MATH 126: 06/24/2014
1. Show that the following is the equation for a sphere:
2x2 + 2y 2 + 2z 2 − 12x + 4z = 40
Solution. We complete the square of the above equation as follows:
2x2 + 2y 2 + 2z 2 − 12x + 4z = 40
⇔x2 + y 2 + z 2 − 6x + 2z = 20
⇔(x2 − 6x) + (y 2 ) + (z 2 + 2z) = 20
⇔(x2 − 6x + 9) + (y 2 ) + (z 2 + 2z + 1) = 30
⇔(x − 3)2 + y 2 + (z + 1)2 = 30
This means that the original
equation represents a 2-dimensional sphere with center
√
(3, 0, −1) and radius 30.
2. Let S be the set of points which are three times as far from the point (3, 2, 1) as
they are from (1, 0, 3). Show that S is a sphere by finding its equation.
Solution. First write down the equation for which all points (x, y, z) in S satisfies:
p
p
3 (x − 1)2 + y 2 + (z − 3)2 = (x − 3)2 + (y − 2)2 + (z − 1)2 .
Square the term on both side and expand out all factors, after some (tedious)
algebra we find out that points in S satisfy the equation
2x2 + 2y 2 + 2z 2 − 3x + y − 13z + 19 = 0.
Now completing square as we did in Problem 1, we find out
2 2 2
1
13
27
3
+ y+
+ z−
= .
x−
4
4
4
16
This
√ means S is a sphere with center (3/4, −1/4, 13/4) and radius
3 3/4.
p
27/16 =
Selected Solution for Review sheet 2.
1.(c) We will use chain rule to find derivative for composite functions. Let w(v) =
a(tan(bv)), then
du
du dw
=
·
,
dv
dw dv
where
du
1
u(w) = ln(w) ⇒
= ,
dw
w
and
dw
w(v) = a(tan(bv)) ⇒
= ab sec2 (bv).
dv
Hence
du
du dw
ab sec2 (bv)
b sec2 (bv)
=
·
=
=
.
dv
dw dv
w
tan(bv)
1
1.(d) The notation
d2 y
dt2
means the second derivative of function y(t). Now
dy
1
=
,
dt
1 + t2
and hence
d
1
−2t
d2 y
d dy
=
=
=
.
2
2
dt
dt dt
dt 1 + t
(1 + t2 )2
where the last equation follows from quotient rule.
2.(a) Let ln x = t, then we have x = et , dx = et dt and the integration over t ranges from
[1, 2]. This gives
( 2
Z 2 t
Z 2
Z e2
ln t1 = ln 2
p = 1;
e dt
dt
dx
=
=
=
t−p+1 2
1
1−p
p
t
p
p
x(ln x)
= 1−p (2
− 1) p 6= 1.
1 e t
1 t
e
−p+1 1
2.(b) We do by integration by part as follows:
Z
Z
Z
1
1
1
1
−2x
−2x
−2x
xe
dx = −
x d(e ) = − xe
− e−2x dx = − xe−2x + e−2x .
2
2
2
2
2