MATH 253 WORKSHEET 13
DIRECTIONAL DERIVATIVES
~ and D~u f at the given point.
(1) In each case nd ∇f
(a) f (x, y) = xey , at (1, 0) in the direction ~u = 35 , − 45 .
~ (x, y) = hey , xey i so ∇f
~ (1, 0) = h1, 1i and
Solution: ∇f
3 4
Dh 3 ,− 4 i f (1, 0) = h1, 1i ·
,−
5
5
5 5
1
3 4
− =− .
=
5 5
5
6 7 6
2
2
2
(b) f (x, y, z) = x + y + z at (1, 2, 3) in the direction ~u = − 11
, 11 , 11 .
~ (x, y, z) = h2x, 2y, 2zi so ∇f
~ (1, 2, 3) = 2 h1, 2, 3i and
Solution: ∇f
6 7 6
Dh− 6 , 7 , 6 i f (1, 2, 3) = 2 h1, 2, 3i · − , ,
11 11 11
11 11 11
2
52
=
h−6 + 14 + 18i =
.
11
11
p
π
(c) f (x, y) = x2 + y 2 + ex at (1, 1) in the direction
4 to the horizontal.
making an angle
E
D
y
x
1
x
~ (x, y) = √
~ (1, 1) = √ + e, √1 and
Solution: ∇f
+ e , √ 2 2 so ∇f
2
2
2
2
x +y
x +y
DD
1
√
, √12
2
E f (1, 1)
=
=
1
1
1
1
√ + e, √
· √ ,√
2
2
2
2
1
e
e
1
+ √ + =1+ √ .
2
2 2
2
(2) You are driving your car towards the northeast at 72km/h along a terrain whose elevation at the
point (x, y) is 8+x12 +y2 (all distances are measured in kilometres). What is your rate of ascent/descent
when your car is at the location (1, 1)? What about
D if theElocation was (1, −1)?
Solution 1: Let f (x, y) = 8+x12 +y2 and Let ~u = √12 , √12 , a unit vector in the northeast direction.
Then the slope in the direction of travel is:
*
~
∇f
(x, y) · ~u =
+ 1
1
√
√
,
·
,
2
2
2
2
(8 + x2 + y 2 ) (8 + x2 + y 2 )
−2x
−2y
which at (1, 1) is
−
1
1
√ h1, 1i · h1, 1i = − √ .
50 2
25 2
In other words, for every unit of horizontal distance in the direction of travel one would lose
units of elevation. We are travelling at 72km/h and are therefore losing altitude at the rate of
√
2 2m
5 s .
1√
25 2
72
√
25 2
If (x, y) = (−1, 1) then the directional derivative would have been
− 501√2 h1, −1i · h1, 1i = 0, so the the car is momentarily level.
D
E
~
Solution 2: Let ~v = 72 √12 , √12 be the velocity vector. Then the rate of change is ∇f
(x, y) · ~v =
D
E
72
1
1
1
− 25√2 . Similarly, − 50 h1, −1i · 72 √2 , √2 = 0.
kilometers per hour, or
Date
: 7/10/2013.
1
(3) An ant is crawling along the curve y = x2 at the rate of vcm/s (distances are measured in cm). The
y
temperature in the xy plane is varying according to T (x, y) = 1+x
2 . What is the rate of change of
the temperature the ant sees when it is located at (x, y)?
Solution 2: A tangent vector to the curve y = x2 at (x, y) is given by h1, 2xi. A unit tangent
1
vector is therefore ~u = 1+4x
2 h1, 2xi. The temperature gradient is
~ (x, y) =
∇T
1
−2xy
,
2
2
(1 + x ) 1 + x2
so, per unit distance travelled, the ant sees a temperature change of
2
2
~ · ~u = −2xy + (1 + x )2x = 2x(1 + x − y) .
∇T
(1 + x2 )2 (1 + 4x2 )
(1 + x2 )2 (1 + 4x2 )
Now the ant is travelling at the rate of v units of distance per unit time, so per unit time the
temperature change is
~ · ~u =
v ∇T
2x(1 + x2 − y)
v.
(1 + x2 )2 (1 + 4x2 )
2