NAME:
Due Date: 10/03/2016
Instructor: Jay Taylor
Math 223 - Vector Calculus (Fall 2016)
Homework 2
1. The Dubois formula relates a person’s surface area, s, measured in m2 , to weight, w , measured
in kg, and height, h, in cm, by
s = f (w , h) = 0.01w 0.25 h0.75 .
Find f (65, 160), fw (65, 160), and fh (65, 160). Interpret your answers in terms of surface area,
height, and weight.
2. Is the following statement True or False: If f is a function with differential df = 2y dx +sin(xy )dy ,
then f changes by approximately −0.4 between the points (1, 2) and (0.9, 2.0002). Give reasons
for your answer.
3. Let h denote the height of a certain hill, measured in metres. The following function models the
height of the hill.
2
2
f (x, y ) = 20e ((1/2)−(x/100) −(y /200) ) .
Here (x, y ) measures the position relative to the summit of the hill; thus (0, 0) is the hill summit,
x measures the distance east of the summit, and y measures distance north of the summit. Note
that x, y , and h are all measured in metres.
(a) Give a linear function which approximates the height h around the point (50, 100).
(b) You start at the point (x, y ) = (50, 100) and set out walking. You want to walk so that
your path is as level as possible, i.e., your altitude doesn’t change. In which direction should
you begin walking? [Note: You should give your answer as a 2-dimensional vector of the
form ai + bj . There are multiple possible correct answers.]
(c) Draw a contour diagram of the surface z = f (x, y ). Your contour diagram should include
a minimum of three contours. One of these contours must contain the point (50, 100),
which should be clearly marked on your diagram. Draw on your diagram the vector you
found in (b). From this diagram, why do you know the vector you found in (b) is correct?
[Note: Remember to label your axes and mark your contours.]
Solutions
1. Evaluating the function we have
f (65, 160) = 0.01 × 650.25 × 1600.75 ≈ 1.2774 m2 .
This is the surface area of a person of weight 65 kg and height 160 cm. Calculating the partial
derivatives with respect to w we have
fw (w , h) = 0.0025w −0.75 h0.75 ⇒ fw (65, 160) ≈ 0.0049 m2 /kg.
This means that the surface area of a person of weight 66 kg and height 160 cm should be
approximately 0.0049 m2 more than that of a person of weight 65 kg and height 160 cm. Finally,
calculating the partial derivative with respect to h we have
fh (w , h) = 0.0075w 0.25 h−0.25 ⇒ fh (65, 160) ≈ 0.006 m2 /cm.
This means that the surface area of a person of weight 65 kg and height 161 cm should be
approximately 0.006 m2 more than that of a person of weight 65 kg and height 160 cm.
2. At the point (1, 2) the differential becomes
df = 4dx + sin(2)dy
The difference between the points (1, 2) and (0.9, 2.0002) in the x-direction is ∆x = −0.1 and
in the y -direction is ∆y = 0.0002. If ∆f = f (0.9, 2.0002) − f (1, 2) then by local linearity we
have
∆f ≈ 4∆x + sin(2)∆y
≈ −4 × 0.1 + sin(2) × 0.0002
≈ −0.3998.
As this is approximately −0.4 the statement is correct.
3. (a) The tangent plane of the surface z = f (x, y ) at the point (50, 100) is a linear function
approximating the height around the point (50, 100). The equation for this plane is
z = f (50, 100) + fx (50, 100)(x − 50) + fy (50, 100)(y − 100).
A quick calculation shows that
2 −(100/200)2 )
f (50, 100) = 20e ((1/2)−(50/100)
= 20e ((1/2)−1/4−1/4) = 20.
Calculating the partial derivatives we see that
fx (x, y ) = 20e
=−
x 1
−2
100 100
4x ((1/2)−(x/100)2 −(y /200)2 )
e
1000
fy (x, y ) = 20e
=−
((1/2)−(x/100)2 −(y /200)2 )
((1/2)−(x/100)2 −(y /200)2 )
y 1 −2
200 200
y
2
2
e ((1/2)−(x/100) −(y /200) )
1000
Evaluating at the point we thus have
4 × 50
1
=−
1000
5
100
1
fy (50, 100) = −
=− .
1000
10
fx (50, 100) = −
Plugging this into the equation for the tangent plane gives
1
1
z = 20 − (x − 50) − (y − 100).
5
10
If we wish, rearranging, we could also write this as
x
y
+
+ z = 40.
5 10
(b) We’re asking that our altitude does not change, so remains constant. In particular, we want
a direction vector u = ai + bj such that the directional derivative fu (50, 100) is 0. If we let
u
v = |u|
be the corresponding unit vector then the directional derivative is simply the dot
product fu (50, 100) = ∇f (50, 100) · v where
∇f (50, 100) = fx (50, 100)i + fy (50, 100)j
1
1
=− i− j
5
10
is the gradient vector of f at (50, 100). Here we used the calculations of fx (50, 100) and
fy (50, 100) from the previous part. We want ∇f (50, 100) · v = 0 but this is the case if and
only if ∇f (50, 100) · u = 0, which is equivalent to
a
b
− −
= 0 ⇒ −2a − b = 0 ⇒ b = −2a.
5 10
If we wanted u to be a unit vector we would need
p
a2 + b2 = 1 ⇒ a2 + (−2a)2 = 1
⇒ 5a2 = 1
√
⇒ a = ±1/ 5.
√
√
√
√
Hence, we must have v is (1/ 5)i −(2/ 5)j or −(1/ 5)i +(2/ 5)j . A non-unit direction
vector could also be obtained by choosing a = 1, so u =
√
5v = i − 2j .
(c) If c is a constant then the level surface c = f (x, y ) has equation
2 −(y /200)2 )
c = 20e ((1/2)−(x/100)
x2
y2
1
−
−
2 1002 2002
2
2
x
y
1
⇒
+
= − ln(c/20)
2
2
100
200
2
⇒ ln(c/20) =
This means that each level surface is an ellipse centred at the origin unless c = 20e 1/2 . If
c = 20e 1/2 then the right hand side of the equation is 0 so the level surface is simply the
origin. The standard equation for an ellipse is of the form
x2 y2
+ 2 = 1.
a2
b
In our case, assuming c 6= 20e 1/2 , we have
r
1
− ln(c/20)
a = 100
2
r
1
− ln(c/20).
b = 200
2
In particular, each ellipse is twice as tall as it is wide. The following is then a contour
diagram of f (x, y ).
y
200
10 100
20
30
-200
-100
(50,100)
x
100
200
-100
-200
On the diagram we’ve marked on the non-unit direction vector u = i − 2j from the previous
part. From the diagram we know that u is correct because it is tangential to the contour
and for such a vector the directional derivative is 0.