Parametric & Implicit Differentiation – definitions, techniques, and applications

1. Parametric Equations – basic concepts

A parametric representation expresses the coordinates of a curve as functions of a third variable (the parameter), usually denoted $t$: $$x=f(t),y=g(t),t\in I\subset \mathbb{R}.$$

The curve is traced as $t$ varies over the interval $I$. Typical parameters are time, angle, or any convenient quantity.

Key properties

2. Derivative of a parametric curve

For a differentiable parametric curve, the slope of the tangent at a point is $$\boxed{\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}}\text{provided }\frac{dx}{dt}\ne 0.$$

Second derivative (useful for curvature, concavity): $$\frac{d^{2}y}{d{x}^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}.$$

3. Worked Example – Projectile motion (real‑world link)

A ball is thrown from ground level with initial speed $20\text{ m/s}$ at an angle $30^{\circ }$ above the horizontal. Ignoring air resistance, its motion can be described parametrically (time $t$ in seconds):

$$\begin{array}{rl}x(t)& =v_0\cos \theta t=20\cos 30^{\circ }t=17.32t,\\ y(t)& =v_0\sin \theta t-\frac{1}{2}g{t}^2=20\sin 30^{\circ }t-4.9t^2=10t-4.9t^2.\end{array}$$

Step‑by‑step derivative

1. Compute $dx/dt=17.32$.

2. Compute $dy/dt=10-9.8t$.

3. Tangent slope: $$\frac{dy}{dx}=\frac{10-9.8t}{17.32}.$$

4. At the apex ($dy/dt=0\Rightarrow t=\frac{10}{9.8}\approx 1.02$ s): $$\frac{dy}{dx}{|}_{t=1.02}=0.$$ The tangent is horizontal, confirming the apex.

5. Second derivative (concavity): $$\frac{d^{2}y}{d{x}^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)/\frac{dx}{dt}=\frac{-9.8}{17.32^2}\approx -0.0327\text{(negative → curve opens downward)}.$$

Figure 1 – Parametric trajectory of the projectile (units in metres).

Real‑world connection – Mobile payment apps like eSewa and Khalti calculate transaction fees based on a piecewise linear function of the amount; the slope of each piece is a derivative of a parametric cost‑vs‑time curve used for dynamic pricing.

4. Implicit Differentiation – basic idea

When a curve is given by an equation $F(x,y)=0$ that cannot be solved for $y$ explicitly, we differentiate both sides with respect to $x$, treating $y$ as an implicit function $y(x)$. Apply the chain rule:

$$\frac{d}{dx}F(x,y)=F_x(x,y)+F_y(x,y)\frac{dy}{dx}=0⟹\boxed{\frac{dy}{dx}=-\frac{F_x}{F_y}}.$$

Higher‑order derivatives follow by differentiating again, always remembering $\frac{dy}{dx}$ is itself a function of $x$ and $y$.

5. Worked Example – Folium of Descartes

The curve $x^3+y^3=3axy$ (with $a>0$) is a classic implicit curve. Find $\frac{dy}{dx}$ at the point $(a,a)$.

1. Differentiate implicitly: $$3x^2+3y^2\frac{dy}{dx}=3a\left(y+x\frac{dy}{dx}\right).$$

2. Collect $\frac{dy}{dx}$: $$3y^2\frac{dy}{dx}-3ax\frac{dy}{dx}=3ay-3x^2.$$

3. Factor: $$\frac{dy}{dx}(3y^2-3ax)=3ay-3x^2.$$

4. Solve: $$\frac{dy}{dx}=\frac{ay-x^2}{y^2-ax}.$$

5. Substitute $(x,y)=(a,a)$: $$\frac{dy}{dx}{|}_{(a,a)}=\frac{a\cdot a-a^2}{a^2-a\cdot a}=\frac{0}{0}.$$ The expression is indeterminate; we must apply L’Hôpital’s rule on the original derivative expression or simplify before substitution. Simplify numerator and denominator:

   $$\frac{dy}{dx}=\frac{a(y-x)}{y^2-ax}.$$

   At $(a,a)$ both numerator and denominator vanish; differentiate numerator and denominator with respect to $x$ (treating $y$ as a function of $x$):

   Numerator derivative: $a(\frac{dy}{dx}-1)$.
   Denominator derivative: $2y\frac{dy}{dx}-a$.

