Rates of Change and Monotonicity — Worked Solutions

Ques­tion 1: Vol­ume of an inflat­ing bal­loon#

The prob­lem#

A bal­loon's radius grows at $0.5$ cm/s. How fast is its vol­ume grow­ing when the radius is $6$ cm?

Under­stand­ing the prob­lem#

The bal­loon is a sphere. You are given $\displaystyle \tfrac{dr}{dt} = 0.5$ cm/s and asked for $\displaystyle \tfrac{dV}{dt}$ at the instant when $r = 6$ cm.

The idea#

Link $V$ and $r$ by $\displaystyle V = \tfrac{4}{3}\pi r^3$, dif­fer­en­ti­ate with respect to $t$ (chain rule), then sub­sti­tute.

Step-by-step solu­tion#

Step 1. Write the for­mula con­nect­ing the quan­ti­ties.

$$\displaystyle V = \frac{4}{3}\pi r^3.$$

Step 2. Dif­fer­en­ti­ate both sides with respect to $t$. Since $r$ depends on $t$, the chain rule gives

$$\displaystyle \frac{dV}{dt} = \frac{4}{3}\pi \cdot 3r^2 \cdot \frac{dr}{dt} = 4\pi r^2\frac{dr}{dt}.$$

Step 3. Sub­sti­tute $r = 6$ and $\displaystyle \tfrac{dr}{dt} = 0.5$.

$$\displaystyle \frac{dV}{dt} = 4\pi (36)(0.5) = 72\pi \text{ cm}^3/\text{s}.$$

Check­ing the answer#

Units: cm² × cm/s = cm³/s, cor­rect for a rate of vol­ume. The rate is pos­i­tive, as it should be for a grow­ing bal­loon. ($72\pi \approx 226$ cm³/s.)

Answer#

The vol­ume grows at $72\pi$ cm³/s.

Ques­tion 2: The slid­ing lad­der#

The prob­lem#

A lad­der $10$ m long leans on a wall. Its foot slides away at $1.5$ m/s. How fast does the top slide down when the foot is $6$ m from the wall?

Under­stand­ing the prob­lem#

Let $x$ be the dis­tance of the foot from the wall and $y$ the height of the top on the wall. The lad­der, wall and ground form a right tri­an­gle with hypotenuse $10$. You know $\displaystyle \tfrac{dx}{dt} = 1.5$ m/s and want $\displaystyle \tfrac{dy}{dt}$ when $x = 6$.

The idea#

Use Pythago­ras, $x^2 + y^2 = 100$, and dif­fer­en­ti­ate with respect to $t$. You will also need $y$ at that instant, which Pythago­ras gives.

Step-by-step solu­tion#

Step 1. The rela­tion between $x$ and $y$ at every moment:

$$x^2 + y^2 = 10^2 = 100.$$

Step 2. Dif­fer­en­ti­ate with respect to $t$ (the $100$ is con­stant).

$$\displaystyle 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0 \quad\Longrightarrow\quad x\frac{dx}{dt} + y\frac{dy}{dt} = 0.$$

Step 3. Find $y$ when $x = 6$.

$$y = \sqrt{100 - 36} = \sqrt{64} = 8.$$

Step 4. Sub­sti­tute $x = 6$, $y = 8$, $\displaystyle \tfrac{dx}{dt} = 1.5$.

$$\displaystyle 6(1.5) + 8\frac{dy}{dt} = 0 \quad\Longrightarrow\quad \frac{dy}{dt} = -\frac{9}{8} = -1.125 \text{ m/s}.$$

Step 5. The minus sign means $y$ is decreas­ing: the top is mov­ing down at $1.125$ m/s.

Check­ing the answer#

The sign is neg­a­tive (the top falls) and the size is rea­son­able: at this moment the foot is closer to the wall than the top is high, so the top falls a lit­tle slower than the foot slides ($\displaystyle \tfrac{6}{8}$ of $1.5$).

Answer#

The top slides down at $1.125$ m/s (that is, $\displaystyle \tfrac{dy}{dt} = -1.125$ m/s).

Com­mon mis­take to avoid#

Do not sub­sti­tute $x = 6$ before dif­fer­en­ti­at­ing. $x$ is chang­ing, so it must stay a vari­able until after you dif­fer­en­ti­ate.

