Chapter 8 — Exercises
38 problems spanning implicit differentiation, tangent lines to implicit curves, higher-order implicit derivatives, inverse-function derivatives, logarithmic differentiation, and related rates (ladders, tanks, expanding figures, approaching vehicles, tracking angles). Tiered ⭐ (routine) to ⭐⭐⭐⭐ (multi-step modeling, applied to a named field).
Work each problem before opening its solution. Selected answers are also collected in
appendices/answers-to-selected.md. Section references point to the chapter's own sections (§8.1–§8.17).
Difficulty Distribution
| Tier | Meaning | Count | Problems |
|---|---|---|---|
| ⭐ | Routine application of one rule | 8 | A1–A4, D1, E1, F1, F2 |
| ⭐⭐ | Two steps or a substitution | 12 | A5–A7, B1–B2, C1, D2, E2, F3–F4, G1, H1 |
| ⭐⭐⭐ | Multi-step setup, careful algebra | 12 | A8, B3, C2–C3, D3, F5–F6, G2–G3, H2, I1–I2 |
| ⭐⭐⭐⭐ | Modeling problem, applied field, synthesis | 6 | F7, G4, H3, I3, I4, I5 |
| Total | 38 |
The applied ⭐⭐⭐⭐ problems span physics/aviation (F7, I3), chemical/civil engineering (G4, I4), economics (I5), and astronomy/tracking (H3).
Part A — Basic Implicit Differentiation
A1 ⭐ Find $dy/dx$ for $x^2 + y^2 = 16$.
Solution
Differentiate: $2x + 2y\,y' = 0$, so $y' = -\dfrac{x}{y}$ (the §8.2 circle result).A2 ⭐ Find $dy/dx$ for $xy = 6$.
Solution
Product rule on the left: $y + x\,y' = 0$, so $y' = -\dfrac{y}{x}$.A3 ⭐ Find $dy/dx$ for $x^3 + y^3 = 8$.
Solution
$3x^2 + 3y^2\,y' = 0 \Rightarrow y' = -\dfrac{x^2}{y^2}$.A4 ⭐ Find $dy/dx$ for $4x^2 - y^2 = 9$.
Solution
$8x - 2y\,y' = 0 \Rightarrow y' = \dfrac{4x}{y}$.A5 ⭐⭐ Find $dy/dx$ for $x^2 + xy + y^2 = 7$, and evaluate at $(1, 2)$.
Solution
$2x + (y + x\,y') + 2y\,y' = 0$. Collect: $y'(x + 2y) = -(2x + y)$, so $y' = -\dfrac{2x + y}{x + 2y}$. At $(1,2)$: $y' = -\dfrac{4}{5}$.A6 ⭐⭐ Find $dy/dx$ for $\sin(xy) = x + y$.
Solution
Differentiate the left with the chain and product rules: $\cos(xy)\,(y + x\,y') = 1 + y'$. Expand: $y\cos(xy) + x\cos(xy)\,y' = 1 + y'$. Collect $y'$: $y'\big(x\cos(xy) - 1\big) = 1 - y\cos(xy)$, so $y' = \dfrac{1 - y\cos(xy)}{x\cos(xy) - 1}$.A7 ⭐⭐ Find $dy/dx$ for $e^{xy} = x + y$ (see §8.2 method, exponential term).
Solution
$e^{xy}(y + x\,y') = 1 + y'$. With $e^{xy} = x+y$ on the curve you may substitute, but symbolically: $y'\big(x e^{xy} - 1\big) = 1 - y e^{xy}$, so $y' = \dfrac{1 - y\,e^{xy}}{x\,e^{xy} - 1}$.A8 ⭐⭐⭐ For the folium $x^3 + y^3 = 6xy$ (§8.2), confirm $y'(3,3) = -1$ and find every point where the tangent is horizontal.
