Mathematics 205
HWK 17b Solutions
Section 2.6 p152
Notation: Dv f (a) denotes the directional derivative of the function f at the point a in the direction
v.
Relevant formula, valid provided f is differentiable at a:
Dv f (a) = ∇f (a) · v
Problem 2b, §2.6, p152. Compute the directional derivative of the function
f (x, y) = log
p
x2 + y 2 =
1
log(x2 + y 2 )
2
1
at the point (x0 , y0 ) = (1, 0) in the direction given by the vector v = √ (2i + j).
5
Solution.
∇f (x, y) = (
x
y
1
,
)= 2
(x, y)
x2 + y 2 x2 + y 2
x + y2
∇f (1, 0) = (1, 0)
√
1
2
2 5
Dv f (1, 0) = ∇f (1, 0) · v = (1, 0) · √ (2, 1) = √ =
5
5
5
To make this problem a bit more concrete, the next page gives a sketch showing the level sets for
1
the function f , the point P (1, 0) and the direction vector v = √ (2i + j), planted at the point P .
5
Although the levels for f are not labeled on the graph, notice that f increases as one moves out
from the smaller circles to the larger circles. Our calculation in this problem is telling us that if
(x, y) were to move a small displacement distance away
√ from P along the direction given by v, then
2 5
the function f would increase by approximately
times the length of the small displacement.
5
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Math 205 HWK 17b Solns continued
Section 2.6
Alternatively, if we were to slice through the graph of z = f (x, y) with a vertical plane that used
the vector
v planted at P as a ”cutting guide”, then the slope at P of the resulting curve would
√
2 5
be
.
5
Problem 4c, §2.6, p152. Find the equation for the plane tangent to the surface xyz = 1 at the
point (1, 1, 1).
Solution.
If we set F (x, y, z) = xyz − 1, then the given surface is the same as the level set
F (x, y, z) = 0. Therefore ∇F (1, 1, 1) will be orthogonal to this surface at the point (1, 1, 1) and
can be used as a normal vector for the desired plane. Differentiating, we have
∇F (x, y, z) = (yz, xz, xy)
so
∇F (1, 1, 1) = (1, 1, 1).
The equation for the specified tangent plane is therefore
(1, 1, 1) · (x − 1, y − 1, z − 1) = 0
or simply
x + y + z = 3.
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Math 205 HWK 17b Solns continued
Section 2.6
Here’s a sketch of the surface and the specified tangent plane.
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Math 205 HWK 17b Solns continued
Section 2.6
Problem 7bc, §2.6, p152.
(1, 1, 1)?
For the function given, what is the direction of fastest increase at
Solution. We know that the direction of fastest increase for a given function at a given point
is the direction pointed out by the gradient for the given function at the given point. So, for each
part of this problem, we’ll find the gradient of the function and evaluate at the point (1, 1, 1).
(b) f (x, y, z) = xy + yz + xz
∇f (x, y, z) = (y + z, x + z, y + x)
∇f (1, 1, 1) = (2, 2, 2)
So the direction of greatest increase is given by the vector (2, 2, 2), or more simply by the vector
(1, 1, 1).
The graph below shows the portion of the level set f (x, y, z) = 3 corresponding to the range
0 ≤ x ≤ 5, 0 ≤ x ≤ 5, 0 ≤ z ≤ 5. Try to see how, at the point (1, 1, 1), the direction of greatest
increase is given by the vector i + j + k.
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Math 205 HWK 17b Solns continued
Section 2.6
(c) f (x, y, z) =
x2
1
= (x2 + y 2 + z 2 )−1
+ y2 + z2
∇f (x, y, z) =
(x2
−1
(2x, 2y, 2z)
+ y 2 + z 2 )2
∇f (1, 1, 1) =
−1
(2, 2, 2)
9
So the direction of greatest increase is given by the vector
−(1, 1, 1).
−1
(2, 2, 2), or more simply by the vector
9
Here’s a sketch of the level set f (x, y, z) = 13 , which passes through the point (1, 1, 1). At the point
(1, 1, 1), the direction of greatest increase is given by the vector −i − j − k.
Note that in (b), the direction of greatest increase at (1, 1, 1) is the same as the direction from the
origin to the point (1, 1, 1). In other words, the gradient vector at the point (1, 1, 1) points in the
same direction as the vector (1, 1, 1) itself.
In part (c), we have just the reverse. The direction of greatest increase for the function, at the
point (1, 1, 1), is exactly opposite to the vector (1, 1, 1) itself. We could have seen this without
computing, by reasoning as follows. The function f (x, y, z) can be interpreted as the reciprocal of
the square of the distance from (x, y, z) to the origin. So to make this function increase as fast
as possible, we need to make the point (x, y, z) move straight toward the origin. That way the
distance from the origin will decrease as fast as possible and the reciprocal of the distance will
increase as fast as possible.
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Math 205 HWK 17b Solns continued
Section 2.6
Problem 9, §2.6, p152. Find a unit normal to the surface cos(xy) = ez − 2 at (1, π, 0).
Solution.
Let F (x, y, z) = cos(xy) − ez + 2. The given surface is the same as the level set
F (x, y, z) = 0. The gradient vector ∇F (1, π, 0) will be normal to this surface at (1, π, 0), so we
find that vector and then normalize it to get a unit vector normal to this surface. Differentiating,
we have
∇F (x, y, z) = (−y sin(xy), −x sin(xy), ez )
so
∇F (1, π, 0) = (0, 0, 1).
This vector already has length one, so it will do. Alternatively, we could use the vector opposite
to it. So the two vectors that meet the conditions of the problem are ±k. The surface is seen to
be horizontally flat at the point in question.
Perhaps you’d like to see what this surface looks like. Here’s a computer-generated sketch corresponding to −4 ≤ x ≤ 4, −4 ≤ y ≤ 4, −2 ≤ z ≤ 2.
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Math 205 HWK 17b Solns continued
Section 2.6
Problem 20, §2.6, p152.
Suppose that a mountain has the shape of an elliptic paraboloid
z = c − ax2 − by 2 , where a, b and c are positive constants, x and y are the east-west and northsouth map coordinates, and z is the altitude above sea level (x, y, z are all measured in meters).
At the point (1, 1), in what direction is the altitude increasing most rapidly? If a marble were
released at (1, 1), in what direction would it begin to roll?
Solution. Think z = f (x, y) = c − ax2 − by 2 = altitude. At the point (1, 1), the altitude will
increase most rapidly in the same direction as the vector ∇f (1, 1). Since ∇f (x, y) = (−2ax, −2by),
we have ∇f (1, 1) = (−2a, −2b). The altitude increases most rapidly toward the southwest in the
direction given by the vector −2ai − 2bj, or more simply the direction given by −ai − bj.
A marble released on the mountain at the map point (1, 1) will roll in the direction in which
altitude decreases the fastest. This direction is the same as that opposite to ∇f (1, 1). In other
words, the marble will begin to roll to the northeast in the direction given by the vector (2a, 2b),
or more simply by the vector (a, b).
For this one, you should be able to sketch your own mountain (after choosing some specific values
for a, b, c.
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