Figure 5.2: Implicit Curve Tangent & Conical Related Rates
1. Implicit Differentiation Fundamentals (Syllabus 5.10)
When an algebraic relation connects $x$ and $y$ implicitly in the form $F(x, y) = 0$ where $y$ cannot be easily isolated as an explicit function $y = f(x)$, we differentiate every term with respect to $x$. Because $y$ depends on $x$, applying the chain rule to any function of $y$ yields a factor of $\frac{dy}{dx}$:
Power & Chain Rule on $y$
$$\frac{d}{dx}[y^n] = n y^{n-1} \frac{dy}{dx}$$ $$\frac{d}{dx}[e^{ky}] = k e^{ky} \frac{dy}{dx}$$ $$\frac{d}{dx}[\ln y] = \frac{1}{y} \frac{dy}{dx}$$Mixed Product & Quotient Rules
$$\frac{d}{dx}[x^m y^n] = m x^{m-1} y^n + n x^m y^{n-1} \frac{dy}{dx}$$ $$\frac{d}{dx}[xy] = y + x\frac{dy}{dx}$$ $$\frac{d}{dx}\left[\frac{x}{y}\right] = \frac{y(1) - x\frac{dy}{dx}}{y^2}$$Trigonometric Terms
$$\frac{d}{dx}[\sin(xy)] = \cos(xy)\left(y + x\frac{dy}{dx}\right)$$ $$\frac{d}{dx}[\arctan(y)] = \frac{1}{1 + y^2} \frac{dy}{dx}$$2. Second Derivatives $\frac{d^2y}{dx^2}$ on Implicit Curves
Finding the second derivative implicitly is a hallmark of IB AA HL Paper 1 exam questions. The systematic procedure is:
- Differentiate the original equation once to find an expression for $\frac{dy}{dx} = \frac{N(x, y)}{D(x, y)}$.
- Differentiate both sides with respect to $x$ using the quotient rule: $$\frac{d^2y}{dx^2} = \frac{d}{dx}\left[\frac{dy}{dx}\right] = \frac{D \cdot \frac{d}{dx}[N] - N \cdot \frac{d}{dx}[D]}{D^2}$$
- The Critical Step: Wherever $\frac{dy}{dx}$ appears in the resulting numerator, substitute the first derivative expression back in.
- Clear compound fractions and simplify using the original curve equation $F(x, y) = 0$ to collapse complicated terms into constants!
3. Horizontal and Vertical Tangents Strategy
Express the first derivative as a single rational fraction: $\frac{dy}{dx} = \frac{N(x, y)}{D(x, y)}$.
- Horizontal Tangents ($\frac{dy}{dx} = 0$): Set the numerator $N(x, y) = 0$ (with $D(x, y) \neq 0$). Solve for one variable in terms of the other, then substitute back into the original curve equation to obtain complete coordinates $(x, y)$.
- Vertical Tangents ($\frac{dy}{dx} \to \pm\infty$): Set the denominator $D(x, y) = 0$ (with $N(x, y) \neq 0$). Substitute this condition into the original curve equation to find coordinate points.
- Singularities / Cusps ($N = 0$ and $D = 0$): The tangent line is undefined or multiple branches intersect (e.g., node at the origin of a folium).
4. Related Rates of Change (Syllabus 5.11)
Related rates problems connect the time derivatives $\frac{dx}{dt}, \frac{dy}{dt}, \frac{dV}{dt}, \frac{d\theta}{dt}$ of geometrically related quantities using the chain rule:
$$\frac{dY}{dt} = \frac{dY}{dX} \cdot \frac{dX}{dt}$$Problem-Solving Method for Related Rates
- Assign variables: Draw a large diagram. Clearly label constant values (e.g., ladder length $L=5$) vs time-dependent variables ($x(t), y(t), \theta(t)$). Never substitute instantaneous numerical values before differentiating!
- State the given and required rates: e.g., Given $\frac{dV}{dt} = +0.4\text{ m}^3/\text{s}$, Find $\left.\frac{dh}{dt}\right|_{h=2}$.
- Formulate the geometric equation: Write a formula relating only the relevant variables (e.g., Pythagoras $x^2 + y^2 = L^2$, Volume $V = \frac{1}{3}\pi r^2 h$, Trig $\tan\theta = \frac{y}{x}$).
