Calculus I Lesson 14: Related Rates
Presentations
Quiz?
Related Rates
We are now ready to start our unit on applications of derivatives. Our first set of applications come from related rates. These are problems where multiple rates of change are related to one another in some equations.
Example: A 10-ft ladder is leaning against a wall. If the top of the ladder slides down the wall at a rate of 2 ft/sec, how fast is the bottom moving along the ground when the bottom of the ladder is 5 ft from the wall?
Solution
- $x$ = horizontal distance from the bottom of the ladder to the wall
- $y$ = distance from the top of the ladder to the floor
- $x^2 + y^2 = 100$.
What do we know? What do we want to know?
- We know that $y^\prime = -2$.
- We know what $x = 5$.
- We can figure out $y$ (Pythagorean theorem).
- We need to figure out $x^\prime$?
There are a few ways we can do this. We can write equations for $y(t)$, the height of the top of the ladder at time $t$, and $x(t)$, the distance from the bottom of the ladder to the wall at time $t$. Then we can compute $x^\prime$. This turns out to be really hard!
A better idea would be to reason implicitly. Use $x^2 + y^2 = 100$, take the derivative with respect to $t$ of both sides. We get \(2xx^\prime + 2yy^\prime = 0.\) Again, we know that $x = 5$ and $y^\prime = -2$. Using the Pythagorean Theorem, we get $y = \sqrt{75}$. So we can fill in: \(2(5)x^\prime + 2(\sqrt{75})(-2) = 0.\) Solving for $x^\prime$, we get $x^\prime = 2\sqrt{3}$ ft / sec.
General Strategy
- Draw a figure, declare variables
- Relate the variables with an equation
- Differentiate both sides of the equation (use chain rule)
- Substitute and solve
YouTube Playlist
Related Rates is a challenging topic, so I suggest the following resource that I found helpful: patrickJMT’s playlist of related rates problems. There are 8 videos in this playlist, going through a few different types of related rates problems.
Example
An airplane is flying overhead at a constant elevation of 4000ft. A man is viewing the plane from a position 3000ft from the base of a radio tower. The airplane is flying horizontally away from the man. If the plane is flying at the rate of 600ft/sec, at what rate is the distance between the man and the plane increasing when the plane passes over the radio tower?
In small groups:
- Draw a picture. Label it with variables.
- Relate your variables with an equation.
- Differentiate implicitly.
Funnel Example
Water is draining from the bottom of a cone-shaped funnel at the rate of 0.05 $\mathrm{ft}^3 / \mathrm{sec}$. The height of the funnel is 2 ft and the radius at the top of the funnel is 1ft. At what rate is the height of the water in the funnel changing when the height of the water is $\frac{1}{2}$ ft?
To declare our variables:
- $r$ = radius at the surface level
- $h$ = height at the surface level
Then we can relate these using some basic geometry. Since the height of the funnel is 2 and the radius at the top of the funnel is 1, we see that $h = 2r$, or $r = \frac{h}{2}$. Moreover we can relate the volume to the height using the formula for volume of a cone: $V = \frac{1}{3} \pi r^2 h$, or $V = \frac{1}{12} \pi h^3$.
Then differentiate both sides: $V^\prime = (\frac{1}{12} \pi) \cdot 3h^2 h^\prime$. Now let’s plug in what we know: \(-0.05 = \frac{\pi}{4} (\frac{1}{2})^2 h^\prime.\)
Now simplify and solve for $h^\prime$:
\[\begin{align} -0.05 = \frac{\pi}{16} h^\prime \\ h^\prime = \frac{(-0.05) \times 16}{\pi} \approx -0.255 \mbox{ ft / sec} \end{align}\]As a follow-up: what about when the funnel is full ($h = 2$)? The steps end up being exactly the same, except with $h = 2$ at the end.
\[\begin{align} V^\prime = (\frac{1}{12} \pi) \cdot 3h^2 h^\prime \\ -0.05 = \frac{\pi}{4} (2^2) h^\prime \\ -0.05 = \pi h^\prime \\ h^\prime = \frac{-0.05}{\pi} \approx -0.016 \mbox{ ft / sec} \end{align}\]So the rate of change of the height is slower when the funnel is full. Does this make sense?
Another Example
(Section 4.1 #10):
A 6-ft-tall person walks away from a 10-ft lamppost at a constant rate of 3ft/sec. What is the rate that the tip of the shadow moves away from the pole when the person is 10ft away from the pole?
In small groups (again):
- Draw a picture. Label it with variables.
- Relate your variables with an equation. Hint: Your picture should have similar triangles. The corresponding side lengths can be put in proportion.
- Differentiate implicitly.
- Solve.
Solution
Let $x$ be the distance from the person to the tip of their shadow, and let $y$ be the distance from the person to the base of the lamppost. Then $y = 10$ and $y^\prime = 3$ at the moment we are looking for.
Set up a proportion: $\frac{x}{6} = \frac{x + y}{10}$. Cross multiply to get $10x = 6(x + y)$.
Now go back to our original equation and differentiate both sides implicitly: \(10x^\prime = 6(x^\prime + y^\prime).\) Plug in $y^\prime = 3$ and solve:
- $10x^\prime = 6x^\prime + 18$.
- $4x^\prime = 18$.
- So $x^\prime = 4.5$ feet per second.
Recall that the question is asking for the speed at which the tip of the shadow moves away from the pole. What is $x^\prime$ in the context of this problem?
Optimization
Continuity is a powerful concept:
- Transforms properties of “domains” $\to$ properties of “ranges”
- Connected domains $\to$ connected ranges
- Closed and bounded domains $\to$ closed and bounded ranges
Instead of just thinking about functions as giving us an output whenever we have some input, we should also think about how the function transforms a domain.
Extreme Value Theorem
Theorem: If $f$ is a continuous function on a closed interval $[a, b]$, then it has a maximum and a minimum in $[a, b]$.
Counterexample?
$f(x) = \tan(x)$ on $(-\pi/2, \pi/2)$:
What happens on $(-\pi/2, \pi/2)$? Is there a maximum value? Minimum value? Why not?
Finding maxima/ minima
Finding maxima / minima is easiest when $f$ is differentiable. Suppose $f(x)$ is differentiable on $(a, b)$ and continuous on $[a, b]$. Then there are 3 options for where maxima / minima can be:
- $x = a$ or $x = b$ (endpoints)
- When $f^\prime(x) = 0$
- When $f^\prime(x)$ DNE
Points satisfying (2) or (3) are called critical points of $f$.
Critical Points
Example
Find the minimum value of $f(x) = x^3 - x^2$ on the interval $[-2, 2]$:
\[\begin{align} f^\prime(x) = 3x^2 - 2x = 0 \\ x(3x - 2) = 0 \\ x = 0 \mbox{ or } x = 2/3 \end{align}\]Points to check: $x = -2, 0, 2/3, 2$
- $f(-2) = -12$
- $f(0) = 0$
- $f(2/3) = -4/27$
- $f(2) = 4$
The minimum value occurs at $x = -2$. The minimum value of the function is -12.
Take a look at the graph:
Local Minima / Maxima
- $x = 0$ is a local maximum of $f(x)$
- $x = 2/3$ is a local minimum of $f(x)$
How can you tell when a point is a local min or local max?
Upcoming
- DeltaMath due tonight
- Topic + Sources for final paper due tonight
- Homework 6, due next Monday (10/27):
- Section 3.9 #332, 334, 343
- Section 4.1 #12, 13
- DeltaMath due next Thursday (10/30)
- Exam 2 on November 3
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