題目 5. (10 points) Find the slope of the tangent line to the polar curve r=1+sin(2θ)r=1+\sin(2\theta)r=1+sin(2θ) at the point specified by θ=π/3\theta=\pi/3θ=π/3. 解答 解法一 極座標曲線的斜率為 dydx=r′sinθ+rcosθr′cosθ−rsinθ.\frac{\mathrm{d}y}{\mathrm{d}x} =\frac{r'\sin\theta+r\cos\theta}{r'\cos\theta-r\sin\theta}.dxdy=r′cosθ−rsinθr′sinθ+rcosθ. 在 θ=π/3\theta=\pi/3θ=π/3，r=1+32r=1+\frac{\sqrt3}{2}r=1+23、r′=2cos(2θ)=−1r'=2\cos(2\theta)=-1r′=2cos(2θ)=−1，所以 dydx=−2−35+23.\boxed{\frac{\mathrm{d}y}{\mathrm{d}x} =-\frac{2-\sqrt3}{5+2\sqrt3}}.dxdy=−5+232−3.
