CALCULUS 4 QUIZ #1 REVIEW / SPRING 09 - Drexel Universityjwd25/Calc4_Winter_09/tests&quizzes... ·...

CALCULUS 4 QUIZ #1 REVIEW / SPRING 09

(1.) Determine the following about the vectors: v→

i→

1( )⋅ j→

1( )⋅+ k→

1−( )⋅+= ,

u→

i→

1( )⋅ j→

1( )⋅+ k→

1( )⋅+= , and w→

i→

4( )⋅ j→

4( )⋅+ k→

2−( )⋅+= .

(a.) Find the norm, v→

, or magnitude of the vector v→

.

v→

1( )2 1( )2+ 1−( )2+= 3=

(b.) Find a unit vector, vunit→⎯⎯

pointing in the direction of v→

.

vunit→⎯⎯ v

→

v→= i

→1( )⋅ j

→1( )⋅+ k

→1−( )⋅+⎡⎣ ⎤⎦

3= i

→ 1

3⎛⎜⎝

⎞⎟⎠

⋅ j→ 1

3⎛⎜⎝

⎞⎟⎠

⋅+ k→ 1

3−⎛⎜⎝

⎞⎟⎠

⋅+⎡⎢⎣

⎤⎥⎦

=

(c.) Find the vector x→

which satisfies this vector equation:

2 v→⋅ 2 u

→⋅+ w

→− x

→+ 0

→=

(d.) Find a vector n→

which is perpendicular to the vectors : u→

, and v→

.

n→

u→

v→×=

i→

1

1

j→

1

1

k→

1

1−

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

= 2−( ) i→⋅ 2( ) j

→⋅+=

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(e.) Decompose n→

in part (d) into 2 vectors, one along i→

j→− , and the other,

z→

, orthogonal to i→

j→− .

Proj n→( )

i→⋅

2−( ) i→⋅ 2( ) j

→⋅+⎡⎣ ⎤⎦ i

→1( )⋅ j

→1( )⋅−⎡⎣ ⎤⎦⋅ i

→1( )⋅ j

→1( )⋅−⎡⎣ ⎤⎦⋅

2( )2= 4−( ) i

→⋅ 4( ) j

→⋅+

2=

z→

n→

Proj n→( )

i→⋅−= 2−( ) i

→⋅ 2( ) j

→⋅+⎡⎣ ⎤⎦ 2−( ) i

→⋅ 2( ) j

→⋅+⎡⎣ ⎤⎦−= 0

→=

n→

Proj n→( )

i→⋅ z

→+= 2−( ) i

→⋅ 2( ) j

→⋅+= 2−( ) i

→⋅ 2( ) j

→⋅+=

(f.) Determine if the vector w→

lies in the plane formed by the vectors u→

and

v→

.The cross product of the vectors u→

and v→

is a vector that is perpendicular to both vectors and the plane that they lie in. If we take the dot product of

that vector with w→

and we get zero, then the vector w→

must lie in the plane.

Otherwise, the vector w→

does not lie in the plane

u→

v→× 2−( ) i

→⋅ 2( ) j

→⋅+=

w→

u→

v→×( )⋅ i

→4( )⋅ j

→4( )⋅+ k

→2−( )⋅+⎡⎣ ⎤⎦ 2−( ) i

→⋅ 2( ) j

→⋅+⎡⎣ ⎤⎦⋅= 8− 8+= 0=

Therefore, the vector w→

lies in the plane formed by the vectors u→

and v→

.

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(g.) Find the magnitude of the resultant vector y→

u→

v→+= and the angle

that the resultant vector makes with the positive x-axis

y→

u→

v→+= i

→1( )⋅ j

→1( )⋅+ k

→1( )⋅+⎡⎣ ⎤⎦ i

→1( )⋅ j

→1( )⋅+ k

→1−( )⋅+⎡⎣ ⎤⎦+=

y→

i→

2( )⋅ j→

2( )⋅+⎡⎣ ⎤⎦=

y→

2 2⋅= θ tan 1− 22⎛⎜⎝⎞⎟⎠

= tan 1− 1( )= π4

=

(h.) Show that the vector AP→⎯

formed from the points P 4 2, 2−,( ) and

A 0 2−, 0,( ) equals the vector w→

.

AP→⎯

4 0−( ) i→⋅ 2 2−( )−[ ] j

→⋅+ 2− 0−( ) k

→⋅+= i

→4( )⋅ j

→4( )⋅+ k

→2−( )⋅+= w

→=

(i.) Show that the vector AB→⎯

formed from the points A 0 2−, 0,( ) and

B 1 1−, 1,( ) equals the vector u→

.

