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(MATH100)midterm100sol.pdf
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1. [10 marks] Divide the positive number a into three parts (i.e., a = x + y + z) such that their product shall be a maximum. Find the three parts as well as the product.
Solution: Let x = .rst part, y = second part. Then z = a . (x + y)= a . x . y = third part, and the function to be examined is
f(x, y)= xy(a . x . y). .f .f
First Step.= ay . 2xy . y2 =0, = ax . 2xy . x2 = 0.
.x .y aa
Solving simultaneously, we get the critical point (x, y)=(, ), x and y are non-zero.
33
.2f .2f .2f
Second Step. .x2 = .2y, .x.y = a . 2x . 2y, .y2 = .2x.
D(x, y)=4xy . (a . 2x . 2y)2 .
aa a2 .2f 2a
Third Step. Since D(, )= > 0 and = . < 0, it follows that
33 3 .x2 3
3
aa
the product = is maximum when x = y = z =.
27 3
2. [10 marks] A bug moves along a path C given by the equations x =2t and y = .t2 . If
2
z = x2 + y, .nd dz along C at the instant when the bug is at the point (2, .1), where s is
ds
the arc length of the path in the xy-plane. [Hint: The arc length is de.ned as ds = p(dx)2 +(dy)2.]
Solution: Recall that
. dx dy
ds = .(dx)2 +(dy)2 = ()2 +()2 dt = .4+4t2 dt.
dt dt
Thus,
ds
=2.1+ t2 .
dt
Since
dz dzdt
=
ds dtds
and
dz .zdx .zdy
3
= +=4x . 4yt =8t +4t,
dt .x dt .y dt
we have
dz 14t +2t3
= (8t +4t3) = . ds 2 1+ t2 1+ t2
Evaluating the derivative dz/ds at the point (2, .1), (i.e., when t = 1), we have
dz 6
.
= =3 2.
ds t=1 2
3. [10 marks] Find parametric equations for the line that contains the point (3, 1, .2) and intersects the line x = .1+ t, y =2+ t, z =1 . t at a right angle.
w
(3, 1, .2)
d
The given line r(t)= r0 + tv
v
r1
r0
The required line r(s)= r1 + sw
O
Solution: [Method 1] The given line has the vector form equation r(t)= r0 + tv, where r0 = ..1, 2, 1., v = .1, 1, .1.. Let the required line have the vector form equation r(s)= r1 + sw, where r1 = .3, 1, .2., w = to be determined. Denote d = r1 . r0 = .4, .1, .3.. Since w v we can determine w by w + projd = d, or
v
w = d . projvd = d . .d v v v . v
= .4, .1, .3. . .4, .1, .3. 3 .1, 1, .1. .1, 1, .1.
= .4, .1, .3. . 2.1, 1, .1.
= .2, .3, .1..
The normal line has the parametric equations
x =3+2s, y =1 . 3s, z = .2 . s.
[Method 2] Recall that d = r1 . r0 = .4, .1, 3. and v = .1, 1, .1.. The required line has the vector form equation r(s)= r1 + sw, where w can be determined by
(d v) v
=
. .
.
i j k
4 .1 .3
1 1 .1
. .
.
v =
(4i + j +5k) v
=
. .
.
i j k
4 1 5
1 1 .1
. .
.
= .6i +9j +3k = .3 (2i . 3j . k) .
Then we take w =2i . 3j . k. The required line has the parametric equations
x =3+2s, y =1 . 3s, z = .2 . s.
[Method 3] Let r(t1)= r0 + t1v = ..1+ t1, 2+ t1, 1 . t1. be the point on the given line that is closest to the point (3, 1, .2) on the required line. To .nd t1, we try to minimize the distance function
D(t)= .r(t) ..3, 1, .2..2 =(t .