- Eliminate secondary variables: Use similar triangles (e.g., $\frac{r}{h} = \frac{R}{H}$) to write the formula in terms of a single variable before differentiating.
- Differentiate implicitly with respect to time $t$: Apply $\frac{d}{dt}$ across both sides, substitute the instantaneous quantities, and solve for the unknown rate with correct physical units.
Worked Examples
Worked Example 1: Tangent & Normal to the Folium of Descartes
Consider the curve $x^3 + y^3 = 6xy$.
(a) Find an expression for $\frac{dy}{dx}$ in terms of $x$ and $y$.
(b) Find the equation of the normal to the curve at the point $(3, 3)$.
Solution:
(a) Differentiate implicitly with respect to $x$:
$$\frac{d}{dx}[x^3] + \frac{d}{dx}[y^3] = 6\frac{d}{dx}[xy]$$ $$3x^2 + 3y^2\frac{dy}{dx} = 6\left(y + x\frac{dy}{dx}\right) \quad \text{[M1]}$$ $$3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx}$$Divide by 3 and collect $\frac{dy}{dx}$ terms:
$$(y^2 - 2x)\frac{dy}{dx} = 2y - x^2 \quad \text{[M1]}$$ $$\frac{dy}{dx} = \frac{2y - x^2}{y^2 - 2x} \quad \text{[A1]}$$(b) Evaluate the tangent gradient at $(3, 3)$:
$$m_T = \left.\frac{dy}{dx}\right|_{(3, 3)} = \frac{2(3) - 3^2}{3^2 - 2(3)} = \frac{6 - 9}{9 - 6} = \frac{-3}{3} = -1 \quad \text{[A1]}$$The normal gradient is perpendicular: $m_N = -\frac{1}{m_T} = -\frac{1}{-1} = 1$. [M1]
Equation of the normal line:
Worked Example 2: Second Derivative Collapse on an Ellipse
Show that for any point $(x, y)$ with $y \neq 0$ on the ellipse $x^2 + 4y^2 = 8$, the second derivative satisfies: $$\frac{d^2y}{dx^2} = -\frac{2}{y^3}$$
Solution:
1. Differentiate $x^2 + 4y^2 = 8$ implicitly with respect to $x$:
$$2x + 8y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{2x}{8y} = -\frac{x}{4y} \quad \text{[M1][A1]}$$2. Differentiate $\frac{dy}{dx} = -\frac{x}{4y}$ with respect to $x$ using the quotient rule:
$$\frac{d^2y}{dx^2} = -\frac{1}{4} \cdot \frac{d}{dx}\left[\frac{x}{y}\right] = -\frac{1}{4} \left[ \frac{y(1) - x\frac{dy}{dx}}{y^2} \right] \quad \text{[M1]}$$3. Substitute $\frac{dy}{dx} = -\frac{x}{4y}$ into the numerator:
$$\frac{d^2y}{dx^2} = -\frac{1}{4y^2} \left( y - x\left(-\frac{x}{4y}\right) \right) = -\frac{1}{4y^2} \left( y + \frac{x^2}{4y} \right) \quad \text{[M1]}$$Combine over common denominator $4y$:
$$= -\frac{1}{4y^2} \left( \frac{4y^2 + x^2}{4y} \right) = -\frac{x^2 + 4y^2}{16y^3} \quad \text{[M1]}$$4. Since $(x, y)$ lies on the ellipse, $x^2 + 4y^2 = 8$. Substitute this into the numerator:
$\frac{d^2y}{dx^2} = -\frac{8}{16y^3} = -\frac{1}{2y^3} \quad \text{[A1][AG]}$Worked Example 3: Angular Rate of Change (Rocket Tracking)
A television camera on the ground is situated 400 m from a vertical rocket launch pad. A rocket ascends vertically at a constant speed of $150\text{ m/s}$. Find the rate of change of the camera's angle of elevation when the rocket is 300 m above the ground.
Solution:
1. Let $h(t)$ be rocket height and $\theta(t)$ be the angle of elevation. The horizontal distance is constant: $D = 400\text{ m}$.
