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Thinking, wishing, or wanting something does not make it real. Dreams are only dreams and nothing more. Life is not fair nor easy. A lie repeated a thousands times is still a lie. Propaganda and narratives cannot change the facts. Either you deal with what reality is, whether you like it or not, or you can be sure that reality is going to deal with you and chances are that it won’t be nice, JustToThePoint, Anawim, #justtothepoint.
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:
The Problem:
A cone with a height of $10\text{ m}$ and a base radius of $3\text{ m}$ is held vertex down and filled with sand. A hole in the bottom allows the sand to escape at a rate of $0.5\text{ m}^3\text{/s}$. Find the rate at which the depth of the sand is decreasing when the depth is exactly $2\text{ m}$.
The Solution:
Understand the problem, draw a diagram, identify Knowns and Unknowns.
Let $V$ be the volume of the sand, $h$ be its height/depth, and $r$ be the radius of the sand’s surface.
Rate of volume change: $\frac{dV}{dt} = -0.5\text{ m}^3\text{/s}$ (Sand is escaping at a rate of $0.5\text{ m}^3\text{/s}$).
Target: Find $\frac{dh}{dt}$ when $h = 2\text{ m}$ (the depth is exactly $2\text{ m}$).
Relate all quantities in a single equation.
Because the sand forms a smaller cone inside the larger one, we use similar triangles to relate the radius to the height.
Their proportions must match the full cone: $\frac{r}{h} = \frac{3}{10} \implies r = \frac{3}{10}h$
Substitute this into the cone volume formula ($V = \frac{1}{3}\pi r^2 h$), and we get: $V = \frac{1}{3}\pi \left(\frac{3}{10}h\right)^2 h = \frac{1}{3}\pi \left(\frac{9}{100}h^2\right) h = \frac{3\pi}{100} h^3$.
Therefore, $\boxed{V = \frac{3\pi}{100} h^3}$
Differentiate with Respect to Time ($t$): $\frac{dV}{dt} = \frac{3\pi}{100} \cdot 3h^2 \frac{dh}{dt} = \frac{9\pi}{100} h^2 \frac{dh}{dt}$
Substitute and Solve. Isolate $\frac{dh}{dt}$: $\frac{dh}{dt} = \frac{\frac{dV}{dt}}{\frac{9\pi}{100} h^2} = \frac{100 \cdot \frac{dV}{dt}}{9\pi h^2}$
Plug in our snapshot values ($h = 2$ and $\frac{dV}{dt} = -0.5$): $\frac{dh}{dt} = \frac{100(-0.5)}{9\pi (2^2)} = \frac{-50}{9\pi (4)} = \frac{-50}{36\pi} = \frac{-25}{18\pi}$
$\frac{dh}{dt} \approx -0.44\text{ m/s}$
Conclusion: When the sand is $2\text{ meters}$ deep, the depth is decreasing at a rate of roughly $0.44\text{ m/s}$.
Problem: A ladder 10 m long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1 m/s, how fast is the top of the ladder sliding down the wall when the bottom of the ladder is 6 m from the wall?
Understand the problem, draw a diagram (Figure 1), identify Knowns and Unknowns.
$x = 6$ m (distance from wall).
$\frac{dx}{dt} = 1$ m/s (rate the bottom of the ladder slides away from the wall).
$z = 10$ m (length of ladder, constant).
Goal: Find $\frac{dy}{dt}$ when $x = 6$ (the bottom of the ladder is 6 m from the wall). \
Relate all quantities in a single equation.
Using the Pythagorean theorem, we get $x^2 + y^2 = 10^2$.
Differentiate with Respect to Time.
$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$
Substitute and Solve.
Calculate $y$ at this specific moment $x = 6$: $x^2 + y^2 = 10^2 \implies 6^2 + y^2 = 100 \implies y = \sqrt{100-36} = \sqrt{64} = 8 \text{ m}$
Plug in our known values to solve for the unknown rate: $2(6)(1) + 2(8)\frac{dy}{dt} = 0$
$12 + 16\frac{dy}{dt} = 0 \implies 16\frac{dy}{dt} = -12 \implies \frac{dy}{dt} = -\frac{3}{4}$
Conclusion. The top of the ladder is sliding down the wall at a rate of -0.75 m/s.
Problem: Car A is traveling west at 50 mph and Car B is traveling north at 60 mph. Both are headed for the intersection of the two roads. At what rate are the cars approaching each other when Car A is 0.3 miles and Car B is 0.4 miles from the intersection? (Note: Units adjusted from meters to miles for consistency with mph).
Understand the problem, draw a diagram (Figure 2), identify Knowns and Unknowns.
