.
Plugging that into the first equation yields
A 80 2
3000 =
2 A
A 6400
3000 = · 2
2 A
3200
,
3000 =
A
16
15 .
Therefore, the plane’s acceleration needs to be at least
2. Evaluate the limit
16
15
m/s2 to ensure takeoff.
1 x
lim
1+
.
x→+∞
x
(Although you derived this as a consequence, some people treat this as the definition of this
number.)
Answer: As x → +∞, the expression 1 + x1 converges to 1, so we have an expression in
the
∞ . To deal with this, let’s take the natural logarithm: ln 1 + 1 x =
indeterminate
form
1
x
x ln 1 + x1 , which is now in the form ∞ · 0. To deal with that, we can re-write as
ln 1 + x1
1
=
x ln 1 +
.
x
1/x
Therefore,
ln 1 + x1
1 x
lim ln
1+
= lim
,
x→+∞
x→+∞
x
1/x
which is now in a form which we can evaluate using L’Hôpital’s Rule. Since
d
1
1
−1
−1
ln 1 +
=
1 · x2 = x2 + x
dx
x
1+ x
and
d 1
−1
= 2,
dx x
x
L’Hôpital’s Rule tells us that
lim ln
x→+∞
1
1+
x
x ln 1 +
= lim
x→+∞
1/x
=
1
x
−1
x2 +x
lim
x→+∞ −1
x2
x2
= lim
x2 + x
x2
1/x2
·
= lim 2
x→+∞ x + x 1/x2
1
= lim
x→+∞ 1 + 1/x
= 1.
x→+∞
2
So we’ve seen that
lim ln
x→+∞
meaning that
1
1+
x
x = 1,
1 x
= e1 = e.
1+
x→+∞
x
lim
3. Use your knowledge of calculus to draw the graph of the function f (x) = x2 ex . In particular,
be sure to label any inflection points, local maxima, and local minima, and if the function
has an absolute minimum and/or an absolute maximum, find them.
Answer: First, we check for asymptotes. The function is defined everywhere, so there are
no vertical asymptotes. As for horizontal asymptotes,
lim x2 ex = +∞,
x→+∞
since both x2 and ex go to +∞. On the other hand,
lim x2 ex
x→−∞
is in the form ∞ · 0 since limx→−∞ ex = 0. Therefore, we can re-write in
using L’Hôpital’s Rule:
∞
∞
form and evaluate
x2
x→−∞ 1/ex
x2
= lim −x
x→−∞ e
d
2
dx x
= lim d
−x
x→−∞
dx (e )
2x
= lim
.
x→−∞ −e−x
lim x2 ex = lim
x→−∞
Notice that this is still in the form
∞
∞,
so we apply L’Hôpital’s Rule again:
2x
= lim
x→−∞
x→−∞ −e−x
lim
= lim
d
dx (2x)
d
−x
dx (−e )
2
x→−∞ e−x
=0
since the denominator goes to +∞. Therefore, the function has a horizontal asymptote at
y = 0.
Now, to look for critical points, we evaluate
f 0 (x) = 2xex + x2 ex = xex (2 + x),
which is defined everywhere, so the critical points occur when f 0 (x) = 0, meaning that either
x = 0 or x = −2. The derivative is positive and negative on the regions shown:
3
+
-
+
-2
0
which we can check by evaluating
3
>0
e3
−1
<0
=
e
f 0 (−3) = −3e−3 (2 + (−3)) = 3e−3 =
f 0 (−1) = −1e−1 (2 + (−1)) = −e−1
f 0 (1) = 1e1 (2 + 1) = 3e > 0
This shows that f (x) is increasing on (−∞, −2) and (0, +∞) and f (x) is decreasing on (−2, 0).
Therefore, f (x) has a local maximum at (−2, f (−2)) = −2, e42 and a local minimum at
(0, f (0)) = (0, 0). Since f (x) shoots off to +∞ as x → +∞, we know that the local maximum
can’t be an absolute maximum. On the other hand, since f (x) is increasing on (−∞, −2) and
approaches 0 as x → −∞, we see that f (x) can never get smaller than 0, so the point (0, 0)
is actually the absolute minimum of the function.
Finally, we can compute the second derivative:
d
(xex (2 + x))
dx
d
d
=
(xex ) (2 + x) + xex (2 + x)
dx
dx
= (1 · ex + xex ) (2 + x) + xex (1)
f 00 (x) =
= 2ex + xex + 2xex + x2 ex + xex
= ex (x2 + 4x + 2).
Therefore, the possible inflection points are when f 00 (x) = 0, meaning when x2 + 4x + 2 = 0,
which happens when
p
√
√
√
−4 ± 42 − 4(1)(2)
−4 ± 16 − 8
−4 ± 2 2
x=
=
=
= −2 ± 2.
2(1)
2
2
Therefore, the second derivative is positive and negative on the regions show:
+
-2 -
+
-2+ 2
2
which we can check by evaluating
2
>0
e4
−1
f 00 (−1) = e−1 ((−1)2 + 4(−1) + 2) =
<0
e
f 00 (0) = e0 (02 + 4(0) + 2) = 2 > 0.
f 00 (−4) = e−4 ((−4)2 + 4(−4) + 2) =
4
√ √
Therefore, the function is concave up on −∞, −2 − 2 and −2 + 2, +∞ and concave
√
√ down on −2 − 2, −2 + 2 , meaning that the inflection points are
√ 2 −2−√2
√ √ 2 −2+√2
√ and
−2 + 2, −2 + 2 e
.
−2 − 2, −2 − 2 e
Putting this all together yields the graph of the function:
2
-2,
J-2 -
2 , I-2 -
2
-2- 2
2M ã
N
4
ã2
1
O
æ
à
J-2+ 2 , H-2+ 2 L2 ã-2+
2
N
à
æ
-2 -
-2
2
5
-2+ 2 H0, 0L