Geometric relationship: $\tan\theta = \frac{h}{400}$. [M1]
2. Differentiate implicitly with respect to time $t$:
$$\sec^2\theta \frac{d\theta}{dt} = \frac{1}{400} \frac{dh}{dt} \quad \text{[M1]}$$3. When $h = 300$, the right triangle has base 400 and opposite side 300. The hypotenuse is:
$$s = \sqrt{400^2 + 300^2} = \sqrt{160000 + 90000} = 500\text{ m}$$ $$\cos\theta = \frac{400}{500} = \frac{4}{5} \implies \sec\theta = \frac{5}{4} \implies \sec^2\theta = \frac{25}{16} \quad \text{[A1]}$$4. Substitute $\sec^2\theta = \frac{25}{16}$ and $\frac{dh}{dt} = 150$:
$$\frac{25}{16} \frac{d\theta}{dt} = \frac{1}{400} (150) = \frac{3}{8} \quad \text{[M1]}$$ $$\frac{d\theta}{dt} = \frac{3}{8} \cdot \frac{16}{25} = \frac{3 \cdot 2}{25} = \frac{6}{25} = 0.24\text{ rad/s} \quad \text{[A1]}$$Practice Exercises (10 Tiered Questions)
Try these 10 exam-style questions, from core skills to harder problems. Open each question to view its solution.
Find $\frac{dy}{dx}$ for the circle $x^2 + y^2 = 100$ at the point $(6, 8)$.
Show Full Solution
$2x + 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y}$. [M1]
At $(6, 8)$: $\frac{dy}{dx} = -\frac{6}{8} = -\frac{3}{4}$. [A1]
Find $\frac{dy}{dx}$ in terms of $x$ and $y$ for $x^2 y + y^3 = 5$.
Show Full Solution
$\frac{d}{dx}(x^2 y) = 2xy + x^2\frac{dy}{dx}$. [M1]
$2xy + x^2\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 0$
$(x^2 + 3y^2)\frac{dy}{dx} = -2xy \implies \frac{dy}{dx} = -\frac{2xy}{x^2 + 3y^2}$. [A1]
The radius of a circle increases at $3\text{ cm/s}$. Find the rate of increase of its area when $r = 10\text{ cm}$.
Show Full Solution
$A = \pi r^2 \implies \frac{dA}{dt} = 2\pi r \frac{dr}{dt}$. [M1]
$\frac{dA}{dt} = 2\pi(10)(3) = 60\pi \approx 188.5\text{ cm}^2/\text{s}$. [A1]
Find the equations of the tangent and normal to the ellipse $x^2 + 4y^2 = 8$ at the point $(2, 1)$.
Show Full Solution
$2x + 8y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{2x}{8y} = -\frac{x}{4y}$. [M1]
At $(2, 1)$: $m = -\frac{2}{4(1)} = -\frac{1}{2}$. [A1]
Tangent: $y - 1 = -\frac{1}{2}(x - 2) \implies x + 2y = 4$. [A1]
Normal: gradient $= 2 \implies y - 1 = 2(x - 2) \implies y = 2x - 3$. [A1]
A spherical balloon is inflated such that its volume increases at $100\text{ cm}^3/\text{s}$. Find the rate at which its surface area is increasing when the radius is $5\text{ cm}$.
Show Full Solution
$V = \frac{4}{3}\pi r^3 \implies \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$. [M1]
$100 = 4\pi (25) \frac{dr}{dt} = 100\pi \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{1}{\pi}\text{ cm/s}$. [A1]
Surface area $S = 4\pi r^2 \implies \frac{dS}{dt} = 8\pi r \frac{dr}{dt}$. [M1]
$\frac{dS}{dt} = 8\pi(5)\left(\frac{1}{\pi}\right) = 40\text{ cm}^2/\text{s}$. [A1]
Find $\frac{d^2y}{dx^2}$ in terms of $y$ for the curve $x^2 - y^2 = 4$.
Show Full Solution
$2x - 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = \frac{x}{y}$. [M1]
Differentiate again using quotient rule: [M1]
$\frac{d^2y}{dx^2} = \frac{1(y) - x\frac{dy}{dx}}{y^2} = \frac{y - x(x/y)}{y^2} = \frac{y^2 - x^2}{y^3}$. [M1]
Since $x^2 - y^2 = 4 \implies y^2 - x^2 = -4$:
$\frac{d^2y}{dx^2} = -\frac{4}{y^3}$. [A1]
Find all points on the curve $x^2 + xy + y^2 = 27$ where the tangent is horizontal.
