# Polar Calculus

This triangle is important for solving many problems related to polar coordinates.

From it we get the following important relationships:

x=r*cos(theta)

y=r*sin(theta)

r^2=x^2+y^2

tan(theta)=y/x

In polar coordinates we write r as a function of theta, r=f(theta).

Now consider the equation x=r*cos(theta). If we differentiate with respect to theta, the we get (dx)/(d theta)=(dr)/(d theta)*cos(theta)-r*sin(theta).

Similarly, (dy)/(d theta)=(dr)/(d theta)*sin(theta)+r*cos(theta).

These relationships will be useful when looking at slope and arc length, which are concepts from rectangular coordinates.

# Simpson’s Rule (Riemann Sums with Parabolas)

int_a^b f(x) dx ~~ (h/3)((y_0+4y_1+y_2)+(y_2+4y_3+y_4)+…+(y_(n-2)+4y_(n-1)+y_n))

int_a^b f(x) dx ~~ (h/3)(y_0+4y_1+2y_2+4y_3+2y_4+…+2y_(n-2)+4y_(n-1)+y_n)

int_a^b f(x) dx ~~ sum_(j=1)^(n/2) (h/3)(f(x_(2j-2))+4f(x_(2j-1))+f(x_(2j)))

where h=(b-a)/n and the error is |E_S|<=(M(b-a)^5)/(180*n^4) and M is an upper bound for |f^((4))| on [a,b].

#### Example 1:

Evaluate int_5^7 5x dx using Simpson’s rule with n=4

x_0 through x_4 are {5, 5.5, 6, 6.5, 7} and h=2/4=1/2

Area ~~ 1/6*(5(5)+4*5(5.5)+2*5(6)+4*5(6.5)+5(7))=60

Since f'(x)=5 and f”(x)=0, then f^((4))(x)=0 and the error is zero here.

#### Example 2:

Consider int_-1^1 (x^2+5) dx with n=4

h=2/4=1/2 so x_k=-1+k/2, which is the set {-1,-1/2,0,1/2,1}

Area ~~1/6(((-1)^2+5)+4((-1/2)^2+5)+2((0)^2+5)+4((1/2)^2+5)+((1)^2+5))

Area=10 2/3

Since f^((4))(x)=0 the error here is also zero.

#### Example 3:

int_0^2 (5t^3+8t) dt with n=4

h=2/4=1/2 and x_k=0+k/2, so the x-values are {0,1/2,1,1 1/2,2}

Area ~~ 1/6((5(0)^3+8(0))+4(5(1/2)^3+8(1/2))+2(5(1)^3+8(1))+4(5(1 1/2)^3+8(1 1/2))+(5(2)^3+8(2)))

Area =36

Since f^((4))(x)=0, the error is zero here.

#### Example 4:

int_3^9 4/s^2 ds with n=4

h=6/4=3/2 and x_k=3+3k/2, so the x-values are {3, 4.5, 6, 7.5, 9}

Area ~~1/2*(4/3^2+4*4/4.5^2+2*4/6^2+4*4/7.5^2+4/9^2)

#### Example 5:

How many intervals are needed to estimate int_1^2 1/x dx to within 0.0001 square units of the actual area?

Recall the error is |E_S|<=(M(b-a)^5)/(180*n^4) and M is an upper bound for |f^((4))| on [a,b].

f(x)=x^(-1)

f'(x)=-1*x^(-2)

f”(x)=2*x^(-3)

f”(x)=-6*x^(-4)

f^((4))(x)=24*x^(-5)

f^((5))(x)=-144*x^(-6)

I found the 5th derivative to show that f^((4))(x) is decreasing, since its derivative is negative on the interval [1,2]. So, f^((4))(1)=24 is the maximum needed for the error estimation.

