Score:
Name:
Worksheet 4 - Section 14.4, 14.5, 14. 6 (Due Tues, Oct 7)
Math 2110Q – Fall 2014
Professor Hohn
You must show all of your work to receive full credit!
1. If R is the total resistance of there resistors, connected in parallel, with resisitances R1 , R2 ,
R3 , then
1
1
1
1
=
+
+
R
R1 R2 R3
If the resistances are measured in ohms as R1 = 25Ω, R2 = 40Ω, R3 = 50Ω, with a possible
error of 0.5% in each case, estimate the maximum error in the calculated value of R.
Solution: First, let’s do a change of variable to clarify what happening. Let x = R1 , y =
R2 , z = R3 , w(x, y, z) = R(R1 , R2 , R3 ). To find that maximum error, we will look at the
differential of w.
dw = wx dx + wy dy + wz dz
We need to find the partial derivatives of w at the point (25, 40, 50), where
1
1 1 1
= + + .
w
x y z
The partial derivative of w w.r.t. x is
−1 ∂w
−1
∂w
w2
·
=
=⇒
=
= (w/x)2 .
w2 ∂x
w2
∂x
x2
At the point (25, 40, 50),
wx =
200
17
!2
= (8/17)2 = 64/289.
25
Now, we find the partial derivative of w w.r.t. y.
−1 ∂w
−1
∂w
w2
=
=
·
=⇒
= (w/y)2 .
w2 ∂y
w2
∂y
y2
At the point (25, 40, 50),
wy =
200
17
40
!2
= (5/17)2 = 25/289.
Page 1 of 5
The partial derivative of w w.r.t. z is
∂w
−1 ∂w
−1
w2
·
=⇒
= (w/z)2 .
=
=
w2 ∂z
w2
∂z
z2
At the point (25, 40, 50),
wz =
200
17
!2
= (4/17)2 = 16/289.
50
We also need to know dx, dy, dz. Recall that x = 25, y = 40 and z = 50. We also know that
the maximum error is 0.5% in x, y, z. Hence, dx = 25 · 0.005 = 0.125, dy = 40 · 0.005 = 0.2,
and dz = 50 · 0.005 = 0.25.
Then,
dw = (64/289) · 0.125 + (25/289) · 0.2 + (16/289) · 0.25 = 1/17
The maximum error in the calculated value for R is then 1/17 ohms.
2. If z = f (x, y), where x = r cos θ, and y = r sin θ, find
(a) ∂z/∂r
Solution: We need to use the chain rule to find ∂z/∂r.
∂z
∂z
=
∂r
∂x
∂z
=
∂x
∂x ∂z ∂y
+
·
∂r
∂y ∂r
∂z
· (cos θ) +
· (sin θ)
∂y
·
(b) ∂z/∂θ
Solution: We need to use the chain rule to find ∂z/∂θ.
∂z
∂z
=
∂θ
∂x
∂z
=
∂x
∂x ∂z ∂y
+
·
∂θ
∂y ∂θ
∂z
· (−r sin θ) +
· (r cos θ)
∂y
·
(c) ∂ 2 z/∂r∂θ
Solution: We are looking for
∂2z
∂ ∂z
∂z
=
· (cos θ) +
· (sin θ)
∂r∂θ
∂θ ∂x
∂y
∂2z
∂z
∂2z
∂z
=
· (cos θ) +
· (− sin θ) +
· (sin θ) +
· (cos θ).
∂θ∂x
∂x
∂y∂θ
∂y
Page 2 of 5
Now, let’s see what
∂2z
is equivalent to.
∂θ∂x
∂2z
∂2z
=
∂θ∂x
∂x∂θ
∂ ∂z
=
∂x ∂θ
∂ ∂z
∂z
=
· (−r sin θ) +
· (r cos θ)
∂x ∂x
∂y
∂2z
∂2z
·
(−r
sin
θ)
+
· (r cos θ)
=
∂x2
∂x∂y
Similarly,
∂2z
is equivalent to.
