Applications: Using Differentiation

The tangent is the line that grazes the mirror at the point of contact, and its gradient is the derivative there. The strut stands at right angles to the surface, so it lies along the normal, whose gradient is the negative reciprocal of the tangent's.

1. The point of contact lies on the curve, so put x = 2 into y = x²8: y = 48 = 0.5. The strut is bolted at the point (2, 0.5).
2. Differentiate for the gradient at any point on the mirror: dydx = 2x8 = x4.
3. At x = 2 the gradient is 24 = 0.5. That is the gradient of the tangent, the line that grazes the mirror there.
4. (a) Write the tangent through (2, 0.5) with gradient 0.5: y − 0.5 = 0.5(x − 2), which tidies to y = 0.5x − 0.5.
5. The strut is at right angles to the tangent, so its gradient is the negative reciprocal: −10.5 = −2. Check: 0.5 × (−2) = −1, as two perpendicular gradients must.
6. (b) Write the normal through (2, 0.5) with gradient −2: y − 0.5 = −2(x − 2), which tidies to y = 4.5 − 2x. Putting x = 0 gives y = 4.5, so the strut's line crosses the axis 4.5 m above the bottom of the mirror.

answer(a) y = 0.5x − 0.5; (b) the normal has gradient −2 and equation y = 4.5 − 2x, crossing the axis at a height of 4.5 m

techniqueTangents and Normals to Curves

examsO-Level · H2

Close to the point of contact a curve and its tangent are almost the same line, so a value of the function nearby can be read off the tangent instead. The gradient at the known point says how much the root changes for each unit of x.

1. Let f(x) = √x. The nearest value whose root is known is x = 100, where f(100) = 10.
2. Differentiate. Written as a power, f(x) = x^0.5, so dydx = 12√x, and at x = 100 that is 12 × 10 = 0.05.
3. Near the point of contact the curve and the tangent almost agree, so f(100 + h) ≈ f(100) + 0.05h. Here the step is h = 1.
4. (a) The estimate is 10 + 0.05 × 1 = 10.05 m.
5. Square it as a rough check: 10.05² = 101.0025 square meters, which is 0.0025 more than the 101 wanted. The estimate is a little too long.
6. (b) The exact side is 10.049876 m, so the estimate is out by 10.05 − 10.049876 = 0.000124 m, about a tenth of a millimeter. It comes out too long because √x bends downwards and the tangent lies above the curve on both sides of the point.

answer(a) about 10.05 m; (b) the estimate is 0.000124 m too long, about 0.12 mm

techniqueThe Tangent-Line Approximation · Tangents and Normals to Curves

examsO-Level · H2

The cameras measure only a distance and a time, which together give an average speed. The mean value theorem turns that average into a statement about one instant: for a smooth journey, some moment has the speedometer reading exactly the average, and that moment is found by setting the derivative equal to it.

1. Check the two ends. At m = 0 the van is at the first camera, s = 0; at m = 8, s = 8 + 0.0625 × 64 = 8 + 4 = 12 km, the second camera.
2. (a) The average speed is the distance divided by the time: 128 = 1.5 km a minute. In an hour that is 1.5 × 60 = 90, so the average speed is 90 km/h.
3. The distance is a smooth function of the time, so the mean value theorem applies: somewhere strictly between the cameras the derivative equals the gradient of the chord, and that gradient is the average of 1.5 km a minute.
4. Differentiate for the speed at any instant: dsdm = 1 + 0.125m km a minute.
5. Set that equal to the average: 1 + 0.125m = 1.5, so 0.125m = 0.5 and m = 4.
6. (b) Four minutes after the first camera the van was traveling at exactly 1.5 km a minute, which is 90 km/h. Check: it passed the first camera at 1 km a minute (60 km/h) and the second at 2 km a minute (120 km/h), so it went through 1.5 on the way.

answer(a) 90 km/h, which is 1.5 km a minute; (b) 4 minutes after the first camera

techniqueThe Mean Value Theorem · Motion in a Straight Line

examsO-Level

The sign of the derivative says which way a quantity is heading: negative for falling, positive for rising. At the stationary point here the second derivative comes out zero and decides nothing, so the change of sign in the first derivative is what names the minimum.

