Calculus/Multivariable and differential calculus:Exercises

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Multivariable and differential calculus:Exercises

Parametric EquationsEdit

1. Find parametric equations describing the line segment from P(0,0) to Q(7,17).
x=7t and y=17t, where 0 ≤ t ≤ 1
x=7t and y=17t, where 0 ≤ t ≤ 1
2. Find parametric equations describing the line segment from to .
3. Find parametric equations describing the ellipse centered at the origin with major axis of length 6 along the x-axis and the minor axis of length 3 along the y-axis, generated clockwise.

Polar CoordinatesEdit

20. Convert the equation into Cartesian coordinates:
Making the substitutions and gives
Making the substitutions and gives
21. Find an equation of the line y=mx+b in polar coordinates.
Making the substitutions and gives
Making the substitutions and gives

Sketch the following polar curves without using a computer.

22.
2-2sin-t.svg
2-2sin-t.svg
23.
R-square-eq-4cos-t.svg
R-square-eq-4cos-t.svg
24.
2sin-5t.svg
2sin-5t.svg

Sketch the following sets of points.

25.
Polar-set-answer-1.svg
Polar-set-answer-1.svg
26.
Polar-set-answer-2.svg
Polar-set-answer-2.svg

Calculus in Polar CoordinatesEdit

Find points where the following curves have vertical or horizontal tangents.

40.

Horizontal tangents occur at points where . This condition is equivalent to .

Vertical tangents occur at points where . This condition is equivalent to .

The condition for a horizontal tangent gives:

Horizontal tangents occur at which correspond to the Cartesian points and .

The condition for a vertical tangent gives:

Vertical tangents occur at which correspond to the Cartesian points and .

Horizontal tangents at (2,2) and (2,-2); vertical tangents at (0,0) and (4,0)

Horizontal tangents occur at points where . This condition is equivalent to .

Vertical tangents occur at points where . This condition is equivalent to .

The condition for a horizontal tangent gives:

Horizontal tangents occur at which correspond to the Cartesian points and .

The condition for a vertical tangent gives:

Vertical tangents occur at which correspond to the Cartesian points and .

Horizontal tangents at (2,2) and (2,-2); vertical tangents at (0,0) and (4,0)
41.
Horizontal tangents occur at points where . This condition is equivalent to .

Vertical tangents occur at points where . This condition is equivalent to .

The condition for a horizontal tangent gives:

Horizontal tangents occur at which correspond to the Cartesian points , , , and . Point corresponds to a vertical cusp however and should be excluded leaving , , and .

The condition for a vertical tangent gives:

Vertical tangents occur at which correspond to the Cartesian points , , and .

Horizontal tangents at (r,θ) = (4,π/2), (1,7π/6) and (1,-π/6); vertical tangents at (r,θ) = (3,π/6), (3,5π/6), and (0,3π/2)
Horizontal tangents occur at points where . This condition is equivalent to .

Vertical tangents occur at points where . This condition is equivalent to .

The condition for a horizontal tangent gives:

Horizontal tangents occur at which correspond to the Cartesian points , , , and . Point corresponds to a vertical cusp however and should be excluded leaving , , and .

The condition for a vertical tangent gives:

Vertical tangents occur at which correspond to the Cartesian points , , and .

Horizontal tangents at (r,θ) = (4,π/2), (1,7π/6) and (1,-π/6); vertical tangents at (r,θ) = (3,π/6), (3,5π/6), and (0,3π/2)

Sketch the region and find its area.

42. The region inside the limaçon
Given an infinitesimal wedge with angle and radius , the area is . The total area is therefore . 9π/2
2-plus-cos-t.svg
Given an infinitesimal wedge with angle and radius , the area is . The total area is therefore . 9π/2
2-plus-cos-t.svg
43. The region inside the petals of the rose and outside the circle
There are 4 petals, as seen in the image below. The area of just one of the petals needs to be computed and the multiplied by .

It is first necessary to compute the angular limits of one of the petals. The petals start and end at points where . The bounds on one of the petals are .

Given an annular wedge with angle , inner radius , and an outer radius of , the area is . The total area of all 4 petals is therefore .

4cos-2t-and-2.svg
There are 4 petals, as seen in the image below. The area of just one of the petals needs to be computed and the multiplied by .

It is first necessary to compute the angular limits of one of the petals. The petals start and end at points where . The bounds on one of the petals are .

Given an annular wedge with angle , inner radius , and an outer radius of , the area is . The total area of all 4 petals is therefore .

4cos-2t-and-2.svg

Vectors and Dot ProductEdit

60. Find an equation of the sphere with center (1,2,0) passing through the point (3,4,5)
The general equation for a sphere is where is the location of the sphere's center and is the sphere's radius.

It is already known that the sphere's center is . The sphere's radius is the distance between (1,2,0) and (3,4,5) which is .

Therefore the sphere's equation is: .
The general equation for a sphere is where is the location of the sphere's center and is the sphere's radius.

It is already known that the sphere's center is . The sphere's radius is the distance between (1,2,0) and (3,4,5) which is .

Therefore the sphere's equation is: .
61. Sketch the plane passing through the points (2,0,0), (0,3,0), and (0,0,4)
Plane-intercepts-2-3-4.svg
Plane-intercepts-2-3-4.svg
62. Find the value of if and

.

Therefore: .

.

Therefore: .

63. Find all unit vectors parallel to
The length of is . Therefore is a unit vector that points in the same direction as , and is a unit vector that points in the opposite direction as . are the unit vectors that are parallel to .
The length of is . Therefore is a unit vector that points in the same direction as , and is a unit vector that points in the opposite direction as . are the unit vectors that are parallel to .
64. Prove one of the distributive properties for vectors in :

.