Plugging into |E_S|<=(M(b-a)^5)/(180*n^4):

0.0001<=(24*(1)^5)/(180*n^4)

Solving for n we get n <=((24)/(180*0.0001))^(1/4)~~6.04

A_(S_6) = 1/3*1/6*(1/1+4*1/(1 1/6)+2*1/(1 2/6)+4*1/(1 3/6)+2*1/(1 4/6)+4*1/(1 5/6)+1/2)~~0.693169

A_(S_4) = 1/3*1/4*(1/1+4*1/(1 1/4)+2*1/(1 2/4)+4*1/(1 3/4)+1/2)~~0.693254

A_(S_2) = 1/3*1/2*(1+4*1/1.5+1/2)~~.694444

ln(2)~~0.69314718

The actual error from the 6th Simpson’s sum is about 0.00002.

A_(S_20) = sum_(n=1)^10 1/(3*20)*(1/(1 + 1/20 (2 n – 2)) + 4/(1 + 1/20 (2 n – 1)) + 1/(1 + (2 n)/20)) = 5555158368718531/8014397185594800~~0.693147375

For comparison, consider 100 terms with right endpoints: sum_(k=1)^100 1/(100 (k/100 + 1))
~~0.690653430481824215252268721472608478928428119821791431288…

Or the Trapezoidal Rule with 50 intervals:
A_(T_50)=sum_(k=1)^50 1/(2*50)*(1/(1+(K-1)/50)+1/(1+k/50))~~0.693172179

#### Example 6:

Here is a picture of a parabola intersecting the reciprocal function at x=1,2,3

The parabola, y=1/6*x^2-1*x+11/6, is shown in red.
The beautiful thing about Simpson’s rule is that we don’t actually need to calculate the coefficients of any of these parabolas. The weighted sum does all of the work for us.

# A number theory story

A friend asked me to investigate how to write algorithmically generated fraction addition problems where the sum would always be reducible or irreducible. Why? It seems unfair, in a testing situation, to have one addition problem lead to a reducible sum while another is irreducible.

# Relatively Prime Linear Combinations

If there exist x and y for which ax+by=GCD(a,b) then GCD(x,y)=1.

Let g=GCD(a,b) and x and y solutions to ax+by=g.

Since g|a and g|b, there exist integers n and m such that a=gn and b=gm.

ax+by=gnx+gmy=g

So, nx+my=1

The GCD(n,m)=1 since g is the GCD(a,b).

Let d be a divisor of x. If d also divides my, then d must divide 1, which isn’t possible. So, GCD(x,y)=1.

# Section 7.3 Hyperbolic Trigonometric Functions

cosh(x) = (e^x+e^-x)/2

which is the shape of a cable strung between two points.

d/dx[cosh(x)]=(e^x-e^-x)/2=sinh(x)

int cosh(x) dx =(e^x-e^-x)/2+c=sinh(x)+c

d/dx[sinh(x)]=(e^x+e^-x)/2=cosh(x)

int sinh(x) dx =(e^x-e^-x)/2+c=cosh(x)+c

tanh(x) = sinh(x)/cosh(x) =(e^x-e^-x)/(e^x+e^-x)

Alternatively, tanh(x)=(e^(2x)-1)/(e^(2x)+1)

#### Examples

Find d/dx[tanh(x)]

Simplify cosh^2(x)-sinh^2(x)

Find int tanh(x) dx

Find cosh^-1(x)

Find tanh^-1(x)

Find d/dx[cosh^-1(x)]

Findd/dx[sinh^-1(cot(x))]

Find int_(-ln(12))^(-ln(2)) 2 e^x cosh(x) dx

# Fraction Sum Divisors

Given a pair of reduced fractions, e.g., 1/30 and 1/42, or 3/10 and 1/5, we add them with the LCD Algorithm.

If the sum reduces, then it will reduce by a factor of the GCD of the denominators.

