Concepts: rate of change of quantities: Application of Derivatives

2.1 Related rates in one variable

A rate of change question connects two quantities that move together in time. The method always has the same shape. Write the formula linking the quantities, differentiate both sides with respect to time using the chain rule, then substitute the values at the instant mentioned in the question.

For a sphere, the volume is four thirds pi r cubed. Differentiating with respect to time gives the volume rate as four pi r squared times the radius rate. When the radius is given, substitution gives the numeric rate with its unit.

For a circle, the area rate is 2 pi r times the radius rate. For a cube, the volume rate is 3a squared times the side rate. Since the formula has a square or cube, the rate grows as the size grows, which is why the numerical answers depend on the instant chosen.

2.2 Two worked examples

Suppose a balloon takes in air at 6 cubic centimetres each second. How fast is its radius changing at the moment the radius measures 8 cm? Differentiating the volume gives 6 equal to 4 pi r squared times dr by dt, so dr by dt equals 6 divided by 256 pi, which simplifies to 3 over 128 pi cm per second. Keeping pi symbolic until the end avoids rounding errors.

A ladder 5 metres long leans on a wall. The foot slides away at 2 metres per second. When the foot is 3 metres from the wall, the top is 4 metres high by Pythagoras. Differentiating x squared plus y squared equals 25 gives 2x dx by dt plus 2y dy by dt equal to zero, so dy by dt equals minus x by y times 2, which is minus 3 by 2 metres per second. The negative sign means the top moves down.

2.3 Exercise 6.1 coverage

Exercise 6.1 is a set of related rate problems: spheres, circles, cubes, ladders, shadows and expanding areas. In each, the formula and the instant are given; the work is the differentiation. Writing the units and the sign in the final line is part of the answer.

2.4 More related rate shapes

A cube whose side grows at 0.1 cm per second has volume rate 3a squared times 0.1. At a side of 5 cm this is 7.5 cubic centimetres per second. The square appears because the volume depends on the cube of the side, so the rate depends on the size at that instant.

A rectangle with a fixed width and a growing length has area rate equal to the width times the length rate. These small examples train the habit of differentiating the formula before substituting numbers.

2.5 Units and signs in the final line

Every related rate answer carries a unit and a direction. A positive value means the quantity grows, a negative value means it shrinks. Statements such as the water level falling or the shadow shortening correspond to negative rates, and the exam accepts the negative sign as part of the answer.

2.6 Two more rate shapes

A cone filling with water has volume one third pi r squared h, and if the radius stays proportional to the height the volume becomes a single-variable function. Differentiating then gives the water level rate in terms of the inflow. Writing the proportionality first is the key step.

The surface area of a sphere has rate 8 pi r times the radius rate. Comparing it with the volume rate shows why the two rates scale differently with the radius, which is exactly the idea behind many one mark conceptual items.

2.7 Rates that depend on geometry

A conical tank filled at a constant rate has a water level that rises faster when the tank is narrow. Writing the volume in terms of the water height using similar triangles gives the exact relation, and differentiating gives the level rate. The same idea applies to a hemispherical bowl and to any vessel whose cross section changes with height.

The surface area and volume of a sphere are linked: differentiating the volume gives the surface area. This is why the volume rate of a sphere equals its surface area times the radius rate. Recognising such links shortens many calculation steps.

2.8 One more related rate numerical

A cubic tank has side a growing at 0.2 cm per second. Its volume is a cubed, so the volume rate is 3a squared times 0.2. At a side of 10 cm this is 3 times 100 times 0.2, which is 60 cubic centimetres per second. The same pattern works for any solid whose volume formula is known: differentiate the formula, then substitute the instant values and finish with the unit.