MATH 125 - FINITE MATHEMATICS Plot and analyze geometric shapes Set 2

MATH 125 - FINITE MATHEMATICS

- Plot and analyze geometric shapes

Question Bank - Set 2

Liberty University

Question 1

Question

Consider the following geometric shape: a circle centered at (3,−2) with a radius

of 5.

(a) Find the equation of the circle in standard form.

(b) Determine the coordinates of the vertices of a square inscribed in this

circle.

Solution

(a) To find the equation of the circle in standard form, we use the general

equation of a circle:

(x−h)2+ (y−k)2=r2,

where (h, k) is the center of the circle and ris the radius. Given that the center

is (3,−2) and the radius is 5, we have:

(x−3)2+ (y+ 2)2= 25.

(b) To determine the coordinates of the vertices of a square inscribed in

this circle, we note that the diagonal of the square will be the diameter of the

circle (which is equal to 10 units). The diagonal of a square divides it into two

congruent right triangles, so the sides of the square would be 10

√2= 5√2 units.

The vertices of the square would lie on the circle, so by considering the distances

from the center of the circle, we find the four vertices: (3 ±5√2,−2±5√2).

In this case, s= 5√2, so the area of the square is:

A= (5√2)2= 50.

Thus, the area of the square inscribed in the given circle is 50 square units.

Question 2

Question

Consider the following parametric equations for a curve in the xy-plane:

x= 2 sin(t), y = 3 cos(t),0≤t≤2π.

(a) Plot the curve generated by the parametric equations. (b) Determine the

equation of any tangent lines to the curve at the points where t=π

2. (c) Find

the area enclosed by the curve.

Solution

(a) To plot the curve, we can eliminate the parameter tand express yin terms

of x:

y= 3 cos arcsin x

2= 3r1−x

22.

This equation represents the top half of an ellipse centered at the origin with

semi-major axis of 3 units along the y-axis and semi-minor axis of 2 units along

the x-axis. The plot of the curve is the top half of this ellipse.

(b) To find the equation of the tangent lines at t=π

2, we need to find the

slope of the curve at this point. The slope of the curve at any point is given by

dx =dy/dt

dx/dt .

dt =−3 sin(t),dx

dt = 2 cos(t).

At t=π

2,dy

dt =−3, dx

dt = 0, so the slope is undefined. This means the tangent

line at this point is vertical and its equation is x= 2.

2y dx over one period

of the curve:

Area = Z2π

2·3 cos(t)·2 cos(t)dt =Z2π

3 cos2(t)dt.

Using the double angle identity cos(2t) = 2 cos2(t)−1, we can rewrite the

integral as:

Area = 3

2Z2π

(1 + cos(2t)) dt =3

2t+1

2sin(2t)

2π

=9π

Therefore, the area enclosed by the curve is 9π

Question 3

Question

Consider the two-dimensional region in the xy-plane bounded by the curves

y=x2and y= 2x−x2. Calculate the area of the region enclosed by these

curves.

Solution

To find the area of the region bounded by the curves y=x2and y= 2x−x2,

we first need to determine the points of intersection of these two curves.

Step 1: Find the points of intersection Setting x2= 2x−x2gives us

2x= 2x2, which simplifies to x(2 −2x) = 0. So, x= 0 or x= 1.

Step 2: Determine the bounds for integration The region is bounded

by the curves y=x2and y= 2x−x2between x= 0 and x= 1.

Step 3: Compute the area The area enclosed by the curves is given by

the definite integral of the difference of the two curves:

A=Z1

((2x−x2)−x2)dx

Step 4: Evaluate the integral

A=Z1

(2x−2x2)dx =x2−2

3x31

= 1 −2

3=1

So, the area of the region enclosed by the curves y=x2and y= 2x−x2is

Question 4

Question

Consider the following geometric shapes in the coordinate plane: a circle cen-

tered at (2,−1) with radius 3, and a line passing through the points (1,5) and

(4,−2). Determine the points of intersection, if any, between these two shapes.

Solution

Step 1: The equation of the circle can be written in the form (x−h)2+(y−k)2=

r2, where (h, k) are the coordinates of the center and ris the radius. For this

circle, we have (x−2)2+ (y+ 1)2= 9.

Step 2: The equation of the line passing through two points (x1, y1) and

(x2, y2) can be found using the point-slope formula: y−y1=y2−y1

x2−x1(x−x1).

Substituting (1,5) and (4,−2) into this formula, we obtain the equation of the

line as y=−7

3x+26

Step 3: To find the points of intersection between the circle and the line, we

substitute the expression for yfrom the equation of the line into the equation

of the circle. This gives us a quadratic equation in terms of x.

Step 4: Solve the quadratic equation to find the possible x-coordinates of

the points of intersection. These x-coordinates can then be substituted back

into the equation of the line to find the corresponding y-coordinates.

Step 5: Finally, check if the points of intersection lie on both the circle and

the line. If they do, then those points are the actual points of intersection.

Otherwise, there are no points of intersection between the circle and the line.

