Download Solutions for Homework 5

M243. Fall 2011. Homework 5. Solutions.
H5.1 (i) Find an equation of a parabola that has curvature 6 at the origin.
(ii) Find all points of the curve y = xex where the curvature is the greatest or the smallest, if such
points exist.
Hint: (Use Maple. )
Solution (i) Let y = ax2 + bx + c be an equation of the parabola. Then, as we know, the curvature
can be computed by the formula:
κ(x) =
|y 00 |
|2a|
=
[1 + (y 0 )2 ]3/2
[1 + (2ax + b)2 ]3/2
|2a|
Substituting x = 0, we get [1+(b)
2 ]3/2 = 6. Since we need just one example of such a parabola, we
can set a = 3 and b = 0. Parabola y = 3x2 passes through the origin. Hence it is an example.
(ii) Using the same formula for the curvature as in part (i), we get
κ(x) =
|(x + 2)ex |
[1 + (x + 1)2 e2x ]3/2
It is clear that the smallest value of κ(x) is 0 and it is attained when x = −2. The corresponding
point on the curve is (−2, −2/e2 ) ≈ (−2, −.2706705664).
To find the largest value of κ(x) we take its derivative and find when it is does not exist or zero.We
understand that at x = −2, the derivative does not exist, but at this point the curvature is the
smallest (at we are looking at th epoint where it is the largest).
We do it by executing the following Maple commands:
> κ := x− > abs(x + 2) ∗ exp(x)/(1 + (x + 1)2 ∗ exp(2 ∗ x))( 3/2);
> dk := dif f (κ(x), x));
> f solve(dk, x);
This gives x0 = −3.014834651 a critical point of κ0 (x). Evaluating
> evalf (subs(x = −3.1, dk)); evalf (subs(x = −3.0, dk));
we obtain that κ0 (x) changes its sign at x0 from − to +. Therefore this is the point of local
maximum.
Will the function attain its absolute maximum at this point?
Lets look at the graph of our function at some interval containing point x0 :
> plot(x ∗ exp(x), x = −4..1); (see Figure 1)
It looks that at some point of [−1, 0] the curvature is is actually greater than at x0 ! Strange...
OK, lets then look at the graph of the curvature function κ(x) on the interval [−6, 4]:
> plot(kappa, x = −6..4); (see Figure 2)
It clearly shows that on this interval κ(x) has two local maxima, and the value at the newly
discovered local maximum is actually greater that at the first! Why then f solve command did not
find it?
Lets try to “help” it by pointing it to the interval where we expect to have the second local
maximum, namely [−2, 0].
> f solve(dk, x = −2..0);
1
2.5
2
1.5
1
0.5
–4
–3
–2
x
0
–1
1
Figure 1: Graph of y = xex
0.8
0.6
0.4
0.2
–6
–4
x
0
–2
2
Figure 2: Graph of y = κ(x)
2
4
This gives x1 = −.291125603. The corresponding point on the curve is (−.291125603, −.217593620).
Now we understand what happened: f solve did not look for all solutions of the equation dk = 0.
Even if we specify an interval that contains both x0 and x1 , say [−4, 0], it still finds only x0 .It is
typical for this command (on polynomials it works better). Maybe it is improved in newer versions
of Maple.
What do we do now? Have we solved the problem? Strictly speaking the answer is “no”. Maybe
there are some critical points for κ(x) outside the interval [−4, 0]. Experimenting with the graph
we see that it is not the case. We can also show (and this is a more rigorous way) that on (−∞, x0 )
and on (x1 , ∞), dk = κ0 (x) is positive and negative, respectively.
Thus the curvature of the curve y = xex is the greatest at x1 = −.291125603 and its value is
κ(x1 ) = .881243008.
H5.2 (a) Find an equation of a line which lies on the surface 9x2 − y 2 − z 2 = 0.
(b) Prove that there is no line lying on the surface 4x3 − y 2 + 3z 4 = 1.
Solution (a) It is easy to see that a line with a parametric equation ht, 3t, 0i is on the surface:
9(t)2 − (3t)2 − (0)2 = 0 for all t. This answers the question.