AB→⎯

1 0−( ) i→⋅ 1− 2−( )−[ ] j

→⋅+ 1 0−( ) k

→⋅+= i

→1( )⋅ j

→1( )⋅+ k

→1( )⋅+= u

→=

(j.) Find the distance between the point P 4 2, 2−,( ) and the line thru the points A 0 2−, 0,( ) and B 3 1, 3−,( ).

AP→⎯

AB→⎯

× AB→⎯

AP→⎯

⋅ sin θ( )⋅=

d AP→⎯

sin θ( )⋅= AP→⎯

AB→⎯

×

AB→⎯= w

→u→×

u→=

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u→

1( )2 1( )2+ 1( )2+= 3=

w→

u→×

i→

4

1

j→

4

1

k→

2−

1

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

= i→ 4

1

2−

1⎛⎜⎝

⎞⎟⎠

⋅ j→ 4

1

2−

1⎛⎜⎝

⎞⎟⎠

⋅− k→ 4

1

4

1⎛⎜⎝

⎞⎟⎠

⋅+=

w→

u→× 6( ) i

→⋅ 6−( ) j

→⋅+=

w→

u→× 6 1( )2 1−( )2+⋅= 6 2⋅=

d6 2⋅

3=

(2.) Determine the following about the given lines.

(a.) Find parametric equations of the line that is parallel to the

vector:v→

7( ) i→⋅ 1−( ) j

→⋅+ 3( ) k

→⋅+= and that passes thru the point: 3− 2, 1,( ).

x 3− 7 t⋅+= y 2 t−= z 1 3 t⋅+=

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(b.) Show that the lines L1: x t( ) 3 2 t⋅−= y t( ) 4 t+= z t( ) 6 t−= and L2: x s( ) 5 4 s⋅−= y s( ) 2− 2 s⋅+= z s( ) 7 2 s⋅−= are parallel.

v1→

2− i→⋅ j→+ k

→−= v2

→4− i→⋅ 2 j

→⋅+ 2k

→−=

v1→

v2→

× 2− i→⋅ j→+ k

→−( ) 4− i

→⋅ 2 j

→⋅+ 2k

→−( )×=

v1→

v2→

× 8 i→⋅ i→× 4 i

→⋅ j→×− 4 i

→⋅ k→×+ 4 j

→⋅ i→×− 2 j

→⋅ j→×+ 2 j

→⋅ k

→×− +=

4 k→⋅ i→⋅ 2 k

→⋅ j

→×− 2 k

→⋅ k

→×+

v1→

v2→

× 0→

4 k→⋅− 4 j

→⋅− 4 k

→⋅+ 0

→+ 2 i

→⋅− 4 j

→⋅+ 2 i

→⋅+ 0→+= 0

→=

v1→

v2→

× v1→

v2→

⋅ sin θ( )⋅= 0=

Thus the angle between the vectors that are parallel to the lines is zero. Accordingly, the lines must be parallel.

Alternatively, note that v2→

2 v1→⋅= and are thus scalar multiples of each

other. Because of that fact, the 2 vectors must be parallel and thus so too must be their corresponding lines.

(c.) Show that the lines L1: x t( ) 3 2 t⋅−= y t( ) 4 4 t⋅+= z t( ) 6 4 t⋅−= and L2: x s( ) 5 4 s⋅−= y s( ) 2− 1 s⋅−= z s( ) 7 s+= are perpendicular to each other but do not intersect each other.

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1 2 3 s⋅+=3 t+ 1 6 s⋅+=1− 4 t⋅+ 13− 12 s⋅+=

(d.) Show that the lines L1: x t( ) 1− 4 t⋅+= y t( ) 3 t+= z t( ) 1= and L2: x s( ) 13− 12 s⋅+= y s( ) 1 6 s⋅+= z s( ) 2 3 s⋅+= intersect and find their point of intersection.

Since we obtained a contradiction, the lines do not intersect.

4 1−≠

4 4 t⋅+ 2− 1+ 4 t⋅+=

s 1− 4 t⋅−=

6 4 t⋅− 7 s+=4 4 t⋅+ 2− 1 s⋅−=3 2 t⋅− 5 4 s⋅−=

Set the corresponding coordinates equal to each other.

