Module MA2E02 (Frolov), Multivariable Calculus
Tutorial Sheet 3
Due: at the end of the tutorial session Tuesday/Thursday, 9/11 February 2016
Name and student number:
1. Consider the function
f (x, y, z) =
p
4y 2 − sin(2x − 3z) ,
and the point P (0, −1, 0) .
(a) Find a unit vector in the direction in which f increases most rapidly at the point P .
(b) Sketch the projection of the vector onto the xz-plane
(c) Find a unit vector in the direction in which f decreases most rapidly at the point
P.
(d) Sketch the projection of the vector onto the yz-plane
(e) Find the rate of change of f at the point P in these directions.
Show the details of your work.
Solution:
(a) f increases most rapidly in the direction of its gradient, so we compute
−2 cos(2x − 3z)
fx (x, y, z) = p
2 4y 2 − sin(2x − 3z)
⇒
1
fx (0, −1, 0) = − ,
2
8y
fy (x, y, z) = p
2
2 4y − sin(2x − 3z)
⇒
fy (0, −1, 0) = −2 ,
3 cos(2x − 3z)
fz (x, y, z) = p
2 4y 2 − sin(2x − 3z)
⇒
fz (0, −1, 0) =
3
.
4
Thus, the gradient and its magnitude are equal to
√
1
3
77
∇f (0, −1, 0) = − , −2 ,
, ||∇f (0, −1, 0)|| =
≈ 2.19374 .
2
4
4
Therefore, the unit vector in the direction of the gradient is
1
4
3
− , −2 ,
≈ (−0.227921, −0.911685, 0.341882) .
u= √
2
4
77
(b) The projection of the vector u onto the xz-plane is the vector
4
3
1
uxz = √
− ,0,
≈ (−0.227921, 0, 0.341882) ,
2
4
77
see the picture.
1
z
0.35
0.30
0.25
0.20
0.15
0.10
0.05
-0.20
-0.15
-0.10
x
-0.05
(c) f decreases most rapidly in the direction opposite to its gradient, so the unit vector is
4
3
1
v = −√
≈ (0.227921, 0.911685, −0.341882) .
− , −2 ,
2
4
77
(d) The projection of the vector v onto the yz-plane is the vector
4
3
vyz = √
0 , −2 ,
≈ (0, 0.911685, −0.341882) .
4
77
It is shown below
z
-0.05
0.2
0.4
0.6
0.8
y
-0.10
-0.15
-0.20
-0.25
-0.30
-0.35
(e) The rate of change of f at P in the direction of u is equal to
√
77
||∇f (0, −1, 0)|| =
≈ 2.19374 ,
4
and the rate of change of f at P in the direction of v is equal to
√
77
−||∇f (0, −1, 0)|| = −
≈ −2.19374 .
4
2
2. Let r =
p
x2 + y 2 .
(a) Show that
∇r =
r
,
r
where r = x i + y j .
Solution: We have
2x
∂r
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
∂r
x
∂f (r)
= 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
3
z = ln
2x3 + y 2 + 2
3
(a) Find an equation for the tangent plane to the surface at the point P (2, −3, z0 ) where
z0 = f (2, −3).
(b) Find points of intersection of the tangent plane with the x-, y- and z-axes.
(c) Sketch the tangent plane.
(d) Find parametric equations for the normal line to the surface at the point P (2, −3, z0 ).
(e) Sketch the normal line to the surface at the point P (2, −3, z0 ).
Show the details of your work.
Solution:
(a) We first simplify
p
3
2x3 + y 2 + 2
1
z = ln
= ln(2x3 + y 2 + 2) − ln 3 .
3
3
and compute z0
1
ln 16 + 9 + 2 − ln 2 = 0 .
3
Then, we compute the partial derivatives at P (2, −3, 0)
z0 = z|x=2,y=−3 =
∂
1
6x2
z=
∂x
3 2x3 + y 2 + 2
3
⇒
∂
8
z|x=2,y=−3 =
.
∂x
27
∂
1
2y
z=
3
∂y
3 2x + y 2 + 2
⇒
∂ 1
2
z|x=2,y=−3 = − .
∂y 2
27
The tangent plane equation is given by
z =0+
8
2
8
2
22
(x − 2) − (y + 3) = x − y −
.
27
27
27
27
27
22
(b) ( 11
, 0, 0) = (2.75, 0, 0), (0, −11, 0) , (0, 0, − 27
) ≈ (0, 0, −0.814815)
4
(c) The tangent plane is the one through the points in (b).
(d) The normal line to the surface (and the tangent plane) is given by
r = 2i − 3j + t −
8
2
i+ j+k .
27
27
(e) The normal line is perpendicular to the plane.
4. Consider the function
f (x, y) = 5 − x4 + 2x2 y − 2y 2 + 2y
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
y=
x2 1
+ ,
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
∂ 2f
1
(0, ) = 2 ,
2
∂x
2
1
D(0, ) = −8 ,
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 maximum.
D(1, 1) = 16 ,
∂ 2f
(1, 2) = −8 ,
∂x2
and therefore (1, 1) is a relative maximum too.
The graph of the function is shown below
6
5
4
1.5
1.0
y
0.5
-1
0
1
x
5
0.0