$A = 0.3$ miles, $\frac{dA}{dt} = -50$ mph (negative because distance to intersection is decreasing)
$B = 0.4$ miles, $\frac{dB}{dt} = -60$ mph
Goal: Find $\frac{dC}{dt}$ (rate of change of distance between them).
Relate all quantities in a single equation. $A^2 + B^2 = C^2$
Differentiate with Respect to Time. $2A\frac{dA}{dt} + 2B\frac{dB}{dt} = 2C\frac{dC}{dt}$
Substitute and Solve.
First, find the current distance $C$ between the cars: $A^2 + B^2 = C^2 \implies 0.3^2 + 0.4^2 = C^2 \implies C = \sqrt{0.25} = 0.5 \text{ miles}$
Plug in the values: $2(0.3)(-50) + 2(0.4)(-60) = 2(0.5)\frac{dC}{dt}$
Simplify, $-30 - 48 = 1\frac{dC}{dt} \implies \frac{dC}{dt} = -78$
Conclusion. The cars are approaching each other at a rate of 78 mph.
Problem: The volume of a sphere is increasing at a constant rate of $12.8\pi$ m³/s. Find the radius of the sphere at the instant when the radius is increasing at a rate of 0.2 m/s.
Understand the problem, draw a diagram, identify Knowns and Unknowns.
$\frac{dV}{dt} = 12.8\pi$
$\frac{dr}{dt} = 0.2$
Goal: Find $r$.
Relate all quantities in a single equation. The volume of a sphere is related to its radius by: $\boxed{V = \frac{4}{3}\pi r^3}$
Differentiate with respect to $t$ using the chain rule: $\frac{dV}{dt} = \frac{4}{3}\pi (3r^2) \frac{dr}{dt} \implies \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$
Substitute the known rates to find $r$: $12.8\pi = 4\pi r^2 (0.2)$
Simplify, $12.8\pi = 0.8\pi r^2$.
Therefore, $r^2 = \frac{12.8}{0.8} = 16$. Since a radius must be positive, $r = 4$.
Conclusion. The radius is 4 m.
Problem: A police car, approaching a right-angled intersection from the north, is chasing a fast sports car that has turned the corner and is now moving straight east. When the police car is 0.6 miles north of the intersection and the car is 0.8 miles to the east, the police determine with a radar gun that the distance between them and the car is increasing at 20 mph. If the police car is moving at 60 mph at the instant of measurement, what is the speed of the car?
Understand the problem, draw a diagram (Figure E), identify Knowns and Unknowns.
$y = 0.6$ miles, $\frac{dy}{dt} = -60$ mph (distance is decreasing).
$x = 0.8$ miles, $\frac{dx}{dt} = ?$
$\frac{dz}{dt} = 20$ mph (distance between them is increasing)
Goal: Find $\frac{dx}{dt}$.
Relate all quantities in a single equation. $x^2 + y^2 = z^2$
Differentiate with respect to $t$: $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2z\frac{dz}{dt} \implies x\frac{dx}{dt} + y\frac{dy}{dt} = z\frac{dz}{dt}$
Substitute and Solve: $x^2 + y^2 = z^2 \implies 0.8^2 + 0.6^2 = z^2 \implies z = \sqrt{0.64 + 0.36} = 1 \text{ mile}$
Substitute the values: $0.8\frac{dx}{dt} + 0.6(-60) = 1(20)$
Simplify, $0.8\frac{dx}{dt} - 36 = 20 \implies 0.8\frac{dx}{dt} = 56 \implies \frac{dx}{dt} = \frac{56}{0.8} = 70$
Conclusion. The sports car is traveling at 70 mph the speed of the car.
Problem: You are observing a rocket launch from an observation post 5,000 meters from the launch pad. When it reaches an altitude of 10,000 meters, it is traveling upward at a rate of 200 m/s. How fast is the rocket moving away from you at this moment?
Understand the problem, draw a diagram (Figure D), identify Knowns and Unknowns.
$a = 5,000$ m (constant distance from pad to observer, $\frac{da}{dt} = 0$).
$b = 10,000$ m (current altitude).
$\frac{db}{dt} = 200$ m/s.
Goal: Find $\frac{dc}{dt}$ (rate at which the distance $c$ from observer to rocket is changing).