But suppose one wants to do much more than what is asked, namely to find equations of (or to
describe) all such lines. Let us show how this can be done.
Let hx0 + at, y0 + bt, z0 + cti be an equation of one of them. Then, substituting the coordinates of
points of this line into equation of the surface and simplifying, we get
(9x20 − y02 − z02 ) + 2(ax0 − by0 − cz0 )t + (9a2 − b2 − c2 )t2 = 0
The expression on the left hand side of this equation is a quadratic polynomial of t. Since it is
zero for more than 2 values of t (actually for all t), all its coefficients must be zero. Therefore
x0 , y0 , z0 , a, b, c satisfy the following system of equations:
9x20 − y02 − z02 = 0, ax0 − by0 − cz0 = 0, 9a2 − b2 − c2 = 0
This is a system of three equations containing six unknowns.
At this moment one may decide to solve this system by using Maple, eg. by executing the command
> solve(9 ∗ x20 − y02 − z02 , a ∗ x0 − b ∗ y0 − c ∗ z0 , 9 ∗ a2 − b2 − c2 );
This is a good idea, and we encourage the reader to do it. The output will be useful, but some
work will be needed to see how it helps.
Here we show how one can proceed without Maple. First we remind ourselves that a line can have
infinitely many distinct parametric equations, since there are infinitely many ways to pick a point
(x0 , y0 , z0 ) on it and a direction vector ha, b, ci. All direction vectors can be obtained from one by
multiplying it by a proper nonzero constant. For example, if we multiply ha, b, ci by 1/a, we get
another direction vector h1, b/a, c/ai. This multiplication can be performed if a is not zero. But at
the moment we do not know that this is the case. Is it?
Suppose a = 0. Then the last equation gives b2 + c2 = 0. Since b and c are real numbers, we
conclude that b = c = 0. Therefore a = b = c = 0. This is impossible, since zero vector can not be
used in a parametric equation of a line. Therefore in our case a 6= 0, and every line that lies on our
surface has a direction vector of the form h1, b0 , c0 i.
Another important conclusion from the fact that a 6= 0 is this: for every x0 one can find a value
of t such that x0 + at = 0. Just take t = −x0 /a. This tells us that each our line contains a point
with zero x-coordinate! What are other coordinates of this point? Substituting x0 = 0 to the first
equation, we get y02 + z02 = 0. Therefore y0 = z0 = 0, and we conclude that
3
every line that lies on our surface passes through the origin and its parametric equation can be
written in the form ht, b0 t, c0 ti.
After this, finding all lines on our surface is very easy. Substituting coordinates of points of such a
line into the system, one gets
0 = 0, 0 = 0, 9(1)2 − (b0 )2 − (c0 )2 = 0,
or just (b0 )2 + (c0 )2 = 9. Conversely, any line with parametric equation ht, b0 t, c0 ti, where b0 , c0 are
real numbers such that (b0 )2 + (c0 )2 = 9, lies on our surface. Hence this is a complete description
of all lines lying on the surface 9x2 − y 2 − z 2 = 0, which is, by the way, a cone.
If one notices that the surface is a cone from the very beginning, one could expect the result: there
are infinitely many lines lying on a cone. For each point of the cone different from the vertex there
exists a unique line lying on the cone and passing through this point.
(b) Suppose that a line hx0 + at, y0 + bt, z0 + cti lies on the surface 4x3 − y 2 + 3z 4 = 1. Note that
ha, b, ci 6= ~0. Then, substituting the coordinates of points of this line in the equation of the surface
and simplifying, we get an equation with respect to t of the form
3c4 t4 +(4a3 +12c3 z0 )t3 +(12x0 a2 +18c2 z02 −b2 )t2 +(−2y0 b+12x20 a+12cz03 )t+(−1−y02 +4x30 +3z04 ) = 0
The expression on the left hand side of this equation is a polynomial of t of degree 4. Since it
has more than four roots (every real number is a root!), all the coefficients must be equal to zero.
Therefore we get the system of equations
3c4 = 4a3 + 12c3 z0 = 12x0 a2 + 18c2 z02 − b2 = −2y0 b + 12x20 a + 12cz03 = −1 − y02 + 4x30 + 3z04 = 0
From the first equation we get c = 0. Substituting c = 0 to 4a3 + 12c3 z0 = 0, we get a = 0.