1. Differentiate: dWdx = 4(x − 12)³1296 = (x − 12)³324 kilograms an hour for each unit of setting.
2. A cube keeps the sign of what is cubed. For x < 12 the bracket x − 12 is negative, so dWdx is negative; for x > 12 it is positive.
3. (a) So the waste falls over the settings from 6 to 12 and rises over the settings from 12 to 18.
4. The only stationary point is where dWdx = 0, and a cube is zero only when its bracket is, so x = 12.
5. Differentiate again: d²Wdx² = 3(x − 12)²324 = (x − 12)²108. At x = 12 this is zero, and a second derivative of zero decides nothing: it leaves a maximum, a minimum and a point of inflection all still possible.
6. (b) The first-derivative test settles it. The gradient is negative just before x = 12 and positive just after, so the curve turns from falling to rising: setting 12 is a minimum, wasting W = 5 kilograms an hour. Check: settings 6 and 18 each waste 5 + 12961296 = 6 kilograms an hour, which is more.

answer(a) the waste falls from setting 6 to setting 12 and rises from setting 12 to setting 18; (b) setting 12, wasting 5 kilograms an hour

techniqueThe First Derivative Test · Increasing and Decreasing Functions · Critical Points and Local Extrema

examsH2

The fencing is fixed, so the two lengths are tied together; that relation turns the area into a function of one length alone. Setting its derivative to zero gives the stationary point, and the sign of the second derivative says whether it is the greatest area or the least.

1. Count the fences. Four of them run from the canal to the far side — the two ends and the two partitions — and each is x meters long. One runs along the far side, parallel to the canal; call it y meters. So 4x + y = 240.
2. Make the area a function of x alone. From the fencing, y = 240 − 4x, so A = xy = x(240 − 4x) = 240x − 4x² square meters.
3. Differentiate and set the result to zero: dAdx = 240 − 8x = 0, so x = 30.
4. (a) Then y = 240 − 4 × 30 = 120, so the plot is 30 m from the canal and 120 m along it. Check the fencing: 4 × 30 + 120 = 240 m, all of it used.
5. Differentiate again: d²Adx² = −8. It is negative, so the curve bends downwards everywhere and the stationary point is a maximum, not a minimum.
6. (b) The greatest area is 30 × 120 = 3600 square meters. Check: x = 29 gives 29 × 124 = 3596 and x = 31 gives 31 × 116 = 3596, both smaller.

answer(a) 30 m from the canal by 120 m along it; (b) 3600 square meters, a maximum because d²Adx² = −8 is negative

techniqueThe Second Derivative Test · Optimization Problems · Finding Stationary Points

examsO-Level · H2

On a closed interval the greatest and least values are either at a stationary point inside it or at one of its two ends. Both ends must therefore be worked out and compared with the stationary point, because at an end the derivative need not be zero.

1. Differentiate: dFdv = −0.2 + 0.0025v liters per 100 km for each km/h.
2. Set it to zero for the stationary point: −0.2 + 0.0025v = 0, so v = 80, which lies inside the allowed range.
3. (a) At v = 80, F = 14 − 16 + 0.00125 × 6400 = 14 − 16 + 8 = 6 liters per 100 km. Here d²Fdv² = 0.0025 is positive, so this stationary point is a minimum, and it is the least fuel the van can use.
4. Now work out both ends. At v = 60, F = 14 − 12 + 0.00125 × 3600 = 14 − 12 + 4.5 = 6.5; at v = 120, F = 14 − 24 + 0.00125 × 14400 = 14 − 24 + 18 = 8.
5. Compare the three values: 6 liters at 80 km/h, 6.5 liters at 60 km/h and 8 liters at 120 km/h.
6. (b) The most fuel is used at 120 km/h, at 8 liters per 100 km. The gradient there is −0.2 + 0.3 = 0.1, which is not zero: the interval simply stops before the curve does, so the greatest value sits at an end.