Therefore, the angle "θ" between the vectors is π2

and thus the lines are

perpendicular to each other.

v1→

v2→⋅ v1

→v2→

⋅ cos θ( )⋅= 0=

v1→

v2→⋅ 2− i

→⋅ 4 j

→⋅+ 4 k

→⋅−( ) 4− i

→⋅ j→− k

→+( )⋅= 8 4− 4−= 0=

v2→

4− i→⋅ j→− k

→+=v1

→2− i→⋅ 4 j

→⋅+ 4 k

→⋅−=

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Thus, this is the intersection point: 54

94,

12,⎛⎜

⎝⎞⎟⎠.

z34⎛⎜⎝⎞⎟⎠

1−64

+= 24

=y34⎛⎜⎝⎞⎟⎠

94

=x34⎛⎜⎝⎞⎟⎠

234

−= 54

=

t34

=12 t⋅ 9=

2 3 t⋅( )⋅ 3 1− 2 t⋅+( )+ 6=

(e.) Where does the line L1: x t( ) 2 t−= y t( ) 3 t⋅= z t( ) 1− 2 t⋅+= intersect the plane 2 y⋅ 3 z⋅+ 6= .

Thus, this is their point of intersection: 17− 1−, 1,( ).

z 1=y 1−=x 17−=

x 4−( ) 1− 4 4−( )⋅+= y 4−( ) 3 4−( )+= z 4−( ) 1=

Therefore, the lines intersect at t 4−= and s 13

−= .

17− 17−=

1− 16− 13− 4−=

t 4−=

3 t+ 1 2−=

s13

−=

(f.) Determine whether the points: A 1 0, 1,( ), B 3 4−, 3−,( ), and C 4 6−, 5−,( ) lie on the same line.

AB→⎯

i→

2( )⋅ j→

4−( )⋅+ k→

4−( )⋅+= AC→⎯

i→

3( )⋅ j→

6−( )⋅+ k→

6−( )⋅+=

AC→⎯ 3

2AB→⎯

⋅=

Since the vectors are parallel, all three points lie on the same line.

(3.) Determine the following about the given planes.

(a.) Find the equation of the plane that is perpendicular to the

vector:n→

1−( ) i→⋅ 7( ) j

→⋅+ 2( ) k

→⋅+= and that passes thru the point: 3− 2, 1,( ).

x 3+( )− 7 y 2−( )⋅+ 2 z 1−( )⋅+ 0=

(b.) Find the equation of the plane that passes thru the points: A 3 2, 1,( ), B 2 1, 1−,( ) and C 1− 3, 2,( ) .

AB→⎯

i→

1−( )⋅ j→

1−( )⋅+ k→

2−( )⋅+= AC→⎯

i→

4−( )⋅ j→

1( )⋅+ k→

1( )⋅+=

n→

AB→⎯

AC→⎯

×=i

→

1−

4−

j→

1−

1

k→

2−

1

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

=

n→

i→

1( )⋅ j→

9( )⋅+ k→

5−( )⋅+=

x 1−( )−[ ] 9 y 3−( )⋅+ 5 z 2−( )⋅− 0=

x 9 y⋅+ 5 z⋅− 16= of 33

(c.) Determine whether the planes: P1: 3 x⋅ 2y− z+ 4= and P2: 6 x⋅ 4y− 3 z⋅+ 7= are parallel, perpendicular or neither.

n1→

i→

3( )⋅ j→

2−( )⋅+ k→

1( )⋅+= n2→

i→

6( )⋅ j→

4−( )⋅+ k→

3( )⋅+=

If the planes are perpendicular, then the dot product of their normal vectors must be zero. Otherwise, the planes are not perpendicular (orthogonal).

n1→

n2→⋅ 18 8+ 3+= 27= 0≠

Thus, the planes are NOT perpendicular to each other.

If the planes are parallel, then their normal vectors must be parallel and thusthey must be scalar multiples and their cross product must be zero. Otherwise, the planes are not parallel.

By inspection of the 2 normal vectors, we conclude that they are NOT scalarmultiples of each other and thus the planes are NOT parallel.

Alternatively, consider their cross product as a test.

n1→

n2→

×i

→

3

6

j→

2−

4−

k→

1

3

⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

= i→

2−( )⋅ j→

3−( )⋅+ k→

0( )⋅+=

The cross product of the normals is NOT the zero vector. Thus, the planes are not parallel.

Since the planes are not parallel, they must intersect; however their intersection is NOT perpendicular.

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(d.) Determine whether the line L: x t( ) 3 t−= y t( ) 2 t+= z t( ) 1 3 t⋅−= andplane: P: 2 x⋅ 2y+ 5− 0= are parallel, perpendicular or neither.

If the plane and line are perpendicular, then the normal to the plane is parallel to the line. if this is the case, then the cross product of the normal vector with a vector parallel to the line must be zero or they are scalar multiples of each other.

n→

i→

2( )⋅ j→

2( )⋅+ k→

0( )⋅+= v→

i→

1−( )⋅ j→

1( )⋅+ k→

3−( )⋅+=

They are NOT scalar multiples. Accordingly, the plane and line are NOT perpendicular.

