Find an equation for the line tangent to the curve at the...

Find an equation for the line tangent to the curve at the point defined by the given value of $t .$ Also, find the value of $d^{2} v / d x^{2}$ at this point.
$$x=\sin 2 \pi t, \quad y=\cos 2 \pi t, \quad t=-1 / 6$$

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Key Concepts

Second Derivative for Parametric Functions

The second derivative d²y/dx² in the context of parametric curves gives the curvature or concavity of the curve and is computed using the formula d²y/dx² = (d/dt (dy/dx)) / (dx/dt). This expression applies the chain rule to differentiate dy/dx with respect to the parameter t, allowing one to understand the behavior of the curve beyond just the immediate tangent.

Tangent Line to a Parametric Curve

The tangent line to a parametric curve at a given point is the line that just touches the curve at that point and has a slope equal to the derivative dy/dx evaluated at that point. Once the slope and point are known, the equation of the tangent line can be formulated using the point-slope form.

Parametric Equations

Parametric equations represent a curve by defining both x and y as functions of a third variable, usually denoted by t. This approach is useful for describing curves where y cannot be written as a function of x and for capturing the dynamic nature of the curve as the parameter changes.

Derivative of Parametric Equations

When a curve is given parametrically, the derivative dy/dx is found using the derivatives with respect to the parameter: dy/dx = (dy/dt) / (dx/dt), provided that dx/dt is nonzero. This method allows one to compute slopes and rates of change along the curve.

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