Calculus · Differential calculus
The derivative was invented, by Newton, to describe motion. Velocity is the derivative of position, acceleration is the derivative of velocity, and the same two ideas, applied to an angle instead of a distance, describe everything that spins. This post shows how these definitions come out of the derivative, and how to use them to solve real problems.
A material point moves along the \(Ox\) axis. Its position at time \(t\) is given by a function
\[ x = f(t), \]called the law of motion. Between the moments \(t\) and \(t + \Delta t\) the point moves by \( \Delta x = f(t + \Delta t) - f(t) \), so its average velocity on that interval is \( \frac{\Delta x}{\Delta t} \). To get the velocity at the instant \(t\), let the time interval shrink.
This is exactly the definition of the derivative from the post on derivatives, with time in the role of \(x\). Everything we learned there now has a physical meaning:
Velocity itself changes with time. Applying the same idea to \(v\) gives the acceleration.
(The letter \(w\) is traditional in older European textbooks; most modern books write \(a\).) So acceleration is the second derivative of position, the physical meaning of \(f''\) mentioned in the post on higher derivatives.
In physics, Newton's notation with dots is common: \( \dot x = \frac{dx}{dt} \), \( \ddot x = \frac{d^2x}{dt^2} \). Newton's second law, \( F = m\ddot x \), says that force determines the second derivative of position. This is why so many laws of physics are differential equations.
\( v = 20 - 4t \), \( w = -4\ \text{m/s}^2 \). The car stops when \( v = 0 \): \( t = 5 \) s. In that time it travels \( x(5) = 100 - 50 = 50 \) m. Here \(v > 0\) and \(w < 0\): opposite signs, so the car is slowing down, as expected.
When a rigid body rotates about a fixed axis, every point moves on a circle, but all points turn through the same angle. So the motion is described by a single function: the angle of rotation \( \varphi = \varphi(t) \), measured in radians.
Replacing distance by angle in the definitions of Sections 1 and 2:
Units: \(\omega\) in rad/s, \(\varepsilon\) in rad/s². In engineering, rotation speed is often given in revolutions per minute (rpm). One revolution is \(2\pi\) rad, so \( n \) rpm corresponds to \( \omega = \frac{2\pi n}{60} = \frac{\pi n}{30} \) rad/s.
The analogy is perfect:
| Motion along a line | Rotation about an axis |
|---|---|
| position \(x(t)\) | angle \(\varphi(t)\) |
| velocity \( v = \frac{dx}{dt} \) | angular velocity \( \omega = \frac{d\varphi}{dt} \) |
| acceleration \( w = \frac{dv}{dt} = \frac{d^2x}{dt^2} \) | angular acceleration \( \varepsilon = \frac{d\omega}{dt} = \frac{d^2\varphi}{dt^2} \) |
| uniform motion: \( x = x_0 + vt \) | uniform rotation: \( \varphi = \varphi_0 + \omega t \) |
| uniformly accelerated: \( x = x_0 + v_0t + \frac{wt^2}{2} \) | \( \varphi = \varphi_0 + \omega_0t + \frac{\varepsilon t^2}{2} \) |
A point at distance \(r\) from the axis travels along an arc of length \( s = r\varphi \) (that's the definition of the radian). Differentiating with respect to time (\(r\) is constant):
\[ v = \frac{ds}{dt} = r\frac{d\varphi}{dt} = \omega r, \qquad w_t = \frac{dv}{dt} = r\frac{d\omega}{dt} = \varepsilon r. \]But even with constant \(\omega\), a point moving on a circle is accelerating, because the direction of its velocity keeps changing. This normal (centripetal) acceleration points towards the axis and equals
\[ w_n = \frac{v^2}{r} = \omega^2 r. \](It follows from differentiating the position vector \( (r\cos\varphi, r\sin\varphi) \) twice; the vector methods of the post on vectors make this a short exercise.) The total acceleration is \( w = \sqrt{w_t^2 + w_n^2} \).