   Set $$\frac{a(\frac{dy}{dx}-1)}{2y\frac{dy}{dx}-a}=\frac{dy}{dx}.$$

   Solve for $\frac{dy}{dx}$ at $(a,a)$: $$a(\frac{dy}{dx}-1)=\frac{dy}{dx}(2a\frac{dy}{dx}-a)$$ $$a\frac{dy}{dx}-a=2a{\left(\frac{dy}{dx}\right)}^2-a\frac{dy}{dx}$$ $$0=2a{\left(\frac{dy}{dx}\right)}^2-2a\frac{dy}{dx}+a$$ $$2{\left(\frac{dy}{dx}\right)}^2-2\frac{dy}{dx}+1=0$$ $${\left(\frac{dy}{dx}-1\right)}^2=0⟹\frac{dy}{dx}=1.$$

Thus the tangent at $(a,a)$ has slope 1 (45° line).

Figure 2 – Folium of Descartes near $(a,a)$ with tangent slope 1.

Real‑world connection – The NEPSE stock‑price chart often exhibits implicit relationships between price and volume (e.g., $P^2+V^2=kPV$). Traders use implicit differentiation to compute instantaneous price sensitivity to volume changes.

6. Comparison of Parametric vs. Implicit Methods

flowchart LR
    A["Start with curve description"]
    B["Parametric: have x(t), y(t)"] --> C["Compute dy/dx = (dy/dt)/(dx/dt)"]
    D["Implicit: have F(x,y)=0"] --> E["Differentiate: F_x + F_y·dy/dx = 0"]
    C --> F["Higher‑order: differentiate again w.r.t t"]
    E --> G["Higher‑order: differentiate again w.r.t x"]
    F --> H["Use for motion, engineering"]
    G --> I["Use for economics, geometry"]
    A --> B
    A --> D

7. Applications

8. Worked Example – Bézier curve tangent (graphics)

A quadratic Bézier curve with control points $P_0(0,0)$, $P_1(2,4)$, $P_2(4,0)$ is defined parametrically: $$\mathbf{B}(t)=(1-t{)}^2{P}_0+2(1-t)t{P}_1+t^2{P}_2,0\le t\le 1.$$

Extract components: $$\begin{array}{rl}x(t)& =2(1-t)t\cdot 2+t^2\cdot 4=4t(1-t)+4t^2=4t,\\ y(t)& =2(1-t)t\cdot 4=8t(1-t)=8t-8t^2.\end{array}$$

Derivative: $$\frac{dx}{dt}=4,\frac{dy}{dt}=8-16t.$$ Tangent slope: $$\frac{dy}{dx}=\frac{8-16t}{4}=2-4t.$$

At $t=0.5$ (mid‑curve): $$\frac{dy}{dx}=2-4(0.5)=0.$$ The curve is horizontal at its apex, a property used by YouTube to smooth thumbnail transitions (the curve controls animation easing).

Figure 3 – Quadratic Bézier curve with tangent slope zero at the midpoint.

9. Common Mistakes & How to Avoid

1. Dividing by zero – Always check that $dx/dt\ne 0$ (parametric) or $F_y\ne 0$ (implicit) before applying the formula.
2. Forgetting the chain rule – When differentiating terms like $y^3$, write $3y^2\frac{dy}{dx}$.
3. Mixing variables – Keep $t$ separate from $x$ and $y$; do not substitute $t$ prematurely.
4. Incorrect second derivative – Remember the outer derivative is with respect to the original independent variable (t for parametric, x for implicit).

10. In the real world

• eSewa uses a parametric cost‑vs‑transaction‑time curve to adjust service fees during peak hours; the derivative tells the system how fast the fee should increase.
• Daraz order‑processing queue can be modeled by an implicit relation between order size $Q$ and waiting time $W$: $Q^2+W^2=kQW$. Implicit differentiation gives $\frac{dW}{dQ}$, helping the logistics team predict how a larger batch will affect delivery time.
• Google Maps computes the curvature of a road segment defined implicitly by $x^2+y^2=R^2$ (a circular arc). Implicit differentiation yields the steering angle needed for navigation instructions.

11. Exam tip

• Identify the form: If the problem gives $x(t),y(t)$ → use $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$. If it gives $F(x,y)=0$ → differentiate implicitly and solve for $\frac{dy}{dx}$.
• Write down $dx/dt$ and $dy/dt$ (or $F_x,F_y$) first; this prevents algebraic slips.
• Check for vertical/horizontal tangents by testing where $dx/dt=0$ (parametric) or $F_y=0$ (implicit).
• For second derivatives, differentiate the first‑derivative expression once more with respect to the original independent variable, not the parameter.
• Mark the domain of the parameter; many marks are lost if the answer is valid only for a restricted interval.

Projectile trajectory illustrating parametric time variable (Image: Py4nf, CC0, via Wikimedia Commons) Implicit curve used in the worked example (Image: Zorgit, CC BY-SA 3.0, via Wikimedia Commons) Bézier curve in computer graphics, showing control points and tangent (Image: Xhungab, CC0, via Wikimedia Commons)