Download M101 Tut4_SolnD

MATHEMATICS 101
TUTORIAL 4 (SOLUTIONS TO DISCUSSIONS PROBLEMS)
MODULE 5 – LINEAR AND MATRIX ALGEBRA 1
Exercise 13.5 (page 867)
Equation of a plane
Question 28
Find the equation of the plane through the point (–1, 6, –5) and parallel to the plane
x  y  z  2  0.
Solution
The vector equation of the plane x  y  z  2  0 is r  1, 1, 1 = –2.
Since the planes are parallel, they have the same normal, that is, the vector equation of the
required plane is
r  1, 1, 1 =  1, 6,  5  1, 1, 1
that is,
r  1, 1, 1 = (1)(1)  (6)(1)  (5)(1)  0 .
Its Cartesian equation is x  y  z  0 .
x yz20
n=i+j+k
x yz 0
Fig. 1
Courtesy Mr. Rajesh Gunesh
Question 35
Find the equation of the plane that passes through the point (6, 0, –2) and contains the line
x = 4 –2t, y = 3 + 5t, z = 7 + 4t.
Solution
n
P (6, 0, –2)
Q (4, 3, 7)
Fig. 2
The vector equation of the line given by parametric equations x = 4 –2t, y = 3 + 5t, z = 7 + 4t is
r = 4, 3, 7  t  2, 5, 4 .
Choose a point Q, for example, (4, 3, 7) on the line.
Then, PQ = 4, 3, 7  6, 0,  2   2, 3, 9 .
It is clear that the normal n to the plane is orthogonal to both the line and PQ (see Fig. 2).
The normal vector to the plane can be easily found by the vector product of the direction vector
of the line,  2, 5, 4 , and  2, 3, 9 , given by
i
j
k
2
5
4
2
3
9
= i
5 4
3 9
 j
2
4
2
9
k
2
5
2
3
= 33i + 10j + 4k.
The equation of the plane is thus
r  33, 10, 4  6, 0,  2  33, 10, 4  190 .
Its Cartesian equation is 33x  10 y  4 z  190 .
Courtesy Mr. Rajesh Gunesh
Question 42
Where does the line through (1, 0, 1) and (4, –2, 2) intersect the plane x + y + z = 6?
Solution
Let P and Q be the points (1, 0, 1) and (4, –2, 2) respectively. One vector equation of line PQ is
therefore OP  t PQ , given by
r = 1, 0, 1  t  4,  2, 2  1, 0, 1  , that is,
r = 1, 0, 1  t 3,  2, 1 ,
The parametric equations of the line are x = 1 + 3t, y = –2t, z = 1 + t.
P (1, 0, 1)
Q (4, –2, 2)
Point of intersection
x+y+z=6
r = 1, 0, 1  t 3,  2, 1
Fig. 3
The point of intersection is contained in the plane and, thus, has to satisfy the equations of both
line and plane (see Fig. 3). Substituting the parametric equations of the line in the equation of
the plane, we have
(1 + 3t) + (–2t) + (1 + t) = 6
 2t  2  6 and t = 2.
The co-ordinates of the point of intersection are (1 + (3)(2), (–2)(2), 1 +(2)), that is, (7, –4, 3).
Courtesy Mr. Rajesh Gunesh
Question 55
Find an equation for the plane consisting of all points that are equidistant from the points
(1, 1, 0) and (0, 1, 1).
Solution
Let A and B be the points (1, 1, 0) and (0, 1, 1) respectively. The plane consisting of all such
points will (naturally) fall ‘between’ the points A and B in space (see Fig. 1.4).
A (1, 1, 0)
d
P
n
d
B (0, 1, 1)
Fig. 4
Its normal vector will not only be parallel to the line segment AB but also contain the midpoint
this line segment (think about it!). In that case, any point P on the plane would be equidistant to
A and B.
The direction vector of line segment AB, that is, the normal to the plane is
0, 1, 1  1, 1, 0   1, 0, 1
 0 1 11 1 0  1
,
,
and the midpoint of AB is 
   , 1,
2
2  2
 2
1
2
.
The vector equation of the plane is therefore
r   1, 0, 1 
1
2
, 1,
1
2
  1, 0, 1  0 .
Its Cartesian equation is –x + z = 0 or x = z.
Courtesy Mr. Rajesh Gunesh
Question 69
Show that the distance
ax  by  cz  d 2  0 is
between
the
D
parallel
d1  d 2
a2  b2  c2
planes
ax  by  cz  d1  0
and
.
Solution
 1 : ax  by  cz  d1  0
P
l1
D
n
O
l2
Q
 2 : ax  by  cz  d 2  0
Fig. 5
The vector equations of planes  1 and  2 (see Fig. 5) are respectively
r  a, b, c   d1 and r  a, b, c   d 2
Since the planes are parallel, they have the same normal vector. If O is the origin, then it should
be clear that the distances l1 and l2 are respectively the scalar projections of OP and OQ onto
the normal vector n, where P and Q are any two points on  1 and  2 respectively. Recall that
these two points have position vectors conventionally denoted by r.
Thus, r 
a, b, c
a2  b2  c2
 l1 
 d1
a2  b2  c2
and r 
a, b, c
a2  b2  c2
  l2 
 d2
a2  b2  c2
.
Note that the second scalar product above is negative since, from the diagram, lies below the
d1  d 2
origin. Therefore, D = l1 – (–l2) = l1 + l2 =
.
a2  b2  c2
Without any loss of cause, we could have taken both planes to be on the same side as the
origin. The result would still have been the same.
Courtesy Mr. Rajesh Gunesh
MODULE 2 – INTRODUCTORY CALCULUS
Exercise 2.5 (page 111)
Continuity
Question 18
Explain why the function
 1
if x  1

