11.2 Calculus with Parametric Curves

Parametric curves add a new dimension to calculus, allowing us to describe complex shapes and motions. By expressing x and y coordinates in terms of a parameter t, we can analyze curves that loop back on themselves or have multiple y-values for a single x.

This section builds on our calculus knowledge, applying it to parametric equations. We'll learn to find derivatives, calculate arc lengths, and determine areas under curves using these new tools. It's a powerful way to tackle problems traditional functions can't handle.

Derivatives of Parametric Curves

Finding Derivatives of Parametric Equations

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• Parametric equations define $x$ and $y$ coordinates in terms of a parameter $t$
  • $x = f(t)$ and $y = g(t)$
• To find $\frac{dy}{dx}$, treat $t$ as the independent variable and differentiate $x$ and $y$ with respect to $t$
  • $\frac{dx}{dt} = f'(t)$ and $\frac{dy}{dt} = g'(t)$
• Divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$ to obtain $\frac{dy}{dx}$
  • $\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{g'(t)}{f'(t)}$
• Example: Given $x = t^2 + 1$ and $y = t^3 - t$, find $\frac{dy}{dx}$
  • $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 3t^2 - 1$
  • $\frac{dy}{dx} = \frac{3t^2 - 1}{2t}$

Applying the Chain Rule to Parametric Equations

• The chain rule is used when finding derivatives of composite functions
• In parametric equations, $x$ and $y$ are both functions of $t$, so the chain rule is needed
• To find $\frac{d^2y}{dx^2}$, differentiate $\frac{dy}{dx}$ with respect to $t$ and divide by $\frac{dx}{dt}$
  • $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$
• Example: Given $x = \cos t$ and $y = \sin t$, find $\frac{d^2y}{dx^2}$
  • $\frac{dx}{dt} = -\sin t$ and $\frac{dy}{dt} = \cos t$
  • $\frac{dy}{dx} = \frac{\cos t}{-\sin t} = -\cot t$
  • $\frac{d}{dt}(\frac{dy}{dx}) = \frac{d}{dt}(-\cot t) = \csc^2 t$
  • $\frac{d^2y}{dx^2} = \frac{\csc^2 t}{-\sin t} = -\csc t \cot t$

Finding Tangent Lines to Parametric Curves

• The tangent line to a curve at a point is the line that passes through the point with the same slope as the curve at that point
• To find the tangent line to a parametric curve at a given point, find the slope using $\frac{dy}{dx}$ and substitute the point's coordinates into the point-slope form of a line
  • Slope $m = \frac{dy}{dx}$ at the given point
  • Tangent line equation: $y - y_1 = m(x - x_1)$
• Example: Find the tangent line to the curve $x = t^2 - 1$, $y = t^3 + 2t$ at the point where $t = 1$
  • When $t = 1$, $x = 0$ and $y = 3$, so the point is $(0, 3)$
  • $\frac{dx}{dt} = 2t$ and $\frac{dy}{dt} = 3t^2 + 2$
  • $\frac{dy}{dx} = \frac{3t^2 + 2}{2t}$, and when $t = 1$, $\frac{dy}{dx} = \frac{5}{2}$
  • Tangent line equation: $y - 3 = \frac{5}{2}(x - 0)$ or $y = \frac{5}{2}x + 3$

Arc Length and Area

Calculating Arc Length of Parametric Curves

• Arc length is the distance along a curve between two points
• For a parametric curve defined by $x = f(t)$ and $y = g(t)$ on the interval $[a, b]$, the arc length is given by the integral:
  • $L = \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt$
• To find the arc length, differentiate $x$ and $y$ with respect to $t$, square and add the results, take the square root, and integrate with respect to $t$
• Example: Find the arc length of the curve $x = \cos t$, $y = \sin t$ for $0 \leq t \leq \frac{\pi}{2}$
  • $\frac{dx}{dt} = -\sin t$ and $\frac{dy}{dt} = \cos t$
  • $L = \int_0^{\frac{\pi}{2}} \sqrt{(-\sin t)^2 + (\cos t)^2} dt = \int_0^{\frac{\pi}{2}} \sqrt{\sin^2 t + \cos^2 t} dt = \int_0^{\frac{\pi}{2}} 1 dt = \frac{\pi}{2}$

Finding Area Under Parametric Curves

• To find the area under a parametric curve, integrate $y$ with respect to $x$ using the formula:
  • $A = \int_a^b y \frac{dx}{dt} dt$
• Substitute the parametric equations for $x$ and $y$, and integrate with respect to $t$
• Example: Find the area under the curve $x = t^2$, $y = t$ for $0 \leq t \leq 1$
  • $\frac{dx}{dt} = 2t$
  • $A = \int_0^1 t \cdot 2t dt = 2 \int_0^1 t^2 dt = 2 \left[\frac{t^3}{3}\right]_0^1 = \frac{2}{3}$

Higher Order Derivatives

Finding Second Derivatives in Parametric Form

• The second derivative of $y$ with respect to $x$, denoted $\frac{d^2y}{dx^2}$, measures the rate of change of the slope of a curve
• To find $\frac{d^2y}{dx^2}$ for a parametric curve, differentiate $\frac{dy}{dx}$ with respect to $t$ and divide by $\frac{dx}{dt}$
  • $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$
• First, find $\frac{dy}{dx}$ by dividing $\frac{dy}{dt}$ by $\frac{dx}{dt}$
• Then, differentiate $\frac{dy}{dx}$ with respect to $t$ to get $\frac{d}{dt}(\frac{dy}{dx})$
• Finally, divide $\frac{d}{dt}(\frac{dy}{dx})$ by $\frac{dx}{dt}$ to obtain $\frac{d^2y}{dx^2}$
• Example: Given $x = e^t$ and $y = e^{2t}$, find $\frac{d^2y}{dx^2}$
  • $\frac{dx}{dt} = e^t$ and $\frac{dy}{dt} = 2e^{2t}$
  • $\frac{dy}{dx} = \frac{2e^{2t}}{e^t} = 2e^t$
  • $\frac{d}{dt}(\frac{dy}{dx}) = \frac{d}{dt}(2e^t) = 2e^t$
  • $\frac{d^2y}{dx^2} = \frac{2e^t}{e^t} = 2$