If the plane and line are parallel, then the normal to the plane is perpendicular to the line. if this is the case, then the dot product of the normal vector with a vector parallel to the line must be zero.

n→

v→⋅ i

→2( )⋅ j

→2( )⋅+ k

→0( )⋅+⎡⎣ ⎤⎦ i

→1−( )⋅ j

→1( )⋅+ k

→3−( )⋅+⎡⎣ ⎤⎦⋅= 2− 2+= 0=

Therefore, the line and plane are parallel.

(e.) Determine whether the line L: x t( ) 3 t⋅= y t( ) 5 t⋅= z t( ) t−= and plane: P: 2 x⋅ y− z+ 1+ 0= intersect and if they do then find their point of intersection.

If the plane and line are parallel, then the normal to the plane is perpendicular to the line. if this is the case, then the dot product of the normal vector with a vector parallel to the line must be zero. Otherwise, theymust intersect.

n→

i→

2( )⋅ j→

1−( )⋅+ k→

1( )⋅+= v→

i→

3( )⋅ j→

5( )⋅+ k→

1−( )⋅+=

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n→

v→⋅ i

→2( )⋅ j

→1−( )⋅+ k

→1( )⋅+⎡⎣ ⎤⎦ i

→3( )⋅ j

→5( )⋅+ k

→1−( )⋅+⎡⎣ ⎤⎦⋅= 6 5− 1−= 0=

Therefore, the line and plane are parallel and do NOT intersect.

Alternatively, we could have arrived at that same conclusion by assuming that they intersect and finding that that assumption leads to a contradiction.

2 x⋅ y− z+ 1+ 0= x t( ) 3 t⋅= y t( ) 5 t⋅= z t( ) t−=

2 3 t⋅( )⋅ 5 t⋅( )− t−( )+ 1+ 0=

1 0≠

Therefore, the line and plane are parallel and do NOT intersect.

(f.) Find the equation of the plane that contains the line L: x t( ) 2 3 t⋅−= y t( ) 1 t+= z t( ) 1− t+= and is perpendicular to the plane P: x y+ 2 z⋅− 1= .

Here is a normal vector for the given plane.

ng→

i→

1( )⋅ j→

1( )⋅+ k→

2−( )⋅+=

Here is a vector parallel to the given line.

v→

i→

3−( )⋅ j→

1( )⋅+ k→

1( )⋅+=

A normal vector for the unknown plane can be found by taking the cross product of the normal vector for the given plane with the direction vector of the line.

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n→

ng→

v→×= i

→1( )⋅ j

→1( )⋅+ k

→2−( )⋅+⎡⎣ ⎤⎦ i

→3−( )⋅ j

→1( )⋅+ k

→1( )⋅+⎡⎣ ⎤⎦×=

n→

i→

3( )⋅ j→

5( )⋅+ k→

4( )⋅+=

Now we choose any point, say 2 1, 1−,( ) on the line to obtain a point on the unknown plane.

3 x 2−( )⋅ 5 y 1−( )⋅+ 4 z 1+( )⋅+ 0=

y

xz

L n→

ng→⎯

Line & Plane Determine Plane

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(g.) Find the equation of the plane thru 1 2, 1−,( ) that is perpendicular to the line of intersection of the planes: P1: 2 x⋅ y+ z+ 2= and P2: x 2 y⋅+ z+ 3= .

n1→

2 i→⋅ j→+ k

→+= n2

→i

→2 j→⋅+ k

→+=

If we take the cross product of the normals of both planes, we obtain a vector that is parallel to both planes and thus the normal of the plane that is perpendicular to the line of intersection.

n→

n1→

n2→

×= 2 i→⋅ j→+ k

→+( ) i

→2 j→⋅+ k

→+( )×= i

→− j

→− 3 k

→⋅+( )=

Here is the equation of the perpendicular plane.

x 1−( )− y 2−( )− 3 z 1+( )⋅+ 0=

x

z

y

Perpendicular Plane to Line of Intersection

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(h.) Find the equation of the plane that contains the line L1: x 3 t⋅= y 1 t+= z 2 t⋅= and is parallel to the intersection of the planes: P1: y z+ 1−= and P2: 2 x⋅ y− z+ 0= .

If we take the cross product of the normals of both planes, we obtain a

vector: n→

that is parallel to both planes.

n→

n1→

n2→

×= j→

k→+( ) 2 i

→⋅ j→− k

→+( )×= 2 i

→j

→+ k

→−( )⋅=

The vector: v→

3 i→⋅ j→+ 2 k

→⋅+( )= is parallel to the line contained by the

unknown plane.