Solution
From §8.2, $y' = \dfrac{2y - x^2}{y^2 - 2x}$. At $(3,3)$: $\dfrac{6 - 9}{9 - 6} = -1$. ✓ Horizontal tangents need numerator $=0$ (denominator $\ne 0$): $2y = x^2$. Substitute $y = x^2/2$ into the curve: $x^3 + x^6/8 = 6x\cdot x^2/2 = 3x^3$, so $x^6/8 = 2x^3 \Rightarrow x^3 = 16$, $x = 16^{1/3} = 2^{4/3}$. Then $y = x^2/2 = 2^{8/3}/2 = 2^{5/3}$. One horizontal tangent at $\big(2^{4/3},\,2^{5/3}\big)\approx(2.52,\,3.17)$ (besides the trivial origin).Part B — Tangent Lines to Implicit Curves
B1 ⭐⭐ Find the tangent line to $x^3 + y^3 = 9$ at $(1, 2)$ (see §8.4).
Solution
$3x^2 + 3y^2\,y' = 0 \Rightarrow y' = -x^2/y^2 = -1/4$ at $(1,2)$. Tangent: $y - 2 = -\tfrac14(x-1)$, i.e. $y = -\tfrac{x}{4} + \tfrac94$.B2 ⭐⭐ Find the tangent line to the ellipse $\dfrac{x^2}{9} + \dfrac{y^2}{16} = 1$ at $(0, 4)$, and at $(3, 0)$.
Solution
$\dfrac{2x}{9} + \dfrac{2y}{16}y' = 0 \Rightarrow y' = -\dfrac{16x}{9y}$. At $(0,4)$: $y' = 0$, horizontal tangent $y = 4$. At $(3,0)$: denominator vanishes, **vertical tangent** $x = 3$ (§8.4 vertical-tangent test).B3 ⭐⭐⭐ Find all points on the circle $x^2 + y^2 = 25$ where the tangent line has slope $\tfrac34$.
Solution
$y' = -x/y = \tfrac34 \Rightarrow x = -\tfrac34 y$. Substitute: $\tfrac{9}{16}y^2 + y^2 = 25 \Rightarrow \tfrac{25}{16}y^2 = 25 \Rightarrow y^2 = 16$, $y = \pm 4$. Points: $(-3, 4)$ and $(3, -4)$.Part C — Higher-Order Implicit Derivatives
C1 ⭐⭐ Find $y''$ for $x^2 + y^2 = 1$.
Solution
$y' = -x/y$. Then $y'' = -\dfrac{y - x\,y'}{y^2} = -\dfrac{y - x(-x/y)}{y^2} = -\dfrac{y^2 + x^2}{y^3} = -\dfrac{1}{y^3}$ (using $x^2+y^2=1$), matching the §8.3 pattern.C2 ⭐⭐⭐ Find $y''$ at $(3, 4)$ for $x^2 + y^2 = 25$ and state the concavity.
Solution
From §8.3, $y'' = -25/y^3$. At $(3,4)$: $y'' = -25/64 < 0$, so the upper semicircle is **concave down** there.C3 ⭐⭐⭐ For $xy = 1$, show $y'' = \dfrac{2}{x^3}$ two ways: implicitly, and from $y = 1/x$.
Solution
Implicit: $y' = -y/x$. Then $y'' = -\dfrac{y'\,x - y}{x^2} = -\dfrac{(-y/x)x - y}{x^2} = -\dfrac{-2y}{x^2} = \dfrac{2y}{x^2}$. With $y = 1/x$: $y'' = \dfrac{2}{x^3}$. Direct: $y = x^{-1}$, $y' = -x^{-2}$, $y'' = 2x^{-3}$. ✓ Agreement.Part D — Inverse-Function Derivatives
D1 ⭐ Find $\dfrac{d}{dx}[\arctan(x^2)]$.
Solution
Chain rule with $\arctan'(u)=1/(1+u^2)$: $\dfrac{1}{1+x^4}\cdot 2x = \dfrac{2x}{1+x^4}$.D2 ⭐⭐ Derive $\dfrac{d}{dx}\arccos x$ by implicit differentiation (§8.5).
Solution
$y = \arccos x \Leftrightarrow x = \cos y$ with $y\in[0,\pi]$, so $\sin y \ge 0$. Differentiate: $1 = -\sin y\, y' \Rightarrow y' = -\dfrac{1}{\sin y} = -\dfrac{1}{\sqrt{1-\cos^2 y}} = -\dfrac{1}{\sqrt{1-x^2}}$.D3 ⭐⭐⭐ Derive $\dfrac{d}{dx}\,\text{arcsec}\,x$ for $x > 1$.
