9.2 Second Derivatives of Parametric Equations

• What does the second derivative of parametric equations, $\frac{d^{2}y}{dx^{2}}$, describe?


  Concavity
• Steps to find the second derivative of parametric equations

  1️⃣ Find the first derivative $\frac{dy}{dx}$ using $\frac{dy}{dx} =$$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$.

  2️⃣ Differentiate $\frac{dy}{dx}$ with respect to $t$ to get $\frac{d}{dt} \left( \frac{dy}{dx} \right)$.

  3️⃣ Divide $\frac{d}{dt} \left( \frac{dy}{dx} \right)$ by $\frac{dx}{dt}$ to obtain $\frac{d^{2}y}{dx^{2}}$.
• The concavity of a parametric curve is determined by the second derivative $\frac{d^{2}y}{dx^{2}}$.
• Finding the second derivative helps identify inflection points of a parametric curve.
• Steps to apply the chain rule

  1️⃣ Identify the inner and outer functions.

  2️⃣ Differentiate each function separately.

  3️⃣ Multiply the derivatives.
• If $y =$$\sin(x^{2})$, then $\frac{dy}{dx} =$$2x \cos(x^{2})$.
• What does the first derivative of parametric equations, $\frac{dy}{dx}$, represent?


  Slope
• The chain rule is used to differentiate composite functions.
• The chain rule formula states that $\frac{dy}{dx} =$$\frac{df}{dg} \cdot \frac{dg}{dx}$
• Steps to apply the chain rule

  1️⃣ Identify the inner and outer functions

  2️⃣ Differentiate each function separately

  3️⃣ Multiply the derivatives
• If $y =$$\sin(x^{2})$, then \frac{dy}{dx} = 2x \cos(x^{2})</latex>, which uses the chain rule.
• What does the first derivative of parametric equations represent?

  Slope
• The formula for the first derivative of parametric equations is $\frac{dy}{dx} =$$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
• Steps to find $\frac{dy}{dx}$ for parametric equations


  1️⃣ Find $\frac{dy}{dt}$ and $\frac{dx}{dt}$

  2️⃣ Divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$
• If $x =$$t^{2}$ and y = 2t^{3}</latex>, then $\frac{dy}{dx} =$$3t$, which represents the slope of the curve.
• The chain rule states that $\frac{dy}{dx} =$$\frac{df}{dg} \cdot \frac{dg}{dx}$
• If $y =$$\sin(x^{2})$, then $\frac{dy}{dx} =$$2x \cos(x^{2})$, which applies the chain rule.
• The chain rule is essential for finding second derivatives of parametric equations.
• The formula for the first derivative of parametric equations is $\frac{dy}{dx} =$$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$, which represents the slope of the curve.
• The first derivative of parametric equations is crucial for analyzing slope and tangents.
• Match the derivative formula with its description:

  Chain rule ↔️ $\frac{dy}{dx} =$$\frac{df}{dg} \cdot \frac{dg}{dx}$

  Parametric first derivative ↔️ $\frac{dy}{dx} =$$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

  Parametric second derivative ↔️ $\frac{d^{2}y}{dx^{2}} =$$\frac{\frac{d}{dt} \left( \frac{dy}{dx} \right)}{\frac{dx}{dt}}$
• If $x =$$\cos(t)$ and $y =$$\sin(t)$, then $\frac{dy}{dx} =$$- \cot(t)$, which uses the chain rule.
• The second derivative \frac{d^{2}y}{dx^{2}}</latex> helps determine the concavity of a curve.
• The first step to find the second derivative of parametric equations is to find the first derivative.
• $\frac{d^{2}y}{dx^{2}}$ helps determine the concavity of the curve and identify inflection points.concavity
• To find the second derivative of parametric equations, we use the chain rule.
• Steps to find $\frac{d^{2}y}{dx^{2}}$ of parametric equations


  1️⃣ Find the first derivative $\frac{dy}{dx}$ using $\frac{dy}{dx} =$$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

  2️⃣ Differentiate $\frac{dy}{dx}$ with respect to $t$

  3️⃣ Divide the derivative by $\frac{dx}{dt}$ to get $\frac{d^{2}y}{dx^{2}}$
• In the example, \frac{dy}{dx} = \frac{\cos(t)}{ - \sin(t)} = - \cot(t)</latex>.-cot(t)
• $\frac{d^{2}y}{dx^{2}}$ is used to analyze the slope of parametric curves.


  False
• The first derivative $\frac{dy}{dx}$ of parametric equations represents the slope of the curve at a given point.slope
• The formula for $\frac{dy}{dx}$ is $\frac{dy}{dx} =$$\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$.
• Steps to calculate $\frac{dy}{dx}$ of parametric equations


  1️⃣ Find \frac{dy}{dt}</latex> and $\frac{dx}{dt}$

  2️⃣ Divide $\frac{dy}{dt}$ by $\frac{dx}{dt}$
• For $x =$$t^{2}$ and $y =$$2t^{3}$, $\frac{dy}{dx} =$$\frac{6t^{2}}{2t} =$$3t$.3t
• The chain rule is used to differentiate composite functions.
• Match the chain rule concepts with their descriptions:

  Composite Function ↔️ Applying one function to the result of another

  Chain Rule Formula ↔️ $\frac{dy}{dx} =$$\frac{df}{dg} \cdot \frac{dg}{dx}$

  Steps to Apply ↔️ Identify inner and outer functions, differentiate separately, multiply derivatives
• For $y =$$\sin(x^{2})$, $\frac{dy}{dx} =$$2x \cos(x^{2})$.2xcos(x^{2})
• The first derivative of parametric equations represents the slope of the curve.
• For $x =$$t^{2}$ and y = 2t^{3}</latex>, $\frac{dy}{dx} =$$3t$.3t
• The first derivative helps in analyzing the slope and tangents of parametric curves.
• What does the first derivative of parametric equations represent?

  Slope of the curve