Ques­tion 3: Area of a spread­ing ring#

The prob­lem#

A stone thrown into a pond makes a cir­cu­lar ring whose radius grows at $3$ cm/s. How fast does the enclosed area grow when the radius is $12$ cm?

Under­stand­ing the prob­lem#

Given $\displaystyle \tfrac{dr}{dt} = 3$ cm/s, find $\displaystyle \tfrac{dA}{dt}$ when $r = 12$ cm.

The idea#

$A = \pi r^2$; dif­fer­en­ti­ate with respect to $t$.

Step-by-step solu­tion#

Step 1. Dif­fer­en­ti­ate $A = \pi r^2$.

$$\displaystyle \frac{dA}{dt} = 2\pi r\frac{dr}{dt}.$$

Step 2. Sub­sti­tute $r = 12$, $\displaystyle \tfrac{dr}{dt} = 3$.

$$\displaystyle \frac{dA}{dt} = 2\pi(12)(3) = 72\pi \text{ cm}^2/\text{s}.$$

Check­ing the answer#

Units cm × cm/s = cm²/s ✓. Pos­i­tive, since the ring is grow­ing.

Answer#

The area grows at $72\pi$ cm²/s.

Ques­tion 4: Mar­ginal rev­enue#

The prob­lem#

The total rev­enue from sell­ing $x$ units is $R(x) = 5x^2 + 30x + 10$. Find the mar­ginal rev­enue at $x = 8$.

Under­stand­ing the prob­lem#

Mar­ginal rev­enue is the rate of change of total rev­enue with respect to the num­ber of units sold, $R'(x)$. It tells you roughly how much extra rev­enue one more unit brings.

The idea#

Dif­fer­en­ti­ate $R(x)$ and sub­sti­tute $x = 8$.

Step-by-step solu­tion#

Step 1. Dif­fer­en­ti­ate term by term.

$$R'(x) = 10x + 30.$$

Step 2. Sub­sti­tute $x = 8$.

$$R'(8) = 80 + 30 = 110.$$

Check­ing the answer#

$R(9) - R(8) = (405 + 270 + 10) - (320 + 240 + 10) = 685 - 570 = 115$, close to $110$, as expected for a mar­ginal value.

Answer#

Mar­ginal rev­enue at $x = 8$ is $110$.

Ques­tion 5: Where a cubic increases and decreases#

The prob­lem#

Find where $f(x) = 2x^3 - 3x^2 - 36x + 7$ increases and decreases.

Under­stand­ing the prob­lem#

You must split the real line into inter­vals and say, for each, whether $f$ is going up or down.

The idea#

Find $f'(x)$, fac­torise it, find where it is zero, and check its sign on each inter­val.

Step-by-step solu­tion#

Step 1. Dif­fer­en­ti­ate.

$$f'(x) = 6x^2 - 6x - 36 = 6(x^2 - x - 6).$$

Step 2. Fac­torise: two num­bers with prod­uct $-6$ and sum $-1$ are $-3$ and $2$.

$$f'(x) = 6(x - 3)(x + 2).$$

Step 3. $f'(x) = 0$ at $x = -2$ and $x = 3$. These split the line into three inter­vals.

Step 4. Test the sign of $f'$ in each inter­val.

• $x < -2$ (try $x = -3$): $6(-6)(-1) = 36 > 0$ → increas­ing.
• $-2 < x < 3$ (try $x = 0$): $6(-3)(2) = -36 < 0$ → decreas­ing.
• $x > 3$ (try $x = 4$): $6(1)(6) = 36 > 0$ → increas­ing.

Check­ing the answer#

$f(-2) = -16 - 12 + 72 + 7 = 51$ and $f(3) = 54 - 27 - 108 + 7 = -74$. The func­tion falls from $51$ to $-74$ on $(-2, 3)$, con­sis­tent with "decreas­ing" there.

Answer#

Increas­ing on $(-\infty, -2)$ and $(3, \infty)$; decreas­ing on $(-2, 3)$.

Ques­tion 6: A func­tion increas­ing every­where#

The prob­lem#

Show that $f(x) = x^3 + 3x$ is increas­ing on $\mathbf{R}$.

Under­stand­ing the prob­lem#

You must prove that $f$ increases on the whole real line. It is enough to show $f'(x) > 0$ for every real $x$.

