Gradients in three dimensions Consider the following functions f, points P, and unit vectors u . a. Compute the gradient of f and evaluate it at P b. Find the unit vector in the direction of maximum increase of f at P. c. Find the rate of change of the function in the direction of maximum increase at P. d. Find the directional derivative at P in the direction of the given vector. 59. f ( x , y , z ) = 1 + sin ( x + 2 y − z ) ; P ( π 6 , π 6 , − π 6 ) ; 〈 1 3 ′ 2 3 ′ 2 3 〉
Gradients in three dimensions Consider the following functions f, points P, and unit vectors u . a. Compute the gradient of f and evaluate it at P b. Find the unit vector in the direction of maximum increase of f at P. c. Find the rate of change of the function in the direction of maximum increase at P. d. Find the directional derivative at P in the direction of the given vector. 59. f ( x , y , z ) = 1 + sin ( x + 2 y − z ) ; P ( π 6 , π 6 , − π 6 ) ; 〈 1 3 ′ 2 3 ′ 2 3 〉
Gradients in three dimensionsConsider the following functions f, points P, and unit vectorsu.
a.Compute the gradient of f and evaluate it at P
b.Find the unit vector in the direction of maximum increase of f at P.
c.Find the rate of change of the function in the direction of maximum increase at P.
d.Find the directional derivative at P in the direction of the given vector.
59.
f
(
x
,
y
,
z
)
=
1
+
sin
(
x
+
2
y
−
z
)
;
P
(
π
6
,
π
6
,
−
π
6
)
;
〈
1
3
′
2
3
′
2
3
〉
Quantities that have magnitude and direction but not position. Some examples of vectors are velocity, displacement, acceleration, and force. They are sometimes called Euclidean or spatial vectors.
Determine whether the lines
L₁ (t) = (-2,3, −1)t + (0,2,-3) and
L2 p(s) = (2, −3, 1)s + (-10, 17, -8)
intersect. If they do, find the point of intersection.
Convert the line given by the parametric equations y(t)
Enter the symmetric equations in alphabetic order.
(x(t)
= -4+6t
= 3-t
(z(t)
=
5-7t
to symmetric equations.
Find the point at which the line (t) = (4, -5,-4)+t(-2, -1,5) intersects the xy plane.
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, calculus and related others by exploring similar questions and additional content below.