Download MATH 203 Lab 6 Solutions Spring 2005 (1) On a certain mountain

MATH 203
Lab 6 Solutions
Spring 2005
(1) On a certain mountain, the elevation z above a point (x, y) in the xy−plane (sea level) is given by
z = 2500 − 4x2 − 3y 2 . The positive x−axis points south, and the positive y−axis points east. Suppose
a climber is at the point (−20, 5, 825), if the climber walks due north, will the climber be ascending or
descending? Justify mathematically. (Hint: partial derivative)
Since the climber walks north, the y coordinate is fixed while the x-coordinate changes. Hence the
partial derivative ∂z/∂x is useful. We have
∂z = −8x|(−20,5,825) = 160.
∂x (−20,5,825)
Note that the positive x−axis is due south, hence the climber is heading to a slope of −160, that is,
descending.
(2) You are given only the following information about a differentiable function f :
f (4, 5) = 20.8, f (4.01, 5) = 21.0, f (4, 5.02) = 20.7
(a) Approximate the equation of the tangent plane to the surface at (4, 5). (Hint: interpret the
formula for the tangent plane)
The equation of the tangent plane is
z − z0 =
∂f
∂f
(x − x0 ) +
(y − y0 )
∂x
∂y
We can approximate the partial derivatives using the information given, from the point (4, 5):
∂f
∂x
∂f
∂y
≈
≈
21.0 − 20.8
= 20
4.01 − 4
20.7 − 20.8
= −5
5.02 − 5
Hence the approximate equation of the plane is
z − 20.8 = 20(x − 4) − 5(y − 5)
(b) Given only the three pieces of information about f , it is impossible to evaluate f (4.01, 5.02). Can
you approximate the value? How?
Use the approximate tangent plane, we get
z ≈ 20.8 + 20(4.01 − 4) − 5(5.02 − 5) = 20.9
(3) Consider the function
2
f (x, y) = yex y .
(a) Find the slope of the curve formed by the intersection of the given surface with a plane perpendicular to the y−axis and passing through (3, 1).
2
∂f
= 2xy 2 ex y
∂x
Slope at (3, 1) is
∂f = 6e9
∂x (3,1)
MATH 203
Lab 6 Solutions
(b) Find the tangent plane at (3, 1).
We have
Spring 2005
2
2
∂f
= ex y + x2 yex y
∂y
and
∂f = 10e9
∂y (3,1)
Furthermore, f (3, 1) = e9 . Hence the equation of the tangent plane is
z − e9 = 6e9 (x − 3) + 10e9 (y − 1)
(4) In a study of frost penetration it was found that the temperature T at time t (measured in days) at a
depth x (measured in feet) can be modeled by the function
T (x, t) = T0 + T1 e−λx sin(ωt − λx)
where ω = 2π/365 and λ is a positive constant.
(a) Find
∂T
∂x .
What is its physical significance?
∂T
= −λT1 e−λx [sin(ωt − λx) + cos(ωt − λx)]
∂x
This quantity represents the rate of change of temperature with respect to depth below the surface,
at a given time t.
(b) Find
∂T
∂t
. What is its physical significance?
∂T
= ωT1 e−λx cos(ωt − λx)
∂t
This quantity represents that rate of change of temperature with respect to time at a fixed depth
x.
(c) Show that T satisfies the heat equation Tt = kTxx for a certain constant k.
Txx =
∂ ∂T
= 2λ2 T1 e−λx cos(ωt − λx)
∂x ∂x
Since
Tt = ωT1 e−λx cos(ωt − λx),
The equation is satisfied with k =
ω
2λ2 .
(d) What is the physical significance of the term −λx in the expression sin(ωt − λx)?
The term −λx represents a phase shift, it takes time for the heat to diffuse as the depth increases.
(5) If R is the total resistance of three resistors in a circuit connected in parallel, with resistances R1 , R2 , R3 ,
then
1
1
1
1
=
+
+
.
R
R1
R2
R3
If the resistances are measured in ohms as R1 = 25Ω, R2 = 15Ω, R3 = 40Ω, with a possible error of
0.2%, 0.3%, 0.5% respectively, estimate the maximum error in the calculated value of R.
MATH 203
Lab 6 Solutions
Substituting R1 , R2 , R3 we get R =
get
79
600 .
Spring 2005
Using the given equation, differentiate with respect to R1 , we
−
1 ∂R
1
=− 2
R2 ∂R1
R1
that is,
R2
∂R
= 2
∂R1
R1
Similarly,
∂R
R2
∂R
R2
= 2,
= 2
∂R2
R2 ∂R3
R3
Then
∆R ≈ dR
=
=
=
=
≈
∂R
∂R
∂R
∆R1 +
∆R2 +
∆R3
∂R1
∂R2
∂R3
R2
R2
R2
(0.002)R
+
(0.003)R
+
(0.005)R3
1
2
R12
R22
R32
0.002 0.003 0.005
R2
+
+
R1
R2
R3
2 0.002 0.003 0.005
600
+
+
79
25
15
40
0.0234Ω
(6) Car A is travelling north on Highway 184 and Car B is travelling west on highway 223. Each car
is approaching the intersection of these highways. At a certain moment, car A is 0.4 miles from the
intersection and travelling at 50 mi/h while car B is 0.3 miles from the intersection and travelling at
60 mi/h. How fast is the distance between the cars changing at that moment?
Let x(t) be the distance function of car A from the intersection, and y(t) be the distance
function of
p
car B from the intersection. Then, the distance between the two cars is f (x, y) = x2 + y 2 .
Since car A is moving at 50 mi/h, then x(t) = 0.4 − 50t. Similarly, y(t) = 0.3 − 60t.
The distance is changing at df /dt when t = 0. By chain rule,
df
dt
=
=
=
Then
∂f dx ∂f dy
+
∂x dt
∂y dt
x
dx
y
dy
p
+p
x2 + y 2 dt
x2 + y 2 dt
−50x
−60y
p
+p
x2 + y 2
x2 + y 2
df −50(0.4)
−60(0.3)
=p
+p
= −76
2
2
dt t=0
(0.4) + (0.3)
(0.4)2 + (0.3)2
That is, the distance is changing at a rate of −76 mi/h.