answer(a) 80 km/h, using 6 liters per 100 km; (b) 120 km/h, using 8 liters per 100 km

techniqueAbsolute Extrema on a Closed Interval · Critical Points and Local Extrema · Finding Stationary Points

examsO-Level

The new cases in a day are the rate at which the total grows, so they are the first derivative. That rate peaks where its own derivative is zero, which is where the curve of totals changes from bending upwards to bending downwards: a point of inflection.

1. The new cases in a day are the rate at which the total grows: dNdx = 120x − 6x² cases a day.
2. That rate is greatest where its own derivative is zero, so differentiate once more: d²Ndx² = 120 − 12x.
3. Set it to zero: 120 − 12x = 0, so x = 10.
4. Check that the bend changes sides. For x < 10, 120 − 12x is positive and the curve of totals bends upwards; for x > 10 it is negative and the curve bends downwards. So x = 10 really is a point of inflection.
5. (a) The new cases peak on day 10, at dNdx = 1200 − 600 = 600 new cases. Check: day 9 gives 1080 − 486 = 594 and day 11 gives 1320 − 726 = 594, both fewer.
6. (b) The total that day is N = 6000 − 2000 = 4000 cases. The outbreak is still growing, but from day 10 onwards it grows by less each day than the day before: that is what the change of bend means, and by day 20 the total has settled at 8000.

answer(a) day 10, when new cases are appearing at 600 a day; (b) 4000 cases in total, where the curve stops bending upwards and starts bending downwards

techniquePoints of Inflection · Curve Sketching with Derivatives

examsO-Level · H2

The area depends on two lengths, but the capacity ties them together, so one can be written in terms of the other. Once the area is a function of the radius alone it can be differentiated, and the stationary point gives the can that uses least metal.

1. Use the capacity to remove one of the two letters. The volume is π r² h = 128π, so h = 128r².
2. Put that into the area: A = 2π r² + 2π r × 128r² = 2π r² + 256πr, which is a function of r alone.
3. Differentiate: dAdr = 4π r − 256πr².
4. Set it to zero: 4π r = 256πr², so 4r³ = 256 and r³ = 64, giving r = 4 cm. A radius cannot be negative, and r³ = 64 has only this one real root, so there is nothing to reject.
5. (a) Then h = 12816 = 8 cm. Check the capacity: π × 16 × 8 = 128π cubic centimeters, as the can must hold.
6. (b) The least metal is A = 2π × 16 + 256π4 = 32π + 64π = 96π ≈ 301.6 square centimeters. Since d²Adr² = 4π + 512πr³ = 12π is positive, this is a minimum, and h = 8 = 2r: the best can is exactly as tall as it is wide.

answer(a) radius 4 cm and height 8 cm; (b) 96π ≈ 301.6 square centimeters, with the height twice the radius

techniqueOptimization Problems · The Second Derivative Test · Finding Stationary Points

examsO-Level · H2

The two coordinates are each given in terms of the time, so each has its own rate. The two rates are at right angles, so the speed comes from Pythagoras' theorem, while dividing one by the other gives the gradient of the path the carriage is on.

1. Differentiate each coordinate with respect to the time: dxdm = 3 meters a second out from the start, and dydm = 4m meters a second upwards.
2. After 1 second the carriage is moving 3 meters a second out and 4 meters a second up, and it is at the point x = 3, y = 2.
3. (a) The two rates are at right angles, so Pythagoras' theorem gives the speed: √3² + 4² = √25 = 5 meters a second.
4. For the steepness of the rail, divide the upward rate by the rate out from the start: dydx = dydm ÷ dxdm = 4m3.
5. (b) After 1 second that gradient is 43, or about 1.333: the rail rises 4 m for every 3 m out.
6. Check it against the rail itself. From x = 3m comes m = x3, so y = 2 × x²9 = 2x²9 and dydx = 4x9. At x = 3 that is 129 = 43, the same gradient by a second route.

answer(a) 5 meters a second; (b) a gradient of 43, about 1.333

techniqueParametric Differentiation · Parametric Equations · Related Rates

examsH2

A Maclaurin series rebuilds a function from its derivatives at zero, and for a small angle the first few terms already give a very close value. Taking one term more and comparing the two answers shows how much the extra term is worth.