Show Full Solution
$2x + \left(y + x\frac{dy}{dx}\right) + 2y\frac{dy}{dx} = 0 \implies (x + 2y)\frac{dy}{dx} = -(2x + y)$. [M1]
$\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$.
Horizontal tangent requires $\frac{dy}{dx} = 0 \implies 2x + y = 0 \implies y = -2x$. [A1]
Substitute into curve equation: [M1]
$x^2 + x(-2x) + (-2x)^2 = 27 \implies x^2 - 2x^2 + 4x^2 = 27 \implies 3x^2 = 27 \implies x^2 = 9 \implies x = \pm 3$.
• $x = 3 \implies y = -6$. Point: $(3, -6)$. [A1]
• $x = -3 \implies y = 6$. Point: $(-3, 6)$. [A1]
A person 1.8 m tall walks away from a 6 m tall streetlight at $1.5\text{ m/s}$. Find: (a) the rate at which the length of the person's shadow is growing, (b) the speed of the tip of the shadow.
Show Full Solution
Let $x$ be the distance from streetlight to person, $s$ be shadow length. [M1]
By similar triangles: $\frac{s}{1.8} = \frac{x + s}{6} \implies 6s = 1.8x + 1.8s \implies 4.2s = 1.8x \implies s = \frac{1.8}{4.2}x = \frac{3}{7}x$. [A1]
(a) Shadow length growth: $\frac{ds}{dt} = \frac{3}{7}\frac{dx}{dt} = \frac{3}{7}(1.5) = \frac{4.5}{7} = \frac{9}{14} \approx 0.643\text{ m/s}$. [A1]
(b) Speed of tip of shadow: Tip position $= x + s = x + \frac{3}{7}x = \frac{10}{7}x$.
Speed $= \frac{10}{7}\frac{dx}{dt} = \frac{10}{7}(1.5) = \frac{15}{7} \approx 2.14\text{ m/s}$. [A1]
Find the gradient of the tangent to $e^y \sin x = x + y$ at the origin $(0, 0)$.
Show Full Solution
Differentiate implicitly: [M1]
$\frac{d}{dx}[e^y \sin x] = e^y \frac{dy}{dx} \sin x + e^y \cos x$.
$\frac{d}{dx}[x + y] = 1 + \frac{dy}{dx}$.
Equation: $e^y \sin x \frac{dy}{dx} + e^y \cos x = 1 + \frac{dy}{dx}$. [A1]
Substitute $(x, y) = (0, 0)$: [M1]
$e^0 \sin(0) \frac{dy}{dx} + e^0 \cos(0) = 1 + \frac{dy}{dx}$
$0 + 1(1) = 1 + \frac{dy}{dx} \implies 1 = 1 + \frac{dy}{dx} \implies \frac{dy}{dx} = 0$. [A1]
A water trough of length 4 m has an equilateral triangular cross-section with sides 2 m pointing downwards. Water is pumped in at $0.2\text{ m}^3/\text{min}$. Find the rate of rise of the water level when the depth is 1 m.
Show Full Solution
Cross section is an equilateral triangle of side 2. Height $H = 2\sin(60^\circ) = \sqrt{3}$. Base $B = 2$. [M1]
At depth $h$, water surface width $b$: $\frac{b}{h} = \frac{2}{\sqrt{3}} \implies b = \frac{2}{\sqrt{3}}h$. [M1]
Cross-sectional area: $A = \frac{1}{2} b h = \frac{1}{2}\left(\frac{2}{\sqrt{3}}h\right)h = \frac{h^2}{\sqrt{3}}$.
Total volume $V = \text{Area} \times \text{length} = 4\left(\frac{h^2}{\sqrt{3}}\right) = \frac{4}{\sqrt{3}}h^2$. [A1]
$\frac{dV}{dt} = \frac{8}{\sqrt{3}}h \frac{dh}{dt}$. [M1]
When $h = 1$ and $\frac{dV}{dt} = 0.2$:
$0.2 = \frac{8}{\sqrt{3}}(1) \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{0.2\sqrt{3}}{8} = \frac{\sqrt{3}}{40} \approx 0.0433\text{ m/min}$. [A1]
IB Formula Booklet References (Topic 5)
Syllabus 5.10, 5.11: Implicit Differentiation & Chain Rule
$$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$$ $$\text{Product Rule: } \frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$$Key Formulas & Identities
1. Standard Implicit Derivative Patterns
$$\frac{d}{dx}[y^n] = n y^{n-1} \frac{dy}{dx}, \quad \frac{d}{dx}[xy] = y + x\frac{dy}{dx}$$2. Related Rates Template
$$\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$$Common Exam Pitfalls
Forgetting $\frac{dy}{dx}$ on $y$-terms
Every time you differentiate a term with $y$ with respect to $x$, you MUST multiply by $\frac{dy}{dx}$. Writing $\frac{d}{dx}(y^2) = 2y$ instead of $2y\frac{dy}{dx}$ is the most common error in HL calculus.
