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Unit 9
10–15% of examThis BC-only unit takes calculus beyond graphs of y = f(x). You work with curves described by parametric equations, vectors and polar coordinates, and find their slopes, lengths, areas and motion. Much of it is the straight-line motion you already know, now moving around a plane.
Longer videos that cover the whole unit. Good for a first pass or a final review.
Parametric equations give x and y separately as functions of a third variable, often t for time. The slope of the curve is dy/dx = (dy/dt)/(dx/dt), as long as dx/dt isn't 0.
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To get d²y/dx², take the derivative of dy/dx with respect to t, then divide by dx/dt. A common mistake is to stop after differentiating dy/dx with respect to t.
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The length of a parametric curve from t = a to t = b is the integral of √((dx/dt)² + (dy/dt)²) dt. When the curve is the path of a moving particle, the same integral gives the total distance it travels.
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A vector-valued function such as r(t) = ⟨x(t), y(t)⟩ gives a position vector for each value of t. You differentiate it one component at a time: r′(t) is the velocity vector and r″(t) is the acceleration vector.
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You integrate a vector-valued function component by component too, with a constant of integration for each component. Given a velocity vector and a starting position, this lets you find the position at any later time.
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For a particle moving in the plane, the velocity is ⟨x′(t), y′(t)⟩, and the speed is that vector's length, √((x′(t))² + (y′(t))²). Integrating speed over a time interval gives the total distance traveled, and adding displacement to a starting point gives the new position.
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Polar coordinates locate a point by its distance r from the origin and its angle θ. For slopes, write x = r cos θ and y = r sin θ, then use dy/dx = (dy/dθ)/(dx/dθ); dr/dθ is different, and when r is positive it tells you whether the curve is moving away from or toward the origin.
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The area swept out by a polar curve from θ = α to θ = β is ½ times the integral of r² dθ. The formula comes from adding up thin wedges, each like a tiny slice of pie.
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To find the area between two polar curves, find the angles where they meet, then integrate ½(R² − r²) dθ, with R the outer curve and r the inner one. A quick sketch shows which curve is outside on each interval.
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