Exponential Growth & Decay (Edexcel A-Level Mathematics): Revision Notes

Exponential growth occurs when a quantity increases at a rate proportional to its current value. The general formula for exponential growth is:

• $y_0$: Initial value (value at $t = 0$)
• $k$: Growth rate constant ($k > 0$ for growth)
• $t$: Time
• $e$: Euler's number (~2.718)

Key Features:

• The rate of change (growth) increases as the quantity grows.
• Commonly seen in population growth, investments, and the spread of diseases.

Exponential decay occurs when a quantity decreases at a rate proportional to its current value. The general formula for exponential decay is:

• $k$: Decay rate constant ( $k > 0$ for decay)

Key Features:

• The rate of change (decay) slows down as the quantity decreases.
• Commonly seen in radioactive decay, depreciation of assets, and cooling processes.

Example 1: $y = 2^{x+3}$

• We can see that every time we add 1 to the power x, we multiply y by 2.

Example 2: $y = 4^{\frac{1}{3}x - 3}$

• The power goes up by $\frac{1}{3}$ every time we add $3$ to $x$.

Example 3: $y = 6^{-x + 2}$

• Every time we increase $x$ by $1$, $y$ is divided by $6$.

Problem 1: A quantity $N$ is increasing exponentially.

Given that at time $t = 0, N = 15$, and that at time $t = 18, N = 30$:

a) Find the value of $N$ when $t = 36$.

b) Find the value of $t$ when $N = 480$.

Solution:

• a) The pattern shows that $N$ doubles every $18$ units of time. $N = 15 \rightarrow 30 \rightarrow 60$

Therefore, $N = 60$ when $t = 36$.

• b) Extending the doubling pattern: $N = 15 \rightarrow 30 \rightarrow 60 \rightarrow 120 \rightarrow 240 \rightarrow 480$

$t$ corresponds to $t = 90$.

Problem 2: A quantity N is decreasing such that at time $t$:

Given the equation $N = 50e^{-0.2t}$:

a) Find the value of $N$ when $t = 10$.

b) Find the value of $t$ when $N = 3$.

c) Find the rate at which $N$ is decreasing when $t = 10$.

Solution:

a) Substitute $t = 10$ into the equation:

$N = 50e^{-0.2(10)} = 50e^{-2} \approx$ 6.767 (to 4 sf)

b) Set $N = 3$ and solve for $t$:

$3 = 50e^{-0.2t}$

$\frac{3}{50} = e^{-0.2t}$

Take the natural logarithm on both sides:

$\ln\left(\frac{3}{50}\right) = -0.2t$

Solve for $t$:

$t \approx \frac{\ln\left(\frac{3}{50}\right)}{-0.2} \approx$ 14.07 (to 4 sf)

c) The rate of change is given by:

$\text{Rate} = \frac{dN}{dt} = -0.2 \times 50e^{-0.2t}$

Substituting $t = 10$:

$\text{Rate} = -0.2 \times 50e^{-2} \approx$ -1.353 (to 4 sf)

Given that when $t = 180, m = 160$, find: a) The value of the constant $k$.

b) The time it takes for the mass of the substance to be halved.

Solution a):

Find the value of $k$:

Given:

Substitute into the equation:

Divide both sides by 240:

Take the natural logarithm on both sides:

Solve for $k$:

Solution b):

Find the time for the mass to be halved:

The initial mass is when $t = 0$:

Half the initial mass is:

Substitute into the equation:

This simplifies to:

Take the natural logarithm on both sides:

Solve for $t$:

Substituting the value of $k$: