Chapter 7: Review

The chapter problem, solved

The open box

A sheet of card measures $16$ cm by $10$ cm. Four equal squares of side $x$ cm are cut from the corners, and the sides are folded up to make an open box.

Step 1. Write the volume. The base measures $(16 - 2x)$ cm by $(10 - 2x)$ cm, and the height is $x$ cm, so

V = x(16 - 2x)(10 - 2x) = 4x^3 - 52x^2 + 160x.

Step 2. Differentiate and solve. Differentiate with the power rule:

\frac{dV}{dx} = 12x^2 - 104x + 160 = 4(3x^2 - 26x + 40).

Set $\dfrac{dV}{dx} = 0$ and solve the quadratic:

3x^2 - 26x + 40 = 0, \qquad x = \frac{26 \pm \sqrt{676 - 480}}{6} = \frac{26 \pm 14}{6}.

So $x = \dfrac{40}{6} = \dfrac{20}{3}$ or $x = \dfrac{12}{6} = 2$ .

Step 3. Reject the impossible root. The card is $10$ cm wide, so $2x$ cannot be larger than $10$ . The root $x = \dfrac{20}{3}$ gives $2x = \dfrac{40}{3} > 10$ , so it is impossible. Only $x = 2$ remains.

Step 4. Confirm the maximum. The second derivative is

\frac{d^2V}{dx^2} = 24x - 104.

At $x = 2$ this is $48 - 104 = -56 < 0$ , so the point is a maximum.

Step 5. Find the greatest volume.

V(2) = 4(8) - 52(4) + 160(2) = 32 - 208 + 320 = 144 \text{ cm}^3.

The box is largest when the squares have side $2$ cm, and the greatest volume is $144$ cm $^3$ .

The rising balloon

A spherical balloon is inflated so that its volume grows at $100$ cm $^3$ per second. Its volume is $V = \dfrac{4}{3}\pi r^3$ , so

\frac{dV}{dr} = 4\pi r^2.

The chain rule gives $\dfrac{dV}{dt} = \dfrac{dV}{dr} \times \dfrac{dr}{dt}$ . Substitute $\dfrac{dV}{dt} = 100$ and rearrange:

\frac{dr}{dt} = \frac{100}{4\pi r^2} = \frac{25}{\pi r^2}.

When $r = 5$ cm:

\frac{dr}{dt} = \frac{25}{25\pi} = \frac{1}{\pi} \approx 0.318 \text{ cm/s}.

When $r = 10$ cm:

\frac{dr}{dt} = \frac{25}{100\pi} = \frac{1}{4\pi} \approx 0.0796 \text{ cm/s}.

The radius grows more slowly as the balloon swells, because the same volume of air spreads over a larger surface. The rate is proportional to $\dfrac{1}{r^2}$ .

Mixed practice — choose the method first

For each question, say which rule or method you will use before you calculate. The items mix this chapter with Chapters 1 to 6.

Differentiate $y = (2x + 3)^4$ .
Differentiate $y = x^2e^{3x}$ .
Differentiate $y = \dfrac{3x - 1}{x + 2}$ .
Find $\dfrac{dy}{dx}$ for $x^2 + y^2 = 100$ .
Find $\dfrac{dy}{dx}$ for the curve $x = t^3$ , $y = t^2$ .
Find the stationary points of $y = x^3 - 12x$ and classify them.
Find the gradient of the tangent to $y = x^2 - 4x$ at $x = 3$ .
Solve $2^x = 7$ , giving $x$ to three significant figures.
Solve $\sin\theta = \dfrac{1}{2}$ for $0 \le \theta < 2\pi$ .
Write $\sqrt{75} + \sqrt{12}$ in the form $k\sqrt{3}$ .
Show that $(n + 1)^2 - n^2$ is always odd.
Find the coefficient of $x^3$ in the expansion of $(2 + x)^6$ .
A balloon has $\dfrac{dV}{dt} = 200$ cm $^3$ /s. Find $\dfrac{dr}{dt}$ when $r = 5$ cm.
A radioactive mass $m$ decays at a rate proportional to $m$ . Write the differential equation and show that $m = m_0e^{-kt}$ satisfies it.