1. Differentiate the cosine at zero, one order at a time: cos 0 = 1, then −sin 0 = 0, then −cos 0 = −1, then sin 0 = 0, then cos 0 = 1. Divide each by the factorial of its order to get the coefficients.
2. (a) So cos x ≈ 1 − x²2 + x⁴24, every odd term vanishing. At x = 0.2: 1 − 0.02 + 0.001624 = 1 − 0.02 + 0.0000667 = 0.980067 to six decimal places.
3. Put that into the height: h = 2(1 − 0.980067) = 2 × 0.019933 = 0.039867 m.
4. (b) So the bob rises about 39.87 mm.
5. Now stop the series at two terms: cos 0.2 ≈ 1 − 0.02 = 0.98, which gives h = 2 × 0.02 = 0.04 m, or 40 mm.
6. The two-term estimate is therefore 40 − 39.87 = 0.13 mm too high. Check: the term left out is x⁴24 = 0.0000667, and 2 × 0.0000667 = 0.000133 m, which is that 0.13 mm.

answer(a) cos 0.2 ≈ 0.980067; (b) the bob rises about 39.87 mm, and the two-term estimate leaves it 0.13 mm too high

techniqueMaclaurin Series · The Tangent-Line Approximation

examsH2

The mean current over an interval that shrinks to nothing is the current at the instant the switch closes, which the circuit certainly has, but the formula turns into zero over zero there. L'Hôpital's rule settles such a limit: when the top and the bottom both tend to zero, the limit of their ratio is the limit of the ratio of their derivatives.

1. Put n = 0 into I = 0.018(1 − e⁻ⁿ)n: the top is 0.018(1 − 1) = 0 and the bottom is 0. The formula gives 00, which decides nothing, so the limit has to be found another way.
2. Both top and bottom tend to zero, so L'Hôpital's rule applies. Differentiate each separately: the top gives 0.018e⁻ⁿ and the bottom gives 1. So lim_{n → 0} I = 0.018e⁰1 = 0.018 A.
3. (a) The mean current closes in on 0.018 A, which is 18 mA. Check: while the capacitor is empty there is no voltage across it, so the whole 9 V lies across the resistor and Ohm's law gives 9500 = 0.018 A.
4. For (b), write the fraction over one line: 0.018 − In = 0.018n − 0.018(1 − e⁻ⁿ)n² = 0.018(n − 1 + e⁻ⁿ)n². At n = 0 the top is 0.018(0 − 1 + 1) = 0 and the bottom is 0. Differentiating both gives 0.018(1 − e⁻ⁿ)2n, which is still 00 at n = 0, so differentiate both again: 0.018e⁻ⁿ2, which tends to 0.0182 = 0.009.
5. (b) The limit is 0.009 amperes a second, or 9 mA a second, so over a short first interval of n seconds the mean current is about 18 − 9n mA. For the first 0.1 seconds that is 18 − 0.9 = 17.1 mA. Check: the formula itself gives 0.018(1 − e^−0.1)0.1 = 0.18 × 0.09516 = 0.01713 A, which is 17.13 mA.

answer(a) 18 mA, the same as the 9500 = 0.018 A Ohm's law gives for the empty capacitor; (b) the limit is 0.009 amperes a second, so the mean current over the first 0.1 seconds is about 18 − 0.9 = 17.1 mA (the formula gives 17.13 mA)

techniqueL'Hôpital's Rule · Indeterminate Forms