Question 5

Question

Consider the geometric shape in the xy-plane defined by the following inequal-

ities:

x2−4x+y2+ 2y≤0

x+y≤4

x≥0

Sketch and analyze the geometric shape formed by the above inequalities.

Solution

To analyze the given geometric shape defined by the system of inequalities, let’s

break it down step by step:

Step 1: First, let’s examine the boundary of the region. To find the boundary

where the inequalities become equalities, we set each inequality to equality and

solve for xand y. For x2−4x+y2+ 2y= 0, we complete the square to rewrite

it as (x−2)2+ (y+ 1)2= 5. This represents a circle centered at (2,−1) with

radius √5. For x+y= 4, the boundary line passes through (4,0) and (0,4).

Step 2: Now, we will sketch the boundary of the region. The circle with

center (2,−1) and radius √5 is sketched, along with the line x+y= 4. This

boundary divides the plane into different regions.

Step 3: Next, we will check the regions defined by the given inequalities to

determine the solution set. Evaluating the inequalities at a test point in each

region allows us to determine the valid solution set. For example, we can test

the region below the circle but above the line by plugging in (0,0) into the

inequalities. We find that all inequalities are satisfied in this region.

Step 4: Finally, we summarize that the geometric shape defined by the given

inequalities is the closed region bounded by the circle (x−2)2+ (y+ 1)2= 5,

the line x+y= 4, the x-axis, and the y-axis. This region includes the center of

the circle and is the solution set.

Therefore, the solution is the region enclosed by the circle (x−2)2+(y+1)2=

5, the line x+y= 4, the x-axis, and the y-axis.

Question 6

Question

Consider the region Din the xy-plane bounded by the curve y=x2and the

line y=x+ 2. Sketch the region Dand calculate the area enclosed by the curve

and the line.

Solution

To find the area enclosed by the curve y=x2and the line y=x+ 2, we first

need to determine the points of intersection.

Step 1: Find the points of intersection.

Setting x2=x+ 2, we have:

x2−x−2=0

Factoring the quadratic equation, we get:

(x−2)(x+ 1) = 0

So, x= 2 or x=−1.

Therefore, the points of intersection are at (−1,1) and (2,4).

Step 2: Sketch the region D

The region Dis enclosed by the curve y=x2and the line y=x+ 2. It lies

between the x-values −1 and 2.

xy2;2y=x+ 2

Step 3: Calculate the area enclosed by the curve and the line.

The area enclosed by the two curves can be found by integrating the differ-

ence of the two functions over the interval [−1,2]:

Area = Z2

−1

(x+ 2 −x2)dx

Now, we compute the integral:

Area = x2

2+ 2x−x3

32

−1

=22

2+ 2(2) −23

3−(−1)2

2+ 2(−1) −(−1)3

3

= (2 + 4 −8

3)−(1

2−2 + 1

=17

Therefore, the area enclosed by the curve y=x2and the line y=x+ 2 is

6square units.

Question 7

Question

Consider the geometric shape defined by the equation x2+y2+z2−x−y−z= 0.

Plot and analyze this geometric shape.

Solution

To begin, we rewrite the given equation in standard form by completing the

square for each variable:

x2−x+1

22

+y2−y+1

22

+z2−z+1

22

=1

22

+1

22

+1

22

Simplifying, we have:

(x−1

2)2+ (y−1

2)2+ (z−1

2)2=3

Step 1: The equation defines a sphere with center (1

2,1

2) and radius

4=√3

Step 2: We now analyze the properties of the sphere. - Center: The

center of the sphere is at (1

2,1

2). - Radius: The radius of the sphere is √3

-Surface Area: The surface area of a sphere is given by 4πr2, where ris the

radius. Thus, the surface area of this sphere is 4π√3

22= 3π. - Volume: The

volume of a sphere is given by 4

3πr3. Therefore, the volume of this sphere is

3π√3

23=π√3

Hence, the given equation represents a sphere with center (1

2,1

2) and radius

√3

2. This sphere has a surface area of 3πsquare units and a volume of π√3

2cubic

units.

Question 8

Question

Consider the following parametric equations:

x= 2 sin(t), y = 3 cos(t)

(a) Sketch the curve formed by these parametric equations in the xy-plane.

(b) Determine the equation of the tangent line to the curve at the point

where t=π

Solution

(a) To sketch the curve, we can eliminate the parameter tby eliminating sine

and cosine. Since sin2(t) + cos2(t) = 1, we have:

x2= 4 sin2(t)

y2= 9 cos2(t)

Now, substituting x2and y2into the equation sin2(t) + cos2(t) = 1 gives us:

4+y2

9= 1

This equation represents an ellipse with major axis along the y-axis and minor

axis along the x-axis.

(b) To find the equation of the tangent line at t=π

4, we first need to find

the coordinates of the point on the curve at this tvalue. Substituting t=π

into the parametric equations gives:

x= 2 sin π

4= 2 1

√2=√2

y= 3 cos π

4= 3 1

√2=3

√2

The point on the curve at t=π

4is (√2,3

√2).

