Post on 04-Aug-2020
CALCULUS 4 QUIZ #2 REVIEW / SPRING 09
(1.) Determine the following about the given quadric surfaces.
(a.) Identify & Sketch the quadric surface:zx2
4y2
9+= .
In planes parallel to the xz-plane and planes parallel to the yz-plane, the traces are parabolic. Thus, this a parabaloid. In planes parallel to the xy-plane, the traces are ellipses. Therefore, this an elliptic paraboloid
This "sketch" is computer-generated.
z
xy
Elliptic Parabaloid
Page 1 of 30
This "sketch" is hand-generated.
(b.) Identify & Sketch the quadric surface the quadric surface:z y2 x2−= .
In planes parallel to the xz-plane and planes parallel to the yz-plane, the traces are parabolic. Thus, this a parabaloid. In planes parallel to the xy-plane, the traces are hyperbolas. Therefore, this a hyperbolic paraboloid
This "sketch" is computer-generated.
Page 2 of 30
xy
z
Hyperbolic Parabaloid
This "sketch" is hand-generated.
Page 3 of 30
(c.) Identify & Sketch the quadric surface the quadric surface:x2 y2+ z2− 16= .
In planes parallel to the xz-plane and planes parallel to the yz-plane, the traces are hyperbolic. Thus, this a hyperboloid. In planes parallel to the xy-plane, the traces are circles. This, nonetheless, is called a hyperboloid of 1 sheet.
This "sketch" is computer-generated.
z
xy
Hyperboloid of One Sheet
This "sketch" is hand-generated.
Page 4 of 30
(d.) Identify & Sketch the quadric surface:x2 y2+ z2− 0= .
In planes parallel to the xz-plane and planes parallel to the yz-plane, the traces are straight lines. Thus, this a cone. In planes parallel to the xy-plane, the traces are circles. Therefore, this is a circular cone, which is a special case of the elliptic cone.
This "sketch" is computer-generated.
Page 5 of 30
x
y
z
Elliptic(Circular) Cone
This "sketch" is hand-generated.
Page 6 of 30
(e.) Identify & Sketch the quadric surface:z2 x2− y2− 1= .
In planes parallel to the xz-plane and planes parallel to the yz-plane, the traces are hyperbolic. Thus, this a hyperboloid. In planes parallel to the xy-plane, the traces are circles. Since z 0>, this a hyperboloid of 2 sheets.
This "sketch" is computer-generated.
xy
z
Hyperboloid of Two Sheets
Page 7 of 30
This "sketch" is hand-generated.
(f.) Identify & Sketch the quadric surface:x2 y2+z2
4+ 1= .
In planes parallel to the xz-plane, planes parallel to the yz-plane, and planes parallel to the xy-plane the traces are all ellipses (circular in planes parallel to the xy-plane which is a specical case of ellitic). Thus, this a specicialized ellipsoid ( prolate sphereoid.).
This "sketch" is computer-generated.Page 8 of 30
xy
z
Ellipsoid(Prolate Sphereoid)
This "sketch" is hand-generated. Page 9 of 30
(a.1.) Find traces in the coordinate planes of the quadric
surface:x2
9y2
25+
z2
4+ 1= and sketch trace in the xyz-coordinate
system.
Page 10 of 30
xy plane− xz plane−
x2
9y2
25+ 1= z 0=
x2
9z2
4+ 1= y 0=
5 4 3 2 1 01 2 3 4 5
54321
12345
x
y
5 4 3 2 1 01 2 3 4 5
54321
12345
x
z
yz plane− x 0=
y2
25z2
4+ 1=
5 4 3 2 1 01 2 3 4 5
54321
12345
y
z
Page 11 of 30
This is an ellipsoid. Here is a sketch.
Page 12 of 30
(b.1.) Find traces in the coordinate planes of the quadric surface:z x2 4 y2⋅+= and sketch trace in the xyz-coordinate system.
xy plane− xz plane−
x2 4 y2⋅+ 0= z 0= z x2=
2 1 0 1 2
1
2
3
4
x
zThe trace is the origin.
yz plane− x 0=
z 4 y2⋅=
5 4 3 2 1 0 1 2 3 4 5123456789
10
y
z
Page 13 of 30
This is an elliptic paraboloid. Here is a sketch.
