Minnesota State University Directional Derivative Calculus III Problems I have to do a homework on web assign. the class is calculus III. see the attachmen

Minnesota State University Directional Derivative Calculus III Problems I have to do a homework on web assign. the class is calculus III. see the attachment for more details. sarour@cua.edu ( sign out )
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Math 221, section 25, Summer 2 2020
INSTRUCTOR
Guoyang Liu
M221_Sum Homework Assignment 8 (14.6, 14.7)
(Homework)
Catholic University of America,
DC
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Current Score
QUESTION
1
2
3
4
5
6
7
8
9
POINTS
–/5
–/1
–/3
–/1

–/1
–/2
–/2
–/3
–/4
TOTAL SCORE
–/66
0.0%
Due Date
MON, JUL 13, 2020
11:56 PM EDT
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1. [–/5 Points]
SCALCET8 14.6.AE.005.
DETAILS
MY NOTES
Video Example
EXAMPLE 5
ASK YOUR TEACHER
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If f(x, y, z) = x sin(yz), (a) find the gradient of f and (b) find the
directional derivative of f at (2, 5, 0) in the direction of v = i + 3j − k.
SOLUTION
+
(a) The gradient of f is
∇f(x, y, z) = fx(x, y, z), fy(x, y, z), fz(x, y, z)
=
​
.
–
Operations
Functions
x
÷
▯
​
▯
▯▯ ​
Symbols
Relations
Sets
▯!
Vectors
. The unit vector in the direction
Trig
(b) At (2, 5, 0) we have ∇f(2, 5, 0) =
of v = i + 3j − k is
Greek
u=
.
Help
Therefore this equation gives
Duf(2, 5, 0) = ∇f(2, 5, 0) · u
= 10k ·
.
=
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2. [–/1 Points]
SCALCET8 14.6.004.
DETAILS
MY NOTES
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MY NOTES
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MY NOTES
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Find the directional derivative of f at the given point in the direction indicated by the angle θ.
f(x, y) = xy3 − x2,
θ = π/3
(1, 3),
Duf(1, 3) =
3. [–/3 Points]
SCALCET8 14.6.009.
DETAILS
Consider the following.
f(x, y, z) = x2yz − xyz5,
P(3, −1, 1),
u = 0,
4
3
,−
5
5
(a) Find the gradient of f.
∇f(x, y, z) =
(b) Evaluate the gradient at the point P.
∇f(3, −1, 1) =
(c) Find the rate of change of f at P in the direction of the vector u.
Duf(3, −1, 1) =
4. [–/1 Points]
SCALCET8 14.6.011.
DETAILS
Find the directional derivative of the function at the given point in the direction of the vector v.
f(x, y) = 3ex sin(y),
(0, π/3),
v = −6, 8
Dvf(0, π/3) =
5. [–/1 Points]
SCALCET8 14.6.015.
DETAILS
MY NOTES
Find the directional derivative of the function at the given point in the direction of the vector v.
f(x, y, z) = x2y + y2z,
(2, 7, 9),
v = 2, −1, 2
Dvf(2, 7, 9) =
6. [–/2 Points]
SCALCET8 14.6.023.MI.
DETAILS
MY NOTES
Find the maximum rate of change of f at the given point and the direction in which it occurs.
f(x, y) = 8 sin(xy),
(0, 7)
maximum rate of change
direction vector
7. [–/2 Points]
SCALCET8 14.6.510.XP.
DETAILS
MY NOTES
Find the maximum rate of change of f at the given point and the direction in which it occurs.
f(x, y, z) =
x2 + y2 + z2 ,
(4, 6, −4)
maximum rate of change
direction vector
8. [–/3 Points]
SCALCET8 14.6.033.
DETAILS
MY NOTES
Suppose that over a certain region of space the electrical potential V is given by the following equation.
V(x, y, z) = 2×2 − 3xy + xyz
(a) Find the rate of change of the potential at P(4, 4, 5) in the direction of the vector v = i + j − k.
(b) In which direction does V change most rapidly at P?
(c) What is the maximum rate of change at P?
9. [–/4 Points]
SCALCET8 14.7.003.
