18.01 Problem Set 1 Solutions
Part II: 15 points
1a) A circle of radius r has area πr 2 and circumference 2πr. The rate of change of
the area with respect to the radius is therefore equal to the circumference. Explain
geometrically why this is so.
If the radius of the circle changes from r to r + ∆r, the circle has added to it a circular band
of thickness ∆r and circumference 2πr. The area of this band is exactly the change ∆A in the area
of the circle. It is approximately equal to 2πr∆r; the reason for the “approximately” is that the
outer circumference of the band is a little more than 2πr, so the area is a little more than this. So
the rate of change in area is approximately
∆A
≈ 2πr.
∆r
As ∆r approaches zero, this approximation gets better and better (the inner and outer circumferences of the band get closer together) so the limit is 2πr.
b) What is the analogous fact for squares?
Write s for the side of the square; then the area is A(s) = s2 . Adding ∆s to a side adds to
the square an inverted “L,” of thickness ∆s and length 2s (two legs, each of length s). (The key
point is that we are not adding a band of thickness ∆s around the entire square.) The area of
this “L” is approximately 2s∆s. (Exactly it’s 2s∆s + (∆s)2 .) So the rate of change in the area is
approximately
∆A
≈ 2s.
∆s
(The error in the approximation is ∆s.) As ∆s approaches zero, this approximation gets better,
so the limit is 2s.
2. The greatest integer in x, written [x], is defined to be the largest integer that is less
than or equal to x. Thus for example
[2.5] = 2,
[π] = 3,
[10] = 10,
[−2/3] = −1.
The fractional part of x, written {x}, is defined by
{x} = x − [x].
For example,
{2.5} = .5,
{π} = .1415926535 . . . ,
{10} = 0,
{−2/3} = 1/3.
a) Graph the function h(x) = {x} − {x}2 on the interval from −2 to 2.
b) Is the function h(x) continuous? Differentiable? Periodic? Even? Odd? Explain
why or why not.
The function {x} is periodic of period 1: in other words {x + 1} = {x} for every x. (That is,
the fractional part of x + 1 is equal to the fractional part of x.) Therefore {x}2 and h(x) are also
periodic of period 1.
The question of even and odd is a little trickier, since {x} is neither even nor odd. These
properties concern what happens when x is replaced by −x, and what’s true is
{−x} =
1 − {x}
0
if x not an integer
if x an integer.
If we plug this fact into the equation h(x) = {x}(1 − {x}), we find h(−x) = h(x), and therefore
that h is even. It is not odd, since the only function that is both even and odd is zero.
The function [x] is constant on each interval between two integers. It is therefore continuous
and differentiable whenever x is not an integer. The function x is continuous and differentiable
everywhere. Consequently {x} = x − [x] is continuous and differentiable whenever x is not an
integer. Since sums and products of continuous and differentiable functions are continuous and
differentiable, h is continuous and differentiable whenever x is not an integer.
Suppose j is an integer. If x = j + ∆x is close to j but slightly larger (so that 0 < ∆x < 1
then {x} = ∆x, and
h(j + ∆x) = (∆x) − (∆x)2 → 0
as ∆x → 0+ . If x is close to j but slightly smaller (so that −1 < ∆x < 0, then {x} = 1 + ∆x, and
h(j + ∆x) = (1 + ∆x) − (1 + ∆x)2 = −∆x − (∆x)2 → 0
as ∆x → 0− . So the left and right limits are both equal to 0, which is h(j); so
lim h(j + ∆x) = h(j),
∆x→0
and h is continuous at j. So h is continuous everywhere.
Exactly the same calculation shows that
∆x − (∆x)2
∆h
=
= 1 − ∆x → 1
∆x
∆x
as ∆x → 0+ ; and
∆h
−∆x − (∆x)2
=
= −1 − ∆x → −1
∆x
∆x
as ∆x → 0− . So the left and right limits are unequal, and the derivative does not exist at j. (This is
pretty clear by looking at the graph as well; the graph takes a sharp bend from pointing 45◦ down
to pointing 45◦ up at each integer, so there is no tangent line.) Therefore h is not differentiable (at
integer values of x).
3a) Graph (together) the three functions f (x) = x, g(x) = −x, and h(x) = x cos(x) on the
interval from x = −π to x = π.
b) Find the slope of the tangent line to the graph of h at x = π/4.
According to the product rule, h′ (x) = 1 · cos(x) + x · (− sin(x)). At x = π/4, this is
1·
√
√
√
2/2 − (π/4) · ( 2/2) = (1 − π/4)( 2/2).
Since π is smaller than 4, the first factor is positive; and of course the second factor is positive. So
the derivative at π/4 is strictly positive.
√
c) When x = π/4, h(x) takes the value π 2/8 (about .555). Is that the largest value it
takes for x between −π and π? (Hint: you don’t need to do any additional calculation.)
I already said that the derivative at π/4 is strictly positive. This means that for sufficiently
small positive ∆x, we must have
(h(π/4 + ∆x) − h(π/4))/∆x > 0.
(These quotients get closer and closer to the derivative as ∆x gets smaller, so they must eventually
be positive.) As soon as the quotient is positive, the numerator must also be positive. That is,
h(π/4 + ∆x) > h(π/4),
(∆x > 0 small enough).
This says that h takes larger values than h(π/4) at everyplace bigger than π/4 but close to it. So
the answer to the question is no.