Solution
$y = \text{arcsec}\,x \Leftrightarrow x = \sec y$. Differentiate: $1 = \sec y\tan y\, y'$, so $y' = \dfrac{1}{\sec y\tan y}$. With $\sec y = x$ and $\tan y = \sqrt{\sec^2 y - 1} = \sqrt{x^2-1}$ (positive on the principal branch for $x>1$): $y' = \dfrac{1}{x\sqrt{x^2-1}}$. (The full-domain formula carries $|x|$: $\dfrac{1}{|x|\sqrt{x^2-1}}$.)Part E — Logarithmic Differentiation
E1 ⭐ Differentiate $y = x^x$ (§8.6).
Solution
$\ln y = x\ln x$, so $y'/y = \ln x + 1$ and $y' = x^x(\ln x + 1)$.E2 ⭐⭐ Differentiate $y = (\sin x)^x$.
Solution
$\ln y = x\ln\sin x$. Differentiate: $\dfrac{y'}{y} = \ln\sin x + x\cdot\dfrac{\cos x}{\sin x} = \ln\sin x + x\cot x$. So $y' = (\sin x)^x\big(\ln\sin x + x\cot x\big)$.Part F — Related Rates: Expanding and Filling Figures
F1 ⭐ A circle's radius increases at $2$ cm/s. Find $dA/dt$ when $r = 5$ cm.
Solution
$A = \pi r^2 \Rightarrow \dfrac{dA}{dt} = 2\pi r\dfrac{dr}{dt} = 2\pi(5)(2) = 20\pi \approx 62.8$ cm²/s.F2 ⭐ A sphere's radius increases at $0.5$ in/s. Find $dV/dt$ when $r = 6$ in.
Solution
$V = \tfrac43\pi r^3 \Rightarrow \dfrac{dV}{dt} = 4\pi r^2\dfrac{dr}{dt} = 4\pi(36)(0.5) = 72\pi \approx 226.2$ in³/s.F3 ⭐⭐ A square's diagonal grows at $3$ cm/s. Find $dA/dt$ when the diagonal is $10$ cm.
Solution
With diagonal $d$, area $A = d^2/2$ (since side $s = d/\sqrt2$, $A = s^2 = d^2/2$). $\dfrac{dA}{dt} = d\dfrac{dd}{dt} = 10(3) = 30$ cm²/s.F4 ⭐⭐ A spherical balloon is inflated so its radius grows at $2$ cm/s. Find $dV/dt$ and $dS/dt$ when $r = 10$ cm (§8.14 expanding balloon).
Solution
$\dfrac{dV}{dt} = 4\pi r^2\dfrac{dr}{dt} = 4\pi(100)(2) = 800\pi \approx 2513$ cm³/s. $\dfrac{dS}{dt} = 8\pi r\dfrac{dr}{dt} = 8\pi(10)(2) = 160\pi \approx 503$ cm²/s.F5 ⭐⭐⭐ Sand pours into a conical pile at $10$ ft³/min; the base radius always equals twice the height. How fast is the height rising when $h = 5$ ft?
Solution
$r = 2h$, so $V = \tfrac13\pi r^2 h = \tfrac13\pi(2h)^2 h = \tfrac43\pi h^3$. Then $\dfrac{dV}{dt} = 4\pi h^2\dfrac{dh}{dt}$. At $h=5$: $10 = 4\pi(25)\dfrac{dh}{dt} = 100\pi\dfrac{dh}{dt}$, so $\dfrac{dh}{dt} = \dfrac{1}{10\pi}\approx 0.0318$ ft/min.F6 ⭐⭐⭐ A trough $10$ ft long has an isosceles-triangle cross-section, vertex down, $2$ ft wide at top and $1$ ft deep. Water fills at $0.5$ ft³/min. How fast is the level rising when the water is $6$ in deep? (Compare §8.12.)
