Harvard-MIT Mathematics Tournament

March 15, 2003

Individual Round: Calculus Subject Test — Solutions

1. A point is chosen randomly with uniform distribution in the interior of a circle of radius
1. What is its expected distance from the center of the circle?
Solution: 2/3
The probability of the point falling between a distance r and r + dr from the center
is the ratio of the area of the corresponding annulus to the area of the whole circle:
π[(r+dr)2 −r2 ]
π
→ 2πrπ dr = 2r dr for small values of dr. Then the expected distance is
R1 2
0 r cot 2r dr = 3 .

2. A particle moves along the x-axis in such a way that its velocity at position x is given
by the formula v(x) = 2 + sin x. What is its acceleration at x = π6 ?
√
Solution: 5 3/4
√
Acceleration is given by a = dv
dt
= dv dx
·
dx dt
= dv
dx
· v = cos x · (2 + sin x) = 5 3/4.
3. What is the area of the region bounded by the curves y = x2003 and y = x1/2003 and
lying above the x-axis?
Solution: 1001/1002
The two curves intersect at (0, 0) and (1, 1), so the desired area is
Z 1³ ´ · ¸1
1/2003 2003 x2004/2003 x2004 1001
x −x dx = − = .
0 2004/2003 2004 0 1002
4. The sequence of real numbers x1 , x2 , x3 , . . . satisfies limn→∞ (x2n + x2n+1 ) = 315 and
limn→∞ (x2n + x2n−1 ) = 2003. Evaluate limn→∞ (x2n /x2n+1 ).
Solution: −1
We have limn→∞ (x2n+1 −x2n−1 ) = limn→∞ [(x2n +x2n+1 )−(x2n +x2n−1 )] = 315−2003 =
−1688; it follows that x2n+1 → −∞ as n → ∞. Then
x2n x2n + x2n+1
lim = lim − 1 = −1,
n→∞ x n→∞ x2n+1
2n+1

since x2n + x2n+1 → 315 while x2n+1 → −∞.

5. Find the minimum distance from the point (0, 5/2) to the graph of y = x4 /8.
√
Solution: 17/2
We want to minimize x2 +(x4 /8−5/2)2 = x8 /64−5x4 /8+x2 +25/4, which is equivalent
to minimizing z 4 /4 − 10z 2 + 16z, where we have set z = x2 . The derivative of this
expression is z 3 − 20z + 16, which √
is seen on inspection
√ to have 4 as a root, leading to
the factorization (z − 4)(z + 2 − 2 2)(z + 2 −√2 2). Since z = x2 ranges over [0, ∞),
the possible minima are√at z = 0, z = −2 + 2 2, and z = 4. However, the derivative
is positive on (0, −2 + 2 2), so this leaves only 0 and 4 to be tried. We find that the
minimum is in fact achieved
q at z = 4, so the closest point on the graph is given by
√
x = ±2, with distance 2 + (24 /8 − 5/2)2 = 17/2.
2

1
6. For n an integer, evaluate
µ ¶
1 1 1
lim √ + √ + · · · + q .
n→∞ n2 − 02 n2 − 12 n2 − (n − 1)2

Solution: π/2
Note that √ 1
n2 −i2
= 1
n
·√ 1
, so that the sum we wish to evaluate is just a Riemann
1−( ni )2
sum. Then,
µ ¶ Z 1
1 n−1
X 1 1 −1 1 π
lim q = √ dx = [sin x]0 = .
n→∞ n i=0 1 − ( i )2 0 1 − x2 2
n

7. For what value of a > 1 is Z a2

1 x−1
log dx
a x 32
minimum?
Solution: 3
R 2
Let f (a) = aa x1 log x−1
32
df
dx. Then we want da = 0; by the Fundamental Theorem of
Calculus and the chain rule, this implies that
à ! à !
1 a2 − 1 1 a−1 d Z a2 1 x−1 Z a
1 x−1
2a 2 log − log = log dx − log dx = 0,
a 32 a 32 da c x 32 c x 32
2 2
where c is any constant with 1 < c < a. Then 2 log a 32−1 = log a−1 32
, so that ( a 32−1 )2 =
a−1
32
. After canceling factors of (a−1)/32 (since a > 1), this simplifies to (a2 −1)(a+1) =
32 ⇒ a3 + a2 − a − 33 = 0, which in turn factors as (a − 3)(a2 + 4a + 11) = 0. The
quadratic factor has no real solutions, so this leaves only a = 3. However, we have that
a > 1, and we can check that f (1) = 0, lima→∞ f (a) > 0, and f (3) < 0, so the global
minimum does occur at a = 3.

8. A right circular cone with a height of 12 inches and a base radius of 3 inches is filled
with water and held with its vertex pointing downward. Water flows out through a
hole at the vertex at a rate in cubic inches per second numerically equal to the height
of the water in the cone. (For example, when the height of the water in the cone is 4
inches, water flows out at a rate of 4 cubic inches per second.) Determine how many
seconds it will take for all of the water to flow out of the cone.
Solution: 9π/2
When the water in the cone is h inches high, it forms a cone similar to the original, so
that its base has radius h/4 and its volume is hence πh3 /48. The given condition then
states that à !
d πh3 πh2 dh dh 32
= −h ⇒ · = −h ⇒ 2h · =− .
dt 48 16 dt dt π
Integrating with respect to t, we get that h2 = −32t/π + C; setting t = 0, h = 12, we
get C = 144. The cone empties when h = 0, so 0 = −32t/π + 144 ⇒ t = 9π/2.

2
 9. Two differentiable real functions f (x) and g(x) satisfy

f 0 (x)
= ef (x)−g(x)
g 0 (x)

for all x, and f (0) = g(2003) = 1. Find the largest constant c such that f (2003) > c
for all such functions f, g.
Solution: 1 − ln 2
d
Rearranging the given equation gives f 0 (x)e−f (x) = g 0 (x)e−g(x) for all x, so dx (e−f (x) −
−g(x) 0 −f (x) 0 −g(x) −f (x) −g(x)
e ) = −f (x)e + g (x)e = 0. Thus, e −e is a constant, and it must
−f (0) −1 −f (2003) −g(2003)
be less than e = e . Thus, e <e +e = 2e−1 = eln 2−1 ⇒ f (2003) >
−1

1 − ln 2. On the other hand, we can find positive-valued functions e−f (x) , e−g(x) that
take on the required values at 0 and 2003 and have constant difference arbitrarily close
to e−1 . For example, for arbitrarily large t, we can set e−f (x) = e−(t(2003−x)+1) + e−1 −
e−(2003t+1) and e−g(x) = e−(t(2003−x)+1) , and we can check that the resulting functions
f, g satisfy the required conditions. Thus, we can make f (2003) arbitrarily close to
1 − ln 2, so this is the answer.

10. Evaluate Z ∞
1 − x2
dx.
−∞ 1 + x4

Solution: 0
R∞
Let S = 0 1/(x4 + 1) dx; note that the integral converges absolutely. Substituting
x = 1/u, so that dx = −1/u2 du, we have
Z ∞ Z 0 Z 0
1 1 du −u2
S= dx = = du
0 1 + x4 ∞ 1 + u−4 −u2 ∞ u4 + 1

Z ∞ Z ∞
u2 x2
= du = dx
0 1 + u4 0 1 + x4
(the
R∞
manipulations are justified by absolute convergence), from which we see that
2 4
0 (1 − x )/(1 + x ) dx = 0. Since the integrand is an even function, it follows that the
integral from −∞ to ∞ is zero as well.