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Perfect is the enemy of the good.
The derivative of a function at a chosen input value, when it exists, is the slope of the tangent line to the graph of the function at that point. It is the instantaneous rate of change —the exact ratio of the instantaneous change in the dependent variable ($y$) to that of the independent variable ($x$).
Definition. A function $f(x)$ is differentiable at a point $a$ of its domain if its domain contains an open interval containing $a$, and the following limit exists: $\boxed{f'(a) = L = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}}$
Formal Definition: More rigorously, for every positive real number $\epsilon > 0$, there exists a positive real number $\delta > 0$, such that for every $h$ satisfying $0 < |h| < \delta$, the difference between the secant slopes at points sufficiently near a are guaranteed to be ε-close to the derivative: $\left| L - \frac{f(a+h) - f(a)}{h} \right| < \epsilon$
Geometrically, $f'(x)$ is the slope of the tangent line to the graph at the point $(x, f(x))$. Physically, $f'(x)$ represents the instantaneous rate of change.
The average rate of change calculates how the output changes over a wide interval of inputs. This can be visualized as the slope of a secant line cutting through two points on a graph. In differential calculus, we shrink that interval to zero to find the instantaneous rate.
Example (Population Dynamics). Average Growth Rate = $\frac{\text{Total Births} - \text{Total Deaths}}{\text{Years}}$.
By taking the limit as the time interval approaches zero, the instantaneous rate of change in population is the derivative of the population function with respect to time, denoted as $\frac{dP}{dt}$.
Whether you are calculating the speed of a falling object, the inflation of a balloon, or the sliding of a ladder, follow these general steps:
During a full-moon night run, an observer -with plenty of time and nothing better to do- is standing 80 meters away from a long, straight fence when he notices a runner running along it, getting closer to him. “What the fork is she doing running so early?”, he wonders.
He points his flashlight at her, tracking her as she runs. When the straight-line distance between him and the runner is exactly 100 meters, she is running at 9 m/s. At this very moment, at what rate is he turning the flashlight to keep her illuminated?
Solution:
Understand and Represent the Problem (Figure C)
To solve a related rates problem, we use the chain rule to relate the rates of change of different variables with respect to time ($t$).
Let’s define our variables:
$\star$ $y = 80$ m, the constant distance from the observer to the closest point on the fence.
$\star$ $d = 100$ m, the hypotenuse, or distance between the observer and the runner at the given instant.
$\star$ $x$ = the distance from the closest point on the fence to the runner.
$\frac{dx}{dt} = -9$ m/s negative because the distance $x$ is decreasing as she gets closer and closer.
Goal: Find $\frac{d\theta}{dt}$ (the rate the angle is changing) when $d = 100$ m.
Relate all quantities in a single equation
Using the Pythagorean theorem ($x^2 + 80^2 = 100^2$), we can determine that at this exact moment, $x = 60$ m.
Using basic trigonometry, we relate the angle to the sides we know: $\boxed{\tan(\theta) = \frac{x}{80}}$
Differentiate the equation. Now, differentiate both sides with respect to time ($t$): $ \frac{d}{dt}(\tan(\theta)) = \frac{d}{dt}\left(\frac{x}{80}\right) $
$ \sec^2(\theta) \frac{d\theta}{dt} = \frac{1}{80} \frac{dx}{dt} $
Solve for the Target Rate.
We know $\frac{dx}{dt} = -9$, but we need to find $\sec(\theta)$ before we can isolate $\frac{d\theta}{dt}$.
Since $\sec(\theta) = \frac{\text{Hypotenuse}}{\text{Adjacent}}$, we get: $ \sec(\theta) = \frac{100}{80} = \frac{5}{4}$
Substitute the known values into our differentiated equation: $ \left(\frac{5}{4}\right)^2 \frac{d\theta}{dt} = \frac{1}{80}(-9)$
Simplify, $\frac{25}{16} \frac{d\theta}{dt} = -\frac{9}{80}$
$ \frac{d\theta}{dt} = -\frac{9}{80} \cdot \frac{16}{25} = -\frac{9 \cdot 16}{80 \cdot 25} = -\frac{144}{2000} = \mathbf{-0.072 \text{ rad/s}}$
Solution. The observer is turning the flashlight at a rate of -0.072 radians per second.
A rocket from Elon Musk’s company SpaceX is launched vertically upward from a point 2 miles west of an observer on the ground. What is the speed of the rocket when the angle of elevation of the observer’s line of sight to the rocket is 50° and is increasing at 5° per second?
Understand and Represent the Problem (Figure 2.c)
Let’s define our variables:
$\star$ Base = 2 miles (constant distance from the observer to the launch pad).
$\star$ $h$ = the height of the rocket.