DETAILS
MY NOTES
ASK YOUR TEACHER
Use the level curves in the figure to predict the location of the critical points of f and whether f has a saddle point or a local maximum or minimum at
each critical point. Then use the Second Derivatives Test to confirm your predictions. (Order your answers by their ordered pairs, from smallest to
largest x.)
f(x, y) = 4 + x3 + y3 − 3xy
(x, y) =
Select Classification
(x, y) =
Select Classification
10. [–/13 Points]
SCALCET8 14.7.AE.003.
DETAILS
MY NOTES
EXAMPLE 3
f(x, y) =
x4
SOLUTION
ASK YOUR TEACHER
Find the local maximum and minimum values and saddle points of
+ y4 − 4xy + 1.
We first locate the critical points:
fy = 4y3 − 4x.
fx =
Setting these partial derivatives equal to 0, we obtain the equations
Video Example
=0
and
y3 − x = 0.
To solve these equations we substitute y = x3 from the first equation into the second
one. This gives
0 = x9 − x = x(x8 − 1)
= x
(x4 + 1)
= x
(x2 + 1)(x4 + 1)
so there are three real roots: x = 0, 1,
. The three critical points are
(0, 0), (1, 1), and (−1, −1). Next we calculate the second partial derivatives and
D(x, y):
fxx =
fxy =
fyy =
D(x, y) = fxxfyy − (fxy)2 =
Since D(0, 0) = −16 0 and fxx(1, 1) = 12 > 0, we see that f(1, 1) = −1 is a
. Similarly, we have D(−1, −1) =
fxx(−1, −1) = 12 > 0, so f(−1, −1) = −1 is also a local
11. [–/3 Points]
SCALCET8 14.7.010.
DETAILS
MY NOTES
> 0 and
—Select—
ASK YOUR TEACHER
.
PRACTICE ANOTHER
Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function
with a domain and viewpoint that reveal all the important aspects of the function. (Enter your answers as a comma-separated list. If an answer does
not exist, enter DNE.)
f(x, y) = 6 − x4 + 2×2 − y2
local maximum value(s)
local minimum value(s)
saddle point(s)
12. [–/3 Points]
(x, y, f) =
SCALCET8 14.7.011.
DETAILS
MY NOTES
ASK YOUR TEACHER
PRACTICE ANOTHER
Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function
with a domain and viewpoint that reveal all the important aspects of the function. (Enter your answers as a comma-separated list. If an answer does
not exist, enter DNE.)
f(x, y) = 3×3 − 9x + 9xy2
local maximum value(s)
local minimum value(s)
saddle point(s)
13. [–/3 Points]
(x, y, f) =
SCALCET8 14.7.504.XP.
DETAILS
MY NOTES
ASK YOUR TEACHER
PRACTICE ANOTHER
Find the local maximum and minimum values and saddle point(s) of the function. If you have three-dimensional graphing software, graph the function
with a domain and viewpoint that reveal all the important aspects of the function. (Enter your answers as a comma-separated list. If an answer does
not exist, enter DNE.)
f(x, y) = 7ey(y2 − x2)
local maximum value(s)
local minimum value(s)
saddle point(s)
(x, y, f) =
14. [–/16 Points]
SCALCET8 14.8.501.XP.MI.SA.
DETAILS
MY NOTES
ASK YOUR TEACHER
PRACTICE ANOTHER
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for the skipped
part, and you will not be able to come back to the skipped part.
Tutorial Exercise
Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given
constraint.
f(x, y) = 7×2 + 7y2;
xy = 1
Click here to begin!
15. [–/2 Points]
DETAILS
SCALCET8 14.8.003.
MY NOTES
ASK YOUR TEACHER
PRACTICE ANOTHER
This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the extreme values of
the function subject to the given constraint.
f(x, y) = x2 − y2;
x2 + y2 = 9
maximum value
minimum value
16. [–/2 Points]
DETAILS
SCALCET8 14.8.007.
MY NOTES
ASK YOUR TEACHER
PRACTICE ANOTHER
This extreme value problem has a solution with both a maximum value and a minimum value. Use Lagrange multipliers to find the extreme values of
the function subject to the given constraint.
f(x, y, z) = 8x + 8y + 4z;
4×2 + 4y2 + 4z2 = 36
maximum value
minimum value
17. [–/2 Points]
DETAILS
SCALCET8 14.8.033.
MY NOTES
ASK YOUR TEACHER
Use Lagrange multipliers to find the points on the given cone that are closest to the following point.
z2 = x2 + y2;
(10, 6, 0)
(x, y, z) =
(smaller z-value)
(x, y, z) =
(larger z-value)
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