Calculus II: Parametric Curves and Calculus

Parametric equations provide a powerful way to describe curves that are difficult or impossible to represent with a single function $y=f(x)$. From tracing the path of a projectile to designing a robotic arm’s motion, parametric forms are essential in engineering and physics. Mastering calculus with parametric curves unlocks the ability to analyze motion, compute geometric properties, and solve complex real-world modeling problems where traditional Cartesian coordinates fall short.

From Parametric Equations to Derivatives

A curve defined parametrically is described by two equations, $x=x(t)$ and $y=y(t)$, where the parameter $t$ (often representing time or an angle) defines a point $(x,y)$ in the plane. To analyze the curve's slope and behavior, we need to find $\frac{dy}{dx}$ without explicitly solving for $y$ in terms of $x$. This is accomplished using the Chain Rule. Assuming $x(t)$ is differentiable and $x^{\prime }(t)\ne 0$, the derivative is given by:

$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{y^{\prime }(t)}{x^{\prime }(t)}$$

This formula makes intuitive sense: the rate of change of $y$ with respect to $x$ is simply the ratio of their individual rates of change with respect to the parameter $t$.

Example: Find the slope of the curve defined by $x=t^2$, $y=t^3-3t$ at the point where $t=2$.

First, compute the derivatives with respect to $t$: $x^{\prime }(t)=2t$ and $y^{\prime }(t)=3t^2-3$. Applying the formula, $\frac{dy}{dx}=\frac{3t^2-3}{2t}$. At $t=2$, the slope is $\frac{3(4)-3}{2(2)}=\frac{12-3}{4}=\frac{9}{4}$.

This slope can then be used to write the equation of a tangent line at the corresponding point $(x(2),y(2))=(4,2)$.

Higher-Order Derivatives and Tangent Line Analysis

The second derivative, $\frac{d^{2}y}{d{x}^2}$, tells us about the concavity of the parametric curve. It is found by differentiating the first derivative $\frac{dy}{dx}$ with respect to $x$, again using the parameter $t$. The formula is:

$$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}$$

This process is essentially applying the derivative-with-respect-to-$x$ formula to the function $\frac{dy}{dx}$ itself.

Example: Find $\frac{d^{2}y}{d{x}^2}$ for the previous curve, $x=t^2$, $y=t^3-3t$. We had $\frac{dy}{dx}=\frac{3t^2-3}{2t}$. Differentiate this with respect to $t$: $$\frac{d}{dt}\left(\frac{3t^2-3}{2t}\right)=\frac{(6t)(2t)-(3t^2-3)(2)}{(2t{)}^2}=\frac{12t^2-6t^2+6}{4t^2}=\frac{6t^2+6}{4t^2}=\frac{3(t^2+1)}{2t^2}$$ Now, divide by $dx/dt=2t$: $$\frac{d^{2}y}{d{x}^2}=\frac{\frac{3(t^2+1)}{2t^2}}{2t}=\frac{3(t^2+1)}{4t^3}$$ This tells us, for instance, that for $t>0$, the second derivative is positive, so the curve is concave up.

Arc Length of a Parametric Curve

How long is a parametric path between two parameter values? The arc length formula generalizes the Cartesian version. For a curve traced exactly once as $t$ increases from $a$ to $b$, the arc length $L$ is the integral of the instantaneous speed along the path:

$$L={\int }_a^b\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt$$

The expression under the radical, $\sqrt{(x^{\prime }(t){)}^2+(y^{\prime }(t){)}^2}$, is the magnitude of the velocity vector if $t$ is time, representing the speed $ds/dt$.

Example: Find the length of one arch of the cycloid defined by $x=t-\sin t$, $y=1-\cos t$ from $t=0$ to $t=2\pi$. Compute derivatives: $x^{\prime }(t)=1-\cos t$, $y^{\prime }(t)=\sin t$. Set up the integrand: $$\sqrt{(1-\cos t{)}^2+(\sin t{)}^2}=\sqrt{1-2\cos t+{\cos }^{2}t+{\sin }^{2}t}$$ Since ${\cos }^{2}t+{\sin }^{2}t=1$, this simplifies to $\sqrt{2-2\cos t}=\sqrt{4{\sin }^2(t/2)}=2|\sin (t/2)|$. On the interval $[0,2\pi ]$, $\sin (t/2)$ is non-negative, so we drop the absolute value. $$L={\int }_0^{2\pi }2\sin (t/2)dt=2{\left[-2\cos (t/2)\right]}_0^{2\pi }=4(\cos 0-\cos \pi )=4(1-(-1))=8$$

Area Under a Parametric Curve

Finding the area under a parametric curve from $x=a$ to $x=b$ requires careful consideration of the direction of tracing. If the curve is traced from left to right as $t$ increases from $\alpha$ to $\beta$ (i.e., $x^{\prime }(t)>0$), then the area under the curve, above the x-axis, is given by:

$$A={\int }_{\alpha }^{\beta }y(t)x^{\prime }(t)dt={\int }_a^{b}ydx$$

The key is to integrate $y$ with respect to $x$ by substituting $dx=x^{\prime }(t)dt$.