∂θ∂y
∂2z
∂2z
=
∂θ∂y
∂y∂θ
∂ ∂z
=
∂y ∂θ
∂ ∂z
∂z
=
· (−r sin θ) +
· (r cos θ)
∂y ∂x
∂y
2
2
∂ z
∂ z
=
· (−r sin θ) + 2 · (r cos θ)
∂y∂x
∂y
Substituting this into our equation, we have
∂2z
∂z
∂2z
∂z
∂2z
=
· (cos θ) +
· (− sin θ) +
· (sin θ) +
· (cos θ)
∂r∂θ
∂θ∂x
∂x
∂y∂θ
∂y
2
∂ z
∂2z
∂z
=
· (−r sin θ) +
· (r cos θ) · (cos θ) +
· (− sin θ)
2
∂x
∂x∂y
∂x
2
∂2z
∂z
∂ z
· (−r sin θ) + 2 · (r cos θ) · (sin θ) +
· (cos θ)
+
∂y∂x
∂y
∂y
3. Show that any function of the form
z = f (x + at) + g(x − at)
is a solution of the wave equation
2
∂2z
2∂ z
=
a
∂t2
∂x2
[Hint: Let u = x + at and v = x − at.]
Solution: First, we rewrite our equation using the hint provided. Let u = x + at and let
v = x − at. Then, our function becomes z(u, v) = f (u) + g(v). We will start with the left
hand side and show that it is equivalent to the right hand side.
Page 3 of 5
Let’s find the partial derivative of z with respect to t.
∂z
∂z ∂u ∂z ∂v
=
·
+
·
∂t
∂u ∂t
∂v ∂t
df
dg
=
·a+
· (−a)
du
dv
df
dg
=a
−
du dv
Now, we find the second partial derivative of z with respect to t.
∂2z
∂
df
dg
=
a
−
∂t2
∂t
du dv
2
d f ∂u d2 g ∂v
=a
− 2
du2 ∂t
dv ∂t
2
d f
d2 g
=a
· a − 2 · (−a)
du2
dv
2
2
d g
2 d f
+
=a
du2 dv 2
∂2z
.
∂x2
The first partial derivative of z with respect to x.
We will show that this is equal to a2
∂z
∂z ∂u ∂z ∂v
=
·
+
·
∂x
∂u ∂x ∂v ∂x
df
dg
=
·1+
·1
du
dv
df
dg
=
+
du dv
The second partial derivative of z w.r.t. x.
∂2z
∂ df
dg
=
+
∂x2
∂t du dv
d2 f ∂u d2 g ∂v
= 2
+ 2
du ∂t
dv ∂t
d2 f
d2 g
= 2 ·1− 2 ·1
du
dv
d2 g
d2 f
= 2+ 2
du
dv
So, indeed
∂2z
= a2
∂t2
d2 f
d2 g
+
du2 dv 2
2
∂ z
= a2 2
∂x
Page 4 of 5
4. Find the maximum rate of change of f (x, y, z) = (x + y)/z at the point (1, 1, −1) and the
direction in which it occurs.
Solution: The maximum rate of change of our function f at our point P is equal to the
magnitude of the gradient vector at our point P (1, 1, −1).
∇f (x, y, z) = h1/z, 1/z, −(x + y)/z 2 i
Therefore,
∇f (P ) = h−1, −1, −(1 + 1)/(−1)2 i = h−1, −1, −2i,
and the magnitude is then
p
√
(−1)2 + (−1)2 + (−2)2 = 6.
√
Thus, the maximum rate of change of our function f at P is 6.
k∇f (P )k =
The direction in which the maximum rate of change occurs is the direction of the gradient
vector. If we specify that we want to know the unit vector that has the same direction as the
gradient vector, we take the gradient vector at our point and divide by the magnitude of the
vector to get the direction
√
√
√
~u = h−1/ 6, −1/ 6, −2/ 6i.
Note: any solution that is a multiple of the gradient vector is in the direction we need. Hence,
anything of the form ~v = kh−1, −1, −2i where k ∈ R is a correct solution.
Page 5 of 5