*Mathematics and the Physical World*is "The Differential Calculus." We apparently can't avoid Calculus forever!

"It's enough if you understand the Propositions with some of the Demonstrations which are easier than the rest." -- Newton's advice to prospective readers of his

*Principia Mathematica*

*Kline begins,*

"Through the study of the motion of projectiles, planets, pendulums, sound, and light the scientific world of the seventeenth century became conscious of the pervasiveness of change. It had also become aware of the usefulness of a function to represent relationships between variables and to deduce new scientific laws."

Kline's opening quote is from one of the co-inventors of Calculus, Isaac Newton. As we can see so far, Calculus deals with functions and change. Kline writes:

"Next to the creation of Euclidean geometry the calculus has proved to be the most original and the most fruitful concept in all of mathematics."

I said that Newton was one of the

*co*-inventors of Calculus. The other was Gottfried Wilhelm Leibniz, who created the subject independently from Newton. Here's a little of what Kline says about Leibniz:

"This man of universal gifts and interests, son and grandson of professors, was born in Leipzig [Germany] in 1646. At the age of fifteen he entered the University of Leipzig with the announced intention of studying law and the unannounced intention of studying everything. His brilliance so excited the jealousy and the envy of his teachers that they never granted him his doctor's degree. During the years in which he acquired his legal training Leibniz was also busily studying mathematics and physics."

Kline's first example of Calculus is to find the instantaneous velocity of a ball three seconds after it has been dropped. He writes the following calculation, starting with the formula Galileo had found for the distance traveled by an object in free fall:

*d*= 16

*t*^2

*d*_3 = 144

*d*_3 +

*k*= 16(3 +

*h*)^2 (That is, in an additional

*h*seconds the ball travels an extra

*k*feet.)

*d*_3 +

*k*= 16(9 + 6

*h*+

*h*^2)

*d*_3 +

*k*= 144 + 96

*h*+ 16

*h*^2

*k*= 96

*h*+ 16

*h*^2 (Here Kline substitutes in the value of

*d*_3 found earlier and subtracts.)

*k*/

*h*= (96

*h*+ 16

*h*^2)/

*h*

*k*/

*h*= 96 + 16

*h*

*Kline writes that to find the instantaneous velocity, we should plug in*

*h*= 0 into the penultimate equation in this list -- but this gives us 0/0. So we plug it in to the last equation instead, and we end up with

*k*/

*h*(distance divided by time, or velocity) equals 96 feet per second. According to Kline, the great insight by Newton and Leibniz is that this is justified using

*limits*-- as

*h*approaches 0, the value of 96 + 16

*h*approaches 96, and so this is taken to be the instantaneous velocity.

The author goes on to demonstrate that at an arbitrary time

*t*=

*x*, we perform the same calculation to

*y*= 16

*x*^2 to obtain 32

*x*as the instantaneous velocity. This is what we now know as the

*derivative*of the expression 16

*x*^2. Kline writes that Leibniz used the notation

*dy*/

*dx*to denote the derivative of

*y*with respect to

*x*, but he prefers Newton's notation,

*y*-dot (that is, a dot above

*y*). This is difficult for me to show in ASCII, so I'll use another commonly used symbol for derivative -- the apostrophe, often pronounced "prime" (just as we do with transformation images in Geometry). So we have:

*y*= 16

*x*^2

*y*' = 32

*x*

Kline also works out the velocity of a bob on a spring, which follows the formula

*y*= sin

*x*, at the initial time

*t*= 0. He shows that this equals the limit of the expression (sin

*h*)/

*h*as

*h*-> 0. He gives the following argument -- when

*h*is a small central angle of a circle, we can let

*r*be the radius of the circle, and then we draw a right triangle with hypotenuse

*r*and opposite leg

*a*(since sine equals opposite over hypotenuse, or

*a*/

*r*). Then

*s*is the arc subtended by the central angle

*h*, and so the measure of

*h*in

*radians*is the arclength

*s*divided by the radius

*r*, or

*s*/

*r*. Then (sin

*h*)/

*h*works out to be

*a*/

*s*, as the

*r*'s cancel. But as

*h*approaches 0,

*a*approaches the arclength

*s*(which is why a regular polygon with many sides approaches a circle), so

*a*/

*s*is approaching 1. Nowadays we would write:

*y*= sin

*x*

*y'*= cos

*x*

*y'*_0 = cos 0

*y'*_0 = 1

but we must actually

*prove*that the derivative of sine is cosine.

Here are a few more amusing examples. Kline begins with the area of a circle and differentiate it with respect to the radius:

*A*= pi

*r*^2

*A'*= 2pi

*r*

*That's funny -- the derivative of the area is the circumference! Kline explains why:*

"This result is intuitively clear, for as the radius increases, one might say that 'successive' circumferences are added to the area."

And of course, we do the same thing with the volume of a sphere:

*V*= (4/3) pi

*r*^3

*V'*= 4pi

*r*^2

which is the surface area of a sphere.

Actually, the volume of a sphere is a great place to segue into PARCC. Question 22 of the PARCC Practice Exam is on the volume of cylinders and -- you guessed it! -- spheres.