#### Example 1/30+1/42

• Find the LCM
30=2 * 3 * 5 and 42=2 * 3 * 7
so the LCM =2 * 3 * 5 * 7 = 210 and the GCD =2* 3=6.
• Modify the first fraction
210/30=7 so we rewrite 1/30 = (7*1)/(7*30) = 7/210
• Modify the second fraction
210/42=5 so 1/42 = (5*1)/(5*42) = 5/210
• Combine the fractions with the common denominator
1/30+1/42 = 7/210+5/210 = 12/210
• Notice if they reduce
12/210= (12 -: 6)/(210 -: 6) = 2/35

#### Example 11/30+1/42

11/30+1/42 = 77/210+5/210 = 82/210= (82 -: 2)/(210 -: 2) = 41/105

#### Proof

Applying the LCD algorithm a/c+b/d=(au+bv)/lcm

lcm=LCM(c,d)
Note that the LCM of two numbers is the product of the highest power of each prime from the prime factorizations of the two numbers.

gcd=GCD(c,d)
Note that the GCD is the lowest power of the primes common to the two numbers.

These sets of prime factors are disjoint and complementary to the product of the two numbers, i.e.,

c* d = lcm * gcd (1)

Alternatively,

lcm/c=d/gcd(2)

To apply the LCD Algorithm we need to find u and v, the complementary factors of the lcm for c and d respectively.

u=lcm/c and v=lcm/d(3)

Since u=lcm/c=d/gcd, then u is made from factors of d that are not common to c. Similarly, v is made from factors of c that are not common to d. Hence u and v have no common factors, i.e., GCD(u,v)=1.

Notice that u*v = lcm/c * lcm/d and gcd=(c*d)/lcm

Hence, lcm = gcd * u * v

so a/c+b/d=(au+bv)/(gcd * u * v)

If this sum is reducible, then there exists an Integer n such that

(au+bv)/n and (gcd * u * v)/n

If n divides u then n must divide bv. However, we know that u is relatively prime to v. Additionally, since b is relatively prime to d, b is also relatively prime to u. Hence n can’t divide bv and then must not divide u,

So if n is to be the GCD of the numerator and denominator of the sum, it must divide the GCD(c,d).

# Fraction Sum Exploration

A common method for adding fractions is to find the lowest common multiple of the denominators. This is called the LCD algorithm, Lowest Common Denominator. Even though we have the smallest possible denominator that allows us to combine the fractions, sometimes the sum will still reduce. I have always wondered why.

Consider 1/3+1/6. The LCD = 6, so the sum becomes 2/6+1/6=3/6 or 1/2. This is reducible.

Now consider 1/3+1/2. The LCD = 6, so the sum becomes 2/6+3/6=5/6. This is clearly not reducible.

I finally found an article explaining the theory behind when a fraction sum is reducible or irreducible.

Biscuits of Number Theory by Arthur T. Benjamin, Ezra Brown, 2006, MAA
URL

Originally published as:
Reducing the Sum of Two Fractions
Shultz, Harris S. and Ray C. Shiflett. Mathematics Teacher. vol. 98, no. 7 (March 2005): pp.486-490.
URL

If the denominators have a commom prime power in the prime factorizations, then some sums are reducible, otherwise, the sums are always irreducible. For denominators 12=2^2*3 and 15=3*5 there exist numerators for which the sum will reduce since they have a common prime power of 3^1, and, for example, 1/12+1/15=9/60 = 3/20. For denominators 8=2^3 and 12=2^2*3 there are no numerators for which the sum reduces, since they do not have a common prime power. Here I am assuming that the numerators are relatively prime to their denominators, i.e., they have a gcd of 1.

Even better, if the common prime power is based on 2, then the sums are always reducible. Otherwise, there exist numerators for which the sum will reduce. The problem then becomes finding those numerators.

10 vs 5 1 2 3 4
1 3/10 1/2 7/10 9/10
3 1/2 7/10 9/10 11/10
7 9/10 11/10 13/10 3/2
9 11/10 13/10 3/2 17/10

# Section 6.5 Physics Applications to Calculus

### Physics Formulas

#### Weight-density

A fluid’s weight-density w is its weight per unit volume. Typical values (lb/ft^3) are listed below.