$\star$ $\theta = 50^\circ$ (the angle of elevation at the given instant).
$\star$ $\frac{d\theta}{dt} = 5^\circ/\text{s}$.
Goal: Find $\frac{dh}{dt}$ (the speed of the rocket) when $\theta = 50^\circ$.
In calculus, all angles and their rates of change must be in radians before you compute them. To convert degrees to radians, multiply by $\frac{\pi}{180}$.
$\star$ $\theta = 50^\circ \cdot \frac{\pi}{180} = \frac{5\pi}{18}$ radians.
$\star$ $\frac{d\theta}{dt} = 5^\circ \cdot \frac{\pi}{180} = \frac{\pi}{36}$ rad/s.
Relate all quantities in a single equation
We relate the height to the angle using the tangent function: $\boxed{\tan(\theta) = \frac{h}{2}}$
Differentiate the equation.
Differentiate both sides with respect to $t$: $ \sec^2(\theta) \frac{d\theta}{dt} = \frac{1}{2} \frac{dh}{dt} $
Multiply by 2 to isolate our goal ($\frac{dh}{dt}$): $ \frac{dh}{dt} = 2 \sec^2(\theta) \frac{d\theta}{dt} $
Solve for the Target Rate.
Substitute the values (using radians for the rate): $ \frac{dh}{dt} = 2 \cdot \sec^2(50^\circ) \cdot \frac{\pi}{36} $
Since $\sec(50^\circ) \approx 1.5557$, then $\sec^2(50^\circ) \approx 2.420$.
$ \frac{dh}{dt} \approx 2 \cdot 2.420 \cdot 0.08726 \approx \mathbf{0.422 \text{ miles per second}} $
Final Solution.
To make this number easier to grasp, we can convert miles per second (m/s) to miles per hour (mph) by multiplying by 3,600 (the number of seconds in an hour): $ 0.422 \cdot 3600 \approx \mathbf{1,520 \text{ mph}} $
The Problem:
A block of ice in the form of a perfect cube with side length $x$ meters is melting at a rate of $9.72\text{ m}^3\text{ per second}$ (because of global warming —so pay your climate taxes!). Assuming the block retains its cube shape as it melts, how fast is the side length shrinking at the exact instant when $x = 12\text{ m}$?
The Solution:
Understand and Represent the Problem.
Let’s define our variables:
$\star$ Let $V$ be the volume of the cube and $x$ be the length of its sides.
$\star$ Rate of volume change: $\frac{dV}{dt} = -9.72\text{ m}^3\text{/s}$ (Negative because it is melting/shrinking).
Target: Find $\frac{dx}{dt}$ when $x = 12\text{ m}$.
Relate all quantities in a single equation
The volume of a cube is: $V = x^3$
Differentiate with Respect to Time ($t$).
Using the chain rule, we differentiate both sides: $\frac{dV}{dt} = 3x^2 \frac{dx}{dt}$
Substitute and Solve:
We want to isolate our target rate, $\frac{dx}{dt}$: $\frac{dx}{dt} = \frac{\frac{dV}{dt}}{3x^2}$
Plug in our known rate and our snapshot value ($x = 12$): $\frac{dx}{dt} = \frac{-9.72}{3(12)^2} = \frac{-9.72}{3(144)} = \frac{-9.72}{432}$
Therefore, $\frac{dx}{dt} = -0.0225\text{ m/s}$
Conclusion: At the instant the side is 12 meters long, it is shrinking at a rate of $0.0225\text{ m/s}$.
The Problem:
A container is shaped like an inverted right circular cone where the angle $\alpha = \tan^{-1}\left(\frac{1}{2}\right)$. Initially empty, water is pumped in at a constant rate of $14\text{ cubic meters per minute}$. If the water has a depth of $x$ meters, find the rate at which the depth $x$ is increasing when $x = 8\text{ m}$ (Figure A).
The Solution:
Understand and Represent the Problem.
Let’s define our variables:
$\star$ Let $V$ be the volume of the water, $x$ be the depth (height), and $r$ be the radius of the water’s surface.
$\star$ Rate of volume change: $\frac{dV}{dt} = 14\text{ m}^3\text{/min}$.
$\star$ Angle relationship: $\alpha = \tan^{-1}\left(\frac{1}{2}\right)$, which means $\tan(\alpha) = \frac{1}{2}$.
Target: Find $\frac{dx}{dt}$ when $x = 8\text{ m}$.
Relate all quantities in a single equation
First, use the angle to relate the radius to the depth. In a right triangle formed by the cone’s cross-section, $\tan(\alpha) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{r}{x}$.