The pattern "limit of \( \frac{\Delta(\text{quantity})}{\Delta t} \)" appears all over physics. Every time a quantity changes, its derivative is the rate of that change:
| Quantity | Derivative | Physical meaning |
|---|---|---|
| electric charge \(q(t)\) | \( i = \frac{dq}{dt} \) | electric current |
| work \(A(t)\) | \( P = \frac{dA}{dt} \) | power |
| momentum \(p(t) = mv\) | \( F = \frac{dp}{dt} \) | force (Newton's second law) |
| heat \(Q(T)\) as a function of temperature | \( C = \frac{dQ}{dT} \) | heat capacity |
| mass \(m(x)\) of a rod from \(0\) to \(x\) | \( \rho = \frac{dm}{dx} \) | linear density |
| amount of a radioactive substance \(N(t)\) | \( \frac{dN}{dt} = -\lambda N \) | decay rate (proportional to \(N\)) |
Note the last two rows: the "variable" need not be time. The derivative measures how fast one quantity changes with respect to another.
| Quantity | Formula | Unit |
|---|---|---|
| Velocity | \( v = \frac{dx}{dt} \) | m/s |
| Acceleration | \( w = \frac{dv}{dt} = \frac{d^2x}{dt^2} \) | m/s² |
| Angular velocity | \( \omega = \frac{d\varphi}{dt} \) | rad/s |
| Angular acceleration | \( \varepsilon = \frac{d\omega}{dt} = \frac{d^2\varphi}{dt^2} \) | rad/s² |
| Linear speed on a circle | \( v = \omega r \) | m/s |
| Tangential / normal acceleration | \( w_t = \varepsilon r \), \( w_n = \omega^2r \) | m/s² |
Recipe: given the law of motion, differentiate once for velocity, twice for acceleration. Set \(v = 0\) to find stops and turning points; compare the signs of \(v\) and \(w\) to see whether the body speeds up or slows down.
Try each one before opening the solution. More solved problems from this topic are in Differential Calculus.
\( v = 6t^2 - 6t \Rightarrow v(2) = 12 \) m/s. \( w = 12t - 6 \Rightarrow w(2) = 18 \) m/s².
\( v = 20 - 10t = 0 \Rightarrow t = 2 \) s; \( h(2) = 40 - 20 = 20 \) m. Acceleration \( w = -10 \) m/s² throughout.
\( v = 2t - 4 \), \( w = 2 > 0 \). For \( 0 \le t < 2 \): \(v < 0\), \(w > 0\), so it slows down. For \( t > 2 \): same signs, so it speeds up. It turns around at \(t = 2\), \(x = -1\).
\( \omega = 6t + 1 \Rightarrow \omega(4) = 25 \) rad/s; \( \varepsilon = 6 \) rad/s². \( v = \omega r = 12.5 \) m/s.
\( \omega = \frac{\pi\cdot 3000}{30} = 100\pi \approx 314 \) rad/s. \( w_n = \omega^2r \approx 314^2\cdot 0.1 \approx 9870 \) m/s², about 1000 g.
\( v = -6\sin 2t \Rightarrow |v|_{\max} = 6 \) (at \(x = 0\)). \( w = -12\cos 2t = -4x \Rightarrow |w|_{\max} = 12 \) (at the extreme points \(x = \pm3\)).
\( \frac{dN}{dt} = -\lambda N_0e^{-\lambda t} = -\lambda N \). The more atoms there are, the faster they decay; the minus sign shows that \(N\) decreases.
Physics gave calculus its first and most natural interpretation: the derivative is a rate of change. Velocity measures how fast position changes, acceleration how fast velocity changes, and the same two definitions with an angle in place of a distance describe every wheel, gear and planet in rotation. Once you see derivatives this way, the laws of physics start to read as statements about rates, and solving them becomes a question of calculus.
Related posts: derivatives (definition and rules), higher derivatives (the meaning of \(f''\)), and the mean value theorems (why average speed equals instant speed at some moment).