6.2 The gradient of e^kx and the exponential model — Practise with help

Work each question closed book first. Look at the exponent and write down the constant $k$ before you differentiate.

Warm-up: remember

Write $\left(e^{2}\right)^3$ as a power of $e$ .
Find the gradient of the straight line through $(1, 2)$ and $(4, 11)$ .
Work out $2^0$ and $2^{-1}$ .

Answers

$\left(e^{2}\right)^3 = e^{6}$ .
The gradient is $\dfrac{11 - 2}{4 - 1} = \dfrac{9}{3} = 3$ .
$2^0 = 1$ and $2^{-1} = \dfrac{1}{2}$ .

If question 2 was hard, revise the gradient of a straight line in Chapter 3.

Practise with help

1. Differentiate $y = e^{5x}$ .

Hint

The constant in the exponent is $k = 5$ , and it comes to the front.

Next step

Keep the exponential $e^{5x}$ and multiply it by $5$ .

Full working

\frac{dy}{dx} = 5e^{5x}.

The curve rises five times as fast as $e^x$ at every point.

2. Differentiate $y = 4e^{-x}$ .

Hint

Here $k = -1$ . The constant $4$ stays in front.

Next step

Multiply $4$ by $-1$ and keep the exponential unchanged.

Full working

\frac{dy}{dx} = 4 \times (-1)e^{-x} = -4e^{-x}.

The derivative is negative everywhere, so the curve falls.

3. Differentiate $y = 3^{2x}$ .

Hint

The base is $3$ , which is not $e$ , so the factor $\ln 3$ appears as well as the constant $2$ from the exponent.

Next step

Write the answer as $2 \times 3^{2x}$ multiplied by $\ln 3$ .

Full working

\frac{dy}{dx} = 2 \times 3^{2x}\ln 3.

Since $\ln 3 \approx 1.099$ , the curve is a little steeper than $e^{2x}$ .

4. Find the gradient of $y = e^{3x}$ at the point where $x = 0$ .

Hint

Differentiate first, then substitute $x = 0$ into the derivative.

Next step

The derivative is $3e^{3x}$ . Substitute $x = 0$ and use $e^0 = 1$ .

Full working

\frac{dy}{dx} = 3e^{3x}, \qquad \left.\frac{dy}{dx}\right|_{x=0} = 3e^{0} = 3.

The tangent at $(0, 1)$ has gradient $3$ .

5. A population is modelled by $N = 50e^{0.2t}$ , where $t$ is in days. Write down $\dfrac{dN}{dt}$ , and find its value at $t = 0$ .

Hint

Use $\dfrac{d}{dt}\left(e^{kt}\right) = ke^{kt}$ with $k = 0.2$ , keeping the constant $50$ .

Next step

Differentiate to get $50 \times 0.2 \times e^{0.2t}$ , then substitute $t = 0$ .

Full working

\frac{dN}{dt} = 10e^{0.2t}, \qquad \frac{dN}{dt}\bigg|_{t=0} = 10e^{0} = 10.

The population starts by growing at about $10$ cells per day.

6. A radioactive mass is modelled by $m = 120e^{-0.5t}$ grams. Find $\dfrac{dm}{dt}$ at $t = 0$ , and explain why the answer is negative.

Hint

Here $k = -0.5$ . The negative value of $k$ is what makes the rate negative.

Next step

Differentiate to get $120 \times (-0.5)e^{-0.5t} = -60e^{-0.5t}$ , then substitute $t = 0$ .

Full working

\frac{dm}{dt} = -60e^{-0.5t}, \qquad \frac{dm}{dt}\bigg|_{t=0} = -60e^{0} = -60.

The rate is negative because the mass is decreasing. It loses about $60$ grams per hour at the start, and the rate becomes less negative as the mass falls.

7. A student says that a quantity satisfying $\dfrac{dy}{dt} = ky$ with $k = 0$ grows without limit. Explain why the statement is wrong.

Hint

Substitute $k = 0$ into the model $y = Ae^{kt}$ and see what happens.

Next step

With $k = 0$ , the equation says $\dfrac{dy}{dt} = 0$ . What kind of quantity has zero rate of change?

Full working

With $k = 0$ the model is $y = Ae^{0} = A$ , a constant. A zero rate of change means the quantity stays the same. Growth needs a positive $k$ , and decay needs a negative $k$ . Only the sign of $k$ decides whether the quantity rises or falls.