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If u(t) = (sin(36), cos(26), e) and v(t) = (t, cos(26), sin(36)), use Formula 4 of this theorem to
find
duce) - Vres]
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-11 points Scalc8 13.2.048.
My Notes
If u(t) = (sin(26), cos(26), e) and v(t) = (t, cos(2), sin(26)), use Formula 5 of this theorem to
find
u(e) x (D)
dc
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1.
+ -13 points Scalc8 14.6.502.XP.
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U=
Consider the following equation.
f(x, y) = sin( 3x + 4y), P(-12,9), - }{/31 -1)
(a) Find the gradient of f.
Vf(x, y) =
(b) Evaluate the gradient at the point P.
vf(-12, 9) =
(c) Find the rate of change of fat P in the direction of the vector u.
DJ-12, 9) =
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2.
-11 points Scalc8 14.6.011.
My Notes
Find the directional derivative of the function at the given point in the
direction of the vector v.
f(x, y) = 7e* sin(y), (0, 1/3),
v = (-6, 8)
DulO, 7/3) -
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3.
+ -/3 points Scalc7 14.6.023.MISA
My Notes
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
Find the maximum rate of change of fat the given point and the
direction in which it occurs.
f(x, y) = sin(xy), (5,0)
Step 1
Recall that the direction in which the maximum rate of change of
f(x, y) occurs at a point (a, b) is given by the vector f(a, b). For
f(x, y) = sin(xy), we have
Vf(x, y) =
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-12 points Scalc8 14.6.510.XP.
My Notes
Find the maximum rate of change of fat the given point and the direction in
which it occurs.
f(x, y, z) = x2 + y2 + 22, (6, 2, -8)
maximum rate of change
direction vector
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5.
+ -12 points Scalc8 14.6.029.MI.SA.
My Notes
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
Find all the points at which the direction of fastest change of the
function rx, y) = x2 + y2 - 10x - 16y is i+j.
Step 1
The direction in which the maximum rate of change of axy)
occurs at a point (a, b) is given by the vector ba, b). For
R(x, ) = x2 + y2 - 10x - 16y, we have
Vf(x,y) -
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6.
+ -13 points Scalc8 14.6.032.
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The temperature at a point (x, y, z) is given by
T(x, y, z) = 200e-x2 - 5y2 - 922
where T is measured in °C and x,y,z in meters.
(a) Find the rate of change of temperature at the point P(4, -1, 5) in the
direction towards the point (5, -5,6).
°C/m
(b) In which direction does the temperature increase fastest at P?
(c) Find the maximum rate of increase at P.
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Theorem Suppose u and v are differentiable vector functions, c is a scalar,
and f is a real-valued function. Then
1. [u(t) + v(t)] = u'(t) + v'(t)
dt
2.
2. [cu(t)] = cu'(t)
dt
3. [f(t)u(t)] = f'(t)u(t) + f(t)u'(t)
d
[u(t) . v(t)] = u'(t) . v(t) + ult) . v'(t)
[u(t) x v(t)] = u'ſt) v(t) + ult) * v'(t)
6. Or u
[u(f(t))] = f'(t)u'(f(t)) (Chain Rule)
dt
4.
dt
5.
dt
d
6.
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If f(x,y) = xy, find the gradient vector v2, 3) and use it to find the tangent
line to the level curve f(x,y) = 6 at the point (2, 3).
gradient vector
tangent line equation
Sketch the level curve, the tangent line, and the gradient vector.
0
KN
AK
7
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8
+ -11 points Scalc8 14.6.512.XP.
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At what point on the paraboloid y = x2 + 22 is the tangent plane parallel to the
plane 7x + 2y + 5z = 6? (If an answer does not exist, enter DNE.)
(x, y, z) =
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9.
+ -/1 points Scalc8 14.6.055.
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Find any points on the hyperboloid x2 - y2 - 72 = 2 where the tangent plane is
parallel to the plane z = 5x + 5y. (If an answer does not exist, enter DNE.)
(X,Y,Z) =
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