The vector: v→

n→× is perpendicular to the unknown plane.

v→

n→× 3 i

→⋅ j→+ 2 k

→⋅+( ) i

→j

→+ k

→−( )×= 3− i

→⋅ 5 j

→⋅+ 2 k

→⋅+( )=

Choose a point: 0 1, 0,( ) on the line that is contained by the unknown plane.

Here is the equation of the desired plane.

3− x 5 y 1−( )⋅+ 2 z⋅+ 0=

3 x⋅ 5 y⋅− 2 z⋅− 5−=

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x

z

y

Parallel Plane to Line of Intersection

(i.) Find an equation of the plane whose points are equidistant from the points: 2 1−, 1,( ) and 3 1, 5,( ). (See graph below.)

xy

z

Equidistant Points to Plane

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If both points are equidistant from all points on the plane, then the points must lie on a line that is perpendicular to the plane. Here is a direction

vector, " v→

", for that line.

v→

1( ) i→⋅ 2( ) j

→⋅+ 4( ) k

→⋅+=

Since the line is perpendicular to the plane, the direction vector for the line is also a normal for the plane.

n→

1( ) i→⋅ 2( ) j

→⋅+ 4( ) k

→⋅+=

Because the points are equidistant from the plane, the line connecting them must intersect the plane mid-way between the two points. Let the point: a b, c,( ) be that point.

a3 2+

2= 5

2= b

1 1−( )+2

= 0= c5 1+

2= 3=

a b, c,( )52

0, 3,⎛⎜⎝

⎞⎟⎠

=

This is the equation of a plane that passes thru the point: a b, c,( ) and has

this normal vector: n→

nx( ) i→⋅ ny( ) j

→⋅+ nz( ) k

→⋅+= .

nx x a−( ) ny y b−( )+ nz z c−( )+ 0=

Make the appropriate associations and find this for the equation of the plane.

1 x52

−⎛⎜⎝

⎞⎟⎠

⋅ 2 y 0−( )⋅+ 4 z 3−( )⋅+ 0=

x 2 y⋅+ 4 z⋅+292

= of 33

(j.) Find parametric equations of the line that passes thru the point P 0 0, 1−,( ) and that is parallel to the planes: x y+ z− 0= and x y− z+ 0= .

If the line is parallel to the two planes, then it must be parallel to the line of intersection.

v→

j→

k→+=

Here is that line.

x 0= y t= z 1− t+=

x

z

y

Line Parallel to both Planes

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of 33 4 x⋅ 13 y⋅− 21 z⋅+ 14−=

4− x 1+( ) 13 y 4−( )⋅+ 21 z 2−( )⋅− 0=

n→

4−( ) i→⋅ 13( ) j

→⋅+ 21−( ) k

→⋅+=

n→

AP→⎯

AB→⎯

×= 1−( ) i→⋅ 11( ) j

→⋅+ 7( ) k

→⋅+⎡⎣ ⎤⎦ 1( ) i

→⋅ 10( ) j

→⋅+ 6( ) k

→⋅+⎡⎣ ⎤⎦×=

AB→⎯

1( ) i→⋅ 10( ) j

→⋅+ 6( ) k

→⋅+=AP

→⎯1−( ) i→⋅ 11( ) j

→⋅+ 7( ) k

→⋅+=

P 1− 4, 2,( )B 1 3, 1,( )A 0 7−, 5−,( )

z 1=y 3=x 1=

z 5−=y 7−=x 0=

Two points are all that is needed.

6 x⋅ z 5+=

4 x⋅ z+ 2− 2− x⋅ 2 z⋅+ 3+=

y 2− x⋅ 2 z⋅+ 3+=y 4 x⋅ z+ 2−=

Determine the line of intersection.

(k.) Find the equation of the plane thru the point: 1− 4, 2,( ) that contains the line of intersection of the planes P1: 4 x⋅ y− z+ 2− 0= and P2: 2 x⋅ y+ 2 z⋅− 3− 0= .

xy

z

Plane contains line of intersection & point

(L.) Find the equation of the plane thru the point: 1− 4, 3−,( ) that is perpendicular to the line x t( ) 2 t+= y t( ) 3− 2 t⋅+= z t( ) t−=

v→

1( ) i→⋅ 2( ) j

→⋅+ 1−( ) k

→⋅+=

n→

v→

= 1( ) i→⋅ 2( ) j

→⋅+ 1−( ) k

→⋅+=

1 x 1+( )⋅ 2 y 4−( )⋅+ z 3+( )− 0=

x 2 y⋅+ z− 10= of 33

x

y

z

Plane with point & perpendicular to line

(M.) Find the equation of the plane thru the points: P1 2− 1, 4,( ) and P2 1 0, 3,( ) that is perpendicular to the plane 4 x⋅ y− 3 z⋅+ 2= .

