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University of South Carolina Math 222: Math for Elementary Educators II Instructor: Austin Mohr Section 002 Fall 2010
Quiz 3 Solutions
- Astronomers use light-years to measure distances between celestial bod- ies. A light-year is the distance light travels in 1 year. The speed of light is (roughly) 300,000 km/sec. 1a. How long is 1 light-year in kilometers?
Solution: Light travels 300,000 kilometers every second, so we need to turn that speed into kilometers per year.
300 , 000 km 1 sec
60 sec 1 min
60 min 1 hour
24 hour 1 day
365 day 1 year
When we multiply it all together, all the units except for kilometers on top and years on bottom will cancel (so we know we’ve set up the solution correctly). All this simplifies to
- 5 ⋅ 1012 km 1 year
So, a lightyear (the distance light travels in a year) is 9. 5 ⋅ 1012 kilometers.
1b. The nearest star (other than the sun) is Alpha Centauri. It is 4. light-years from Earth. How far is that in kilometers? Solution: We know from part a how long a single lightyear is, so we just multiply that answer by 4.34 to get 4. 1 ⋅ 1013 kilometers.
1c. How many years will it take a rocket traveling 60,000 km/hr to reach Alpha Centauri? Solution: We use the fact that distance equals speed times travel time. In other words, travel time equals distance divided by speed. So, we get
time =
- 1 ⋅ 1013 km 60000 kmhr = 684331200 hr.
So now we know how many hours it takes, but we want to convert this to years. 684331200 hr 1
1 day 24 hr
1 year 365 days which comes out to 78,120 years.
1d. This question was omitted.
- The figure below is a square with its corners rounded into perfect quarter-circles. Give your answers to the following in inches, square inches, or cubic inches, as appropriate.
2a. Find the perimeter of the shape.
Solution: The key to these kinds of problems is breaking the shape up into pieces you know about. This perimeter can be viewed as four 2-foot line segments and four quarter-circles of radius 3 inches. So, our preliminary formula might look like
perimeter = 4 ⋅ line segment + 4 ⋅ quarter-circle. Since our answers are supposed to be in inches, we should convert the 2-foot line segment into 24 inches. The perimeter (usually called circumference) of a circle is given by 2πr, which is 6π inches in this case. Since we are only interested in quarter -circles, we divide this by 4 to get 32 π inches.
2d. Find the volume of a 6 inch tall “cone” having this shape as its base.
Solution: The volume of a cone is one-third the volume of the related cylinder (by “related”, I mean they have exactly the same base).
volume =
(volume of related cylinder)
=
( 5184 + 54 π cubic inches) = 1728 + 18 π cubic inches.
2e. Find the surface area of the “cylinder” in part c.
Solution: The surface area can be viewed as two bases plus the lateral face, which is a rectangle. This means
surface area = 2 ⋅ (bases) + 1 ⋅ (lateral face).
We know the area of the bases from part b. The lateral face is a rectangle whose height is the same as the height of the cylinder and its width is the perimeter of the base. (Imagine making a cylinder out of paper. When you unroll it, you’ll have a rectangle with these dimensions). As scratch work, we would compute
area of lateral face = (height of cylinder) ⋅ (perimeter of base) = (6 inches) ⋅ ( 96 + 6 π inches) = 576 + 36 π square inches.
Returning to our formula,
surface area = 2 ⋅ (bases) + 1 ⋅ (lateral face) = 2 ⋅ ( 864 + 9 π square inches) + 1 ⋅ ( 576 + 36 π square inches) = ( 1728 + 18 π square inches) + ( 576 + 36 π square inches) = 2304 + 54 π square inches.
- Suppose you have a cylinder of height h and radius R and you bore out a centrally-located cylinder of radius r. The result is called a cylindrical shell.
3a. Write an equation for the volume of the cylindrical shell in the picture.
Solution: The volume of the shape if nothing was removed would be πR^2 h (area of the circular base times the height of the cylinder). The vol- ume of the removed cylinder is πr^2 h. So, the remaining area is πR^2 h − πr^2 h, which you might rewrite as πh(R^2 − r^2 ).
3b. Write an equation for the surface area of the cylindrical shell in the picture. Solution: The surface area is made up of two bases, an outside lateral face, and an inside lateral face. So,
surface area = 2 ⋅(area of base)+ 1 ⋅(area of outside lateral face)+ 1 ⋅(area of inside lateral face).
If the base had nothing removed, the area would be πR^2. The area of the removed portion is πr^2. So, the remaining area is πR^2 − πr^2. The outside lateral face is a rectangle with height h and width 2πR (the circumference of the larger circle). So, the total area is 2πRh. The inside lateral face is a rectangle with height h and width 2πr (the circumference of the smaller circle). So, the total area is 2πrh. Putting all the pieces together,
surface area = 2 ⋅ (area of base) + 1 ⋅ (area of outside lateral face) + 1 ⋅ (area of inside lateral face)
= 2 ⋅ (πR^2 − πr^2 ) + 1 ⋅ ( 2 πRh) + 1 ⋅ ( 2 πrh) = 2 πR^2 − 2 πr^2 + 2 πRh + 2 πrh.
- Repeated application of the Pythagorean Theorem can be used to find the distance between points in higher dimensions.
4c. Use this idea to find the distance between the origin and the point (2,5,3,7) in four-dimensional space. Solution: We have limited power to visualize four-dimensional space, but algebra lets us explore it, nonetheless. The triangle perpendicular to all of three-dimensional space (whatever that means) is a right triangle whose legs are length
38 (this is the leg contained in the first three dimensions) and 7 (this is the leg poking out into the fourth dimension). The hypotenuse is the distance we are interested in finding. If we let c be this distance, then the Pythagorean Theorem tells us
(
38 )^2 + 72 = c^2 38 + 49 = c^2 87 = c^2 √ 87 = c.
The beauty of mathematics is that it lets you explore what you can never personally experience. You can take that as a sign of impracticality, or you can take it as the beginnings of understanding a reality much richer than the one we experience day-to-day.
- A circle has the points (-1, 2) and (4, -3) as opposite enpoints of one of its diameters. 5a. What is the radius of the circle?
Solution: The distance between the given points will give us the diam- eter of the circle. Using the distance formula, the diameter has length » (x 2 − x 1 )^2 + (y 2 − y 1 )^2 =
( 4 − (− 1 ))^2 + (− 3 − 2 )^2
52 + (− 5 )^2
Since the radius is just half the diameter, the radius has length
√ 50
5b. What is the center of the circle?
Solution: The center of the circle is the midpoint of the two endpoints of the diameter (the end points of any diameter would have worked). The midpoint formula tells us the center is at
x 1 + x 2 2
y 1 + y 2 2
5c. What is the equation of the circle?
Solution: Now that we have the radius and the center, we just put all this information in the generic equation of a circle.
(x − h)^2 + (y − k)^2 = r^2
(x −
)^2 + (y − (−
))^2 =
2
(x −
)^2 + (y +
)^2 =
(x −
)^2 + (y +
)^2 =
5d. Use your equation in part c to determine if the point (1, -1) lies on the circle. Solution: The equation in part c describes all points (x, y) that lie on the circle. If we plug in values for x and y and get a false statement, then they do not lie on the circle. If we get a true statement, then they do.
(x −
)^2 + (y +
)^2 + (− 1 +
)^2 + (−
=^? 25
=^? 25
The last line is clearly false, so we conclude that the point ( 1 , − 1 ) does not lie on the circle.