Page 14 of 30
(c.1.) Find traces in the coordinate planes of the quadric
surface:x2
9y2
16+
z2
4− 1= and sketch trace in the xyz-coordinate
system.
xy plane− xz plane−
x2
9y2
16+ 1= z 0= x2
9z2
4+ 1= y 0=
5 4 3 2 1 01 2 3 4 5
54321
12345
x
y
5 4 3 2 1 01 2 3 4 5
54321
12345
x
z
yz plane− x 0=
5 4 3 2 1 01 2 3 4 5
54321
12345
y
z
y2
25z2
4+ 1=
Page 15 of 30
This is a hyperboloid of one sheet. Here is a sketch.
(2.) Determine the following about the given partial derivatives.
(a.) Find fx 1 1,( ) and fy 1 1,( ), where f x y,( ) x2 y⋅ ex y⋅⋅= .
fx 2 x⋅ y⋅ ex y⋅⋅ x2 y2⋅ ex y⋅⋅+= fy x2 ex y⋅⋅ x3 y⋅ ex y⋅⋅+=
fx 1 1,( ) 2 e⋅ e+= fy 1 1,( ) e e+=
fy 1 1,( ) 2 e⋅=fx 1 1,( ) 3 e⋅=
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(b.) A point moves along the intersection of the given elliptic paraboloid z x y,( ) x2 3 y2⋅+= and plane y 1= . At what rate is " z " changing with "x" when the point is at 1 1, 4,( ) ? Also, find the equation of the tangent line at that point.
y
z
x
Paraboloid & Plane
z x y,( ) x2 3 y2⋅+= y 1=
xz∂
∂2 x⋅=
xz 1 1,( )∂
∂2=
The tangent line lies in the plane y 1= . Therefore, "y" is fixed. Accordingly, these are the equations of the tangent line to z x y,( ) at the point 1 1, 4,( ) in the "x" direction.
x t( ) 1 t+= y t( ) 1= z t( ) 4 2 t⋅+=Page 17 of 30
(c.) Calculate x
z x y,( )∂
∂ using implicit differentiation where
ln 2 x2⋅ y+ z3−( ) x= .
xln 2 x2⋅ y+ z3−( )( )∂
∂ xx( )∂
∂=
1
2 x2⋅ y+ z3−( ) x2 x2⋅ y+ z3−( )∂
∂⋅ 1=
4 x⋅ 3 z2⋅x
z∂
∂⋅−
2 x2⋅ y+ z3−( ) 1=
4 x⋅ 3 z2⋅x
z∂
∂⋅− 2 x2⋅ y+ z3−=
xzd
d2− x2⋅ y−⋅ z3+ 4 x⋅+
3 z2⋅=
(d.) Let f x y,( ) x cos y( )⋅= . Find fxx, fyy, fxy, and fyx.
fx1
2 x⋅cos y( )⋅= fy x− sin y( )⋅=
Page 18 of 30
fxy1
2 x⋅− sin y( )⋅= fyx
12 x
− sin y( )⋅=
fyy x− cos y( )⋅=fxx1
4 x
32⋅
− cos y( )⋅=
(3.) Determine the following about differentials.
(a.) Find the local linear approximation "L x y,( )" to
f x y,( ) x2 y2+= at the point P 3 4, 5,( ) and compare the error in approximating "f" by "L" at Q 4 3, 5,( ) with the distance between "P" & "Q" .
