5c-14.6.math

14.6 5
Find the directional derivative of f at the given point in the direction indicated by the angle θ.
f(x, y) = y cos(xy),    (0, 1),    θ = π/6

fx=-y^2sin(xy)
fx(0 1)= -1^2sin(0*1) =0

fy= cos(xy)-sin(xy)y^2
fy= 1

Duf(0 1)= sin(pi/6)
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14.6 9
Consider the following.
f(x, y, z) = x^2yz − xyz^3,    P(2, −1, 1),    u = <0, 4/5, −3/5>

a) gradient

∇f= <2xyz-yz^3,x^2z-xz^3,x^2y-xy3z^2>
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14.6 17
Find the directional derivative of the function at the given point in the direction of the vector v.
h(r, s, t) = ln(3r + 6s + 9t),    (2, 2, 2),    v = 14i + 42j + 21k

∇h=<3/(3r + 6s + 9t),6/(3r + 6s + 9t),9/(3r + 6s + 9t)>
∇h(2 2 2)=<3/(6 + 12 + 18),6/(18 + 18),9/(36)>
∇h(2 2 2)=<1/12,1/6,1/4>

|v|= sqrt(14^2+42^2+21^2)=49

u= <14/49,42/49,21/49>
u=<2/7,6/7,3/7>

Duh(2 2 2)= 1/42+1/7+3/28 =23/84
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14.6 19
Find the directional derivative of f(x, y) = sqrt(xy) at P(8, 8) in the direction from P to Q(11, 4).
PQ = <11-8,4-8>
PQ = <3,-4>
|PQ| = sqrt(9+16) = 5
u= <3/5,-4/5>
Duf(8, 8) = <y/2sqrt(xy),x/2sqrt(xy)>.u
Duf(8 8) = <8/2*8,1/2>.u
Duf(8 8) = 3/10-4/10 = -1/10
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14.6 23
Find the maximum rate of change of f at the given point and the direction in which it occurs.
f(x, y) = 6 sin(xy),    (0, 4)

∇f = <6ycos(xy), 6x(cos(xy)>

∇f(0 4) = <6(4)cos(0), 0>
∇f(0 4) = <24, 0>
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14.6 45
Find equations of the tangent plane and the normal line to the given surface at the specified point.
x + y + z = 4e^(xyz),    (0, 0, 4)

x+y+z-4e^(xyz)= 0
F(x y z) = x+y+z-4e^(xyz)=0
∇F=<1-4yze^(xyz),1-4xze^(xyz),1-4xye^(xyz)>
∇F(0 0 4)=<1,1,1>

tangent plane
x+y+z = 4
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14.6 41
Find equations of the following.
2(x − 1)^2 + (y − 1)^2 + (z − 8)^2 = 10,    (2, 3, 10)
a) tangent plane
Fx=4(x-1)
Fy=2(y-1)
Fz=2(z-8)

Fx(2,3,10)= 4(2-1)=4
Fy(2,3,10)= 2(3-1)=4
Fz(2,3,10)= 2(10-8)=4

4(x-2)+4(y-3)+4(z-10)=0
4x-8+4y-12+4z-40=0
4(x+y+z)=60
x+y+z=15

b) normal line
(2+4t,3+4t,10+4t)