Thursday, May 10, 2012

Calculus Revisited #21: The Catch All

Welcome to Part 21 of our 21 part series on Calculus Series. Here are some interesting facts about calculus that were not discussed in the first 20 parts, but are worth paying attention to:

Implicit Differentiation

The key here is that when you are differentiating, you differentiate with respect to the specific variable. All other variables are treated as constants.

So for the function f(x,y):

d/dx f(x,y) means differentiate with respect to x, y is held constant

d/dy f(x,y) means differentiate with respect to y, x is held constant

Example: Let f(x,y) = ln(x + 2y) + y^2

d/dx f(x,y)
= d/dx ln(x + 2y) + d/dx y^2
= 1/(x + 2y) + 0
= 1/(x + 2y)

d/dx f(x,y)
= d/dy ln(x + 2y) + d/dy y^2
= 2/(x + 2y) + 2y

Arc Length

The arc length of a function f(x) is given by the integral:

b
∫ √( 1 + (f'(x))^2 ) dx
a

Usually, you will be finding the arc length by numerical methods and/or the calculator.

A numerical example: The arc length of f(x) = sin x from x = 0 to x = π/2 (to seven decimal places)

f(x) = sin x
f'(x) = cos x
(f'(x))^2 = cos^2 x

Then:

π/2
∫ (1 + cos^2 x) dx ≈ 1.9100989 (by calcualtor)

A Numerical Method of Integration: Simpson's Rule

For n partitions (n is even):

b
∫ f(x) dx
a

≈ (b - a)/3n * (f(a) + 4*f(x1) + 2*f(x2) + 4*f(x3) + 2*f(x4) + ... + 4*f(x_2k+1) + 2*f(x_2k+2) + .... + 4*f(x_n-1) + f(b))

where x_k = a + (b - a)/n * k

Error:

At maximum (b - a)^5/(180 * (2n)^4) * max|f''''(x)|

In general, the more complex the integral, the more terms needed

Example:
Use the Simpsons rule to calculate

2
∫ e^x dx with n = 4 (7 decimal places)
1

(b - a)/n = (2 - 1)/4 = 1/4

The integral is approximately:

≈ (2 - 1)/(3 * 4) * (f(1) + 4 * f(1.25) + 2 * f(1.5) + 4 * (1.75) + f(2))
≈ 4.6708749

(actual value is about 4.6707743)

Area Between Curves

Let f(x) and g(x) be two functions where f(x) ≥ g(x). The area between curves f(x) and g(x) is:

b
∫ f(x) - g(x) dx
a

Often a and b will be intersection points of f(x) and/or g(x).

Example:

Let f(x) = x^2 and g(x) = x. Find the area between the curves from x = 0 to x = 5.

When 0 ≤ x ≤ 5, f(x) ≥ g(x). Then the area between the curves is:

5
∫ x^2 - x dx = 175/6 ≈ 29.1666667
0

Volume of a Solid - Disk Method

The volume of a solid of revolution - method of discs revolving around the x-axis is:

b
∫ π * (f(x))^2 dx
a

Like arc length, you may be finding the volumes numerically and/or by calculator (like I will do by this example).

Example: Let f(x) = sin x from x = 0 to x = π

Then the volume is:

π
∫ π * sin^2 x dx ≈ 4.9348022 = π^2/2
0

If you are finding a volume with discs revolving around the y-axis then the volume would be:

b
∫ π * (g(y))^2 dy
a

where a and b are y-values.

Polar Equations and Polar Integral

To convert functions to their polar form and back, use the following:

x = r cos θ
y = r sin θ

r^2 = x^2 + y^2
θ = atan (y/x)

The Polar Integral is:

θ2
∫ 1/2 * (r(θ))^2 dθ
θ1

Example: Let r(θ) = 2 θ . Find the polar integral from θ = 0 to θ = 2π

Area:


∫ 1/2 * (2 θ)^2 dθ = 16 π^3/ 3 ≈ 165.3668090
0


Parametric Equations

Parametric equations are in the form of ( x(t), y(t) ), where t is the independent variable and x and y are dependent variables. Each part can be separately differentiated and integrated.

Arc length of a parametric curve from t = a to t = b:

b
∫ √( (x'(t))^2 + (y'(t))^2 ) dt
a

Let:
x(t) = 2t
y(t) = t^3

Find (x, y) at t = 0 and t = 1. Find the arc length.

x(0) = 0, y(0) = 0
x(1) = 2, y(1) = 1

Arc length:

x'(t) = 2
y'(t) = 3t^2

(x'(t))^2 = 4
(y'(t))^2 = 9t^4

1
∫ √(4 + 9t^4) dt ≈ 2.3650656
0

Tips for Sketching a Curve y = f(x)

Find out where the extrema of f(x) are. Recall the critical points are found when:

f'(x) = 0

Let c be the critical points.

If f''(c) > 0, then x = c is strict minimum.

If f''(c) < 0, then x = c is a strict maximum.

If f'(c) > 0 when both f'(x) > 0 for x < c and x > c OR
f'(c) < 0 when both f'(x) < 0 for x < c and x > c,
then x = c is an inflection point.

When f'(x) > 0, f(x) is increasing.

When f'(x) < 0, f(x) is decreasing.

When f''(x) > 0, f(x) is concave up (holding water).

When f''(x) < 0, f(x) is concave down (spilling water).

If f(x) → +∞ or -∞ when x → a, then f(x) has a vertical asymptote at x = a.

If f(x) → b when x → +∞ or -∞, then f(x) has a horizontal asymptote at y = b.

Thank you as always. I hope you enjoyed the Calculus Revisited Series.

To all the students who are in calculus, good luck in finals!

Until next time,

Eddie

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