problem 3. consider the curve described by the equation $(x^{2}+y^{2})^{2}=4(x^{2}-y^{2})$. (a) 6pts. use…

problem 3. consider the curve described by the equation $(x^{2}+y^{2})^{2}=4(x^{2}-y^{2})$. (a) 6pts. use the method of implicit differentiation to give a formula for $y$ in terms of $x$ and $y$. after you differentiate, it might help your algebra to temporarily let $r^{2}=(x^{2}+y^{2})$. (b) 4pts. find the four points where the tangent line to the curve is horizontal. none of them have $x$-coordinate equal to 0; the only point on the curve whose $x$-coordinate is equal to 0 is $(0,0)$; your formula for $y$ is undefined here.

problem 3. consider the curve described by the equation $(x^{2}+y^{2})^{2}=4(x^{2}-y^{2})$. (a) 6pts. use the method of implicit differentiation to give a formula for $y$ in terms of $x$ and $y$. after you differentiate, it might help your algebra to temporarily let $r^{2}=(x^{2}+y^{2})$. (b) 4pts. find the four points where the tangent line to the curve is horizontal. none of them have $x$-coordinate equal to 0; the only point on the curve whose $x$-coordinate is equal to 0 is $(0,0)$; your formula for $y$ is undefined here.

Answer

Explanation:

Step1: 对等式两边求导

设(r^{2}=x^{2}+y^{2})，原方程为(r^{4}=4(x^{2}-y^{2}))。 对左边求导：(\frac{d(r^{4})}{dx}=4r^{3}\cdot\frac{dr^{2}}{dx}=4r^{3}(2x + 2y\cdot y'))。 对右边求导：(\frac{d}{dx}[4(x^{2}-y^{2})]=4(2x-2y\cdot y'))。 则(4r^{3}(2x + 2y\cdot y')=4(2x-2y\cdot y'))，将(r^{2}=x^{2}+y^{2})代回，即(4(x^{2}+y^{2})^{\frac{3}{2}}(2x + 2y\cdot y')=4(2x-2y\cdot y'))，化简得((x^{2}+y^{2})^{\frac{3}{2}}(x + y\cdot y')=x-y\cdot y')，进一步展开((x^{2}+y^{2})^{\frac{3}{2}}x+(x^{2}+y^{2})^{\frac{3}{2}}y\cdot y'=x-y\cdot y')，移项可得(y'=\frac{x-(x^{2}+y^{2})^{\frac{3}{2}}x}{(x^{2}+y^{2})^{\frac{3}{2}}y + y}=\frac{x(1-(x^{2}+y^{2})^{\frac{3}{2}})}{y((x^{2}+y^{2})^{\frac{3}{2}} + 1)})。

Step2: 求水平切线的点

因为水平切线处(y' = 0)，由(y'=\frac{x(1-(x^{2}+y^{2})^{\frac{3}{2}})}{y((x^{2}+y^{2})^{\frac{3}{2}} + 1)} = 0)，且(x\neq0)，则(1-(x^{2}+y^{2})^{\frac{3}{2}}=0)，即(x^{2}+y^{2}=1)。 将(x^{2}+y^{2}=1)代入原方程((x^{2}+y^{2})^{2}=4(x^{2}-y^{2}))，得(1 = 4(x^{2}-y^{2}))，又(y^{2}=1 - x^{2})，代入(1 = 4(x^{2}-(1 - x^{2})))，即(1 = 4(2x^{2}-1))，(8x^{2}-4 = 1)，(8x^{2}=5)，(x^{2}=\frac{5}{8})，(x=\pm\frac{\sqrt{10}}{4})。 当(x=\frac{\sqrt{10}}{4})时，(y^{2}=1 - x^{2}=1-\frac{5}{8}=\frac{3}{8})，(y=\pm\frac{\sqrt{6}}{4})；当(x=-\frac{\sqrt{10}}{4})时，(y^{2}=1 - x^{2}=1-\frac{5}{8}=\frac{3}{8})，(y=\pm\frac{\sqrt{6}}{4})。 所以四个点为((\frac{\sqrt{10}}{4},\frac{\sqrt{6}}{4}))，((\frac{\sqrt{10}}{4},-\frac{\sqrt{6}}{4}))，((-\frac{\sqrt{10}}{4},\frac{\sqrt{6}}{4}))，((-\frac{\sqrt{10}}{4},-\frac{\sqrt{6}}{4}))。

Answer:

(a) (y'=\frac{x(1-(x^{2}+y^{2})^{\frac{3}{2}})}{y((x^{2}+y^{2})^{\frac{3}{2}} + 1)}) (b) ((\frac{\sqrt{10}}{4},\frac{\sqrt{6}}{4}))，((\frac{\sqrt{10}}{4},-\frac{\sqrt{6}}{4}))，((-\frac{\sqrt{10}}{4},\frac{\sqrt{6}}{4}))，((-\frac{\sqrt{10}}{4},-\frac{\sqrt{6}}{4}))