Substituting a = c = 0 to 12x0 a2 + 18c2 z02 − b2 = 0, we get b = 0. We can stop at this moment,
since we already showed that the direction vector ha, b, ci is zero vector. The obtained contradiction
implies that no line lies on the surface 4x3 − y 2 + 3z 4 = 1.
Comment To determine the coefficient at ti , i = 0, 1, 2, 3, 4, in the 4th degree polynomial, one can
use the following Maple command (to avoid doing it by hand):
> coeff(subs({x = x0 + a ∗ t, y = y0 + b ∗ t, z = z0 + c ∗ t}, 4 ∗ x3 − y 2 + 3 ∗ z 4 − 1), t, i);
H5.3 Consider the curve f~(t) = h2t2 − t, e2t + 3t − 2, 2e2t − 6t2 + 9t − 4i, t is any real number.
(a) Does the curve intersect itself ? Justify your answer.
(b) Is the curve planar? (i.e., is there a plane such that all points of the curve are in it). Justify
your answer.
Solution (a) Note that the second component of f~ is y(t) = e2t + 3t − 2 is increasing function on
(−∞, ∞). This can be seen from its derivative: y 0 (t) = 2e2t + 3 > 0 for all t. Therefore there exist
no two distinct values of t for which y(t) takes the same values. This alone proves that the curve
does not cross itself.
Another, and a more general approach would be to try to find two distinct numbers a, b such that
f~(a) = f~(b). Though it is easy to see from the first component that we must have 2a + 2b − 1 = 0,
it is not that easy to see that for these a and b we cannot have equalities of the other components.
(b) Let x(t) = 2t2 − t, y(t) = e2t + 3t − 2, and z(t) = 2e2t − 6t2 + 9t − 4. First we wish to say that by
just looking for a minute or two at the components of f~, one may note that 3x(t) − 2y(y) + z(t) = 0
for all t. Hence, the curve is in the plane α : 3x − 2y + z = 0, and the question is answered.
4
If one does not note the above, here is a more general way to proceed. The curve is planar if there
is a plane α containing it. Let α : ax + by + cz + d = 0 be a general equation of the plane. Then
points (x(t), y(t), z(t)) must belong to α for each t. Substitution gives
a(2t2 − t) + b(e2t + 3t − 2) + c(2e2t − 6t2 + 9t − 4) + d = 0,
which is equivalent to
(b + 2c)e2t + (2a − 6c)t2 + (−a + 9c)t + (−2b − 4c + d) = 0.
Therefore if there are numbers a, b, c, d, not all zero, satisfying the system
b + 2c = 0, 2a − 6c = 0, −a + 3b + 9c = 0, and − 2b − 4c + d = 0,
the curve is in α. Solving the system, we obtain a = 3c, b = −2c and d = 0. Taking c = 1, we
obtain an equation for the plane: α : 3x − 2y + z = 0.
We wish to mention about one more simple approach. Let 0, 1, 2, c be four distinct values of t.
Consider three vectors:
~ = f~(1) − f~(0), B
~ = f~(2) − f~(0), C
~ = f~(c) − f~(0).
A
~ B,
~ C
~ are coplanar for each c distinct
It is clear that the curve is planar if and only if the vectors A,
from 0, 1, 2. The last condition is equivalent to the mixed product of these vectors being zero:
~ · (A
~ × B)
~ = 0. We leave it for the reader to perform this verification. (Using Maple can, certainly,
C
help.)
H5.4 If you present solution for part (ii), you do not have to do it for part (i).
(i) Let A : (0, 0, 0), B : (3, 0, 0). Consider the set all points C in the space such that AC/BC = 2.
Prove that the set is a sphere. Find the coordinates of its center and its radius length.
Hint: Let C : (x, y, z).
(ii) Let A and B be two distinct points in space. For a fixed real number k, 0 < k 6= 1, consider the
set all points C in the space such that AC/BC = k. Prove that the set is a sphere. Describe the
position of the center and its radius length as functions of k.