22. Hank is putting jelly candies into two containers. One container is a cylindrical jar with a height of 33.3 centimeters and a diameter of 8 centimeters. The other container is spherical. Hank determines that the candies are cylindrical in shape and that each candy has a height of 2 centimeters and a diameter of 1.5 centimeters. He also determines that air will take up 20% of the volume of the containers. The rest of the space will be taken up by the candies.

Part A

After Hank fills the cylindrical jar with candies, what will be the volume, in cubic centimeters, of the air in the cylindrical jar? Round your answer to the nearest cubic centimeter.

Part B

What is the maximum number of candies that will fit in the cylindrical jar?

Part C

The spherical container can hold a maximum of 260 candies. Approximate the length of the radius, in centimeters, of the spherical container. Round your answer to the nearest tenth.

Part D

Hank is filling the cylindrical candy container using bags of candy that have a volume of 150 cubic centimeters. Air takes up 10% of the volume of each bag, and the rest of the volume is taken up by candy. How many bags of candy are needed to fill the cylindrical container with 260 candies?

To answer this question, we obviously have to calculate volume. For Part A, we calculate the volume of the cylinder as pi(4)^2 (33.3) = 1673.8 cm^3. Only 20% of this is air, so this gives us an air volume of 334.8 cm^3, which rounds off to 335 cm^3.

For Part B, we see that 1339 cm^3 is left for the candy. Now the volume of a single candy is pi(0.75)^2 (2) = 3.53 cm^3, and so we divide the volumes to give a value of 378.88 candies. (This last value is

*exact*-- the factors of pi cancel out to leave a rational number.) Notice that there isn't quite enough room for 379 candies -- so the answer is 378 candies.

For Part C, we begin by multiplying 3.53 cm^3 for each candy by 260 to obtain 918.92 cm^3. But notice that this

*doesn't*include the air. The candy only takes us 80% of the space, so we must divide this volume by 0.8 to obtain the total volume of the sphere as 1148.64 cm^3. This is the volume of a sphere and we want to know the radius, so we use the sphere volume formula that we found in Kline:

(4/3) pi

*r*^3 = 1148.64

*r*^3 = 274.21875 (exact rational value)

*r*= 6.497, which rounds to 6.5 cm

For Part D, first of all, I'm wondering, why does the question have Hank put 260 candies in the

*cylinder*when we already know that exactly 260 candies fit in the

*sphere*? Wouldn't it have been more logical to ask, "How many bags of candy are needed to fill the spherical container?" (Then there wouldn't have been any need to say "260 candies" again -- "fill the sphere" would've been sufficient.)

At this point, it's probably easier to notice that since air takes up only 10% of the 150 cm^3-bag, the remaining 135 cm^3 is candy. We already calculated the volume of 260 candies earlier (when trying to find the radius of the

*sphere*) as 918.92 cm^3, so we only need to divide this by 135 to obtain an answer of 6.8 bags. Six bags aren't enough -- we need 7 bags.

This question has several places where student error can creep in. This includes how to handle the air factor in both cases. Students must know the formulas for both the cylinder and the sphere without mixing the two up, as well as plug in the radius (not the

*diameter*) correctly and solve for it properly (for example, don't take the square root instead of the cube root of

*r*^3).

As for rounding error, according to the PARCC answer key, both 334 and 335 are acceptable for Part A, and both 6.4 and 6.5 are acceptable for Part C. However, both Part B and Part D have only one acceptable answer, as there isn't enough room for the 379th candy, and six candy bags aren't enough to fill the sphere.

One problem with questions that include many values that need to be rounded off is that rounding error may accumulate throughout the problem. (I assume this is the reason that Parts A and C have two acceptable answers.) But notice that since both the cylinder and sphere have pi in their formulas, the irrational value of pi cancelled out. This means that for Parts A, B, and C, we could have simply ignored pi altogether (or used a crude value like 3.14) and obtained the correct answer. Only Part D requires a value for pi as it doesn't cancel out (since the volume of the bag is 150 cm^3, not anything in terms of pi). But even a crude value like 3.14 still produces seven bags as the correct answer. (In fact, the approximation pi = 3 gives an answer of exactly six and a half bags.)

With so many possibilities for error, I expect most students to get at least one part, if not several parts, of this question wrong.

**PARCC Practice EOY Question 22**

**U of Chicago Correspondence: Lessons 10-5 and 10-8, Volumes of Cylinders and Spheres**

**Key Theorem: Volume Formulas**

**The volume**

*V*of any prism or cylinder is the product of its height*h*and the area*B*of its base.

*V*=*Bh*

**The volume**

*V*of any sphere is (4/3)pi times the cube of its radius*r*.

*V*= (4/3)pi*r*^3

**Common Core Standard:**

CCSS.MATH.CONTENT.HSG.GMD.A.3

Use volume formulas for cylinders, pyramids, cones, and spheres to solve problems.

**Commentary: There are no problems just like this one in the U of Chicago text. Instead, I combine parts of two questions -- one from the SPUR section of Chapter 10 about the volume of a cylindrical cup, the other from Lesson 10-8 on the volume of a spherical tank -- and ask about how many cupfuls of water fit in the tank.**

## No comments:

## Post a Comment