Solution
At depth $h$, the surface width by similar triangles is $w = 2h$ (full triangle: width $2$ at depth $1$). Cross-sectional area $A = \tfrac12 w h = \tfrac12(2h)(h) = h^2$. Volume $V = 10h^2$, so $\dfrac{dV}{dt} = 20h\dfrac{dh}{dt}$. At $h = 0.5$ ft: $0.5 = 20(0.5)\dfrac{dh}{dt} = 10\dfrac{dh}{dt}$, so $\dfrac{dh}{dt} = 0.05$ ft/min $= 0.6$ in/min.F7 ⭐⭐⭐⭐ (Chemical engineering.) A hemispherical bowl reactor of radius $R = 1$ m drains so that the volume of liquid of depth $h$ is $V = \pi\big(R h^2 - \tfrac13 h^3\big)$. Liquid leaves at a constant $0.02$ m³/min. How fast is the depth falling when $h = 0.5$ m?
Solution
Differentiate: $\dfrac{dV}{dt} = \pi\big(2Rh - h^2\big)\dfrac{dh}{dt} = \pi h(2R - h)\dfrac{dh}{dt}$. With $\dfrac{dV}{dt} = -0.02$, $R=1$, $h=0.5$: $\pi(0.5)(2 - 0.5) = \pi(0.5)(1.5) = 0.75\pi$. So $\dfrac{dh}{dt} = \dfrac{-0.02}{0.75\pi} \approx -0.00849$ m/min — the level falls about $8.5$ mm/min. (Negative sign: draining.)Part G — Related Rates: The Sliding Ladder and Kin
G1 ⭐⭐ A $13$-ft ladder leans on a wall; the base slides away at $2$ ft/s. How fast does the top descend when the base is $5$ ft out? (§8.8 method.)
Solution
$x^2 + y^2 = 169$; at $x=5$, $y = 12$. Differentiate: $2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 0$, so $\dfrac{dy}{dt} = -\dfrac{x}{y}\dfrac{dx}{dt} = -\dfrac{5}{12}(2) = -\dfrac{5}{6}\approx -0.833$ ft/s (descending).G2 ⭐⭐⭐ For the $13$-ft ladder of G1, how fast is the area of the triangle formed by the ladder, wall, and ground changing at that instant?
Solution
$A = \tfrac12 xy$. $\dfrac{dA}{dt} = \tfrac12\big(\dfrac{dx}{dt}y + x\dfrac{dy}{dt}\big) = \tfrac12\big(2\cdot 12 + 5\cdot(-\tfrac56)\big) = \tfrac12\big(24 - \tfrac{25}{6}\big) = \tfrac12\cdot\tfrac{119}{6} = \dfrac{119}{12}\approx 9.92$ ft²/s (increasing).G3 ⭐⭐⭐ A $6$-ft person walks at $4$ ft/s away from an $18$-ft lamppost. How fast does (a) the shadow lengthen, (b) the shadow tip move? (§8.11.)
Solution
Similar triangles: $\dfrac{s}{6} = \dfrac{x+s}{18} \Rightarrow s = \tfrac12 x$. (a) $\dfrac{ds}{dt} = \tfrac12\dfrac{dx}{dt} = 2$ ft/s. (b) Tip position $x + s = \tfrac32 x$, so $\dfrac{d}{dt}(x+s) = \tfrac32(4) = 6$ ft/s.G4 ⭐⭐⭐⭐ (Civil engineering — reservoir.) A water tank is an inverted cone, height $12$ m, top radius $3$ m. During a flood, inflow is $5$ m³/min while a spillway releases $1.2$ m³/min. How fast is the water level rising when the depth is $4$ m?