Gasoline  42
Mercury  849
Milk  64.5
Molasses  100
Olive oil  57
Seawater  64
Freshwater  62.4

### Homework

1. It took 1800 J of work to stretch a spring from its natural length of 2 m to a length of 5 m. Find the spring’s force constant.
2. It takes a force of 21,714 lb to compress a coil spring assembly on a New York City Transit Authority subway car from its free height of 8 in. to its fully compressed height of 5 in.
a. What is the assembly’s force constant?
b. How much work does it take to compress the assembly the first half inch? the second half inch? Answer to the nearest in.-lb.
3. Lifting an elevator cable An electric elevator with a motor at the top has a multistrand cable weighing 4.5 lb > ft. When the car is at the first floor, 180 ft of cable are paid out, and effectively 0 ft are out when the car is at the top floor. How much work does the motor do just lifting the cable when it takes the car from the first floor to the top?
4. The rectangular tank shown here, with its top at ground level, is used to catch runoff water. Assume that the water weighs 62.4 lb/ft^3 .
a. How much work does it take to empty the tank by pumping the water back to ground level once the tank is full?
b. If the water is pumped to ground level with a (5 / 11)-horsepower (hp) motor (work output 250 ft-lb/sec), how long will it take to empty the full tank (to the nearest minute)?
c. Show that the pump in part (b) will lower the water level 10 ft (halfway) during the first 25 min of pumping.
d. The weight of water What are the answers to parts (a) and (b) in a location where water weighs 62.26 lb/ft^3 ? 62.59 lb/ft^3 ?
5. How much work would it take to pump oil from the tank in Example 5 to the level of the top of the tank if the tank were completely full?
6. a. Pumping milk Suppose that the conical container in Example 5 contains milk (weighing 64.5 lb/ft 3 ) instead of olive oil. How much work will it take to pump the contents to the rim?
b. Pumping oil How much work will it take to pump the oil in Example 5 to a level 3 ft above the cone’s rim?
7. Softball How much work has to be performed on a 6.5-oz softball to pitch it 132 ft / sec (90 mph)?
8. On June 11, 2004, in a tennis match between Andy Roddick and Paradorn Srichaphan at the Stella Artois tournament in London, England, Roddick hit a serve measured at 153 mi/h. How much work was required by Andy to serve a 2-oz tennis ball at that speed?
9. Drinking a milkshake The truncated conical container shown here is full of strawberry milkshake that weighs 4/9 oz/in^3 . As you can see, the container is 7 in. deep, 2.5 in. across at the base, and 3.5 in. across at the top (a standard size at Brigham’s in Boston). The straw sticks up an inch above the top. About how much work does it take to suck up the milkshake through the straw (neglecting friction)? Answer in inch-ounces.
10. Calculate the fluid force on one side of the plate in Example 6 using the coordinate system shown here.
11. A semicircular plate 2 ft in diameter sticks straight down into freshwater with the diameter along the surface. Find the force exerted by the water on one side of the plate.
12. The end plates of the trough shown here were designed to withstand a fluid force of 6667 lb. How many cubic feet of water can the tank hold without exceeding this limitation? Round down to the nearest cubic foot. What is the value of h?

# Section 6.4 Surface Areas

 int_{x=a}^b 2\pi f(x) sqrt{1+[dy/dx]^2} dx

#### Homework int_{x=0}^b f(x)dx

Graph each function in the given region and note the axis of rotation. Find and reduce the integrand. Evaluate numerically and exactly with an antiderivative if possible.

1. y=tan(x), 0 <= x <= pi/4 about the x-axis.
2. xy=1, 1 <= y <= 2 about the y-axis.
3. x=int_0^y tan(t)dt, 0 <= y <= pi/3 about the y-axis.
4. Find the lateral (side) surface area of the cone generated by revolving the line segment y = x/2, 0 \le x \le 4 , about the x-axis. Check your answer with the geometry formula
Lateral Surface Area = 1/2\timesbase circumference\timesslant height
5. y=x^3/9, 0 \le x \le 2 about the x-axis.
6. y=\sqrt{2x-x^2}, 0.5 \le x \le 1.5 about the x-axis.
7. x=y^3/3, 0 \le y \le 1 about the y-axis.
8. x=2\sqrt{4-y}, 0 \le y \le 15/4 about the y-axis.
9. x=(e^y+e^{-y})/2, 0 \le y \le \ln(2) about the y-axis.
10. Write an integral for the area of the surface generated by revolving the curve y = \cos(x), -\pi/2\le x\le\pi/2, about the x-axis.
11. Suppose the semicircle shown to the right is revolved about the​ x-axis to generate a sphere. Let AB be an arc of the semicircle that lies above an interval of length h on the​ x-axis. Show that the area swept out does not depend on the location of the interval.​ Therefore, it does not depend on the location of the slice.
12. Here is a schematic drawing of the 90-ft dome used by the U.S. National Weather Service to house radar in Bozeman, Montana. How much outside surface is there to paint (not counting the bottom)? Express the answer to the nearest square foot.