$\frac{r}{x} = \frac{1}{2} \implies r = \frac{x}{2}$
Now, substitute this into the volume formula for a cone ($V = \frac{1}{3}\pi r^2 x$): $V = \frac{1}{3}\pi \left(\frac{x}{2}\right)^2 x = \frac{1}{3}\pi \left(\frac{x^2}{4}\right) x = \frac{\pi}{12} x^3$. Therefore, $\boxed{V = \frac{\pi}{12} x^3}$
Differentiate with Respect to Time ($t$): $\frac{dV}{dt} = \frac{\pi}{12} \cdot 3x^2 \frac{dx}{dt} = \frac{\pi}{4} x^2 \frac{dx}{dt}$
Substitute and Solve: Isolate $\frac{dx}{dt}$: $\frac{dx}{dt} = \frac{\frac{dV}{dt}}{\frac{\pi}{4} x^2} = \frac{4 \cdot \frac{dV}{dt}}{\pi x^2}$
Plug in our values ($x = 8$ and $\frac{dV}{dt} = 14$): $\frac{dx}{dt} = \frac{4(14)}{\pi (8^2)} = \frac{56}{64\pi} = \frac{7}{8\pi}$
$\frac{dx}{dt} \approx 0.278\text{ m/min}$
Note: The rate is in minutes, matching the input volume rate.
Conclusion: The water level is rising at approximately $0.28\text{ m/min}$.
The Problem:
Police are parked $30\text{ feet}$ off the side of the road, salivating over their strict daily ticket quota. Their radar gun spots your car approaching, showing the distance between your car and the radar gun shrinking at $80\text{ feet per second}$ when you are exactly $50\text{ feet}$ away from the gun. The speed limit is $65\text{ mph}$. Are you speeding?
The Solution:
Understand and Represent the Problem (Figure 2.a).
$\star$ Let’s define our variables:
$\star$ Let $x$ be your distance down the road from the point directly across from the police.
$\star$ Let $D$ be the straight-line distance from your car to the radar gun (the hypotenuse).
$\star$ Distance from road = $30\text{ ft}$ (Police are parked 30 feet off the side of the road).
$\star$ Rate of change of distance to radar: $\frac{dD}{dt} = -80\text{ ft/s}$ (It’s negative because you are getting closer).
Target: Find your actual speed on the road, $\left|\frac{dx}{dt}\right|$, when $D = 50\text{ ft}$.
Evaluate Missing Variables for the “Snapshot”:
Using the Pythagorean theorem ($x^2 + 30^2 = D^2$), find $x$ when $D = 50$: $x = \sqrt{50^2 - 30^2} = \sqrt{2500 - 900} = \sqrt{1600} = 40\text{ ft}$
Relate all quantities in a single equation
Using the Pythagorean theorem: $x^2 + 30^2 = D^2$
Differentiate with Respect to Time ($t$): $2x \frac{dx}{dt} = 2D \frac{dD}{dt}$
$x \frac{dx}{dt} = D \frac{dD}{dt}$
Substitute and Solve: $40 \frac{dx}{dt} = 50 (-80)$
Simplify, $40 \frac{dx}{dt} = -4000$
Therefore, $\frac{dx}{dt} = -100\text{ ft/s}$
Your actual speed is $100\text{ ft/s}$.
Solution. Let’s convert that to mph to see if you get a ticket: $100\text{ ft/s} \cdot \left(\frac{3600\text{ s}}{1\text{ hr}}\right) \cdot \left(\frac{1\text{ mile}}{5280\text{ ft}}\right) \approx 68.18\text{ mph}$
Conclusion: Yes, you are speeding, but not for much. $68.2\text{ mph} > 65\text{ mph}$. The police got their quota for the day. You are making them happy -Just saying'!
The Problem:
The population of a city is tripling every $5\text{ years}$. If its current population is $10,000$, what will be its approximate population $2\text{ years}$ from now?
The Solution:
Understand and Represent the Problem.
Let $P(t)$ be the population (in thousands) $t$ years from now.
Current population: $P(0) = 10$
Population in 5 years: $P(5) = 30$ (since it triples)
Find the Average Rate of Change:
To estimate the current growth rate $P'(0)$, we can use the slope of the secant line between $t=0$ and $t=5$: $P'(0) \approx \frac{P(5) - P(0)}{5 - 0} = \frac{30 - 10}{5} = \frac{20}{5} = 4$
This means the population is growing by roughly $4,000\text{ people per year}$.
Apply Linear Approximation:
For small time steps ($h$), a function can be estimated using its tangent line: $f(a + h) \approx f(a) + f'(a) \cdot h$
In this scenario, our starting point $a = 0$ and our time step $h = 2$:
$P(2) \approx P(0) + P'(0) \cdot 2$
$P(2) \approx 10 + 4(2) = 10 + 8 = 18$
Conclusion: Using linear approximation, the population will be roughly $18,000$ in two years.
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