L x y,( ) f x0 y0,( ) fx x0 y0,( ) x x0−( )⋅+ fx x0 y0,( )⋅ y y0−( )⋅+=
L x y,( ) x0( )2 y0( )2+x0
x0( )2 y0( )2+x x0−( )⋅+ +=
y0
x0( )2 y0( )2+y y0−( )⋅
L x y,( ) 535
x 3−( )⋅+45
y 4−( )⋅+=
L 4 3,( ) 535
4 3−( )⋅+45
3 4−( )⋅+=245
= f 4 3,( ) 5=
Page 19 of 30
PQ→⎯
12 12+ 02+= 2=
L 4 3,( ) f 4 3,( )−
PQ→⎯
15 2⋅
=
(b.) A function f x y,( ) x2 y⋅= and its Linear Approximation at some point P x0 y0,( ) is L x y,( ) 4 y⋅ 4 x⋅− 8.+= Determine the point "P".
L x y,( ) x0( )2 y0⋅ 2 x0⋅ y0⋅ x x0−( )⋅+ x0( )2 y y0−( )+=
x0( )2 y0⋅ 2 x0( )2⋅ y0⋅− x0( )2 y0⋅− 8= x0( )2 y0⋅ 4−=
2 x0⋅ y0⋅ 4−= x0( )2 4= x0 2−= 2,
y0 1−= x0 2=
P 2 1−,( )
(c.) According to the ideal gas law, the pressure, temperature,
and volume of a confined gas are related by Pk T⋅V
= where "k"
is a constant. Use differentials to approximate the percentage change in pressure if the temperature of the gas is increased by 3% and the volume is increased 5%.
Page 20 of 30
dPT
P∂
∂
⎛⎜⎝
⎞⎟⎠
dT⋅V
P∂
∂
⎛⎜⎝
⎞⎟⎠
dV⋅+=
TP∂
∂
kV
=V
P∂
∂
k T⋅
V2−=
PV
−=
kP V⋅T
=
VP∂
∂
PV
−=T
P∂
∂
P V⋅TV
=PT
=
dPPT
⎛⎜⎝
⎞⎟⎠
dT⋅PV
−⎛⎜⎝
⎞⎟⎠
dV⋅+=
ΔPPT
⎛⎜⎝
⎞⎟⎠ΔT⋅
PV
−⎛⎜⎝
⎞⎟⎠ΔV⋅+=
ΔPP
ΔTT
ΔVV
−=0.03 T⋅
T0.05 V⋅
V−= 0.02−=
ΔPP
0.02−=
ΔPP
100% 2− %=
Page 21 of 30
Page 22 of 30 tz t 10−=( )d
d0=
tzd
d10− t−=
yz
vy
tvd
d⋅∂
∂⋅∂
∂3− t⋅=
yz
uy
tud
d⋅∂
∂⋅∂
∂6−=
xz
vx
tvd
d⋅∂
∂⋅∂
∂2 t⋅=
xz
ux
tud
d⋅∂
∂⋅∂
∂4−=
tvd
dt=
tud
d2−=
vy∂
∂1−=
uy∂
∂1=
yz∂
∂3=
vx∂
∂1=
ux∂
∂1=
xz∂
∂2=
tzd
d xz
ux
tud
d⋅∂
∂ vx
tvd
d⋅∂
∂+
⎛⎜⎝
⎞⎟⎠
⋅∂
∂ yz
uy
tud
d⋅∂
∂ vy
tvd
d⋅∂
∂+
⎛⎜⎝
⎞⎟⎠
⋅∂
∂+=
z z x u t( ) v t( ),( ) y u t( ) v t( ),( ),( )=
(a.) Find tzd
d and then find the value of
tzd
d at t 10−= where
z 2 x⋅ 3 y⋅+= and x u v+= , y u v−= , u 2− t⋅ 3+= , vt2
22+= .
(4.) Determine the following about the Chain Rule.
