The Main Idea 

When working with parametric curves, you often need to find the slope at any point. Instead of the tedious process of eliminating the parameter, there’s a direct formula that uses the chain rule.

For parametric equations [latex]x = x(t)[/latex] and [latex]y = y(t)[/latex], the slope is: [latex]\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{y'(t)}{x'(t)}[/latex]

This formula works for ANY parametric curve—even loops, cusps, and curves that cross themselves. You don’t need to worry about whether the curve can be written as [latex]y = f(x)[/latex].

Critical Points: The derivative is zero when [latex]y'(t) = 0[/latex] (horizontal tangent) and undefined when [latex]x'(t) = 0[/latex] (vertical tangent). These give you the critical points of the parametric curve.

Finding Tangent Lines: Once you have the slope at parameter value [latex]t_0[/latex], find the point [latex](x(t_0), y(t_0))[/latex] and use point-slope form: [latex]y - y(t_0) = \frac{dy}{dx}\big|_{t=t_0}(x - x(t_0))[/latex].

The Main Idea 

For parametric equations, finding the second derivative follows naturally from the first derivative formula. Since we know [latex]\frac{dy}{dx} = \frac{y'(t)}{x'(t)}[/latex], we can find the second derivative by treating this ratio as a function of [latex]t[/latex] and applying the chain rule again.

The formula is: [latex]\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}[/latex]

Problem-Solving Strategy:

• Find the first derivative [latex]\frac{dy}{dx}[/latex] in terms of [latex]t[/latex]
• Differentiate that expression with respect to [latex]t[/latex]
• Divide by [latex]\frac{dx}{dt}[/latex]

The second derivative formula comes from applying the chain rule to [latex]\frac{d}{dx}\left(\frac{dy}{dx}\right)[/latex]. Since both [latex]x[/latex] and [latex]\frac{dy}{dx}[/latex] depend on [latex]t[/latex], we need to account for how [latex]x[/latex] changes with [latex]t[/latex].

This approach is much more efficient than converting parametric equations to rectangular form first, especially when the elimination process would create complex expressions.

The Main Idea 

Finding the area under a parametric curve requires a different approach than regular integration. Instead of integrating with respect to [latex]x[/latex], we integrate with respect to the parameter [latex]t[/latex].

The formula comes from using rectangle approximations. For each small rectangle, the height is [latex]y(t)[/latex] and the width is the change in [latex]x[/latex] over a small time interval. As the intervals get smaller, this width approaches [latex]x'(t) \cdot dt[/latex].

The Area Formula: For a parametric curve [latex]x = x(t), y = y(t)[/latex] where [latex]a \leq t \leq b[/latex]: [latex]A = \int_a^b y(t) \cdot x'(t) , dt[/latex]

Key Requirements:

• The curve must not cross itself
• [latex]x(t)[/latex] should increase as [latex]t[/latex] increases from [latex]a[/latex] to [latex]b[/latex]
• [latex]x(t)[/latex] must be differentiable

Why This Works: The Mean Value Theorem guarantees that [latex]\frac{x(t_i) - x(t_{i-1})}{t_i - t_{i-1}}[/latex] equals [latex]x'(c)[/latex] for some point [latex]c[/latex] in the interval. As intervals shrink, this becomes [latex]x'(t)[/latex].

The Main Idea 

Arc length measures the actual distance along a curve—if you walked along the path, this tells you how far you’d travel. For parametric curves, we can’t use the usual arc length formula, so we need a different approach.

The key insight comes from approximating the curve with tiny line segments. Each segment has length [latex]\sqrt{(\Delta x)^2 + (\Delta y)^2}[/latex] using the distance formula. As we make the segments smaller and smaller, this approximation becomes exact.

The Arc Length Formula: For a parametric curve [latex]x = x(t), y = y(t)[/latex] from [latex]t_1[/latex] to [latex]t_2[/latex]: [latex]s = \int_{t_1}^{t_2} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt[/latex]

The Main Idea 

Think of a parametric curve as a flexible wire that you can bend into any shape you want. When you spin that wire around the x-axis, it sweeps out a 3D surface—like how spinning a jump rope creates a cylinder-like shape in the air. Finding the surface area of that 3D shape is what we’re after here.

The key insight: We’re essentially wrapping tiny circumferences around each point on our curve and adding them all up. For each point [latex](x(t), y(t))[/latex] on our parametric curve, we create a circle with circumference [latex]2\pi y(t)[/latex] when we rotate around the [latex]x[/latex]-axis. But here’s the catch—we need to account for how stretched out our curve is as we move along it.

The key formula: [latex]S = 2\pi \int_a^b y(t) \sqrt{[x'(t)]^2 + [y'(t)]^2} dt[/latex]

Critical requirement: [latex]y(t) \geq 0[/latex] throughout your interval. This just means your curve stays on or above the x-axis—which makes sense since negative radii don’t exist in the real world.

Make sure you’re using the correct derivatives—[latex]x'(t)[/latex] and [latex]y'(t)[/latex] are derivatives with respect to the parameter [latex]t[/latex], not [latex]x[/latex]. This is different from the regular surface area formula you might remember from earlier.