The idea#

Dif­fer­en­ti­ate and show the deriv­a­tive can never be zero or neg­a­tive.

Step-by-step solu­tion#

Step 1. Dif­fer­en­ti­ate.

$$f'(x) = 3x^2 + 3.$$

Step 2. For every real $x$, $x^2 \ge 0$, so $3x^2 \ge 0$.

Step 3. There­fore

$$f'(x) = 3x^2 + 3 \ge 0 + 3 = 3 > 0 \quad \text{for all } x \in \mathbf{R}.$$

Step 4. Since $f'(x) > 0$ every­where, $f$ is increas­ing on $\mathbf{R}$.

Check­ing the answer#

$f(-1) = -4$, $f(0) = 0$, $f(1) = 4$: the val­ues rise as $x$ rises. ✓

Answer#

$f'(x) = 3x^2 + 3 \ge 3 > 0$ for all real $x$, so $f$ is increas­ing on $\mathbf{R}$.

Ques­tion 7: Increas­ing inter­vals of $x^2 e^{-x}$#

The prob­lem#

Find the inter­vals on which $f(x) = x^2 e^{-x}$ increases.

Under­stand­ing the prob­lem#

You need the inter­vals where $f'(x) > 0$. The func­tion is a prod­uct, so use the prod­uct rule.

The idea#

Dif­fer­en­ti­ate with the prod­uct rule, fac­torise, and remem­ber that $e^{-x} > 0$ always, so it never affects the sign.

Step-by-step solu­tion#

Step 1. Prod­uct rule with $u = x^2$, $v = e^{-x}$ ($v' = -e^{-x}$).

$$f'(x) = 2x e^{-x} + x^2(-e^{-x}) = 2xe^{-x} - x^2e^{-x}.$$

Step 2. Take out the com­mon fac­tor $xe^{-x}$.

$$f'(x) = xe^{-x}(2 - x).$$

Step 3. Since $e^{-x} > 0$, the sign of $f'$ is the sign of $x(2 - x)$. This is zero at $x = 0$ and $x = 2$.

Step 4. Test each inter­val.

• $x < 0$ (try $x = -1$): $(-1)(3) < 0$ → decreas­ing.
• $0 < x < 2$ (try $x = 1$): $(1)(1) > 0$ → increas­ing.
• $x > 2$ (try $x = 3$): $(3)(-1) < 0$ → decreas­ing.

Check­ing the answer#

$f(0) = 0$, $f(1) = e^{-1} \approx 0.37$, $f(2) = 4e^{-2} \approx 0.54$: ris­ing on $(0, 2)$ ✓. $f(3) = 9e^{-3} \approx 0.45$: falling after $2$ ✓.

Answer#

$f$ is increas­ing on $(0, 2)$ (and decreas­ing on $(-\infty, 0)$ and $(2, \infty)$).

Ques­tion 8: Monot­o­nic­ity of $\log(\sin x)$#

The prob­lem#

Show that $f(x) = \log(\sin x)$ is increas­ing on $\displaystyle \left(0, \tfrac{\pi}{2}\right)$ and decreas­ing on $\displaystyle \left(\tfrac{\pi}{2}, \pi\right)$.

Under­stand­ing the prob­lem#

On $(0, \pi)$, $\sin x > 0$, so $\log(\sin x)$ is defined. You must show $f' > 0$ on the first inter­val and $f' < 0$ on the sec­ond.

The idea#

Dif­fer­en­ti­ate using the chain rule: $\displaystyle \tfrac{d}{dx}\log u = \tfrac{1}{u}\cdot\tfrac{du}{dx}$. Then study the sign of $\cot x$.

Step-by-step solu­tion#

Step 1. Dif­fer­en­ti­ate.

$$\displaystyle f'(x) = \frac{1}{\sin x}\cdot\cos x = \cot x.$$

Step 2. On $\displaystyle \left(0, \tfrac{\pi}{2}\right)$: both $\cos x > 0$ and $\sin x > 0$, so $\cot x > 0$. Hence $f$ is increas­ing there.

Step 3. On $\displaystyle \left(\tfrac{\pi}{2}, \pi\right)$: $\cos x < 0$ and $\sin x > 0$, so $\cot x < 0$. Hence $f$ is decreas­ing there.

