A fielder picks up the ball at the long-on boundary and must return it to the keeper. Throw it too flat and it hits the grass early. Throw it too steeply and it climbs high but comes down short. Coaches say that the longest throw leaves the hand at about \(45°\). Why that angle, and not \(30°\) or \(60°\)?
Where the tangent goes flat
Throw a ball at \(20\,\text{m/s}\) from level ground, with \(g = 10\,\text{m/s}^2\). Motion in Two Dimensions shows that it lands at the range \(R = (u^2/g)\sin 2\theta = 40\sin 2\theta\) metres. Fig 5.1 sweeps the launch angle and plots \(R\) against \(\theta\) below the path. A tangent slides along the \(R\)-\(\theta\) curve as the angle grows.
Watch the tangent in Fig 5.1. While the range is still growing, its slope is positive. Once the range starts to shrink, the slope is negative. At the peak, between the two, the slope must pass through zero. So a smooth maximum sits where the tangent is flat. The same holds at the bottom of a valley, a minimum. A point where \(dy/dx = 0\) is called a stationary point.
If a smooth function has a maximum or a minimum at a point inside its allowed range, its derivative is zero there. The converse is not true: a zero derivative marks a candidate that must still be tested.
Maximum or minimum: two tests
The first test watches the slope on the two sides of the flat point. If \(dy/dx\) changes from positive to negative, the curve climbs and then falls: a maximum. If it changes from negative to positive, a minimum. If it keeps the same sign on both sides, the point is neither. Fig 5.2 shows all three cases.
The second test is quicker. At a maximum the slope goes from positive to negative, so the slope is falling: \(d^2y/dx^2 < 0\). At a minimum the slope is rising: \(d^2y/dx^2 > 0\). In the language of lesson 4, a maximum sits on a dome and a minimum in a cup. If \(d^2y/dx^2 = 0\), this test says nothing, and the first test must decide.
Take \(f(x) = x^3 - 3x^2 + 1\). Its derivative \(3x^2 - 6x = 3x(x - 2)\) is zero at \(x = 0\) and \(x = 2\). The second derivative is \(6x - 6\). At \(x = 0\) it is \(-6\), so \(f\) has a maximum there, \(f(0) = 1\). At \(x = 2\) it is \(+6\), so \(f\) has a minimum there, \(f(2) = -3\).
For \(f(x) = x^3 - 3x^2 + 1\), the derivative is zero at \(x = 0\) and the second derivative there is \(-6\). What is \(x = 0\)?
For \(y = x^3\), \(dy/dx = 3x^2\) is zero at \(x = 0\). What kind of point is \(x = 0\)?
A stone is thrown up with \(h = 30t - 5t^2\) metres, \(t\) in seconds. Use \(dh/dt = 0\) to find its greatest height, in metres.
Where the rule is silent
The rule \(dy/dx = 0\) finds only peaks and valleys that are smooth and inside the allowed range. Physics often limits the range. Suppose a cell of emf \(6\,\text{V}\) and internal resistance \(2\,\Omega\) may drive only a load between \(5\,\Omega\) and \(20\,\Omega\). The load power \(P = 36R/(R + 2)^2\) falls all the way across that range. There is no flat tangent inside it. The greatest power, \(3.67\,\text{W}\), sits at the end \(R = 5\,\Omega\).
A sharp corner is the other place the rule misses. The height of a ball bouncing on a floor is least at the instant of each bounce. There the height-time graph has a sharp corner, and no derivative exists. So a complete search checks three kinds of candidates: the flat points, the ends of the range and the corners.
A zero derivative is necessary for a smooth extreme inside the range, but it is not enough. The flat point of \(y = x^3\) is neither a maximum nor a minimum. A maximum found this way is also only the greatest nearby: another peak, or an end of the range, may be higher. Always test the candidate, and compare it with the ends.
A cell of \(6\,\text{V}\) and \(2\,\Omega\) may drive only a load between \(5\,\Omega\) and \(20\,\Omega\). Its load power \(P = 36R/(R + 2)^2\) peaks at \(R = 2\,\Omega\). Within the allowed range, where is \(P\) greatest?
Relative Velocity in Two Dimensions finds the least distance between two ships by making \(s^2\), not \(s\), least. Why does this give the right instant?
Back to physics
Motion in Two Dimensions finds the angle of the longest throw from the fact that a sine is never more than 1. The derivative finds the same angle without that fact, and it would work for any formula.
This is a check of Motion in Two Dimensions, lesson 5, done with the tool. A ball is thrown at \(20\,\text{m/s}\) on level ground; take \(g = 10\,\text{m/s}^2\). The range is \(R = (u^2/g)\sin 2\theta\). Find the angle of the greatest range, and that range.
\(R = (400/10)\sin 2\theta = 40\sin 2\theta\) metres, for \(\theta\) between \(0°\) and \(90°\). At both ends \(R = 0\).
\(dR/d\theta = 80\cos 2\theta\), by the chain rule, with \(\theta\) in radians. Set it equal to zero.
\(\cos 2\theta = 0\) gives \(2\theta = 90°\), so \(\theta = 45°\). There \(d^2R/d\theta^2 = -160\sin 2\theta = -160 < 0\), a maximum.
The greatest range is \(R = 40\sin 90° = 40\,\text{m}\), at \(45°\). This is \(u^2/g\), the answer of the physics chapter.
A cell of emf \(6\,\text{V}\) and internal resistance \(2\,\Omega\) drives a load \(R\) that you may choose freely. The load power is \(P = 36R/(R + 2)^2\), and \(dP/dR = 36(2 - R)/(R + 2)^3\). Find the greatest power the load can take, in watts.
Back to the boundary
The fielder's throw is a range problem. On level ground the derivative of \(R = (u^2/g)\sin 2\theta\) is zero at \(45°\), and the second derivative there is negative. So \(45°\) gives the longest throw, \(u^2/g\), and nearby angles lose very little because the tangent is flat. A real fielder releases the ball above the ground and meets air resistance. Both effects make the best angle a little lower than \(45°\).
Pierre de Fermat was a lawyer in Toulouse who did mathematics in his spare time. Around 1636 he described a way to find the greatest and least values of a quantity. He compared its value at a point with its value a tiny step away, and demanded that the two be equal in the limit. That is our rule \(dy/dx = 0\), decades before Newton and Leibniz. In 1662 he used the same idea on light. He supposed that light takes the path of least time, and from that he derived the law of refraction.
Maxima and minima answer every "greatest" or "least" of the book. The range of a projectile peaks at \(45°\) in Motion in Two Dimensions. The closest approach of two moving bodies is a minimum in Relative Velocity in Two Dimensions. A cell gives most power to a load equal to its internal resistance in Cells, Bridges and Measuring Instruments. The field on the axis of a charged ring, or of a ring of mass, is greatest at a definite distance. The deviation of a prism is least at one angle in the optics unit, and the amplitude of a forced oscillator peaks at resonance. Stable equilibrium is a minimum of the potential energy.
- Calling every point with \(dy/dx = 0\) a maximum.A flat tangent is only a candidate. Test it with the sign of \(d^2y/dx^2\) or the change of sign of \(dy/dx\).
- Putting the stationary value of \(x\) back into \(dy/dx\) to find the greatest value.\(dy/dx\) is zero there by construction. Put it into \(y\) itself.
- Forgetting the ends of the allowed range.When the range of \(x\) is limited, compare the values at the ends with the values at the flat points.