To find the slope of the tangent line, we differentiate xand ywith respect

to t:dx

dt = 2 cos(t)

dt =−3 sin(t)

At t=π

dt t=π

= 2 cos π

4=√2

dt t=π

=−3 sin π

4=−3

√2

Therefore, the slope of the tangent line at t=π

4is dy

dx =−3/√2

√2=−3

Using the point-slope form of the equation of a line, the equation of the

tangent line is:

y−3

√2=−3

2x−√2

Question 9

Question

Consider the following parametric equations for a curve in the xy-plane:

x(t) = 2 cos(t), y(t) = sin(t), t ∈[0,2π].

Plot the curve represented by these parametric equations and determine the

points of intersection with the x-axis.

Solution

To plot the curve represented by the parametric equations x(t) = 2 cos(t) and

y(t) = sin(t), we will vary the value of tfrom 0 to 2π.

Step 1: Plotting the curve Let’s compute the values of xand yfor various

values of t.

t0π

3π

4π5π

3π

7π

x2√2 0 −√2 -2 −√20√2

y0 1 1 1 0 -1 -1 -1

Plotting these points will give us the curve.

Step 2: Plotting the curve

Now, let’s plot the curve represented by the parametric equations x(t) =

2 cos(t) and y(t) = sin(t).

parametric_curve.png

Step 3: Determining points of intersection with the x-axis

The curve intersects the x-axis at the points where y(t) = 0. Thus, we need

to solve sin(t) = 0 for tin the interval [0,2π].

The solutions to sin(t) = 0 are t= 0, π. Therefore, the curve intersects the

x-axis at the points (2,0) and (−2,0).

In conclusion, the curve intersects the x-axis at the points (2,0) and (−2,0).

Question 10

Question

Consider the following functions: 1. f(x) = x3−3x2−4x+ 12 2. g(x) =

x2−4x−5

(a) Plot f(x) and g(x) on the same set of axes.

(b) Determine the points of intersection of f(x) and g(x).

and x= 5.

Solution

(a) To plot f(x) and g(x) on the same set of axes, we will first graph each

function separately and then overlay the plots.

Step 1: Graph f(x): Since f(x) = x3−3x2−4x+ 12, we first find the

critical points by setting f′(x) = 0:

f′(x)=3x2−6x−4

Setting f′(x) = 0 gives us:

3x2−6x−4=0

Solving for x, we find x=−1 and x=4

3as critical points.

Step 2: Graph g(x): Since g(x) = x2−4x−5, we can easily graph this

quadratic function.

Step 3: Overlay the plots of f(x) and g(x):

Now, we overlay the plots of f(x) and g(x) on the same set of axes to visualize

their intersection points.

(b) To determine the points of intersection, we solve f(x) = g(x):

x3−3x2−4x+ 12 = x2−4x−5

Rearranging the terms gives us:

x3−4x2+x+ 7 = 0

One of the roots of this equation is x= 1.

To find the other roots, we divide the cubic polynomial by (x−1) using

polynomial long division or synthetic division. This will give us a quadratic

equation that can be solved to find the other roots.

x= 1 and x= 5, we calculate the definite integral of |f(x)|between x= 1 and

x= 5. This will give us the total area enclosed by the curve.

Question 11

Question

Consider the following geometric figure: a square with side length sand an

inscribed circle. Let Asand Acdenote the areas of the square and circle,

respectively. Express the ratio Ac

Asin terms of s.

Solution

Step 1: Find the area of the square.

The area of a square with side length sis given by As=s2.

Step 2: Find the area of the circle inscribed in the square.

The circle is inscribed in the square, so its diameter is equal to the side length

of the square, which is s. The radius of the circle is thus r=s

2. The area of the

circle is Ac=πr2=πs

22=πs2

Step 3: Determine the ratio of the areas.

Now we can express the ratio of the areas as:

πs2

s2=π

Therefore, the ratio Ac

Asis π

4in terms of s.

Question 12

Question

Consider the geometric shape defined by the equation x2+ 4y2+z2= 16.

Determine the type of shape represented by this equation and sketch its graph.

Identify any important features of the shape, such as its center, axes, and asymp-

totes.

Solution

Step 1: To determine the type of shape represented by the equation x2+ 4y2+

z2= 16, we can rewrite it in standard form. Since the coefficients of x2,y2, and

z2are positive, this suggests that the shape is an ellipsoid.

Step 2: We rewrite the equation by dividing each term by 16:

16 +y2

4+z2

16 = 1

Step 3: Next, we identify the center of the ellipsoid. The center is at (h, k, l)

where h= 0, k= 0, and l= 0. Thus, the center of the ellipsoid is at the origin

(0,0,0).

Step 4: We find the semi-axes lengths of the ellipsoid. The semi-axes lengths

along the x-axis, a, along the y-axis, b, and along the z-axis, c, are given by

√16 = 4, √4 = 2, and √16 = 4 respectively.