Check­ing the answer#

$\sin x$ itself rises from $0$ to $1$ on $\displaystyle \left(0, \tfrac{\pi}{2}\right)$ and falls back on $\displaystyle \left(\tfrac{\pi}{2}, \pi\right)$; $\log$ is increas­ing, so $\log(\sin x)$ fol­lows the same pat­tern. ✓

Answer#

$f'(x) = \cot x$, which is pos­i­tive on $\displaystyle \left(0, \tfrac{\pi}{2}\right)$ and neg­a­tive on $\displaystyle \left(\tfrac{\pi}{2}, \pi\right)$; so $f$ increases on the first inter­val and decreases on the sec­ond.

Ques­tion 9: A para­me­ter that makes a func­tion increas­ing#

The prob­lem#

For what $a$ is $f(x) = ax + \cos x$ increas­ing on $\mathbf{R}$?

Under­stand­ing the prob­lem#

You must find all val­ues of the con­stant $a$ for which $f$ never goes down, what­ever $x$ is.

The idea#

Require $f'(x) = a - \sin x \ge 0$ for every $x$. Since $\sin x$ can be as large as $1$, $a$ must be at least $1$.

Step-by-step solu­tion#

Step 1. Dif­fer­en­ti­ate.

$$f'(x) = a - \sin x.$$

Step 2. For $f$ to increase on all of $\mathbf{R}$, we need $a - \sin x \ge 0$, i.e. $a \ge \sin x$, for every $x$.

Step 3. The largest value of $\sin x$ is $1$. So the con­di­tion holds for all $x$ exactly when

$$a \ge 1.$$

Step 4. If $a > 1$, then $f'(x) \ge a - 1 > 0$ every­where. If $a = 1$, then $f'(x) = 1 - \sin x \ge 0$, and it is zero only at iso­lated points $\displaystyle x = \tfrac{\pi}{2} + 2n\pi$; the func­tion still keeps ris­ing between them, so it is increas­ing. If $a < 1$, then at $\displaystyle x = \tfrac{\pi}{2}$, $f' = a - 1 < 0$, so $f$ decreases near there.

Check­ing the answer#

Take $a = 0.5$: $\displaystyle f'\left(\tfrac{\pi}{2}\right) = 0.5 - 1 < 0$, so it fails ✓. Take $a = 2$: $f' \ge 1 > 0$ ✓.

Answer#

$a \ge 1$

Ques­tion 10: A melt­ing snow­ball#

The prob­lem#

A spher­i­cal snow­ball melts so that its vol­ume decreases at $10$ cm³/min. How fast does the radius shrink when the radius is $5$ cm?

Under­stand­ing the prob­lem#

"Decreases at $10$ cm³/min" means $\displaystyle \tfrac{dV}{dt} = -10$ (neg­a­tive, because the vol­ume is going down). You need $\displaystyle \tfrac{dr}{dt}$ when $r = 5$ cm.

The idea#

Dif­fer­en­ti­ate $\displaystyle V = \tfrac{4}{3}\pi r^3$ with respect to $t$ and solve for $\displaystyle \tfrac{dr}{dt}$.

Step-by-step solu­tion#

Step 1. Dif­fer­en­ti­ate.

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

Step 2. Sub­sti­tute $\displaystyle \tfrac{dV}{dt} = -10$ and $r = 5$.

$$\displaystyle -10 = 4\pi(25)\frac{dr}{dt} = 100\pi\frac{dr}{dt}.$$

Step 3. Solve.

$$\displaystyle \frac{dr}{dt} = -\frac{10}{100\pi} = -\frac{1}{10\pi} \text{ cm/min}.$$

The radius is shrink­ing at $\displaystyle \tfrac{1}{10\pi} \approx 0.032$ cm/min.

Check­ing the answer#

The sign is neg­a­tive (radius decreas­ing) ✓, and units cm³/min ÷ cm² = cm/min ✓.

Answer#

$\displaystyle \frac{dr}{dt} = -\frac{1}{10\pi}$ cm/min: the radius shrinks at $\displaystyle \frac{1}{10\pi}$ cm/min.

Com­mon mis­take to avoid#

Using $\displaystyle \tfrac{dV}{dt} = +10$ gives a pos­i­tive answer, which would mean the snow­ball is grow­ing. A decrease is a neg­a­tive rate.