Solution
Net inflow $\dfrac{dV}{dt} = 5 - 1.2 = 3.8$ m³/min. Similar triangles: $\dfrac{r}{h} = \dfrac{3}{12} = \tfrac14$, so $r = h/4$ and $V = \tfrac13\pi r^2 h = \tfrac13\pi\dfrac{h^2}{16}h = \dfrac{\pi h^3}{48}$. Differentiate: $\dfrac{dV}{dt} = \dfrac{\pi h^2}{16}\dfrac{dh}{dt}$. At $h = 4$: $3.8 = \dfrac{\pi(16)}{16}\dfrac{dh}{dt} = \pi\dfrac{dh}{dt}$, so $\dfrac{dh}{dt} = \dfrac{3.8}{\pi}\approx 1.21$ m/min. The spillway calculation determines whether the tank overtops — exactly the §8.9 engineering point.Part H — Related Rates: Angles and Tracking
H1 ⭐⭐ A balloon rises vertically at $5$ ft/s; an observer stands $100$ ft from the launch point. How fast is the angle of elevation changing when the balloon is $50$ ft up? (§8.13.)
Solution
$\tan\theta = h/100$. Differentiate: $\sec^2\theta\dfrac{d\theta}{dt} = \dfrac{1}{100}\dfrac{dh}{dt} = \dfrac{5}{100} = 0.05$. At $h=50$: $\tan\theta = 0.5$, so $\sec^2\theta = 1 + 0.25 = 1.25$. Thus $\dfrac{d\theta}{dt} = \dfrac{0.05}{1.25} = 0.04$ rad/s.H2 ⭐⭐⭐ A lighthouse beacon $2$ km offshore rotates at $3$ rev/min ($= 6\pi$ rad/min). How fast does the spot of light move along the straight shoreline at the point nearest the lighthouse?
Solution
Let $x$ be the distance along shore from the nearest point and $\theta$ the beam angle from the perpendicular: $x = 2\tan\theta$ (km). $\dfrac{dx}{dt} = 2\sec^2\theta\dfrac{d\theta}{dt}$. At the nearest point $\theta = 0$, $\sec^2\theta = 1$: $\dfrac{dx}{dt} = 2(1)(6\pi) = 12\pi \approx 37.7$ km/min.H3 ⭐⭐⭐⭐ (Astronomy/tracking.) A radar dish tracks a rocket launched straight up from a pad $4$ km away (horizontal ground distance). When the rocket is $3$ km high it is climbing at $0.5$ km/s. Find (a) the rate the straight-line range is increasing, and (b) the rate the elevation angle is increasing.
Solution
Let $h$ be altitude, $D$ the range, $\theta$ the elevation. (a) $D^2 = 4^2 + h^2 = 16 + h^2$. At $h=3$: $D = \sqrt{16+9} = 5$ km. Differentiate: $2D\dfrac{dD}{dt} = 2h\dfrac{dh}{dt}$, so $\dfrac{dD}{dt} = \dfrac{h}{D}\dfrac{dh}{dt} = \dfrac{3}{5}(0.5) = 0.3$ km/s. (b) $\tan\theta = h/4$. Differentiate: $\sec^2\theta\dfrac{d\theta}{dt} = \dfrac14\dfrac{dh}{dt}$. At $h=3$: $\tan\theta = 3/4$, $\sec^2\theta = 1 + 9/16 = 25/16$. So $\dfrac{d\theta}{dt} = \dfrac{16}{25}\cdot\dfrac14(0.5) = \dfrac{16}{25}\cdot 0.125 = 0.08$ rad/s.Part I — Multi-Rate, Theory, and Field Modeling
I1 ⭐⭐⭐ A rectangle's length grows at $2$ cm/s while its width shrinks at $1$ cm/s. When length $= 8$, width $= 5$, find the rate of change of (a) area, (b) perimeter, (c) diagonal.
Solution
(a) $A = LW$: $\dfrac{dA}{dt} = W\dfrac{dL}{dt} + L\dfrac{dW}{dt} = 5(2) + 8(-1) = 2$ cm²/s. (b) $P = 2L + 2W$: $\dfrac{dP}{dt} = 2(2) + 2(-1) = 2$ cm/s. (c) $D = \sqrt{L^2 + W^2} = \sqrt{89}$: $D\dfrac{dD}{dt} = L\dfrac{dL}{dt} + W\dfrac{dW}{dt} = 16 - 5 = 11$, so $\dfrac{dD}{dt} = \dfrac{11}{\sqrt{89}}\approx 1.166$ cm/s.I2 ⭐⭐⭐ Prove the inverse-derivative formula: show $(f^{-1})'(x) = \dfrac{1}{f'(f^{-1}(x))}$ (§8.5).
