91Ó°ÊÓ

Differentiation is a fundamental concept in calculus that involves finding the rate at which a function is changing. It can be applied to parametric equations, where each variable is expressed in terms of a parameter, usually denoted as \( t \). In our exercise, we have two parametric equations: \( x = t^2 \) and \( y = 2t^3 + 4t - 1 \). Differentiating each function with respect to \( t \), we find:

• \( \frac{dx}{dt} = 2t \)
• \( \frac{dy}{dt} = 6t^2 + 4 \)

Differentiation here helps us understand how the curve behaves as the parameter \( t \) changes. It's crucial for determining the slope of the tangent line, which we access through the derivative \( \frac{dy}{dx} \), calculated using the chain rule.

Tangent lines are straight lines that touch a curve at a single point, sharing the same slope as the curve at that point. They're useful for approximating the curve locally. To find the slope of a tangent line given parametric equations, we use the derivative \( \frac{dy}{dx} \), which is the ratio of \( \frac{dy}{dt} \) to \( \frac{dx}{dt} \). In our specific problem, this results in:\[ \frac{dy}{dx} = \frac{6t^2 + 4}{2t} = \frac{3t^2 + 2}{t} \]This formula tells us how the slope of the tangent line changes with \( t \). Horizontal tangent lines occur when the numerator of \( \frac{dy}{dx} \) is zero, indicating that there are no real values of \( t \) that satisfy this for our curve. Vertical tangent lines occur when \( 2t = 0 \), which results in \( t = 0 \), indicating a vertical tangent line at the corresponding point on the curve.

The chain rule is a powerful tool in calculus for differentiating composite functions. When dealing with parametric equations, it allows us to connect the derivatives \( \frac{dy}{dt} \) and \( \frac{dx}{dt} \) to find \( \frac{dy}{dx} \), the rate of change of \( y \) concerning \( x \). Here's how we applied it:\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{6t^2 + 4}{2t} \]This results in:\[ \frac{dy}{dx} = \frac{3t^2 + 2}{t} \]The chain rule ensures a smooth transition between differentiating with respect to the parameter \( t \) and moving to derivatives concerning \( x \). It is vital for ensuring the correct calculation of derivatives for functions defined in terms other than \( x \). By understanding the chain rule, students can proficiently handle derivatives of parametric equations and interpret their geometrical implications on the curve.

The second derivative \( \frac{d^2 y}{dx^2} \), or the derivative of \( \frac{dy}{dx} \) with respect to \( x \), provides valuable information about the curvature and concavity of a graph. Our exercise involves calculating this second derivative for a parametric curve.First, differentiate \( \frac{dy}{dx} = \frac{3t^2 + 2}{t} \) with respect to \( t \):\[ \frac{d}{dt}\left(\frac{3t^2 + 2}{t}\right) = 3 - \frac{2}{t^2} \]Then, divide this result by \( \frac{dx}{dt} = 2t \) to find \( \frac{d^2 y}{dx^2} \):\[ \frac{d^2 y}{dx^2} = \frac{3 - \frac{2}{t^2}}{2t} = \frac{3t^2 - 2}{2t^3} \]The second derivative helps us determine the concavity of the curve. If \( \frac{d^2 y}{dx^2} > 0 \), the curve is concave up, and if \( \frac{d^2 y}{dx^2} < 0 \), it's concave down. This information is crucial in understanding the shape and behavior of the curve beyond just the slope of tangent lines.