F: Exponentials and logarithms: Know that the gradient of e^(kx) is equal to k e^(kx) and hence understand why the exponential model is suitable in many applications.

The idea

Differentiating y = e^(kx) gives dy/dx = k e^(kx), so the gradient at any point equals k times the current value. The rate of change is therefore proportional to the quantity present, with constant of proportionality k: growth when k > 0, decay when k < 0. This self-replicating property makes e^(kx) suitable whenever the rate of change is proportional to the amount present, such as populations, radioactive decay and compound interest. For example, if N = N0 e^(0.05t), then dN/dt = 0.05N, so the rate is 5% of the current value per unit time. Newton's law of cooling also fits this model: defining y as the temperature difference from the surroundings gives dy/dt = -ky, an exponential decay. To use the result, differentiate term by term, apply the chain rule if needed, and interpret k as the continuous rate constant.

What earns the marks

• Differentiate e^(kx) correctly as k e^(kx), applying the chain rule where the exponent is a function of x.
• Explain that the derivative is proportional to the original function, with constant of proportionality k.
• Interpret the sign of k as indicating growth (k > 0) or decay (k < 0), and state its meaning and units in context.
• Apply the result to contextual problems, such as finding the rate of change from a given exponential model or the gradient at a specific point.
• Recognise that the exponential model is appropriate when the rate of change is proportional to the current value, and use the derivative to form the corresponding differential equation.
• Explain that Newton's law of cooling fits this model when the variable is defined as the temperature difference from the surroundings, since then dy/dt = -ky.

Common mistakes and how to put them right

• Differentiating e^(kx) as e^(kx) without multiplying by k.

  Put it right by applying the chain rule to get k e^(kx).

• Thinking that the gradient is constant.

  Put it right by noting that the gradient changes with x and equals k times the function value.

• Misinterpreting k as the growth rate per unit time in discrete terms.

  Put it right by explaining that k is the continuous rate constant, and the actual rate depends on the current value.

• Claiming Newton's law of cooling is not an exponential model.

  Put it right by defining y as the temperature difference from the surroundings, so that dy/dt = -ky and y decays exponentially.

• Assuming any dependence of the rate of change on the amount present makes an exponential model suitable.

  Put it right by checking that the rate is proportional to the amount, i.e. dy/dt = ky.

Exam tips

• When asked to show that a model is suitable, differentiate the given function and show that the derivative is a constant multiple of the function.
• In application questions, state the meaning of k clearly in the context of the problem, including its sign and units.
• Use the derivative to find the rate of change at a given time by substituting the time into k e^(kx).
• If the model is N = N0 e^(kt), remember that N0 is the initial value and k is the continuous rate constant.