Answers and concise solutions

1. The derivative

Answer: A function of a function, so use the chain rule. The outside gives $4(2x + 3)^3$ and the inside gives $2$ , so $\dfrac{dy}{dx} = 8(2x + 3)^3$ .

2. The derivative

Answer: A product, so use the product rule: $2xe^{3x} + 3x^2e^{3x} = e^{3x}(3x^2 + 2x)$ after taking out the common factor.

3. The derivative

Answer: A quotient, so use the quotient rule: $\dfrac{3(x + 2) - (3x - 1)}{(x + 2)^2} = \dfrac{7}{(x + 2)^2}$ . The top simplifies to the constant $7$ .

4. The derivative

Answer: An implicit circle, so differentiate both sides and apply the chain rule to $y^2$ : $2x + 2y\dfrac{dy}{dx} = 0$ , giving $\dfrac{dy}{dx} = -\dfrac{x}{y}$ .

5. The derivative

Answer: A parametric curve, so divide $\dfrac{dy}{dt}$ by $\dfrac{dx}{dt}$ : $\dfrac{2t}{3t^2} = \dfrac{2}{3t}$ .

6. The sign of the second derivative at each point

Answer: $\dfrac{dy}{dx} = 3x^2 - 12 = 0$ gives $x = 2$ and $x = -2$ . The second derivative is $6x$ : at $x = -2$ it is $-12 < 0$ , a maximum; at $x = 2$ it is $12 > 0$ , a minimum.

7. The gradient

Answer: $\dfrac{dy}{dx} = 2x - 4$ , which is $2$ at $x = 3$ . The point on the curve is $(3, -3)$ , so the tangent is $y = 2x - 9$ .

8. The value of $x$

Answer: Take logarithms of both sides: $x = \dfrac{\ln 7}{\ln 2} \approx 2.81$ to three significant figures.

9. The solutions

Answer: The principal value is $\dfrac{\pi}{6}$ . The second solution in the interval is $\pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6}$ .

10. The value of $k$

Answer: $\sqrt{75} = 5\sqrt{3}$ and $\sqrt{12} = 2\sqrt{3}$ , so the sum is $7\sqrt{3}$ . The value of $k$ is $7$ .

11. The simplified difference

Answer: Expand: $(n + 1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n + 1$ . This is one more than an even number, so it is always odd.

12. The coefficient

Answer: The term in $x^3$ is $\binom{6}{3}2^3x^3 = 20 \times 8x^3 = 160x^3$ , so the coefficient is $160$ .

13. The rate

Answer: $\dfrac{dV}{dr} = 4\pi r^2 = 100\pi$ at $r = 5$ . Then $200 = 100\pi \dfrac{dr}{dt}$ , so $\dfrac{dr}{dt} = \dfrac{2}{\pi} \approx 0.637$ cm/s.

14. The differential equation

Answer: The mass falls, so $\dfrac{dm}{dt} = -km$ with $k > 0$ . Substituting $m = m_0e^{-kt}$ gives $\dfrac{dm}{dt} = -km_0e^{-kt} = -km$ .

What you can now do

estimate a gradient from a shrinking chord, and explain the tangent as a limit;
differentiate a whole-number power of $x$ from first principles;
differentiate powers, exponentials, trigonometric functions and logarithms;
find and interpret a second derivative, and test a point of inflection;
find stationary points, classify them, and solve an optimisation problem;
find the equation of a tangent and of a normal;
decide where a function is increasing and where it is decreasing;
use the product, quotient and chain rules;
connect two rates of change, and differentiate an inverse function;
differentiate an implicit relation and a parametric curve;
build a differential equation from a written description.

Your relevance reflection

Return to the prompt in the chapter opener: a quantity that grows or shrinks. Write down the quantity again. Now add three things you could not have said before this chapter: how to measure its rate of change at an instant, how to find when that rate is zero, and how to connect its rate to the rate of something it depends on. Finish with one sentence about a rate you would measure from now on.