Calculus with Parametric Curves
Applying differential and integral calculus directly to parametric curves
Tangents to Parametric Curves
By the Chain Rule, $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$. Solving for $\frac{dy}{dx}$:
- Kinematic View: $\frac{dy}{dx}$ is the ratio of vertical velocity $y'(t)$ to horizontal velocity $x'(t)$. The tangent line points along the velocity vector $\vec{v}(t) = \langle x'(t), y'(t) \rangle$.
- Direct computation: no need to convert to explicit Cartesian $y = F(x)$ first.
- Horizontal Tangents: $\frac{dy}{dt} = 0$ (vertical velocity) and $\frac{dx}{dt} \neq 0$ (horizontal velocity), slope $m = 0$.
- Vertical Tangents: $\frac{dx}{dt} = 0$ (horizontal velocity) and $\frac{dy}{dt} \neq 0$ (vertical velocity), slope $m = \pm\infty$.
- If both $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} = 0$, evaluate $\lim_{t \to t_0} \frac{y'(t)}{x'(t)}$ (e.g. via L'HΓ΄pital's Rule).
Second Derivatives & Concavity
Let $y' = \frac{dy}{dx}$. Differentiating with respect to $x$:
- 1st Derivative: $\frac{dy}{dx} = \frac{y'(t)}{x'(t)} = \frac{2t}{1} = 2t$
- Differentiate wrt $t$: $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{d}{dt}(2t) = 2$
- Divide by $x'(t) = 1$: $$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{x'(t)} = \frac{2}{1} = 2$$
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Concavity:
β’ $\frac{d^2y}{dx^2} = 2 > 0$ for all $t \implies$ Everywhere Concave Up βͺ
β’ Matches Cartesian derivative: $y = x^2 \implies \frac{d^2y}{dx^2} = 2$. - Pitfall check: $\frac{y''(t)}{x''(t)} = \frac{2}{0}$ is undefined, whereas $\frac{d^2y}{dx^2} = 2$.
Area Under a Parametric Curve
- Substitution: Replace $y = y(t)$ and differential $dx = x'(t)\, dt$.
- Limits: Match boundaries by setting $x(\alpha) = a$ and $x(\beta) = b$.
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Orientation & Sign:
β’ Left-to-right ($x' > 0$): $\alpha < \beta \implies A > 0$ directly.
β’ Right-to-left ($x' < 0$): $\alpha > \beta \implies$ Negative $x'(t)$ flips limits: $$\int_\alpha^\beta y(t)\, x'(t)\, dt = -\int_\beta^\alpha y(t)\, x'(t)\, dt > 0$$
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Parametrization: $x = t^2, \; y = t \quad (0 \le t \le 1)$.
This traces the curve $y = \sqrt{x}$ from $(0, 0)$ to $(1, 1)$. -
Differential & Range:
β’ Differential: $dx = x'(t)\, dt = 2t\, dt$.
β’ When $x = 0 \implies \mathbf{t = 0}$; when $x = 1 \implies \mathbf{t = 1}$ ($t$ increases $0 \to 1$). -
Parametric Area Integral:
$$A = \int_0^1 y(t)\, x'(t)\, dt = \int_0^1 (t)(2t\, dt) = \int_0^1 2t^2\, dt = \left[ \frac{2}{3}t^3 \right]_0^1 = \mathbf{\frac{2}{3}}$$Cartesian check: $\int_0^1 \sqrt{x}\, dx = \left[\frac{2}{3}x^{3/2}\right]_0^1 = \frac{2}{3} \quad \checkmark$
Arc Length & Particle Speed
The differential arc length element is $ds = \sqrt{dx^2 + dy^2} = \sqrt{(x'(t))^2 + (y'(t))^2}\, dt$.
- No Retracing: The curve must be traversed exactly once; retracing or looping multiplies the measured travel distance, not arc length.