Step 5: Now we sketch the graph of the ellipsoid. The ellipsoid is centered

at the origin with semi-axes lengths 4, 2, and 4 along the x-, y-, and z-axes,

respectively. The ellipsoid has no asymptotes and is symmetric about each axis.

Question 13

Question

Consider the following set of points in the coordinate plane:

S={(x, y)|x2+y2= 9, x ≥0, y ≥0}

Plot the set Sin the coordinate plane and analyze the geometric shape

formed by these points.

Solution

1. To plot the set S, we first note that the equation x2+y2= 9 represents a circle

with radius 3 centered at the origin. However, since we are only considering

points where both xand yare non-negative, we are only interested in the top-

right quarter of the circle.

2. We can now proceed to plot the set Sin the coordinate plane.

xy33

3. The geometric shape formed by the set Sis a quarter circle in the first

quadrant of the coordinate plane, with radius 3 and the x-axis and y-axis as its

boundaries.

4. This shape is often referred to as a quarter circle.

Question 14

Question

Consider the following piecewise-defined function:

f(x) = (x2if x < 0,

2x+ 1 if x≥0.

Plot the graph of the function f(x) and analyze the behavior of the func-

tion for both x < 0 and x≥0. Identify any critical points, intervals of in-

crease/decrease, local extrema, and concavity.

Solution

To plot the graph of the function f(x), we need to consider two cases: x < 0

and x≥0.

Case 1: x < 0 For x < 0, the function is given by f(x) = x2. This is a

simple parabolic graph which opens upwards. The critical point for this part of

the graph occurs at x= 0 where the minimum of the parabola is located.

Case 2: x≥0 For x≥0, the function is given by f(x)=2x+ 1. This is

a linear function with a slope of 2 and a y-intercept of 1. There are no critical

points for this linear function as it is a straight line.

Now, let’s plot the graphs for both pieces of the function on the same coor-

dinate system to get a complete picture of f(x).

xf(x)2; [domain = 0 : 4, samples = 100, color =red, ]2 ∗x+ 1;

From the graph, we can see that the function f(x) behaves differently for

x < 0 and x≥0: - For x < 0, f(x) is a parabola opening upwards with a

minimum at x= 0. - For x≥0, f(x) is a straight line with a positive slope.

Therefore, the function f(x) has a local minimum at x= 0 and is increasing

for x < 0 and x≥0. It is concave up for x < 0 and has no concavity for x≥0.

Question 15

Question

Consider the following system of equations:

(x2+y2= 4

y=x2

Plot and analyze the geometric shapes represented by these equations.

Solution

Step 1: To analyze the shapes represented by the equations, let’s first plot them

on the coordinate plane.

Step 2: The first equation x2+y2= 4 represents a circle with center at the

origin and radius 2.

Step 3: The second equation y=x2represents a parabola opening upwards

with vertex at the origin.

Step 4: Now, let’s plot both the circle and the parabola on the same coor-

dinate plane to analyze their intersection points.

Step 5: From the plot, we can see that the circle intersects the parabola at

two points: (−1,1) and (1,1).

Step 6: Therefore, the geometric shapes represented by the given system of

equations are a circle with center at the origin and radius 2, and a parabola

opening upwards with vertex at the origin. They intersect at the points (−1,1)

and (1,1).

Question 16

Question

Consider the region bounded by the curves y=x3and y=x2in the first

quadrant. Plot the region and find the area enclosed by these curves.

Solution

Step 1: To find the points of intersection between the curves y=x3and y=x2,

we set x3=x2and solve for x.

x3=x2

x3−x2= 0

x2(x−1) = 0

x= 0 or x= 1

Step 2: Next, we find the area enclosed by the curves by integrating the

difference x3−x2from 0 to 1.

Area = Z1

(x3−x2)dx

=x4

4−x3

31

=1

4−1

3−(0 −0)

Therefore, the area enclosed by the curves y=x3and y=x2in the first

quadrant is 1

12 .

Question 17

Question

Plot and analyze the geometric shape defined by the equation x2+y2−6x+

4y−12 = 0.

Solution

Step 1: Complete the square for xand y.

We rearrange the terms in the equation to group xand yterms together:

(x2−6x)+(y2+ 4y) = 12

To complete the square for x, we add (6/2)2= 9 inside the parentheses:

(x2−6x+ 9) + (y2+ 4y) = 12 + 9

To complete the square for y, we add (4/2)2= 4 inside the parentheses:

(x2−6x+ 9) + (y2+ 4y+ 4) = 12 + 9 + 4

This simplifies to:

(x−3)2+ (y+ 2)2= 25

Step 2: Identify the center and radius.

Comparing the equation to the standard form of a circle (x−h)2+(y−k)2=

r2, we see that the center is at (3,−2) and the radius is 5.

Step 3: Plot the circle.

The circle has center (3,−2) and radius 5.

Step 4: Analyze the circle.

The circle has center (3,−2) and radius 5. It represents all points in the

plane that are equidistant from the center (3,−2).