Substituting Numbers TOO EARLY in Related Rates
Never substitute the numerical snapshot (e.g. $h = 3$) BEFORE differentiating! If you substitute early, the quantity becomes a constant and its derivative erroneously vanishes to zero.
Topic Diagnostic Test
Complete this timed test to check your understanding and exam technique.
Section A: Diagnostic Multiple Choice Questions
What is $\frac{d}{dx}(x^2 y)$?
At what points does the circle $x^2 + y^2 = 25$ have vertical tangents?
If $V = x^3$ and $\frac{dx}{dt} = 2$, what is $\frac{dV}{dt}$ when $x = 4$?
The tangent to an implicit curve is horizontal if and only if:
Section B: Examination-Style Free Response Questions
A curve is defined implicitly by $x^2 + 2xy + 3y^2 = 11$.
(a) Show that $\frac{dy}{dx} = -\frac{x + y}{x + 3y}$. [3 marks]
(b) Find the coordinates of the two points on the curve where the tangent is parallel to the $x$-axis. [3 marks]
Show Full Markscheme
(a) Implicit differentiation: [3 marks]
$2x + 2\left(y + x\frac{dy}{dx}\right) + 6y\frac{dy}{dx} = 0$ [M1]
$2x + 2y + (2x + 6y)\frac{dy}{dx} = 0$
Divide by 2: $(x + 3y)\frac{dy}{dx} = -(x + y) \implies \frac{dy}{dx} = -\frac{x + y}{x + 3y}$ [A1].
(b) Horizontal tangents: [3 marks]
Parallel to $x$-axis $\implies \frac{dy}{dx} = 0 \implies x + y = 0 \implies y = -x$ [M1].
Substitute into curve equation:
$x^2 + 2x(-x) + 3(-x)^2 = 11 \implies x^2 - 2x^2 + 3x^2 = 11 \implies 2x^2 = 11 \implies x = \pm\sqrt{\frac{11}{2}}$ [A1].
Points: $\left(\sqrt{5.5}, -\sqrt{5.5}\right)$ and $\left(-\sqrt{5.5}, \sqrt{5.5}\right)$ [A1].
A sand timer consists of sand falling from an upper container onto a conical pile. The height of the cone is always equal to its base radius ($h = r$). Sand falls at a rate of $12\text{ cm}^3/\text{s}$.
(a) Express the volume $V$ in terms of $h$ only. [1 mark]
(b) Find the rate at which the height of the sand pile is increasing when $h = 6\text{ cm}$. [3 marks]
(c) Find the rate at which the base area of the cone is increasing at this instant. [4 marks]
Show Full Markscheme
(a) Volume formula: [1 mark]
$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi h^3$ [A1].
(b) Rate of height increase: [3 marks]
$\frac{dV}{dt} = \pi h^2 \frac{dh}{dt}$ [M1].
Substitute $h = 6$ and $\frac{dV}{dt} = 12$:
$12 = \pi (36) \frac{dh}{dt} \implies \frac{dh}{dt} = \frac{12}{36\pi} = \frac{1}{3\pi} \approx 0.106\text{ cm/s}$ [A1].
(c) Rate of base area increase: [4 marks]
Base area $A = \pi r^2 = \pi h^2$ [M1].
$\frac{dA}{dt} = 2\pi h \frac{dh}{dt}$ [M1].
Substitute $h = 6$ and $\frac{dh}{dt} = \frac{1}{3\pi}$:
$\frac{dA}{dt} = 2\pi(6)\left(\frac{1}{3\pi}\right) = 12\pi \left(\frac{1}{3\pi}\right) = 4\text{ cm}^2/\text{s}$ [A1].