- Instantaneous Speed: Length of velocity vector $\vec{v}(t)$: $$v(t) = \frac{ds}{dt} = |\vec{v}(t)| = \sqrt{(x'(t))^2 + (y'(t))^2}$$
- Total Distance Traveled: Integrating speed over time: $$\text{Distance} = \int_\alpha^\beta v(t)\, dt = L$$
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Sanity Check (Circle of Radius $r$):
$x = r\cos t, y = r\sin t \implies (x')^2 + (y')^2 = r^2$. $$L = \int_0^{2\pi} \sqrt{r^2}\, dt = 2\pi r \quad \checkmark$$
Deriving Arc Length via the Pythagorean Theorem
We approximate the curved path by connecting consecutive sample points with straight chord segments $\Delta s_i$.
Surface Area
General principle: $\text{Surface Area} = \int 2\pi (\text{Radius})\, ds$.
Each point on the 2D curve sweeps a 3D circle of circumference $2\pi R$ around the axis. Multiplying this circumference by arc length $ds$ gives the ribbon's surface area $dS = 2\pi R\, ds$.
Rotate semicircle $x = r\cos t, y = r\sin t$ ($0 \le t \le \pi$) about $x$-axis:
Exercises
Exercise: Tangents & Concavity
$$x'(t)=6t, \qquad y'(t)=3t^2-3=3(t^2-1)$$ $$\frac{dy}{dx}=\frac{y'(t)}{x'(t)}=\frac{3(t^2-1)}{6t} =\boxed{\frac{t^2-1}{2t}}, \qquad t\neq 0$$
$$P=(x(2),y(2))=(12,2), \qquad m=\left.\frac{dy}{dx}\right|_{t=2}=\frac34$$ $$y-2=\frac34(x-12) \quad\Longrightarrow\quad \boxed{y=\frac34x-7}$$
Horizontal: $y'(t)=0$ and $x'(t)\neq 0$, so $t=\pm1$. $$t=-1:\ (3,2),\ \boxed{y=2} \qquad t=1:\ (3,-2),\ \boxed{y=-2}$$ Vertical: $x'(t)=0$ and $y'(t)\neq 0$, so $t=0$. $$t=0:\ (0,0),\ \boxed{x=0}$$
$$\frac{d}{dt}\left(\frac{dy}{dx}\right) =\frac{d}{dt}\left(\frac{t}{2}-\frac{1}{2t}\right) =\frac{t^2+1}{2t^2}$$ $$\frac{d^2y}{dx^2} =\frac{1}{x'(t)}\frac{d}{dt}\left(\frac{dy}{dx}\right) =\boxed{\frac{t^2+1}{12t^3}}$$ Because $t^2+1>0$, the curve is concave down for $t\in(-\infty,0)$ and concave up for $t\in(0,\infty)$.
Exercise: Area & Exact Arc Length
Find the area between the curve and the $x$-axis.
$$x'(t)=6t, \qquad dx=x'(t)\,dt=6t\,dt$$
For $0\le t\le1$, $y(t)\le0$ and $x'(t)\ge0$. Hence $$A=\int_0^1 |y(t)|x'(t)\,dt =\int_0^1(3t-t^3)(6t)\,dt =6\int_0^1(3t^2-t^4)\,dt.$$
$$A=6\left[t^3-\frac{t^5}{5}\right]_0^1 =6\left(1-\frac15\right) =\boxed{\frac{24}{5}}.$$
Find the arc length of the traced curve.
$$[x'(t)]^2+[y'(t)]^2 =(6t)^2+[3(t^2-1)]^2$$ $$=36t^2+9(t^4-2t^2+1) =\boxed{9(t^2+1)^2}.$$
Since $t^2+1>0$, $$\frac{ds}{dt}=\sqrt{[x'(t)]^2+[y'(t)]^2} =\sqrt{9(t^2+1)^2}=\boxed{3(t^2+1)}.$$
$$L=\int_0^2\frac{ds}{dt}\,dt =\int_0^2 3(t^2+1)\,dt =\left[t^3+3t\right]_0^2 =\boxed{14}.$$