Hint: Introduce a Cartesian coordinate system in the space such that A becomes the origin and
B : (1, 0, 0). Let C : (x, y, z).
(iii) Let (a, b, c) be the coordinates of the center of the sphere from part (ii), when the coordinate
system was chosen as was suggested in the hint, and r be the length of the radius. Then a = a(k)
and r = r(k).
(i) What can be said about
lim a(k),
and
lim a(k),
and
k→1
lim r(k)?
k→1
(ii) What can be said about
k→0+
lim r(k)?
k→0+
Solution
(i) We know from the distance formula that AC/BC = 2 if and only if
p
x2 + y 2 + z 2
p
=2
(3 − x)2 + y 2 + z 2
5
which is equivalent to
p
x2 + y 2 + z 2 = 2
p
(3 − x)2 + y 2 + z 2
x2 + y 2 + z 2 = 4(3 − x)2 + 4y 2 + 4z 2
3x2 − 24x + 3y 2 + 3z 2 = −36
x2 − 8x + y 2 + z 2 = −12
(x − 4)2 + y 2 + z 2 = 4.
When squaring, we used the fact that for positive a, b: a = b ⇔ a2 = b2 . Since this is the equation
for a sphere in standard form, the set of points C such that AC/BC = 2 must be the sphere with
center (4, 0, 0) and radius 2.
(ii) As the hint suggests, we choose to use a Cartesian coordinate system such that A : (0, 0, 0) and
B : (1, 0, 0), and we let C(x, y, z) as in part (i). Using the distance formula, we have AC/BC = k
if and only if
p
x2 + y 2 + z 2
p
=k
(x − 1)2 + y 2 + z 2
and so
p
x2 + y 2 + z 2 = k
2
2
2
p
(x − 1)2 + y 2 + z 2
2
2
2 2
⇔
(as both sides are nonnegative)
2 2
x + y + z = k (x − 1) + k y + k z
2
2k
k2
2
2
x
+
y
+
z
=
−
⇔ (as k 2 6= 1)
k2 − 1
k2 − 1
¶2
µ
k2
k4
k2
+ y2 + z2 = − 2
+ 2
x− 2
k −1
k − 1 (k − 1)2
µ
¶
2
k2
k2
x− 2
+ y2 + z2 = 2
.
k −1
(k − 1)2
³ 2
´
Again, the set is a sphere. The center is at k2k−1 , 0, 0 and the radius is
x2 −
(iii) From part (ii), we have b(k) = c(k) = 0, a(k) =
following.
k2
k2 −1 ,
and r(k) =
k
|k2 −1| ,
k
|k2 −1| .
as k > 0.
So we obtain the
k2
does not exist. More precisely,
k→1 k 2 − 1
lim a(k) = lim
k→1
lim− a(k) = lim−
k→1
k→1
lim r(k) = lim
k→1
k→1
k2
k2
= −∞, and lim+ b(k) = lim+ 2
= ∞.
−1
k→1
k→1 k − 1
k2
k
= ∞.
|k 2 − 1|
As we see, when k → 1, the center of the sphere moves to −∞ or ∞ along x-axis, depending on
whether k < 1 or k > 1. At the same time the radius length is becoming “infinitely large”. When
k = 1, the set of all points C such that AC/BC = 1 is the set of points C such that AC = BC.
Now it is clear from geometry that it is the plane which is perpendicular to the segment AB and
that passes through the midpoint of it. This coincides with our intuition that a plane is a “sphere
of infinite radius” with the “center at infinity”.
6
Investigating what happens when k → 0+ , i.e., AC is becoming much smaller than BC, we obtain
lim a(k) = 0;
k→0+
k2
= 0;
−1
k→0
k→0
k
lim+ r(k) = lim+ 2
= 0.
k→0
k→0 |k − 1|
lim+ b(k) = lim+
k2
Hence, the sphere “shrinks” to point A, which again coincides with our intuition.
This sphere is often referred as the Sphere of Apollonius, due to Apollonius from Perga (c. 262 – 190
B.C. ), who was a great Greek mathematician, and who contributed a great deal to our understanding of conic sections.
7