(b.) Find u
z∂
∂ and
vz∂
∂ where z esin x y⋅( )= , x u2 v⋅= , y v2= .
z z x u v,( ) y v( ),( )=
uz∂
∂ xz
ux∂
∂⋅∂
∂ yz
uy∂
∂⋅∂
∂+=
uy∂
∂0=
uz∂
∂ xz
ux∂
∂⋅∂
∂=
vz∂
∂ xz
vx∂
∂⋅∂
∂ yz
vyd
d⋅∂
∂+=
xz∂
∂esin x y⋅( ) cos x y⋅( )⋅ y( )⋅=
yz∂
∂esin x y⋅( ) cos x y⋅( )⋅ x( )⋅=
xz∂
∂esin x y⋅( ) cos x y⋅( )⋅ y( )⋅= v2 cos u2 v3⋅( )⋅ esin u2 v3⋅( )
⋅=
yz∂
∂esin x y⋅( ) cos x y⋅( )⋅ x( )⋅= u2v cos u2 v3⋅( )⋅ esin u2 v3⋅( )
⋅=
uy∂
∂0=
vy∂
∂ vyd
d= 2 v⋅=
ux∂
∂2 u⋅ v⋅=
vx∂
∂u2=
Page 23 of 30
uz∂
∂3 u⋅ v3⋅ cos u2 v3⋅( )⋅ esin u2 v3⋅( )
⋅=
(c.) Find u
z∂
∂ and
vz∂
∂ where z ln cos x2( )(= , x v tan u( )⋅= .
z z x u v,( )( )=
uz∂
∂ xz
ux∂
∂⋅d
d=
vz∂
∂ xz
vx∂
∂⋅d
d=
xzd
d1
cos x2( ) sin x2( )−( )⋅ 2 x⋅( )⋅= 2− x⋅ tan x2( )⋅=
xzd
d2− v⋅ tan u( )⋅ tan v4 tan u( )( )2⎡⎣ ⎤⎦⋅=
ux∂
∂v sec u( )( )2⋅=
vx∂
∂tan u( )=
uz∂
∂2− v2⋅ tan u( )⋅ sec u( )( )2⋅ tan v4 tan u( )( )2⋅⎡⎣ ⎤⎦⋅=
vz∂
∂2− v⋅ tan u( )( )2⋅ tan v4 tan u( )( )2⋅⎡⎣ ⎤⎦⋅=
Page 24 of 30
(4.) Determine the following about Directional Derivatives.
(a.) You are given the equation of the sphere: x2 y2+ z2+ 4= and the point: 1 1, 2,( ).
xy
z
f(x,y) & Steepest Tangent
(a.1.) Find a unit vector in the direction in which " z " increases most rapidly at the point: 1 1, 2,( )
Since the point is located in the upper hemisphere,
f x y,( ) z x y,( )= 4 x2− y2−= .
Δf x y,( )→⎯⎯⎯
xf x y,( )∂
∂
⎛⎜⎝
⎞⎟⎠
i→⋅
xf x y,( )∂
∂
⎛⎜⎝
⎞⎟⎠
j→⋅+=
Page 25 of 30
We can calculate the partial derivatives by differentiating
f x y,( ) z x y,( )= 4 x2− y2−= explicitely or implicitely. I chose to do it implicitly.
xx2 y2+ z2+( )∂
∂ x4∂
∂=
yx2 y2+ z2+( )∂
∂ y4∂
∂=
xz∂
∂
xz
−=y
z∂
∂
yz
−=
Δf x y,( )→⎯⎯⎯ x
z−⎛⎜⎝
⎞⎟⎠
i→⋅
yz
−⎛⎜⎝
⎞⎟⎠
j→⋅+=
Δf 1 1,( )→⎯⎯⎯ 1
2−⎛⎜⎝
⎞⎟⎠
i→⋅
12
−⎛⎜⎝
⎞⎟⎠
j→⋅+=
Δf 1 1,( )→⎯⎯⎯
Δf 1 1,( )→⎯⎯⎯
12
−⎛⎜⎝
⎞⎟⎠
i→⋅
12
−⎛⎜⎝
⎞⎟⎠
j→⋅+=
(a.2.) Find the rate of change of " f x y,( ) " in that direction at the point: 1 1, 2,( )
sf 1 1,( )d
d12
−⎛⎜⎝
⎞⎟⎠
i→⋅
12
−⎛⎜⎝
⎞⎟⎠
j→⋅+⎡
⎢⎣
⎤⎥⎦
12
−⎛⎜⎝
⎞⎟⎠
i→⋅
12
−⎛⎜⎝
⎞⎟⎠
j→⋅+⎡
⎢⎣
⎤⎥⎦
⋅=
sf 1 1,( )d
d1=
Page 26 of 30
Page 27 of 30
u0→⎯
i→ 1
2−⎛⎜⎝
⎞⎟⎠
j→ 1
2⎛⎜⎝
⎞⎟⎠
⋅+=oru0→⎯
i→ 1
2⎛⎜⎝
⎞⎟⎠
j→ 1
2−⎛⎜⎝
⎞⎟⎠
⋅+=
xr
→1( )⋅d
di
→1( )⋅ j
→1−( )⋅+=
xr
→x( )d
di
→1( )⋅ j
→ x
2 x2−
⎛⎜⎝
⎞⎟⎠
−⎡⎢⎣
⎤⎥⎦
⋅+=
r→
x( ) i→
x( )⋅ j→
2 x2−( )⋅+ 2 k→⋅+=
Since x and y are both positive in the neighborhood of the
given point, the trace of the level curve is given by r→
x( ).