Question 18

Question

Consider the following geometric shapes:

Shape A: A circle with radius 3 centered at the origin.

Shape B: A square with side length 6 centered at the point (2,2).

Shape C: An ellipse centered at the origin with semi-major axis length 4

along the x-axis and semi-minor axis length 2 along the y-axis.

For each shape, sketch the shape on the coordinate plane and analyze its

key properties (e.g., symmetry, intercepts, asymptotes).

Solution

To analyze each shape, we will consider the characteristics of circles, squares,

and ellipses.

Shape A: Circle with radius 3 centered at the origin



Sketch: This circle is centered at the origin and has a radius of 3.



Key Properties:

– Symmetry: The circle is symmetric with respect to both the x-axis

and the y-axis.

– Intercepts: The circle intersects the x-axis at (-3, 0) and (3, 0), and

the y-axis at (0, -3) and (0, 3).

Shape B: Square with side length 6 centered at the point (2, 2)



Sketch: This square is centered at the point (2, 2) and has side length 6.



Key Properties:

– Symmetry: The square is symmetric with respect to both the x-axis

and the y-axis.

– Intercepts: The square intersects the x-axis at (-1, 2) and (5, 2),

and the y-axis at (2, -1) and (2, 5).

Shape C: Ellipse centered at the origin with semi-major axis of

length 4 along the x-axis and semi-minor axis of length 2 along the

y-axis



Sketch: This ellipse is centered at the origin with semi-major axis length

4 along the x-axis and semi-minor axis length 2 along the y-axis.



Key Properties:

– Symmetry: The ellipse is symmetric with respect to both the x-axis

and the y-axis.

– Intercepts: The ellipse intersects the x-axis at (-4, 0) and (4, 0),

and the y-axis at (0, -2) and (0, 2).

– Asymptotes: There are no asymptotes for an ellipse.

Question 19

Question

Consider the following parametric equations:

x(t) = sin(t) + 2 cos(t), y(t) = sin(t)−2 cos(t)

Plot and analyze the geometric shape represented by these parametric equa-

tions.

Solution

To plot and analyze the geometric shape represented by the parametric equa-

tions, we need to eliminate the parameter and find the Cartesian equation of

the curve.

Step 1: Eliminate the parameter tTo eliminate the parameter t, we can

square both equations and add them together to get rid of the trigonometric

functions:

(x(t))2+ (y(t))2= (sin(t) + 2 cos(t))2+ (sin(t)−2 cos(t))2

x2+y2= sin2(t) + 4 sin(t) cos(t) + 4 cos2(t) + sin2(t)−4 sin(t) cos(t) + 4 cos2(t)

x2+y2= 2(sin2(t) + cos2(t)) + 8 cos2(t)

x2+y2= 2 + 8 cos2(t)

Step 2: Analyze the Cartesian equation The Cartesian equation we

found is x2+y2= 2 + 8 cos2(t). This is the equation of a circle centered at the

origin with radius p2 + 8(1) = √10.

Step 3: Plot the geometric shape The geometric shape represented by

the parametric equations is a circle centered at the origin with radius √10.

Therefore, the plot of the geometric shape is a circle centered at the origin

with radius √10.

Question 20

Question

Consider the geometric shape defined by the following equation:

4x2+ 9y2−16x+ 36y−36 = 0

Determine the type of shape the equation represents and plot the shape.

Analyze the characteristics of the shape including the center, major axis, minor

axis, foci, and eccentricity.

Solution

Step 1: Rearrange the equation to determine the type of shape.

4x2−16x+ 9y2+ 36y= 36

4(x2−4x) + 9(y2+ 4y) = 36

4(x2−4x+ 4) + 9(y2+ 4y+ 4) = 36 + 4(4) + 9(4)

4(x−2)2+ 9(y+ 2)2= 100

Comparing the equation to the standard form of the ellipse:

(x−h)2

a2+(y−k)2

b2= 1

We can see that this equation represents an ellipse.

Step 2: Identify the center and radii of the ellipse. The center of the ellipse

is (2, -2). The major axis has length 2a= 20 and the minor axis has length

2b= 18. Therefore, a= 10 and b= 9.

Step 3: Determine the major and minor axes. The major axis is along the

x-axis, which means the ellipse is wider horizontally. The endpoints of the major

axis are (2 ±10,−2) = (12,−2) and (−8,−2). The length of the major axis is

20.

The minor axis is along the y-axis. The endpoints of the minor axis are

(2,−2±9) = (2,7) and (2,−11). The length of the minor axis is 18.

Step 4: Find the foci. The distance from the center to the foci is c=

√a2−b2=√100 −81 = √19.

The foci are located at (2 ±√19,−2) ≈(6.36,−2) and (−2.36,−2).

Step 5: Calculate the eccentricity. The eccentricity is given by e=c

√19

10 ≈0.44.

Thus, the given equation represents an ellipse centered at (2, -2) with a

major axis of length 20, a minor axis of length 18, foci at approximately (6.36,

-2) and (-2.36, -2), and an eccentricity of approximately 0.44.

