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Module MA2E02 (Frolov), Multivariable Calculus, 2017
Homework Sheet 3
Due: by 12:00 on the first Friday after the corresponding tutorial session
Name and student number:
1. Consider the function
p
f (x, y, z) = −2 + 2y + x4 + yz − sin(3y − 2z) ,
and the point P = (−1, 2, 3) .
(a) Calculate f (−1, 2, 3), fx (−1, 2, 3), fy (−1, 2, 3), fz (−1, 2, 3).
(b) Find a unit vector in the direction in which f increases most rapidly at the point P .
(c) Sketch the projection of the vector onto the xz-plane
(d) Find a unit vector in the direction in which f decreases most rapidly at the point
P.
(e) Sketch the projection of the vector onto the xy-plane
(f) Find the rate of change of f at the point P in these directions.
Show the details of your work.
Solution:
(a) We find
f (−1, 2, 3) =
p
−2 + 2y + x4 + yz − sin(3y − 2z) = 3 ,
P
fx (x, y, z) =
fy (x, y, z) =
−2 cos(2x − 3z)
2f (x, y, z)
2 + z − 3 cos(3y − 2z)
2f (x, y, z)
fz (x, y, z) =
y + 2 cos(2z − 3x)
2f (x, y, z)
2
fx (−1, 2, 3) = − ,
3
⇒
⇒
⇒
fy (−1, 2, 3) =
fz (−1, 2, 3) =
1
,
3
2
.
3
(b) f increases most rapidly in the direction of its gradient. The gradient and its magnitude are equal to
2 1 2
∇f (−1, 2, 3) = − , ,
, ||∇f (−1, 2, 3)|| = 1 .
3 3 3
Therefore, the unit vector in the direction of the gradient is
∇f (−1, 2, 3)
2 1 2
u=
= − , ,
≈ (−0.666667, 0.333333, 0.666667) .
||∇f (−1, 2, 3)||
3 3 3
1
z
0.7
0.6
0.5
0.4
0.3
0.2
0.1
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
x
-0.1
(c) The projection of the vector u onto the xz-plane is the vector
2
2
uxz = − , 0 ,
≈ (−0.666667, 0, 0.666667) ,
3
3
see the picture.
(d) f decreases most rapidly in the direction opposite to its gradient, so the unit vector
is
2
1
2
∇f (−1, 2, 3)
=
,− ,−
v=−
≈ (0.666667, −0.333333, −0.666667) .
||∇f (−1, 2, 3)||
3
3
3
(e) The projection of the vector v onto the xy-plane is the vector
1
2
, − , 0 ≈ (0.666667, −0.333333, 0) .
vxy =
3
3
It is shown below
y
0.1
0.2
0.3
0.4
0.5
0.6
0.7
x
-0.05
-0.10
-0.15
-0.20
-0.25
-0.30
-0.35
(f) The rate of change of f at P in the direction of u is equal to
||∇f (−1, 2, 3)|| = 1 ,
and the rate of change of f at P in the direction of v is equal to
−||∇f (−1, 2, 3)|| = −1 .
2
2. Let r =
p
x2 + y 2 .
(a) Show that
∇r =
r
,
r
where r = x i + y j .
Solution: We have
∂r
2x
x
= p
= ,
∂x
r
2 x2 + y 2
∂r
2y
y
= p
= ,
∂y
r
2 x2 + y 2
which proves the formula.
(b) Show that
∇f (r) = f 0 (r)∇r =
f 0 (r)
r.
r
Solution: We use the chain rule to get
∂f (r)
∂r
x
= f 0 (r)
= f 0 (r) ,
∂x
∂x
r
∂f (r)
∂r
y
= f 0 (r)
= f 0 (r) ,
∂y
∂y
r
which proves the formula.
3. Consider the surface
p
13 3
2
2x − y + 10
z = f (x, y) = ln
2
(a) Calculate z0 ≡ f (1, −2), fx (1, −2), fy (1, −2).
(b) Find an equation for the tangent plane to the surface at the point P = (1, −2, z0 )
where z0 = f (1, −2).
(c) Find points of intersection of the tangent plane with the x-, y- and z-axes.
(d) Sketch the tangent plane.
(e) Find parametric equations for the normal line to the surface at the point P (1, −2, z0 ).
(f) Sketch the normal line to the surface at the point P (1, −2, z0 ).
Show the details of your work.
Solution:
(a) We first simplify
p
13 3
1
2
z = ln
2x − y + 10 = ln(2x3 − y 2 + 10) − ln 2 .
2
3
and compute z0
1
ln 2 − 4 + 10 − ln 2 = 0 .
3
Then, we compute the partial derivatives at P (1, −2, 0)
z0 = z|x=1,y=−2 =
∂
1
6x2
z=
∂x
3 2x3 − y 2 + 10
3
⇒
∂
1
z|x=1,y=−2 = .
∂x
4
∂
1
−2y
z=
3
∂y
3 2x − y 2 + 10
⇒
∂
1
z|x=1,y=−2 = .
∂y
6
(b) The tangent plane equation is given by
1
1
1
1
1
z = 0 + (x − 1) + (y + 2) = x + y +
.
4
6
4
6
12
1
(c) (− 31 , 0, 0) = (−0.33333, 0, 0), (0, − 21 , 0) , (0, 0, 12
) ≈ (0, 0, 0.08333)
(d) The tangent plane is the one through the points in (c).
(e) The normal line to the surface (and the tangent plane) is given by
1
1
r = i − 2j + t − i − j + k .
4
6
(f) The normal line is perpendicular to the plane.
4. Consider the function
f (x, y) = x4 − 2x2 y + 2y 2 − 2y − 5
Locate all relative maxima, relative minima, and saddle points, if any.
Solution: We first find all critical points
fx (x, y) = 4x3 − 4xy = 0 ,
fy (x, y) = −2x2 + 4y − 2 = 0 .
From the second equation we find y in terms of x
x2 1
+ ,
y=
2
2
and substituting it to the first equation, we derive the following equation for x
2x − 2x3 = 0 .
There are three solutions to this equation
x = 0 , x = −1 , x = 1 ,
and, therefore, three critical points
1
(x = 0 , y = ) ,
2
(x = −1 , y = 1) ,
(x = 1 , y = 1) .
Computing the values of f at critical points, we get
1
11
f (0, ) = − ,
2
2
f (−1, 1) = −6 ,
4
f (1, 1) = −6 .
To find out if they are maximum, minimum or saddle points we use the second derivative
test. To this end we compute
∂ 2f
∂ 2f
∂ 2f
2
(x, y) = −4x ,
(x,
y)
=
12x
−
4y
,
(x,
y)
=
4
,
∂x2
∂y 2
∂x∂y
and
∂ 2f ∂ 2f
D(x, y) =
−
∂x2 ∂y 2
Computing D and
∂2f
∂x2
∂ 2f
∂x∂y
2
= 32x2 − 16y ,
for the three critical points, we get
1
D(0, ) = −8 ,
2
∂ 2f
1
(0, ) = −2 ,
2
∂x
2
and therefore (0 , 12 ) is a saddle point.
D(−1, 1) = 16 ,
∂ 2f
(−1, 1) = 8 ,
∂x2
and therefore (−1, 1) is a relative minimum.
D(1, 1) = 16 ,
∂ 2f
(1, 2) = 8 ,
∂x2
and therefore (1, 1) is a relative minimum too.
The graph of the function is shown below
5