Calculus II Lesson 11: Arc Length / Differential Equations Intro

1. Upcoming
2. Volumes
   1. Activity
3. Arc Length
   1. Example
   2. Example
   3. Questions?
4. Differential Equations
   1. Example
   2. Solution (if time)
5. Overview
   1. Solutions to differential equations
   2. Order
   3. Example
   4. Solution
   5. Wrap-up

Upcoming

• Today:
  • wrap up volumes and arc length
  • Differential equations intro
• Monday: HW (MyOpenMath)
• Next Friday: Presentation 2 (VoiceThread)
• Next week: quiz
• Exam 2: 2 weeks (March 16)
  • Bring a graphing calculator!

Volumes

• Disk method
• Washer method
• Shell method

More complicated regions:

• Can convert from $y = f(x)$ to $x = g(y)$
• Then use disk / shell / washer?

So we have actually 6 possible formulas:

1. $V = \int_a^b \pi (f(x)^2) dx$.
2. $V = \int_a^b \pi (g(y)^2) dy$.
3. $V = \int_a^b 2\pi x f(x) dx$.
4. $V = \int_a^b \pi (f(x)^2 - g(x)^2) dx$.
5. $V = \int_a^b \pi (g(y)^2 - h(y)^2) dy$.
6. $V = \int_a^b 2\pi y g(y) dy$.

Activity

Consider the following problems:

1. Revolve the region bounded by $y = x$, $y = x^2$, $x = 0$ and $x = 1$ around the $x$-axis.
2. Revolve the region bounded by $y = x^2$, $x = 0$, $x = 2$ and $y = 4$ around the $y$-axis.
3. Revolve the region bounded by $y = x^2$, $x = 0$, $x = 2$ and $y = 0$ around the $y$-axis.

In small groups: For each problem:

• Pick the method you will use for this problem.
• Pick the formula you will use for the problem.
• Solve one of the problems.
• Pick one student to present your problem on one of the boards.

Arc Length

Formula:

\[s = \int_a^b \sqrt{1 + (f^\prime)^2 } dx\]

• These integrals are tricky.
• Know how to do the algebra to set it up
• Then use a calculator (graphing or online)

First: review the notes / examples from last time.

• In particular, check out the notes on catenary arches.

Example

The path of a rock thrown off a 100 meter cliff (approximately) follows the curve $f(t) = 100 - 5t^2$, from $t = 0$ to $t = \sqrt{20}$ seconds. Find the length of the path the rock travels from $t = 0$ to $t = \sqrt{20}$. Round your answer to the nearest hundredth of a meter.

Example

Arc length of $f(t) = 100 - 5t^2$ from $t = 0$ to $t = \sqrt{20}$. Round your answer to the nearest hundredth of a meter.

\[\begin{align} f^\prime(x) = -10x \\ (f^\prime(x))^2 = 100x^2 \\ \int_0^{\sqrt{20}} \sqrt{1 + 100x^2}dx \end{align}\]

WolframAlpha: $\approx 100.25$ meters.

Questions?

Before we move on:

• Questions about volumes?
• Questions about arc length?
• Catenary arch example?

Differential Equations

• Scientific experiments: can be easier to observe changes to data
• Might not know the actual formula describing your data
• How a quantity changes may be related to the quantity itself
  • Ex: the more money you have, the more it earns in interest / investment

Sometimes:

• Can be easier to find the relationship between the change and the quantity
• Set up a differential equation

Example

• Suppose we are growing a bacteria culture.
• The larger the culture, the bigger the growth of the culture at any given time.
• Let $m(t)$ be the mass of the culture at time $t$. (Can easily measure its mass.)
• We notice: at any given $t$ (in seconds?), the culture grows at a rate $.05m(t)$
  • Idea: take lots of measurements, compare mass to previous mass. Look for a pattern.
• Initial mass: 2 grams. Mass in 10 minutes?

We have a differential equation:

\[m^\prime(t) = .05 m(t)\]

What functions are proportional to their own derivatives? Exponential functions! Guess $m(t) = Ae^{kt}$, and use $m^\prime$ and $m(0)$ to find these constants $A$ and $k$.

Solution (if time)

\[\begin{align} m(t) &= Ae^{kt} \\ m^\prime(t) &= Ake^{kt} \\ &= .05 m(t) \\ &= .05 Ae^{kt} \\ k &= .05 \end{align}\]

So we have:

• $m(t) = Ae^{.05t}$
• $m(0) = 2$

Find $A$?

\[\begin{align} Ae^{0} = 2 \\ A = 2 \end{align}\]

Solution:

• $m(t) = 2e^{.05t}$
• $m(10) = 2e^{.5} \approx 3.3$ grams.

Overview

• Any equation involving the derivative of a quantity is known as a differential equation.
• Example: $y^\prime = x$.
• A function $y = f(x)$ satisfying that equation is called a solution to the differential equation.
• In general: differential equations do not have unique solutions:
  • $y = \frac{x^2}{2}$ is a solution to $y^\prime = x$.
  • $y = \frac{x^2}{2} + 1$ is another solution.

Solutions to differential equations

• We look for a general form for a solution:
  • $y = \frac{x^2}{2} + C$.
• If we have an initial condition, we can find an exact solution.
  • Suppose $y^\prime = x$ and $y(0) = -1$.
  • Then $y = \frac{x^2}{2} + C$.
  • $-1 = \frac{0^2}{2} + C$
  • $C = -1$.
  • So: $y = \frac{x^2}{2} - 1$.

Order

The order of a differential equation is the highest derivative that appears in the equation. Exercise: Identify the orders of the following:

1. $y^\prime = y + x$
2. $(y^\prime)^2 = (y - x)^2$
3. $y^{\prime\prime} = -y$
4. $y^{\prime\prime\prime} + (y^\prime)^5 = y$

We will mostly focus on first-order differential equations, but some simple higher order differential equations can be solved.

Example

A ball is thrown straight into the air with an initial velocity of 5 meters per second. It is acted on by a constant, downward force of gravity, causing an acceleration of $-9.8$ $m/s^2$.

1. Find a formula $v(t)$ for the velocity at time $t$ seconds after the ball is thrown.
2. How many seconds after the ball is thrown does it reach its maximum height?
3. If the ball is thrown at a height of 2 meters above the ground, what is the maximum height reached?

Solution

Since $v^\prime(t) = a(t)$, and $a(t) = -9.8$, we know:

• $v^\prime(t) = -9.8$
• $v(0) = 5$.

So: $v(t) = -9.8t + C$. Since $v(0) = 5$, we can compute $5 = -9.8(0) + C$, so $C = 5$.

So our formula for the velocity is $v(t) = -9.8t + 5$. Going back to Calc I:

• the ball reaches its maximum height when its upward velocity is 0
• $v(t) = -9.8t + 5 = 0$
• Solve for $t$: $t = \frac{-5}{-9.8} \approx .51$ seconds.
• Derivative of height: velocity.
• $h^\prime(t) = -9.8t + 5$.
• So: $h(t) = -4.9t^2 + 5t + C$.
• $h(0) = 2$ means $-4.9(0) + 5(0) + C = 2$
• So $C = 2$

Therefore $h(t) = -4.9t^2 + 5t + 2$. Then $h(\frac{5}{9.8}) \approx 3.28$ meters. This is basically the exam 1 extra credit problem.

Wrap-up

• We really solved a second-order differential equation:
• $h^{\prime\prime}(t) = -9.8$
• Initial condition: $h(0) = 2$ and $h^\prime(0) = 5$.
• We did this by splitting it up into two steps:
  • First find $h^\prime(t)$
  • Then find $h(t)$.