Question 21

Question

Plot and analyze the geometric shape defined by the equation 4x2+ 9y2−8x+

36y−4 = 0.

Solution

Step 1: Rewrite the equation in standard form by completing the square for

both xand yterms.

Step 2: Begin with the xterms.

4x2−8x= 4(x2−2x)

= 4(x2−2x+ 1 −1)

= 4((x−1)2−1)

= 4(x−1)2−4

Step 3: Now work on the yterms.

9y2+ 36y= 9(y2+ 4y)

= 9(y2+ 4y+ 4 −4)

= 9((y+ 2)2−4)

= 9(y+ 2)2−36

Step 4: Substitute these results back into the original equation.

4(x−1)2−4 + 9(y+ 2)2−36 −4=0

4(x−1)2+ 9(y+ 2)2= 44

Step 5: Divide by 44 to get the standard form of the equation.

(x−1)2

√44

√42+(y+ 2)2

√44

√92= 1

(x−1)2

11 +(y+ 2)2

4= 1

Step 6: Compare the resulting equation with the standard form of the ellipse,

(x−h)2

a2+(y−k)2

b2= 1. This ellipse has a center at (1,−2), a horizontal axis of

length 2a= 2√11, and a vertical axis of length 2b= 2 ·2 = 4.

Step 7: Thus, the plot is an ellipse centered at (1,−2) with a horizontal axis

of length 2√11 and a vertical axis of length 4.

Question 22

Question

Consider the geometric shape defined by the following equations:

x= 2 cos(t), y = 3 sin(t), z =t

where 0 ≤t≤2π. Plot this shape in 3D space and analyze its properties.

Solution

To plot this shape and analyze its properties, we first need to parameterize the

given equations. Let P(t) = (2 cos(t),3 sin(t), t) for 0 ≤t≤2π.

Step 1: Plotting the Shape

We can plot this shape by varying the parameter tfrom 0 to 2π. This will

give us a curve in 3D space. Let’s plot this curve using a graphing software.

Step 2: Analyzing the Shape

From the plot, we observe that the shape defined by the given equations is

a helix. It spirals upwards along the z-axis as tincreases from 0 to 2π.

Step 3: Calculating Length of the Curve

To calculate the length of the curve, we use the formula for the arc length

of a curve in 3D space:

L=Zt2

t1sdx

dt 2

+dy

dt 2

+dz

dt 2

Here, dx

dt =−2 sin(t), dy

dt = 3 cos(t), and dz

dt = 1.

Thus, the length of the curve P(t) from t= 0 to t= 2πis:

L=Z2π

0p(−2 sin(t))2+ (3 cos(t))2+ 12dt

After evaluating this integral, we can find the total length of the helix.

Question 23

Question

Consider the following parametric equations:

x(t) = 2 cos(t)

y(t) = 3 sin(t)

Determine the shape of the curve described by these equations and analyze its

properties.

Solution

Step 1: To find the shape of the curve, we can eliminate the parameter tby

expressing yin terms of x.

y(t) = 3 sin(t)

⇒sin(t) = y

Similarly,

x(t) = 2 cos(t)

⇒cos(t) = x

Step 2: Since sin2(t) + cos2(t) = 1, we can substitute the expressions for

sin(t) and cos(t) into this equation.

y

32+x

22= 1

9+x2

4= 1

4x2+ 9y2= 36

Step 3: The equation 4x2+ 9y2= 36 is the equation of an ellipse centered

at the origin with major axis along the y-axis and minor axis along the x-axis.

Step 4: The major axis has length 2a= 6 and the minor axis has length

2b= 4. Thus, the semi-major axis a= 3 and the semi-minor axis b= 2.

Step 5: The ellipse has foci (0,±√5).

Therefore, the curve described by the parametric equations x(t) = 2 cos(t)

and y(t) = 3 sin(t) is an ellipse centered at the origin with major axis along the

y-axis, minor axis along the x-axis, foci at (0,±√5), semi-major axis a= 3, and

semi-minor axis b= 2.

Question 24

Question

Consider the following geometric shape in the xy-plane:

16x2−24xy + 9y2+ 32x−36y+ 36 = 0

Determine the type of conic section represented by this equation, sketch the

shape, and identify its key features.

Solution

Step 1: Identify the Conic Section

To determine the type of conic section represented by the given equation, we

examine the coefficients of x2,xy, and y2. Since the coefficients of x2and y2

are both positive and the coefficient of xy is negative, the conic section is an

ellipse.

Step 2: Find the Center of the Ellipse

To find the center of the ellipse, we complete the square for both xand yterms.

16(x2+ 2x)−24xy + 9(y2−4y) = −36

16(x+ 1)2+ 9(y−2)2= 36

The center of the ellipse is (−1,2).

Step 3: Determine the Major and Minor Axes

The major axis of the ellipse corresponds to the larger of the 16 and 9 coefficients.

