Mathematics

Mathematics

Problem

The answers are in Chapter 24. 2. Find the equation of the tangent line to x=4sin⁡2tx=4 \sin ^{2} t and y=2cos⁡ty=2 \cos t at t=π3t=\frac{\pi}{3}. (A) y−1=−143(x−3)y-1=-\frac{1}{4 \sqrt{3}}(x-3) (B) y−3=−143(x−1)y-\sqrt{3}=-\frac{1}{4 \sqrt{3}}(x-1) (C) y−3=−12(x−1)y-\sqrt{3}=-\frac{1}{2}(x-1) (D) y−1=−12(x−3)y-1=-\frac{1}{2}(x-3) (A) y−1=−143(x−3)y-1=-\frac{1}{4 \sqrt{3}}(x-3) (B) y−3=−143(x−1)y-\sqrt{3}=-\frac{1}{4 \sqrt{3}}(x-1) (C) y−3=−12(x−1)y-\sqrt{3}=-\frac{1}{2}(x-1) (D) y−1=−12(x−3)y-1=-\frac{1}{2}(x-3)
Intuitive explanation

For parametric curves, the tangent line's slope is dy/dtdx/dt\frac{dy/dt}{dx/dt}, not dydx\frac{dy}{dx} directly. Find both derivatives with respect to tt, evaluate them at t=π3t=\frac{\pi}{3} to get the slope and the point (x,y)(x, y) on the curve, then use point-slope form. The key is remembering that parametric differentiation requires the ratio of rates of change along each coordinate.

Intuitive explanation

For parametric curves, the tangent line's slope is dy/dtdx/dt\frac{dy/dt}{dx/dt}, not dydx\frac{dy}{dx} directly. Find both derivatives with respect to tt, evaluate them at t=π3t=\frac{\pi}{3} to get the slope and the point (x,y)(x, y) on the curve, then use point-slope form. The key is remembering that parametric differentiation requires the ratio of rates of change along each coordinate.

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