Relate all quantities in a single equation. $a^2 + b^2 = c^2$
Differentiate with respect to $t$: $a^2 + b^2 = c^2 \implies 2a\frac{da}{dt} + 2b\frac{db}{dt} = 2c\frac{dc}{dt} (\star)$
Substitute and Solve. First, calculate $c$ at the moment when the rocket reaches an altitude of 10,000 meters: $c = \sqrt{5000^2 + 10000^2} = \sqrt{25,000,000 + 100,000,000} = \sqrt{125,000,000} = 5000\sqrt{5} \text{ m}$
Since $a$ is constant, $\frac{da}{dt} = 0$, the first term drops out $(\star)$: $2(5000)(0) + 2(10000)(200) = 2(5000\sqrt{5})\frac{dc}{dt}$
Simplify, $4,000,000 = 10000\sqrt{5}\frac{dc}{dt}$
$\frac{dc}{dt} = \frac{4,000,000}{10000\sqrt{5}} = \frac{400}{\sqrt{5}} = 80\sqrt{5}$
Conclusion. The rocket is moving away from you at $80\sqrt{5}$ m/s (approx. 178.9 m/s).
Problem: The volume of water in a reservoir is given by $V = 3(2 + \sqrt{h})^6 - 192$, where $V$ is volume in cubic meters and $h$ is the depth of water in meters. Water is flowing into the reservoir at a constant rate of 150 m³/hr. Find the rate at which the depth of the water is increasing at the instant when the depth is 1.4 m.
Understand the problem, draw a diagram, identify Knowns and Unknowns.
$\frac{dV}{dt} = 150$ m³/hr
$h = 1.4$ m
Goal: Find $\frac{dh}{dt}$.
Relate all quantities in a single equation. It is already given. $V = 3(2 + \sqrt{h})^6 - 192$
Differentiate the volume equation with respect to $t$ using the chain rule: $\frac{dV}{dt} = 18(2 + \sqrt{h})^5 \cdot \frac{d}{dt}(2 + \sqrt{h}) = 18(2 + \sqrt{h})^5 \cdot \left(\frac{1}{2\sqrt{h}}\right) \frac{dh}{dt}$
Simplify, $\frac{dV}{dt} = \frac{9(2 + \sqrt{h})^5}{\sqrt{h}} \frac{dh}{dt}$
Substitute and Solve. Isolate $\frac{dh}{dt}$ and substitute the known values: $150 = \frac{9(2 + \sqrt{1.4})^5}{\sqrt{1.4}} \frac{dh}{dt}$
$\frac{dh}{dt} = \frac{150\sqrt{1.4}}{9(2 + \sqrt{1.4})^5}$
Conclusion. The depth is increasing at a rate of roughly 0.06 m/hr.
Problem: An airplane is flying horizontally at 480 mph, exactly 3 miles above the ground, when it passes directly over a boy on the ground. How fast is the distance between the boy and the plane increasing half a minute later?
Understand the problem, draw a diagram (Figure 2.b), identify Knowns and Unknowns.
$a = 3$ miles (constant altitude of the plane, so $\frac{da}{dt} = 0$).
$\frac{db}{dt} = 480$ mph (horizontal speed of the plane).
$t = 30$ seconds (time elapsed since passing overhead).
Goal: Find $\frac{dc}{dt}$ (the rate at which the straight-line distance $c$ between the boy and the plane is changing) at exactly $t = 30$ seconds.
We need to convert between different units.
480 miles ———- 1h = 60*60 seconds
b miles ———— 30 seconds
The horizontal distance b the plane has traveled in half a minute is b = $\frac{480·30}{60·60} = 4 miles$
Relate all quantities in a single equation. $a^2 + b^2 = c^2$
Differentiate with respect to $t$: $a^2 + b^2 = c^2 \implies 2a\frac{da}{dt} + 2b\frac{db}{dt} = 2c\frac{dc}{dt} (\star)$
Substitute and Solve. Now, use the Pythagorean theorem to find the hypotenuse $c$ (the distance between the boy and the plane) at this exact moment (half a minute later the plane passes over the boy): $a^2 + b^2 = c^2 \implies 3^2 + 4^2 = c^2 \implies 9 + 16 = c^2 \implies c = \sqrt{25} = 5 \text{ miles}$
Since the altitude $a$ is constant, $\frac{da}{dt} = 0$, causing the first term to cancel out entirely $(\star)$: $2b\frac{db}{dt} = 2c\frac{dc}{dt} \implies b\frac{db}{dt} = c\frac{dc}{dt}$
Finally, substitute the known values to solve for $\frac{dc}{dt}$: $4(480) = 5\frac{dc}{dt}$
$1920 = 5\frac{dc}{dt} \implies \frac{dc}{dt} = \frac{1920}{5} = 384$
Conclusion. Half a minute later, the distance between the boy and the plane is increasing at a rate of 384 mph.
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