1. No category

Name:
Math 1321
Score:
Week 8 Lab Worksheet
/8
Due Thursday 10/30
1. Partial Derivatives:
(a) (1 point) Suppose z = (1 + xy)y . Find first partial derivatives
∂z ∂z
,
.
∂x ∂y
Solution:
∂z
= y(1 + xy)y−1 · y = y 2 (1 + xy)y−1 .
∂x
When differentiating with respect to y, we rewrite z as z = ey ln(1+xy) . Therefore,
∂z
xy
y ln(1+xy)
0
y
=e
[y ln(1 + xy)] = (1 + xy) ln(1 + xy) +
.
∂y
1 + xy
∂ 2z ∂ 2z ∂ 2z
y
,
,
.
(b) (1 point) Suppose z = arctan . Find second partial derivatives
x
∂x2 ∂x∂y ∂y 2
Solution:
− xy2
y
∂z
=
.
2 = − 2
y
∂x
x + y2
1+ x
Therefore,
2xy
∂ 2z
= 2
,
2
∂x
(x + y 2 )2
and
∂ 2z
∂
=
∂x∂y
∂y
∂z
∂x
=
y 2 − x2
2y · y − (x2 + y 2 )
=
.
(x2 + y 2 )2
(x2 + y 2 )2
Similarly,
1
x
∂z
x
=
,
2 = 2
y
∂y
x + y2
1+ x
and thus
∂ 2z
2xy
=− 2
.
2
∂y
(x + y 2 )2
(c) (1 point) Show that u = z arctan
x
∂ 2u ∂ 2u ∂ 2u
satisfies the Laplace’s equation 2 + 2 + 2 = 0
y
∂x
∂y
∂z
Solution: Similar to part (b),
∂ 2u
2xyz
=− 2
,
2
∂x
(x + y 2 )2
and
∂ 2u
2xyz
= 2
.
2
∂y
(x + y 2 )2
Also since
we have
∂ 2u
∂
=
2
∂z
∂z
x
arctan
y
= 0,
∂ 2u ∂ 2u ∂ 2u
2xyz
2xyz
+ 2 + 2 =− 2
+ 2
+ 0 = 0.
2
2
2
∂x
∂y
∂z
(x + y )
(x + y 2 )2
2. Directional Derivatives and Gradient Vector:
Suppose f (x, y) = x2 − xy + y 2 .
(a) (1 point) Find all the directional derivatives at (1, 1), i.e. Du f (1, 1), with u =
(cos α, sin α).
Solution: Compute the directional derivative using gradient,
Du f (x, y) = ∇f ·u = (2x−y, 2y−x)·(cos α, sin α) = (2x−y) cos α+(2y−x) sin α.
Plugging in (x, y) = (1, 1), we have
Du f (1, 1) = cos α + sin α.
(b) (1 point) When does the directional derivative get to its maximum? its minimum?
When is the directional derivative 0?
Solution: We can do this in two ways. First, we can take the derivative of
Du f (1, 1) with respect to α, i.e. the standard process of how to find maximum
and minimum values in single variable calculus. Or we can use the proposition
in the text book, that the directional derivative gets to its maximum when the
directional vector has the same direction as the gradient, and that the directional derivative gets to its minimum when the directional vector has the inverse
direction as to the gradient. In this specific case, the directional
get
√ derivative
√
2
2
π
to its maximum when α = 4 and the directional vector u = ( 2 , 2 ), and the
directional derivative
get to its maximum when α = 5π
and the directional
4
√
√
vector u = (− 22 , − 22 ). The directional derivative is 0 when
cos α + sin α = 0,
or equivalently,
tan α = −1.
Therefore, α =
3π
4
or
7π
,
4
√
and u = (
√
2
2
,
−
)
2
2
√
or (−
√
2
2
,
).
2
2
3. Wind-Chill The wind-chill index W is the perceived temperature when the actual
temperature is T and the wind speed is v so, we can write W = f (T, v).
(a) (2 points) The following table of values is an excerpt from Table 1 in Section 11.1.
Use the table to find a linear approximation to the wind-chill index function when
T is near −15o C and v is near 50km/h.
(b) (1 point) Estimate the wind-chill index when the temperature is −17o C and the
wind speed is 55km/h.
Solution:
(a) From the table, f (−15, 50) = −29.
By definition,
f (T + h, v) − f (T, v)
h→0
h
f (T, v + h) − f (T, v)
fv (T, v) = lim
h→0
h
To estimate fT (−15, 50) and fv (−15, 50), we follow a procedure similar to that
given in Section 11.3 of your textbook. Based on the values given in the table,
we approximate
fT (T, v) = lim
f (−15 + h, 50) − f (−15, 50)
h→0
h
fT (−15, 50) = lim
using h = ±5. Thus,
f (−10, 50) − f (−15, 50)
−22 − (−29)
7
=
=
5
5
5
f
(−20,
50)
−
f
(−15,
50)
−35
−
(−29)
6
fT− (−15, 50) ≈
=
=
−5
−5
5
fT+ (−15, 50) ≈
Averaging these values, gives
fT (−15, 50) ≈
fT+ (−15, 50) + fT− (−15, 50)
13
=
= 1.3
2
10
Similarly, we approximate
f (−15, 50 + h) − f (−15, 50)
h→0
h
fv (−15, 50) = lim
using h = ±10. Thus,
f (−15, 60) − f (−15, 50)
−30 − (−29)
−1
=
=
10
10
10
−27
−
(−29)
−2
f
(−15,
40)
−
f
(−15,
50)
=
=
fv− (−15, 50) ≈
−10
−10
10
fv+ (−15, 50) ≈
Averaging these values, gives
fv (−15, 50) ≈
fv+ (−15, 50) + fv− (−15, 50)
−3
=
= −0.15
2
20
The linear approximation to the wind-chill function, then, is
f (T, v) ≈ f (−15, 50) + fT (−15, 50)(T − (−15)) + fv (−15, 50)(v − 50)
f (T, v) ≈ −29 + 1.3(T + 15) − 0.15(v − 50)
f (T, v) ≈ −2 + 1.3T − 0.15v
Remark: In the process of approximating partial derivatives, it is
also acceptable to use only one sided approximation of derivatives,
i.e. only use h > 0 or h < 0.
(b) Using the linear approximation found in part a, we estimate the windchill index when the temperature is −17o C and the wind speed is
55km/h as
f (−17, 55) ≈ −2 + 1.3T − 0.15v = −2 + 1.3(−17) − 0.15(55) ≈ −32.35o C