P1P2→⎯⎯

3( ) i→⋅ 1−( ) j

→⋅+ 1−( ) k

→⋅+= ng

→4( ) i→⋅ 1−( ) j

→⋅+ 3( ) k

→⋅+=

n→

P1P2→⎯⎯

ng→

×= 3( ) i→⋅ 1−( ) j

→⋅+ 1−( ) k

→⋅+⎡⎣ ⎤⎦ 4( ) i

→⋅ 1−( ) j

→⋅+ 3( ) k

→⋅+⎡⎣ ⎤⎦×=

n→

4−( ) i→⋅ 13−( ) j

→⋅+ 1( ) k

→⋅+=

4− x 2+( )⋅ 13 y 1−( )⋅− 1 z 4−( )⋅+ 0=

4 x⋅ 13 y⋅+ z− 1=

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Plane with 2 points & perpendicular to plane

(4.) Determine the following about the vector-valued functions.

(a.) Find the domain of r t( )→⎯

i→

3 t⋅ 1+( )⋅ j→

t2( )⋅+= and the value of r 1( )→⎯

3 t⋅ 1+ 0≥

t13

−≥

r 1( )→⎯

i→

3 1( )⋅ 1+⎡⎣ ⎤⎦⋅ j→

1( )2⎡⎣ ⎤⎦⋅+= i→

2( )⋅ j→

1( )⋅+=

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z 3=y4

cos t( )=x2

sin t( )=

z 3=y 4 cos t( )⋅=x 2 sin t( )⋅=

(d.) Describe the graph of r t( )→⎯

i→

2 sin t( )⋅( )⋅ j→

4 cos t( )⋅( )⋅+ k→

3( )⋅+= .

This is a circle in the xy-plane centered at the origin and with radius 2.

x2 y2+ 2( )2=

x2 y2+ 4 sin 3 t⋅( )( )2⋅ 4 cos 3 t⋅( )( )2⋅+= 4=

y 2− cos 3 t⋅( )⋅=x 2 sin 3 t⋅( )⋅=

(c.) Describe the graph of r t( )→⎯

i→

2 sin 3 t⋅( )⋅( )⋅ j→

2− cos 3 t⋅( )⋅( )⋅+= .

This is a straight line in the xy-plane of slope 52

− and

y-intercept 152

.

y x( )152

52

x⋅−=

x 3 2y5⋅−=

y 5 t⋅=x 3 2 t⋅−=

(b.) Describe the graph of r t( )→⎯

i→

3 2 t⋅−( )⋅ j→

5 t⋅( )⋅+= .

x2⎛⎜⎝⎞⎟⎠

2 y4⎛⎜⎝⎞⎟⎠

2+ sin t( )( )2 cos t( )( )2+= 1=

x( )2

4y( )2

16+ 1= z 3=

This is an ellipse in the plane z 3= with major axis of 8 parallel to the y-axis and minor axis of 4 parallel to the x-axis.

xy

z

Elliptical Trace in z=3 Plane

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(e.) Sketch the graph of r t( )→⎯

i→

3 cos t( )⋅( )⋅ j→

4 sin t( )⋅( )⋅+ k→

t( )⋅+= .

x3⎛⎜⎝⎞⎟⎠

2 y4⎛⎜⎝⎞⎟⎠

2+ 1= z t=

x( )2

9y( )2

16+ 1= z t=

This is an elliptical spiral that spirals upward along the positive z-axis with major axis of 8 parallel to the y-axis and minor axis of 6 parallel to the x-axis.

xy

z

Elliptical Spiral advancing along positive z-axis

r

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(f.) Show that the graph of r t( )→⎯

i→

t cos t( )⋅( )⋅ j→

t sin t( )⋅( )⋅+ k→

t( )⋅+= lies on the

cone: z x2 y2+= .

x t cos t( )⋅= y t sin t( )⋅= z t=

x2 y2+ t2= z2=

This is the upper branch of the cone.

z x2 y2+=

xy

z

Curved Line lies on Cone Surface

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(5.) Determine the following about the calculus of vector-valued functions.

