MATH 1320 : Spring 2014
Lab Instructor : Kurt VanNess
Lab 11 Solutions
Write all your solutions on a separate sheet of paper.
1. (a) Take the partial derivatives to obtain
x
fx (x, y) = (1/2)(x2 + 3y 2 )−1/2 · 2x = p
2
x + 3y 2
3y
fy (x, y) = (1/2)(x2 + 3y 2 )−1/2 · 6y = p
2
x + 3y 2
and evaluate
fx (1, 4) = √
1
1
=
7
1 + 48
fy (1, 4) = √
12
12
=
7
1 + 48
(b) The equation of the tangent plane is equivalent to the linearization at the given point, so we have
z−7=
1
12
(x − 1) + (y − 4)
7
7
or
z=
x + 12y
7
(c) We use the tangent plane to make an approximation for the function at the point, so we have
z=
1.1 + 46.8
= 6.842857 . . .
7
(d) The actual value is
f (1.1, 3.9) =
√
1.21 + 45.63 = 6.843975 . . .
(e) The difference between the approximation and the actual value is
6.843975 . . . − 6.842857 . . . = 0.0011183 . . .
2. We first obtain a value for R from
1
1
1
31
1
=
+
+
=
R
25 35 50
350
=⇒
R=
350
≈ 11.29032
31
(1)
Now we can use differentials to find the maximum error in this calculation since we have errors for the
three resistors. We use
∂R
∂R
∂R
dR1 +
dR2 +
dR3
dR =
∂R1
∂R2
∂R3
Now we need to find the partial derivatives. If we differentiate equation (1) implicitly with respect to
R1 , we have
−1 ∂R
−1
∂R
R2
=
=⇒
=
R2 ∂R1
R12
∂R1
R12
We have similar results for R2 and R3 . Now we have percent errors for R1 , R2 , and R3 , so we need to
convert those to actual difference errors. This can be done as follows:
dR1 = (0.005)R1
Page 1 of 3
MATH 1320 : Spring 2014
Lab Instructor : Kurt VanNess
Putting all of this together, we have
R2
R2
R2
(0.005)R1 + 2 (0.005)R2 + 2 (0.005)R3
2
R1
R2
R3
2
2
2
0.005R
0.005R
0.005R
+
+
=
R1
R2
R3
1
1
1
= 0.005R2
+
+
R1
R2
R3
1
= 0.005R2
R
dR =
= 0.005R
7
350
=
≈ 0.05645
= 0.005
31
124
3. (a) The negative sign in ∂W/∂T means that wheat production decreases with an increase in the
average temperature at current production levels. The positive sign in ∂W/∂R means that wheat
production increases with an increase in annual rainfall at current production levels.
(b) We note that units aren’t given on the partial derivatives, so we make the assumption that the
units match up correctly. Using the chain rule, we can estimate the current rate of change of
wheat production using the changes in average temperature and annual rainfall
dW
∂W dT
∂W dR
=
+
= (−2)(0.15) + (6)(−0.1) = −0.9
dt
∂T dt
∂R dt
4. (a) Using the chain rule, we have
∂z ∂x ∂z ∂y
∂z
∂z
∂z
=
+
=
cos θ +
sin θ
∂r
∂x ∂r
∂y ∂r
∂x
∂y
∂z ∂x ∂z ∂y
∂z
∂z
∂z
=
+
=
(−r) sin θ +
r cos θ
∂θ
∂x ∂θ
∂y ∂θ
∂x
∂y
(b) To show this equation, we start with the right-hand side since we have solved for those terms
already. We then can simplify as follows:
2
2
2
2 ∂z
∂z
1 ∂z
∂z
1 ∂z
∂z
cos θ +
sin θ + 2
(−r) sin θ +
r cos θ
+ 2
=
∂r
r
∂θ
∂x
∂y
r
∂x
∂y
2
2
∂z
∂z ∂z
∂z
=
cos2 θ + 2
sin θ cos θ +
sin2 θ
∂x
∂x ∂y
∂y
" #
2
2
1
∂z
∂z
2
2
2 ∂z ∂z
2
2
sin θ − 2r
sin θ cos θ + r
cos θ
+ 2 r
r
∂x
∂x ∂y
∂y
2
2
∂z
∂z ∂z
∂z
2
=
cos θ + 2
sin θ cos θ +
sin2 θ
∂x
∂x ∂y
∂y
2
2
∂z
∂z ∂z
∂z
+
sin2 θ − 2
sin θ cos θ +
cos2 θ
∂x
∂x ∂y
∂y
2
2
2
2
∂z
∂z
∂z
∂z
2
2
2
=
cos θ +
sin θ +
sin θ +
cos2 θ
∂x
∂y
∂x
∂y
2
2
∂z
∂z
=
cos2 θ + sin2 θ +
sin2 θ + cos2 θ
∂x
∂y
2 2
∂z
∂z
=
+
∂x
∂y
Page 2 of 3
MATH 1320 : Spring 2014
Lab Instructor : Kurt VanNess
5. Let us introduce the variables v(x, t) = x + at and w(x, t) = x − at. Then we have
∂u
∂u ∂v
∂u ∂w
∂u
∂u
=
+
=
(a) +
(−a)
∂t
∂v ∂t
∂w ∂t
∂v
∂w
∂ ∂u
∂2u
∂ ∂u
∂u
∂ 2 u ∂v
∂ 2 u ∂w
=
=
(a)
+
(−a)
=
a
−
a
∂t2
∂t ∂t
∂t ∂v
∂w
∂v 2 ∂t
∂w2 ∂t
2
2
2
2
∂ u
∂ u
∂ u
∂ u
= a2 2 + a2 2 = a2
+
2
∂v
∂w
∂v
∂w2
∂u ∂v
∂u ∂w
∂u
∂u
∂u
=
+
=
(1) +
(1)
∂x
∂v ∂x ∂w ∂x
∂v
∂w
∂2u
∂ ∂u
∂
=
=
∂x2
∂x ∂x
∂x
∂2u
∂2u
=
+
∂v 2
∂w2
∂u
∂u
+
∂v
∂w
=
∂ 2 u ∂v
∂ 2 u ∂w
+
∂v 2 ∂x ∂w2 ∂x
From these expressions, we can see that
∂2u
= a2
∂t2
∂2u
∂2u
+
2
∂v
∂w2
= a2
∂2u
∂x2
Page 3 of 3