x2 y2+ 2=
x2 y2+ 2( )2+ 4=
A zero slope is in the direction of the level curve at the given point.
(a.4.) Find a unit vector, u0→⎯
, in the direction in which " z " has a slope of
zero at the point: 1 1, 2,( ).
z t( ) 2 2 t⋅−=y t( ) 1 1 t⋅+=x t( ) 1 1 t⋅+=
z t( ) 2 1 t⋅+=y t( ) 12
2t⋅−=x t( ) 1
22
t⋅−=
(a.3.) Find parametric equations of the tangent line to " f x y,( ) " in that direction at the given point: 1 1, 2,( ).
Δf 1 1,( )→⎯⎯⎯
u0→⎯⋅ 0=
Alternatively, we could have determined u0→⎯
as follows.
u0→⎯ u
→
u→=
Δf 1 1,( )→⎯⎯⎯
u→⋅
12
−⎛⎜⎝
⎞⎟⎠
i→⋅
12
−⎛⎜⎝
⎞⎟⎠
j→⋅+⎡
⎢⎣
⎤⎥⎦
i→
ux⋅ j→
uy⋅+( )⋅= 0=
uy ux−=
u→
ux i→
1( )⋅ j→
1−( )⋅+⎡⎣ ⎤⎦⋅=
u0→⎯ u
→
u→=
ux i→
1( )⋅ j→
1−( )⋅+⎡⎣ ⎤⎦⋅
2 ux⋅= i
→ 12
⎛⎜⎝
⎞⎟⎠
⋅ j→ 1
2−⎛⎜⎝
⎞⎟⎠
⋅+= or
u0→⎯
i→ 1
2−⎛⎜⎝
⎞⎟⎠
⋅ j→ 1
2⎛⎜⎝
⎞⎟⎠
⋅+=
(a.5.) Find parametric equations of the tangent line to " f x y,( ) " in that direction at the given point: 1 1, 2,( ).
x t( ) 1 1 t⋅+= y t( ) 1 t−= z t( ) 2=
Page 28 of 30
xy
z
Steepest & Shallowest Tangent Lines
(a.6.) Find the rate of change of " f x y,( ) " in the direction
a→
1 i→⋅ 3 j
→⋅+= at the point: 1 1, 2,( ).
sf 1 1,( )d
dΔf 1 1,( )
→⎯⎯⎯ a→
a→⋅=
12
−⎛⎜⎝
⎞⎟⎠
i→⋅
12
−⎛⎜⎝
⎞⎟⎠
j→⋅+⎡
⎢⎣
⎤⎥⎦
1 i→⋅ 3 j
→⋅+( )
10⋅=
Page 29 of 30
sf 1 1,( )d
d25
−=
(a.7.) Find the parametric equations of the tangent line to " f x y,( ) " in the direction a
→1 i→⋅ 3 j
→⋅+= at the point: 1 1, 2,( ).
x t( ) 1 1 t⋅+= y t( ) 1 3 t⋅+= z t( ) 2 2 2⋅ t⋅−=
xy
z
Steepest, Shallowest & Representative Tangent Lines
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