Solution
Let $y = f^{-1}(x)$, so $x = f(y)$. Differentiate both sides w.r.t. $x$: $1 = f'(y)\,y'$. Hence $y' = \dfrac{1}{f'(y)} = \dfrac{1}{f'(f^{-1}(x))}$, valid where $f'(y)\ne 0$.I3 ⭐⭐⭐⭐ (Physics/aviation — closing speed.) Two aircraft fly at the same altitude toward a common waypoint along perpendicular straight tracks. Jet A is $40$ km from the waypoint closing at $600$ km/h; Jet B is $30$ km from it closing at $800$ km/h. How fast is the distance between them changing at this instant? Are they getting closer or farther apart?
Solution
Let $a$, $b$ be distances to the waypoint, $D = \sqrt{a^2 + b^2}$. Closing means $\dfrac{da}{dt} = -600$, $\dfrac{db}{dt} = -800$. At $a=40$, $b=30$: $D = \sqrt{1600 + 900} = 50$ km. Differentiate $D^2 = a^2 + b^2$: $D\dfrac{dD}{dt} = a\dfrac{da}{dt} + b\dfrac{db}{dt} = 40(-600) + 30(-800) = -24000 - 24000 = -48000$. So $\dfrac{dD}{dt} = \dfrac{-48000}{50} = -960$ km/h — closing at $960$ km/h. (A genuine collision-avoidance computation.)I4 ⭐⭐⭐⭐ (Chemical engineering — ideal gas.) A gas obeys $PV = nRT$ with $nR = 8.314$ J/K. A piston compresses the gas so that volume decreases at $0.01$ m³/s while temperature is held constant at $T = 300$ K. When $V = 0.5$ m³, how fast is the pressure rising? (Use §8.15 / §8.17 constant-$T$ relation.)
Solution
With $T$ constant, $PV = nRT = 8.314(300) = 2494.2$ J (constant). Differentiate $PV = \text{const}$: $\dfrac{dP}{dt}V + P\dfrac{dV}{dt} = 0$. At $V = 0.5$: $P = 2494.2/0.5 = 4988.4$ Pa. With $\dfrac{dV}{dt} = -0.01$: $\dfrac{dP}{dt} = -\dfrac{P}{V}\dfrac{dV}{dt} = -\dfrac{4988.4}{0.5}(-0.01) = 99.77$ Pa/s. Pressure rises as volume shrinks — Boyle's law in rate form.I5 ⭐⭐⭐⭐ (Economics — isoquant.) A firm's output obeys the Cobb–Douglas relation $Q = 100\,L^{0.5}K^{0.5}$, held fixed at $Q_0 = 1000$. Using implicit differentiation (§8.15), find the marginal rate of technical substitution $-\dfrac{dK}{dL}$ at the point $L = 100$, $K = 100$, and interpret it.
Solution
Along $Q = Q_0$, differentiate $100L^{0.5}K^{0.5} = 1000$ w.r.t. $L$ (with $K = K(L)$): $100\big(0.5L^{-0.5}K^{0.5} + 0.5L^{0.5}K^{-0.5}\dfrac{dK}{dL}\big) = 0$. Solve: $\dfrac{dK}{dL} = -\dfrac{L^{-0.5}K^{0.5}}{L^{0.5}K^{-0.5}} = -\dfrac{K}{L}$. At $L = K = 100$: $\dfrac{dK}{dL} = -1$, so MRTS $= 1$. Interpretation: at this balanced bundle, one extra unit of labor lets the firm shed exactly one unit of capital while holding output at $1000$. (Check: this is the §8.15 result $\dfrac{dK}{dL} = -\dfrac{\alpha K}{\beta L}$ with $\alpha=\beta=0.5$.)Reflection
Implicit differentiation extends the chain rule to relations that refuse to be solved for $y$; related rates points the same chain rule at time, turning a static geometric truth into a relationship among rates. Every problem above — circle, folium, ladder, cone, radar dish, isoquant — is one gesture: differentiate the connecting equation, then substitute the instant. Master that order (§8.7, step 6) and the rest is algebra.