Quiz 9
1. (a) Approach the origin along the x-axis (i.e., y = 0):
|x|
= lim 1 = 1.
x→0
(x,0)→(0,0) |x| + |y|
lim
Approach the origin along the y-axis (i.e., x = 0):
|x|
= lim 0 = 0.
|x| + |y| y→0
lim
(0,y)→(0,0)
Since the limit changes with the different directions, the limit does not exist.
(b) Recall that for every x, y:
π
−1
1
≤ ,
0 ≤ tan
x2 + y 2 2
so
π
1
2
2
−1
≤ tan x2 + y 2 .
0 ≤ tan x + y tan
x2 + y 2 2
Apply limits:
0 ≤ lim
tan x2 + y 2 tan−1
(x,y)→(0,0)
π
1
≤
2
2
x +y 2
2
2 lim
tan
x
+
y
(x,y)→(0,0)
= 0.
By the squeeze theorem,
lim
2
tan x + y
2
tan
−1
(x,y)→(0,0)
1
2
x + y2
= 0.
(c) Let x = r cos θ and y = r sin θ. Then:
r2
r→0
r2 + 1 − 1
2r
= lim+ 2r
√
r→0
−0
2 r 2 +1
p
= lim+ 2 r2 + 1
x2 + y 2
p
(x,y)→(0,0)
x2 + y 2 + 1 − 1
=
lim
lim+ √
r→0
=
2.
2. We want to find suitable a, b, c, d of ax + by + cz = d that satisfies the given conditions. Since
the trace on the xy-plane (i.e., when z = 0) is −x + y = 4, let’s be easy on ourselves and set
a = −1, b = 1, and d = 4. Then to find c.
The normal vector of this plane is h−1, 1, ci, and it should form an angle of π/3 with the angle
of the normal vector of P, h0, 1, −1i, so:
kh−1, 1, cikkh0, 1, −1ik cos (π/3) = 1 − c
p
√
2 + c2 =
2(1 − c)
2
2 + c = 2 − 4c + 2c2
= c(c − 4),
√
2 + c2 = 2(1 − c)), so our plane is −x + y = 4.
0
√
so c = 0 (c = 4 is extraneous, doesn’t satisfy
Any nonzero constant multiple of this works.
3. Note that
uxx (x, y) = −a2 cos(ax)eby
and
uyy (x, y) = b2 cos(ax)eby .
We need
∆u = (b2 − a2 ) cos(ax)eby = 0.
This must hold for every x, y, so b2 − a2 = 0. Thus, the set of a, b that makes u harmonic is
{(a, b) : a = |b|}.
4. Notice that the gradient of g is
∇gP = e−yz , −xze−yz , −xye−yz (x,y,z)=P
and the unit vector in the direction of v is
1
1
1
.
ev = √ , √ , √
3
3
3
Hence, the directional derivative is
Dev g(P )
=
=
∇gP · ev
1
√ e−yz − xze−yz − xye−yz .
3
Therefore,
Dev g(1, 2, 0)
=
=
1
√ (1 − 0 − 2)
3
1
−√ .
3
5. Let f (x, y) = xy 3 + 8y −1 , and note that
fx (x, y) = y 3
and
fy (x, y) = 3xy 2 − 8y −2 .
At a point a, b, the tangent plane is
z
=
f (a, b) + fx (a, b)(x − a) + fy (a, b)(y − b)
=
ab3 + 8b−1 + b3 (x − a) + (3ab2 − 8b−2 )(y − b)
=
ab3 + 8b−1 + b3 x − ab3 + (3ab2 − 8b−2 )y − 3ab3 + 8b−1 ,
so the equation of the plane is
b3 x + (3ab2 − 8b−2 )y − z = 3ab3 − 16b−1 .
To be parallel to 2x + 7y + 2z = 0, we need λ 6= 0 such that
hb3 , 3ab2 − 8b−2 , −1i = λh2, 7, 2i.
That is, the normal vectors of the plane are parallel. We can also think of this as having a function F (x, y, z) = k, where F (x, y, z) = b3 x + (3ab2 − 8b−2 )y − z, and be solving ∇F = λh2, 7, 2i.
From the third components, we see that λ = −1/2. It follows from the first components that
b3 = −1, so b = −1. Using the second components
gives a = 3/2. Therefore, the only point
that satisfies the given condition is 23 , −1, − 19
2 .
6. We’re in the form F (x, y, z) = k, where F (x, y, z) = x2 + z 2 ey−x and k = 13. At a point
P = (a, b, c), the equation for the tangent plane level to the surface is
∇FP · hx − a, y − b, z − ci
=
2x − z 2 ey−x , z 2 ey−x , 2zey−x (x,y,z)=P · hx − a, y − b, z − ci
= 2a − c2 eb−a , c2 eb−a , 2ceb−a · hx − a, y − b, z − ci.
When P = 2, 3, √3e , we have
0
=
0
=
=
=
=
9 1 9 1 6 1
3
· e , · e , √ e · x − 2, y − 3, z − √
e
e
e
e
√ 3
−5, 9, 6 e · x − 2, y − 3, z − √
e
√ √ 3
−5, 9, 6 e · hx, y, zi + −5, 9, 6 e · −2, −3, − √
e
√ −5, 9, 6 e · hx, y, zi − 35.
4−
Write the equation of this plane in whatever form is your favourite.
7. First, the partial derivative:
∂h
∂q
∂h ∂u ∂h ∂v
+
∂u ∂q
∂v ∂q
= (ev ) 3q 2 + (uev ) r2
= ev 3q 2 + ur2
=
Notice that when (q, r) = (3, 2), (u, v) = (27, 12), so
∂h = e12 (27 + 27 · 4)
∂q (q,r)=(3,2)
=
135e12 .
8. Let t ∈ R. We are considering f (x1 , x2 , x3 ), where x1 = tx, x2 = ty, and x3 = tz.
∂
f (x1 , x2 , x3 )
∂t
=
=
∂f ∂x1
∂f ∂x2
∂f ∂x3
+
+
∂x1 ∂t
∂x2 ∂t
∂x3 ∂t
∂f
∂f
∂f
x
+y
+z
.
∂x1
∂x2
∂x3
Now, since our x, y, z are independent of t,
∂ n
[t f (x, y, z)]
∂t
∂ n
[t ]
∂t
n−1
= nt
f (x, y, z).
= f (x, y, z) ·
Since f (tx, ty, tz) = tn f (x, y, z) for every t ∈ R (and n is the same for every t), the derivatives
will match:
∂f
∂f
∂f
x
+y
+z
= ntn−1 f (x, y, z)
∂x1
∂x2
∂x3
for every t ∈ R. In particular, this is true for t = 1, so
x
which is the desired result.
∂f
∂f
∂f
+y
+z
= nf (x, y, z),
∂x
∂y
∂z