Example: Find the area under one arch of the cycloid from the previous example. For the cycloid $x=t-\sin t$, $y=1-\cos t$ from $t=0$ to $t=2\pi$, we have $x^{\prime }(t)=1-\cos t$. The curve starts at $(0,0)$, goes up, and returns to $(2\pi ,0)$, staying above the x-axis. $$A={\int }_0^{2\pi }(1-\cos t)(1-\cos t)dt={\int }_0^{2\pi }(1-2\cos t+{\cos }^{2}t)dt$$ Using the identity ${\cos }^{2}t=\frac{1+\cos 2t}{2}$: $$A={\int }_0^{2\pi }\left(1-2\cos t+\frac{1}{2}+\frac{1}{2}\cos 2t\right)dt={\int }_0^{2\pi }\left(\frac{3}{2}-2\cos t+\frac{1}{2}\cos 2t\right)dt$$ Integrate term by term: $={\left[\frac{3}{2}t-2\sin t+\frac{1}{4}\sin 2t\right]}_0^{2\pi }=3\pi$.

Surface Area of Revolution for Parametric Curves

To find the surface area generated by revolving a smooth parametric curve about the x-axis, we integrate the circumference of a circular slice $(2\pi y)$ times the arc length element $ds$. The formula for revolution about the x-axis is:

$$S={\int }_a^{b}2\pi yds={\int }_{\alpha }^{\beta }2\pi y(t)\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2}dt$$

For revolution about the y-axis, the formula uses $2\pi x$ instead of $2\pi y$.

Example: Revolve the cycloid arch from the previous examples about the x-axis and find the surface area. We have $y(t)=1-\cos t$ and $ds=2\sin (t/2)dt$ as before. The limits are $t=0$ to $t=2\pi$. $$S={\int }_0^{2\pi }2\pi (1-\cos t)\cdot 2\sin (t/2)dt$$ Use the identity $1-\cos t=2{\sin }^2(t/2)$. Substitute: $$S={\int }_0^{2\pi }2\pi \cdot 2{\sin }^2(t/2)\cdot 2\sin (t/2)dt={\int }_0^{2\pi }8\pi {\sin }^3(t/2)dt$$ Let $u=t/2$, so $du=dt/2$ and when $t=2\pi$, $u=\pi$. The integral becomes: $$S={\int }_0^{\pi }8\pi {\sin }^{3}u\cdot (2du)=16\pi {\int }_0^{\pi }{\sin }^{3}udu$$ Since ${\sin }^{3}u=\sin u(1-{\cos }^{2}u)$, and using symmetry or a standard integral, ${\int }_0^{\pi }{\sin }^{3}udu=\frac{4}{3}$. Therefore, $S=16\pi \cdot \frac{4}{3}=\frac{64\pi }{3}$.

Common Pitfalls

1. Misapplying the derivative formula: A common error is to try to find $\frac{dy}{dx}$ by dividing the original equations ($y(t)/x(t)$). Remember, you must always differentiate first with respect to the parameter $t$ before taking the ratio: $\frac{dy}{dx}=\frac{y^{\prime }(t)}{x^{\prime }(t)}$.
2. Incorrect limits for area and arc length: When computing area under a curve or arc length, your limits of integration must be in the parameter $t$. If the problem gives x-limits like $x=a$ to $x=b$, you must solve $x(t)=a$ and $x(t)=b$ to find the corresponding $t$-values. Using the x-limits directly in the parametric integral formulas will give an incorrect result.
3. Forgetting the $x^{\prime }(t)$ factor in the area integral: The formula for area is $\int y(t)x^{\prime }(t)dt$, not simply $\int y(t)dt$. The $x^{\prime }(t)dt$ term is the critical substitution for $dx$.
4. Confusing $\frac{d^{2}y}{d{x}^2}$ with the derivative of $\frac{dy}{dx}$ with respect to $t$: The second derivative requires an extra division by $dx/dt$. A frequent mistake is to stop after computing $\frac{d}{dt}\left(\frac{dy}{dx}\right)$, forgetting the final step: $\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}$.

Summary

• The derivative of a parametric curve is $\frac{dy}{dx}=\frac{y^{\prime }(t)}{x^{\prime }(t)}$, which is used to find slopes and equations of tangent lines.
• The second derivative, indicating concavity, is found by differentiating $\frac{dy}{dx}$ with respect to $t$ and then dividing by $x^{\prime }(t)$: $\frac{d^{2}y}{d{x}^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}$.
• Arc length is computed by integrating the speed along the path: $L={\int }_a^b\sqrt{(x^{\prime }(t){)}^2+(y^{\prime }(t){)}^2}dt$.
• The area under a parametric curve (above the x-axis) is given by $A={\int }_{\alpha }^{\beta }y(t)x^{\prime }(t)dt$, where the integration is performed in the parameter $t$.
• The surface area generated by revolving a parametric curve about the x-axis is $S={\int }_{\alpha }^{\beta }2\pi y(t)\sqrt{(x^{\prime }(t){)}^2+(y^{\prime }(t){)}^2}dt$, which integrates the circumference of a revolving slice over the arc length element.