(a.) Let r t( )→⎯

i→

2 sin t( )⋅( )⋅ j→

3 cos t( )⋅( )⋅+= in 2-space. Sketch the graph of r t( )→⎯

and draw tr

π6

⎛⎜⎝

⎞⎟⎠

→⎯⎯dd

.

r t( )→⎯

i→

2 sin t( )⋅( )⋅ j→

3 cos t( )⋅( )⋅+= rπ6

⎛⎜⎝

⎞⎟⎠

→⎯⎯

i→

1( )⋅ j→

33

2⋅

⎛⎜⎝

⎞⎟⎠

⋅+=

tr t( )→⎯d

di

→2 cos t( )⋅( )⋅ j

→3− sin t( )⋅( )⋅+=

tr

π6

⎛⎜⎝

⎞⎟⎠

→⎯⎯dd

i→

3( )⋅ j→ 3

2−⎛⎜⎝

⎞⎟⎠

⋅+=

x2

4y2

9+ 1=

4 3 2 1 0 1 2 3 4

4

3

2

1

1

2

3

4Trace and Tangent Vector

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(b.) Find parametric equations of the line tangent to the graph of

r t( )→⎯

cos t( )( ) i→⋅ sin t( )( ) j

→⋅+ t( ) k

→⋅+= at the point where t

π2

= . [The graph of r t( )→⎯

and

its tangent line are provided as a guide.]

xy

z

Curve & Tangent

tr t( )→⎯d

dsin t( )−( ) i

→⋅ cos t( )( ) j

→⋅+ 1( ) k

→⋅+=

rπ2

⎛⎜⎝

⎞⎟⎠

→⎯⎯

cosπ2

⎛⎜⎝

⎞⎟⎠

⎛⎜⎝

⎞⎟⎠

i→⋅ sin

π2

⎛⎜⎝

⎞⎟⎠

⎛⎜⎝

⎞⎟⎠

j→⋅+

π2

⎛⎜⎝

⎞⎟⎠

k→⋅+= 0( ) i

→⋅ 1( ) j

→⋅+

π2

⎛⎜⎝

⎞⎟⎠

k→⋅+=

tr

π2

⎛⎜⎝

⎞⎟⎠

→⎯⎯dd

sinπ2

⎛⎜⎝

⎞⎟⎠

−⎛⎜⎝

⎞⎟⎠

i→⋅ cos

π2

⎛⎜⎝

⎞⎟⎠

⎛⎜⎝

⎞⎟⎠

j→⋅+ 1( ) k

→⋅+= 1−( ) i

→⋅ 0( ) j

→⋅+ 1( ) k

→⋅+=

x t( ) 0 t−= y t( ) 1 0 t⋅+= z t( )π2

t+=

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(c.) Find a vector equation of the line tangent to the graph of

r t( )→⎯

t2( ) i→⋅

1t 1+( )

⎡⎢⎣

⎤⎥⎦

−⎡⎢⎣

⎤⎥⎦

j→⋅+ 4 t2−( ) k

→⋅+= at the point 4 1, 0,( ).

1t0 1+( )

⎡⎢⎣

⎤⎥⎦

− 1=t0( )2 4= 4 t0( )2−⎡⎣ ⎤⎦ 0=

Only t0 2−= satisfies all three constraints.

r 2−( )→⎯⎯

4( ) i→⋅ 1( ) j

→⋅+ 0( ) k

→⋅+=

tr t( )→⎯d

d2 t⋅( ) i

→⋅

1

t 1+( )2⎡⎢⎣

⎤⎥⎦

j→⋅+ 2− t⋅( ) k

→⋅+=

tr 2−( )

→⎯⎯dd

4−( ) i→⋅ 1( ) j

→⋅+ 4( ) k

→⋅+=

r t( )tan→⎯⎯

r 2−( )→⎯⎯

ttr 2−( )

→⎯⎯dd⎛⎜⎝

⎞⎟⎠

⋅+=

r t( )tan→⎯⎯

4( ) i→⋅ 1( ) j

→⋅+ 0( ) k

→⋅+ t 4−( ) i

→⋅ 1( ) j

→⋅+ 4( ) k

→⋅+⎡⎣ ⎤⎦⋅+=

(d.) Evalute tr t( )→⎯⌠

⎮⎮⌡

d where r t( )→⎯

3 t2⋅( ) i→⋅ 2− t⋅( ) j

→⋅+

1t

⎛⎜⎝⎞⎟⎠

k→⋅+=

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tr t( )→⎯⌠

⎮⎮⌡

d t3 t2⋅⌠⎮⎮⌡

d C1+⎛⎜⎜⎝

⎞⎟⎟⎠

i→⋅ t2− t⋅

⌠⎮⎮⌡

d C2+⎛⎜⎜⎝

⎞⎟⎟⎠

j→⋅+ t

1t

⌠⎮⎮⌡

d C3+⎛⎜⎜⎜⎝

⎞⎟⎟⎟⎠

k→⋅+=

tr t( )→⎯⌠

⎮⎮⌡

d t3 C1+( ) i→⋅ t2− C2+( ) j

→⋅+ ln t( ) C3+( ) k

→⋅+=

tr t( )→⎯⌠

⎮⎮⌡

d t3( ) i→⋅ t2−( ) j

→⋅+ ln t( )( ) k

→⋅+ C

→+=

(e.) Solve this initial- value problem: ty t( )→⎯d

dcos t( )( ) i

→⋅ sin t( )( ) j

→⋅+=

where y 0( )→⎯

i→

j→−= .