# Section 6.3 Arc Length Problems

Arc Length = $$\displaystyle \int_{x=a}^b \sqrt{1+\left[f'(x)\right]^2}\,dx$$

Arc Length = $$\displaystyle \int_{y=c}^d \sqrt{1+\left[g'(y)\right]^2}\,dy$$

#### Steps

1. find $$f'(x)$$
2. find $$\big[f'(x)\big]^2$$
3. find $$1+\big[f'(x)\big]^2$$
4. find $$\sqrt{1+\big[f'(x)\big]^2}$$
5. $$\displaystyle \int_{x=a}^b{\sqrt{1+\left[f'(x)\right]^2}\,dx}$$
1. hope to find an antiderivative
2. or evaluate numerically with a computer program

#### Homework

1. Sketch and find the length of the curve $$y=\frac{1}{3}(x^2+2)^\frac{3}{2}$$ from $$x=0$$ to $$x=3$$.
1. $$y’=\frac{1}{3}\frac{3}{2} 2x (x^2+2)^\frac{1}{2}=x (x^2+2)^\frac{1}{2}$$
2. $$(y’)^2 = x^2 (x^2+2)$$
3. $$1+(y’)^2 =1+ x^2 (x^2+2)$$
$$= x^4 + 2x^2 +1 = (x^2+1)^2$$
4. $$\sqrt{1+(y’)^2} = x^2+1$$
5. $$\int_0^3 \left(x^2+1\right) \,dx=\bigg[\frac{x^3}{3}+x\bigg]_0^3=12$$
2. Sketch and find the length of the curve $$y=x^\frac{3}{2}$$ from $$x=0$$ to $$x=4$$.
3. Sketch and find the length of the curve $$x=\frac{y^3}{3}+\frac{1}{4y}$$ from $$y=1$$ to $$y=3$$.
4. Sketch and find the length of the curve $$x=\frac{y^4}{4}+\frac{1}{8y^2}$$ from $$y=1$$ to $$y=2$$.
5. Sketch and find the length of the curve $$x=\int_0^x \sqrt{\sec^4(t)-1}\,dt$$, $$-\frac{\pi}{4}\le x \le\frac{\pi}{4}$$
1. $$x’=\sqrt{\sec^4(x)-1}$$
2. $$(x’)^2=\sec^4(x)-1$$
3. $$1+(x’)^2 =\sec^4(x)$$
4. $$\sqrt{1+(x’)^2} =\sec^2(x)$$
5. $$\displaystyle \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sqrt{1+(x’)^2}\,dx =\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\sec^2(x)\,dx=\bigg[\tan(x)\bigg]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}=2$$
6. $$y=x^2$$ , $$-1 \le x \le 2$$
Sketch the graph and set up the integral for the length of the curve. Use a calculator or program to approximate the curve’s length.
7. 13
8. 14
9. 15
10. 21
11. Find the arc length of an astroid. $$\displaystyle x^\frac{2}{3}+y^\frac{2}{3}=a^\frac{2}{3}$$ is a curve called an astroid, star-shaped. Solving for $$y$$ we get $$\displaystyle y=\left(a^\frac{2}{3}-x^\frac{2}{3}\right)^\frac{3}{2}$$. We will start by finding the arclength of half of the section in the first quadrant $$\frac{a\sqrt{2}}{4}$$.
In the graph to the right a = 3. You can see there that Geogebra had a difficult time drawing the cusps on each axis.
Solution
12. 23