.
65. Find all unit vectors orthogonal to in

Rotating counterclockwise gives . is orthogonal to , and the normalization of and its negative are the only unit vectors that are orthogonal to .

The magnitude of is so the only unit vectors that are orthogonal to are .

Rotating counterclockwise gives . is orthogonal to , and the normalization of and its negative are the only unit vectors that are orthogonal to .

The magnitude of is so the only unit vectors that are orthogonal to are .
66. Find all unit vectors orthogonal to in

All vectors that are orthogonal to must satisfy .

The set of possible values of is . The restriction that becomes .

The set of possible and is an ellipse with radii and . One possible parameterization of and is and where . This parameterization yields where as the complete set of unit vectors that are orthogonal to .

Re-parameterizing by letting gives the set

All vectors that are orthogonal to must satisfy .

The set of possible values of is . The restriction that becomes .

The set of possible and is an ellipse with radii and . One possible parameterization of and is and where . This parameterization yields where as the complete set of unit vectors that are orthogonal to .

Re-parameterizing by letting gives the set
67. Find all unit vectors that make an angle of with the vector

The angle that makes with the x-axis is counterclockwise.

Making a both a clockwise and a counterclockwise rotation of gives

The angle that makes with the x-axis is counterclockwise.

Making a both a clockwise and a counterclockwise rotation of gives

Cross ProductEdit

Find and

80. and
81. and

Find the area of the parallelogram with sides and .

82. and

The cross product of vectors and is a vector with length where is the angle between and . is the area of the parallelogram with sides and , so this area is found by computing .

The cross product of vectors and is a vector with length where is the angle between and . is the area of the parallelogram with sides and , so this area is found by computing .

83. and

The cross product of vectors and is a vector with length where is the angle between and . is the area of the parallelogram with sides and , so this area is found by computing .

The cross product of vectors and is a vector with length where is the angle between and . is the area of the parallelogram with sides and , so this area is found by computing .


84. Find all vectors that satisfy the equation
The cross product is orthogonal to both multiplicand vectors. should be orthogonal to both and . However, so and are not orthogonal. The equation is never true, and therefore the set of vectors that satisfy the equation is "None".
The cross product is orthogonal to both multiplicand vectors. should be orthogonal to both and . However, so and are not orthogonal. The equation is never true, and therefore the set of vectors that satisfy the equation is "None".
85. Find the volume of the parallelepiped with edges given by position vectors , , and

The volume of a parallelepiped with edges defined by the vectors , , and is the absolute value of the scalar triple product: .

The volume of a parallelepiped with edges defined by the vectors , , and is the absolute value of the scalar triple product: .

86. A wrench has a pivot at the origin and extends along the x-axis. Find the magnitude and the direction of the torque at the pivot when the force is applied to the wrench n units away from the origin.
The moment arm is , so the torque applied is The magnitude of the torque is . The torque's direction is .
The moment arm is , so the torque applied is The magnitude of the torque is . The torque's direction is .

Prove the following identities or show them false by giving a counterexample.

87.
False:
False:
88.
True: Once expressed in component form, both sides evaluate to
True: Once expressed in component form, both sides evaluate to
89.
True:
True:

Calculus of Vector-Valued FunctionsEdit

100. Differentiate .
101. Find a tangent vector for the curve at the point .
so a possible a tangent vector at is
so a possible a tangent vector at is
102. Find the unit tangent vector for the curve .
so the unit tangent vector is
so the unit tangent vector is
103. Find the unit tangent vector for the curve at the point .

so the unit tangent vector is

At :

so the unit tangent vector is

At :
104. Find if and .
For an arbitrary the position can be computed by the integral .

For an arbitrary the position can be computed by the integral .

105. Evaluate

Motion in SpaceEdit

120. Find velocity, speed, and acceleration of an object if the position is given by .
, ,
, ,
121. Find the velocity and the position vectors for if the acceleration is given by .

Length of CurvesEdit

Find the length of the following curves.

140.
For an infinitesimal step , the length traversed is approximately

.

The total length is therefore:

For an infinitesimal step , the length traversed is approximately

.

The total length is therefore:

141.
For an infinitesimal step , the length traversed is approximately

.

The total length is therefore:

For an infinitesimal step , the length traversed is approximately

.

The total length is therefore:

Parametrization and Normal VectorsEdit

142. Find a description of the curve that uses arc length as a parameter:
For an infinitesimal step , the length traversed is approximately

Given an upper bound of , the arc length swept out from to is:

The arc length spans a range from to . For an arc length of , the upper bound on that generates an arc length of is , and the point at which this upper bound occurs is:
For an infinitesimal step , the length traversed is approximately

Given an upper bound of , the arc length swept out from to is:

The arc length spans a range from to . For an arc length of , the upper bound on that generates an arc length of is , and the point at which this upper bound occurs is:
143. Find the unit tangent vector T and the principal unit normal vector N for the curve Check that TN=0.
A tangent vector is . Normalizing this vector to get the unit tangent vector gives:

A vector that has the direction of the principal unit normal vector is

Normalizing gives the principal unit normal vector:

A tangent vector is . Normalizing this vector to get the unit tangent vector gives:

A vector that has the direction of the principal unit normal vector is

Normalizing gives the principal unit normal vector:

Equations of Lines And PlanesEdit

160. Find an equation of a plane passing through points
Let denote a plane that contains points , , and . Let denote an arbitrary vector that is orthogonal to , and denote the position vector of an arbitrary point contained by . A point at position vector is contained by if and only if the displacement from is orthogonal to . This yields the equation .

The displacement from to , which is , and the displacement from to , which is