ty t( )→⎯d

dcos t( )( ) i

→⋅ sin t( )( ) j

→⋅+=

y t( )→⎯

sin t( )( ) i→⋅ cos t( )−( ) j

→⋅+ C

→+=

y 0( )→⎯

i→

j→−= 0( ) i

→⋅ 1−( ) j

→⋅+ C

→+=

C→

i→=

y t( )→⎯

sin t( ) 1+( ) i→⋅ cos t( )−( ) j

→⋅+=

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of 33θ2 71degrees=θ1 76degrees=

θ2 π 1.904−=θ1 1.328radians=

cos θ2( ) 3−

84=cos θ1( ) 3

156=

cos θ2( )n→

tr 1( )→⎯d

d⎛⎜⎝

⎞⎟⎠

⋅

n→

tr 1( )→⎯d

d⋅

=cos θ1( )n→

tr 2−( )

→⎯⎯dd⎛⎜⎝

⎞⎟⎠

⋅

n→

tr 2−( )

→⎯⎯dd

⋅

=

n→

2( ) i→⋅ 1−( ) j

→⋅+ 1( ) k

→⋅+=

tr 1( )→⎯d

d1( ) i→⋅ 2( ) j

→⋅+ 3−( ) k

→⋅+=

tr 2−( )

→⎯⎯dd

1( ) i→⋅ 4−( ) j

→⋅+ 3−( ) k

→⋅+=

tr t( )→⎯d

d1( ) i→⋅ 2 t⋅( ) j

→⋅+ 3−( ) k

→⋅+=

The corresponding points of intersection are 2− 4, 6,( ) and 1 1, 3−,( ).

t 2−= 1,

t2 t+ 2− 0=

2 t( )⋅ t2( )− 3− t⋅( )+ 2−=

(f.) Find the points where the curve r t( )→⎯

t( ) i→⋅ t2( ) j

→⋅+ 3− t⋅( ) k

→⋅+= intersects

the plane 2 x⋅ y− z+ 2−= . Then find the acute angle that the tangent line to the curve makes with a line normal to the plane at each point of intersection.

xy

z

Curve Intersects Plane @ 2 Points

(6.) Determine the following about functions of multi-variables.

(a.) Find g u x y,( ) v x y,( ),( ) if g x y,( ) y sin x2 y⋅( )⋅= , u x y,( ) x2 y3⋅= , and v x y,( ) π x⋅ y⋅= .

g x y,( ) y sin x2 y⋅( )⋅=

g u v,( ) v sin u2 v⋅( )⋅=

g u x y,( ) v x y,( ),( ) π x⋅ y⋅ sin x2 y3⋅( )2π x⋅ y⋅( )⋅

⎡⎣

⎤⎦⋅=

g u x y,( ) v x y,( ),( ) π x⋅ y⋅ sin π x5⋅ y7⋅( )⋅= of 33

(b.) Let f x y, z,( ) z x⋅ y⋅ x+= . Find f x y+( ) x y−( ), x2( ),⎡⎣ ⎤⎦.

f x y+( ) x y−( ), x2( ),⎡⎣ ⎤⎦ x2 x y+( )⋅ x y−( )⋅ x y+( )+=

f x y+( ) x y−( ), x2( ),⎡⎣ ⎤⎦ x4 x2 y2⋅− x+ y+=

(c.) Sketch the domain of f x y,( ) x2 y2+ 4−= .

x2 y2+ 4− 0≥

x2 y2+ 2( )2≥

4 3 2 1 0 1 2 3 4

4

3

2

1

1

2

3

4Domain is on and outside circle boundary

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(d.) Sketch the graph of f x y,( ) y 1+= .

z y 1+=

x 0, 1,( ) x 1−, 0,( )

This is a plane whose projection onto the yz-plane is a line of slope 1 and intercept 1.

x

y

z

f(x,y)=y+1

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CALCULUS 4 QUIZ #1 REVIEW / SPRING 09 - Drexel Universityjwd25/Calc4_Winter_09/tests&quizzes... ·...

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