The minor axis corresponds to the smaller coefficient. So, the major axis is along

the x-axis with length 2a= 2√6 and the minor axis is along the y-axis with

length 2b= 2√4.

Step 4: Find the Foci of the Ellipse

The foci of an ellipse are located at a distance cfrom the center, where c2=

a2−b2. Here, c2= 6 −4 = 2, so c=√2. The foci are −1−√2,2and

−1 + √2,2.

Step 5: Plot and Analyze the Ellipse

With the center, major and minor axes, and foci identified, we can sketch the

ellipse in the xy-plane. The ellipse has center (−1,2), major axis along the

x-axis with length 2√6, minor axis along the y-axis with length 2√4, and foci

at −1−√2,2and −1 + √2,2.

Question 25

Question

Consider the graph of the region defined by x2+y2≤1. Calculate the area of

the region and find the centroid of the region.

Solution

Step 1: To find the area of the region, we will calculate the area of the circle

with radius 1.

Area = π·(radius)2=π·12=π

Step 2: Next, to find the centroid of the region, we will use the formulas:

¯x=Rx·dA

Area and ¯y=Ry·dA

Area

where dA =dx ·dy.

Step 3: For the given region, we will switch to polar coordinates, with x=

r·cos(θ) and y=r·sin(θ). The integral for centroid calculation then becomes:

¯x=R2π

0R1

0(r·cos(θ)) ·r dr dθ

¯y=R2π

0R1

0(r·sin(θ)) ·r dr dθ

Step 4: Evaluating the integrals, we get:

¯x=R2π

01

3·cos(θ)dθ

¯y=R2π

01

3·sin(θ)dθ

Step 5: Solving the integrals, we find:

¯x=1

3π·Z2π

cos(θ)dθ =1

3π·[sin(θ)]2π

0= 0

¯y=1

3π·Z2π

sin(θ)dθ =1

3π·[−cos(θ)]2π

0= 0

Step 6: Therefore, the centroid of the given region is (0,0).

Question 26

Question

Let Abe the point (2,3), and let Bbe the point (5,7). Determine the equations

of:

1. the line passing through Aand B,

2. the perpendicular bisector of the line segment AB.

Solution

To determine the equations of the lines passing through the points Aand B, we

first need to find the slope of the line segment AB using the formula:

m=y2−y1

x2−x1

Step 1: Find the slope of line segment AB.Given points A(2,3) and

B(5,7), the slope mof the line passing through Aand Bis:

m=7−3

5−2=4

Step 2: Determine the equation of the line passing through Aand

B.Since the equation of a line with slope mpassing through point (x1, y1) is

given by y−y1=m(x−x1), the equation of the line passing through points A

and Bis:

y−3 = 4

3(x−2).

Expanding and simplifying:

y−3 = 4

3x−8

y=4

3x−8

3+ 3

y=4

3x−8

3+9

y=4

3x+1

Therefore, the equation of the line passing through Aand Bis y=4

3x+1

Step 3: Find the midpoint of line segment AB.To find the midpoint

of the line segment AB, we use the midpoint formula:

x1+x2

2,y1+y2

2.

For points A(2,3) and B(5,7), the midpoint Mof the line segment AB is:

2+5

2,3+7

2=7

2,10

2=7

2,5.

Step 4: Determine the slope of the perpendicular bisector. The

slope of a line perpendicular to a line with slope mis −1

m. Thus, the slope of

the perpendicular bisector of the line segment AB is:

−1

=−3

Step 5: Find the equation of the perpendicular bisector. Since the

perpendicular bisector passes through the midpoint M7

2,5and has a slope of

−3

4, its equation is:

y−5 = −3

4x−7

2.

Expanding and simplifying:

y−5 = −3

4x+21

y=−3

4x+21

8+ 5

y=−3

4x+21

8+40

y=−3

4x+61

Therefore, the equation of the perpendicular bisector of the line segment AB

is y=−3

4x+61

Question 27

Question

Consider the set of points in R2defined by the equation 3x2+ 4xy + 3y2= 1.

(a) Sketch the graph of this equation. (b) Classify the shape of the graph

and explain your reasoning.

Solution

(a) To sketch the graph of the equation 3x2+ 4xy + 3y2= 1, let’s first rewrite

the equation in matrix form:

x y3 2

2 3x

y= 1

This is a quadratic equation in two variables, which represents a conic sec-

tion. To sketch the graph, we will analyze its level curves.

(b) To classify the shape of the graph, we will analyze the matrix 3 2

2 3.

The determinant of this matrix is 3 ×3−2×2 = 5, and the trace is 3 + 3 = 6.

Since the determinant is positive and the trace is greater than zero, we conclude

that the graph represents an ellipse.

Therefore, the graph of the equation 3x2+ 4xy + 3y2= 1 is an ellipse in R2.

Question 28

Question

Consider the curve defined by the polar equation r= 2 sin(3θ).

(a) Plot the graph of the curve.

(b) Determine the number of petals in the graph and provide a geometric

interpretation for this number.

