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( )⋅=

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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=

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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− %=

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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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