AP Calculus BC 9.1 Defining and Differentiating Parametric Equations Study Notes - New Syllabus
AP Calculus BC 9.1 Defining and Differentiating Parametric Equations Study Notes- New syllabus
AP Calculus BC 9.1 Defining and Differentiating Parametric Equations Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Derivatives allow us to solve real-world problems involving rates of change.
Key Concepts:
- Defining and Differentiating Parametric Equations
Defining and Differentiating Parametric Equations
Defining and Differentiating Parametric Equations
Methods for calculating derivatives of real valued functions extend naturally to parametric curves. A parametric curve in the plane is described by functions \( x = x(t) \) and \( y = y(t) \) for \( t \) in an interval. The variable \( t \) is called a parameter.
Parametric form:
\( x = x(t) \), \( y = y(t) \)
Slope of the tangent:
if \( \dfrac{dx}{dt} \neq 0 \), then \( \displaystyle \dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} \)
Second derivative:
if \( \dfrac{dx}{dt} \neq 0 \), then \( \displaystyle \dfrac{d^{2}y}{dx^{2}} = \dfrac{\dfrac{d}{dt}\!\left( \dfrac{dy}{dx} \right)}{\dfrac{dx}{dt}} \)
Horizontal tangent at \( t = t_{0} \):
\( \dfrac{dy}{dt}(t_{0}) = 0 \) and \( \dfrac{dx}{dt}(t_{0}) \neq 0 \)
Vertical tangent at \( t = t_{0} \):
\( \dfrac{dx}{dt}(t_{0}) = 0 \) and \( \dfrac{dy}{dt}(t_{0}) \neq 0 \)
Stationary or cusp type issues:
when both \( \dfrac{dx}{dt} = 0 \) and \( \dfrac{dy}{dt} = 0 \), further analysis is needed to classify the behavior
Arc Length and Speed
Speed:
\( \displaystyle v(t) = \sqrt{\left( \dfrac{dx}{dt} \right)^{2} + \left( \dfrac{dy}{dt} \right)^{2}} \)
Arc length from \( t_{1} \) to \( t_{2} \):
\( L = \displaystyle \int_{t_{1}}^{t_{2}} \sqrt{\left( \dfrac{dx}{dt} \right)^{2} + \left( \dfrac{dy}{dt} \right)^{2}} \, dt \)
Eliminating the Parameter
- Sometimes it is possible to write \( y \) as a function of \( x \) by eliminating \( t \). This can simplify finding intercepts, domains, or checking symmetry.
Example
Compute \( \dfrac{dy}{dx} \) and \( \dfrac{d^{2}y}{dx^{2}} \) for \( x = t^{2} \), \( y = t^{3} \) at a general \( t \), and classify horizontal or vertical tangents.
▶️ Answer/Explanation
\( \dfrac{dx}{dt} = 2t \), \( \dfrac{dy}{dt} = 3t^{2} \)
\( \displaystyle \dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} = \dfrac{3t^{2}}{2t} = \dfrac{3}{2} t \) for \( t \neq 0 \).
Second derivative: \( \displaystyle \dfrac{d^{2}y}{dx^{2}} = \dfrac{\dfrac{d}{dt}\!\left( \dfrac{dy}{dx} \right)}{\dfrac{dx}{dt}} = \dfrac{\dfrac{d}{dt}\!\left( \dfrac{3}{2} t \right)}{2t} = \dfrac{\dfrac{3}{2}}{2t} = \dfrac{3}{4t} \) for \( t \neq 0 \).
Horizontal tangent when \( \dfrac{dy}{dt} = 0 \Rightarrow 3t^{2} = 0 \Rightarrow t = 0 \). But at \( t = 0 \), \( \dfrac{dx}{dt} = 0 \) too, so this is an indeterminate case that requires deeper analysis. The curve has a cusp at \( t = 0 \) since both derivatives vanish.
Vertical tangent when \( \dfrac{dx}{dt} = 0 \) and \( \dfrac{dy}{dt} \neq 0 \). Here \( \dfrac{dx}{dt} = 2t = 0 \Rightarrow t = 0 \) but \( \dfrac{dy}{dt}(0) = 0 \) as well, so there is no regular vertical tangent away from the cusp.
Example
Find the slope of the tangent line and the equation of the tangent at the point corresponding to \( t = \dfrac{\pi}{4} \) for \( x = \cos t \), \( y = \sin 2t \).
▶️ Answer/Explanation
\( \dfrac{dx}{dt} = -\sin t \), \( \dfrac{dy}{dt} = 2\cos 2t \).
\( \displaystyle \dfrac{dy}{dx} = \dfrac{2\cos 2t}{-\sin t} = -\dfrac{2\cos 2t}{\sin t} \).
At \( t = \dfrac{\pi}{4} \): \( x = \cos\!\left( \dfrac{\pi}{4} \right) = \dfrac{\sqrt{2}}{2} \), \( y = \sin\!\left( \dfrac{\pi}{2} \right) = 1 \).
Slope \( m = -\dfrac{2\cos\!\left( \dfrac{\pi}{2} \right)}{\sin\!\left( \dfrac{\pi}{4} \right)} = -\dfrac{2 \cdot 0}{\dfrac{\sqrt{2}}{2}} = 0 \). Tangent is horizontal.
Tangent line at \( \left( \dfrac{\sqrt{2}}{2}, 1 \right) \): \( y – 1 = 0 \cdot \left( x – \dfrac{\sqrt{2}}{2} \right) \Rightarrow y = 1 \).
Example
Eliminate the parameter and find \( \dfrac{dy}{dx} \) in terms of \( x \) when \( x = 1 + t^{2} \), \( y = t + t^{3} \).
▶️ Answer/Explanation
From \( x = 1 + t^{2} \) we get \( t = \pm \sqrt{x – 1} \) where \( x \ge 1 \). The curve is not single valued in \( y \) as a function of \( x \) because both signs can occur.
Differentiate parametrically: \( \dfrac{dx}{dt} = 2t \), \( \dfrac{dy}{dt} = 1 + 3t^{2} \).
\( \displaystyle \dfrac{dy}{dx} = \dfrac{1 + 3t^{2}}{2t} = \dfrac{1}{2t} + \dfrac{3}{2} t \).
In terms of \( x \), use \( t^{2} = x – 1 \). If a branch is chosen with \( t = \sqrt{x – 1} \) for \( t \ge 0 \), then \( \displaystyle \dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x – 1}} + \dfrac{3}{2} \sqrt{x – 1} \).
If the branch \( t = -\sqrt{x – 1} \) is chosen, the first term changes sign accordingly. Always be clear about the branch when eliminating parameters.
Checklist for Parametric Differentiation
- Compute \( \dfrac{dx}{dt} \) and \( \dfrac{dy}{dt} \) carefully.
- Use \( \displaystyle \dfrac{dy}{dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} \) only where \( \dfrac{dx}{dt} \neq 0 \).
- For concavity, compute \( \displaystyle \dfrac{d^{2}y}{dx^{2}} = \dfrac{\dfrac{d}{dt}\!\left( \dfrac{dy}{dx} \right)}{\dfrac{dx}{dt}} \).
- Classify horizontal and vertical tangents using signs of \( \dfrac{dy}{dt} \) and \( \dfrac{dx}{dt} \).
- When both \( \dfrac{dx}{dt} \) and \( \dfrac{dy}{dt} \) are zero, analyze higher derivatives or reparametrize to understand local behavior.