Solution

(a) To plot the graph of the curve defined by the polar equation r= 2 sin(3θ),

we can first consider different values of θin the interval 0 ≤θ≤2πand compute

the corresponding values of r. Then we can convert the polar coordinates to

Cartesian coordinates using the formulas x=rcos(θ) and y=rsin(θ).

When θ= 0, r= 0, so the point is at the origin.

Let’s now consider θ=π

6,π

3,π

2,2π

3,5π

6, π, 7π

6,4π

3,3π

2,5π

3,11π

6,2π:

For θ=π

6,r= 2 sinπ

2= 2 ⇒x= 2 cosπ

6=√3, y= 2 sinπ

6= 1.

For θ=π

3,r= 2 sin(π) = 0 ⇒x= 0, y= 0.

For θ=π

2,r= 2 sin3π

2=−2⇒x=−2 cosπ

2= 0, y=−2 sinπ

2=−2.

For θ=2π

3,r= 2 sin(2π)=0⇒x= 0, y= 0.

For θ=5π

6,r= 2 sin3π

2=−2⇒x=−2 cos5π

6=−√3, y=−2 sin5π

6=

−1.

For θ=π,r= 2 sin(3π)=0⇒x= 0, y= 0.

For θ=7π

6,r= 2 sin5π

2= 2 ⇒x= 2 cos7π

6=−√3, y= 2 sin7π

6=−1.

For θ=4π

3,r= 2 sin(3π)=0⇒x= 0, y= 0.

For θ=3π

2,r= 2 sin9π

2=−2⇒x=−2 cos3π

2= 0, y=−2 sin3π

2= 2.

For θ=5π

3,r= 2 sin15π

2= 0 ⇒x= 0, y= 0.

For θ=11π

6,r= 2 sin33π

6= 2 ⇒x= 2 cos11π

6=√3, y= 2 sin11π

6= 1.

Finally, for θ= 2π,r= 2 sin(6π)=0⇒

Question 29

Question

Consider the following geometric shape in the xy-plane: - Circle centered at

(2,3) with radius 4. - Line passing through points (5,−1) and (1,5).

Plot the circle and the line on the same set of axes and find the points of

intersection, if any.

Solution

To plot the circle and the line on the same set of axes, we first need to find their

equations.

Step 1: Find the equation of the circle The equation of a circle with

center (h, k) and radius ris given by:

(x−h)2+ (y−k)2=r2

In this case, the center of the circle is (2,3) and the radius is 4, so the

equation of the circle is:

(x−2)2+ (y−3)2= 16

Step 2: Find the equation of the line The equation of a line passing

through two points (x1, y1) and (x2, y2) can be found using the point-slope form

given by:

y−y1=y2−y1

x2−x1·(x−x1)

Using the points (5,−1) and (1,5), we have:

y−(−1) = 5−(−1)

1−5·(x−5)

y+ 1 = 6

−4·(x−5)

y+ 1 = −3

2(x−5)

y=−3

2x+13

Step 3: Plot the circle and the line Now, we can plot the circle with

equation (x−2)2+ (y−3)2= 16 and the line y=−3

2x+13

2on the same set of

axes.

Step 4: Find the points of intersection To find the points of intersection,

we need to solve the system of equations formed by the circle and the line.

Substituting y=−3

2x+13

2into the equation of the circle:

(x−2)2+−3

2x+13

2−32

= 16

Question 2

Question

Consider the following parametric equations for a curve in the xy-plane:

x= 2 sin(t), y = 3 cos(t),0≤t≤2π.

(a) Plot the curve generated by the parametric equations. (b) Determine the

equation of any tangent lines to the curve at the points where t=π

2. (c) Find

the area enclosed by the curve.

Solution

(a) To plot the curve, we can eliminate the parameter tand express yin terms

of x:

y= 3 cos arcsin x

2= 3r1−x

22.

This equation represents the top half of an ellipse centered at the origin with

semi-major axis of 3 units along the y-axis and semi-minor axis of 2 units along

the x-axis. The plot of the curve is the top half of this ellipse.

(b) To find the equation of the tangent lines at t=π

2, we need to find the

slope of the curve at this point. The slope of the curve at any point is given by

dx =dy/dt

dx/dt .

dt =−3 sin(t),dx

dt = 2 cos(t).

At t=π

2,dy

dt =−3, dx

dt = 0, so the slope is undefined. This means the tangent

line at this point is vertical and its equation is x= 2.

2y dx over one period

of the curve:

Area = Z2π

2·3 cos(t)·2 cos(t)dt =Z2π

3 cos2(t)dt.

Using the double angle identity cos(2t) = 2 cos2(t)−1, we can rewrite the

integral as:

Area = 3

2Z2π

(1 + cos(2t)) dt =3

2t+1

2sin(2t)

2π

=9π

Therefore, the area enclosed by the curve is 9π

Question 3

Question

Consider the two-dimensional region in the xy-plane bounded by the curves

y=x2and y= 2x−x2. Calculate the area of the region enclosed by these

curves.