Using a Function In Exercises 67 and 68, (a) find the gradient of the function at P, (b) find a unit normal vector to the level curve f ( x , y ) = c at P, (c) find the tangent line to the level curve f ( x , y ) = c at P, and (d) sketch the level curve, the unit normal vector, and the tangent line in the xy -plane. f ( x , y ) = 9 x 2 − 4 y 2 c = 65 , P ( 3 , 2 )
Using a Function In Exercises 67 and 68, (a) find the gradient of the function at P, (b) find a unit normal vector to the level curve f ( x , y ) = c at P, (c) find the tangent line to the level curve f ( x , y ) = c at P, and (d) sketch the level curve, the unit normal vector, and the tangent line in the xy -plane. f ( x , y ) = 9 x 2 − 4 y 2 c = 65 , P ( 3 , 2 )
Solution Summary: The author explains how the formula for the gradient of a function f(x,y) is given by: 54i-16j.
Using a Function In Exercises 67 and 68, (a) find the gradient of the function at P, (b) find a unit normal vector to the level curve
f
(
x
,
y
)
=
c
at P, (c) find the tangent line to the level curve
f
(
x
,
y
)
=
c
at P, and (d) sketch the level curve, the unit normal vector, and the tangent line in the xy -plane.
f
(
x
,
y
)
=
9
x
2
−
4
y
2
c
=
65
,
P
(
3
,
2
)
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.
(I) Determine the gradient of f.
(II) Calculate the gradient at point P.
(III) Determine a rate of change of f in P in the direction of vector u.
a) f (x, y, z) xe2yz , P(3, 0, 2), u = (2/3, - 2/3, 1/3)
b) f (x, y, z) √x + yz, P(1, 3, 1), u = (2/7, 3/7, 6/7)
Using Properties of the Derivative In Exercise 26, use the properties of the derivative to find the following. (a) r′(t)
(b) d dt [u(t) − 2r(t)] (c) d dt [(3t)r(t)]
(d) d dt [r(t) ∙ u(t)] (e) d dt [r(t) × u(t)]
(f) d dt [u(2t)]
26. r(t) = sin ti + cos tj + tk, u(t) = sin ti + cos tj + 1 t k
* only d ,e, f *
Using Green's Theorem, find the outward flux of F across the closed curve C.F = (x - y) i + (x + y) j; C is the triangle with vertices at (0, 0), (6, 0), and (0, 6)
a)
216
b)
72
c)
0
d)
36
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