題目
Problem
(3) (10 pts) Let f:(−δ,δ)→R be a function so that f(0)=21 for δ>0. Assume that f(x) satisfies the equation
x2+(f(x))2=(2x2+2(f(x))2−x)2
for all x∈(−δ,δ). Suppose we know that f is differentiable at 0. Compute the tangent line to the curve y=f(x) at the point (0,21).
解答
解法一
令 y=f(x),並把右側括號記作
A=2x2+2y2−x.
對隱式關係 x2+y2=A2 兩邊對 x 微分:
2x+2y\frac{\mathrm{d}y}{\mathrm{d}x}
=&\,2A\left(4x+4y\frac{\mathrm{d}y}{\mathrm{d}x}-1\right).
在 (0,21) 處,A=21,所以
\frac{\mathrm{d}y}{\mathrm{d}x}
=&\,2\frac{\mathrm{d}y}{\mathrm{d}x}-1,
\qquad
\frac{\mathrm{d}y}{\mathrm{d}x}=1.
切線經過 (0,21),斜率為 1,故為
y=x+21.