f ( x)   x  1
2
if x  1
is discontinuous at x = 1. Sketch the function.
Solution
(i)
Note that lim f ( x)   and lim f ( x)   . Since the left-hand and right-hand limits
x 1
x 1
are not the same, we conclude that lim f ( x) does not exist and deduce that the function
x1
is discontinuous at x = 1.
(ii)
y
2
0
2
x
Courtesy Mr. Rajesh Gunesh
Exercise 3.1 (page 132)
The derivative
Question 4
If the tangent line to y  f (x) at (4, 3) passes through the point (0, 2), find f (4) and f (4) .
Solution
It is too obvious that f (4)  3 since the point (4, 3) lies on the curve!
f (4) is just the gradient of y  f (x) at the point (4, 3), that is, the gradient of the tangent line.
Gradient of tangent =
Therefore, f (4) =
32 1
 .
40 4
1
.
4
Question 19
(1  h)10  1
represents the derivative of some function f at some point a. State such an f
h 0
h
and the value of a.
lim
Solution
Consider the derivative of the function f ( x)  x10 from first principles.
f ( x)  lim
h 0
f ( x  h)  f ( x )
( x  h)10  x10
.
 lim
h 0
h
h
(1  h)10  1(10)
(1  h)10  1
.
 lim
h 0
h 0
h
h
At x = 1, it is clear that f (1)  lim
Therefore, f ( x)  x10 and a = 1.
Courtesy Mr. Rajesh Gunesh
Exercise 3.2 (page 144)
The derivative
Question 25
Find the derivative of g ( x)  1  2 x using the definition of derivative. State the domain of the
function and the domain of its derivative.
Solution
Differentiating from first principles,
g ( x)  lim
h0
= lim
h 0
1  2( x  h)  1  2 x
g ( x  h)  g ( x)
 lim
h0
h
h
 1  2( x  h)  1  2 x  1  2( x  h) 
h 1  2( x  h)  1  2 x 
1  2x

(rationalising numerator)
= lim
h 0
= lim
h 0
= lim
h 0
=
1  2( x  h)  (1  2 x)

h 1  2( x  h)  1  2 x
2h

h 1  2( x  h)  1  2 x
2
 1  2( x  h) 
2
 1  2x 
1  2x

1  2x


= lim
h 0
2h

h 1  2( x  h)  1  2 x

 since h  0, h  0
2
  1  2( x  0) 
=
2
2 1  2x

1
1  2x
1  2x

.
Df : { x | x   12 }
Df’ : { x | x   12 }
Courtesy Mr. Rajesh Gunesh
Question 41
Show that the function f ( x)  x  6 is not differentiable at 6. Find a formula for f  and
sketch its graph.
Solution
Using differentiation from first principles,
f ( x)  lim
( x  h)  6  x  6
h 0
h
Using the fact that x  6  ( x  6) for x > 6, we can use h small enough so that (x + h) – 6 > 0.
In that case, ( x  h)  6  x  h  6 . Thus,
f ( x)  lim
( x  h)  6  x  6
h 0
 lim
h 0
h
( x  h)  6  ( x  6)
h
 lim  lim 1  1 .
h

0
h
h h 0
Therefore, f is differentiable for any x > 6.
Similarly, x  6  ( x  6) for x < 6. Again, we can use h small enough so that (x + h) – 6 < 0.
This time,
f ( x)  lim
( x  h)  6  x  6
h 0
 lim
h 0
h
 ( x  h)  6  ( x  6)
h
 lim
 lim (1)  1 .
h 0
h 0
h
h
Therefore, f is differentiable for any x < 6.
However, since
lim
( x  h)  6  x  6
h 0
 lim
( x  h)  6  x  6
h 0
h
h
,
f (6) does not exist. Thus, f is differentiable at all x except 6.
Note that
d
x  d
dx
dx
 x   dxd x 
2
2
1
2

 
1 2
x
2
 12
(2 x) 
x
x 
2
1
2

x
.
x
This result can generalised for the function f ( x)  x  a so that f ( x) 
( x  a)
.
xa
Courtesy Mr. Rajesh Gunesh
( x  a) for x  a
From definition, x  a  
 ( x  a) for x  a
It is therefore easy to conclude that for x < a, f ( x)  1 and that, x > a, f ( x)  1 .
The graph of y  f (x) (a = 6) is given below.
y
1
0
4
8
x
–1
Fig. 3
Courtesy Mr. Rajesh Gunesh