题目

求抛物线y=x2 与直线x+y+2=0 之间的最短距离.2、 lim _(xarrow y)+dfrac (ln (1-x-y))(sin (2x+2y)cdot cos xy)=-|||-__ .+-|||-3、设 (x,y)=(x)^2-4xy+3(y)^2, 则 lim _(harrow 0)dfrac (f(1.2+h)-f(1.2))(h)= __-|||-4、已知函数 (xy,x+y)=(x)^2+(y)^2+xy, 则 dfrac (partial f(x:y))(partial x)+dfrac (partial f(x,y))(partial y)= __-|||-5、设 =ln (2y-x) 则 dfrac ({partial )^2z}(partial xpartial y)(|)_(x=1)^x-1= __ _·-|||-6、设 =((x-y))^2 则 /(1,0,1)= __-|||-7、若 =f((x)^2-(y)^2,(e)^xv) 则 dfrac (du)(partial x)= __ dfrac (partial u)(partial y)= __ ---|||-____-|||-若 u=f(2x,xy,xyz) 则 (int )_(x)^1= __ _(y)^1= __ _(2)^1= __-|||-8、二元函数 z=f(x,y) 在点(x0,y0)处满足的关系为 () +-|||-A)可微→可导→连续 B)可微→可导,或可微→连续,-|||-但可导不一定连续-|||-C)可微→可导→连续 D)可导→连续,但可导不一定可微2、 lim _(xarrow y)+dfrac (ln (1-x-y))(sin (2x+2y)cdot cos xy)=-|||-__ .+-|||-3、设 (x,y)=(x)^2-4xy+3(y)^2, 则 lim _(harrow 0)dfrac (f(1.2+h)-f(1.2))(h)= __-|||-4、已知函数 (xy,x+y)=(x)^2+(y)^2+xy, 则 dfrac (partial f(x:y))(partial x)+dfrac (partial f(x,y))(partial y)= __-|||-5、设 =ln (2y-x) 则 dfrac ({partial )^2z}(partial xpartial y)(|)_(x=1)^x-1= __ _·-|||-6、设 =((x-y))^2 则 /(1,0,1)= __-|||-7、若 =f((x)^2-(y)^2,(e)^xv) 则 dfrac (du)(partial x)= __ dfrac (partial u)(partial y)= __ ---|||-____-|||-若 u=f(2x,xy,xyz) 则 (int )_(x)^1= __ _(y)^1= __ _(2)^1= __-|||-8、二元函数 z=f(x,y) 在点(x0,y0)处满足的关系为 () +-|||-A)可微→可导→连续 B)可微→可导,或可微→连续,-|||-但可导不一定连续-|||-C)可微→可导→连续 D)可导→连续,但可导不一定可微2、 lim _(xarrow y)+dfrac (ln (1-x-y))(sin (2x+2y)cdot cos xy)=-|||-__ .+-|||-3、设 (x,y)=(x)^2-4xy+3(y)^2, 则 lim _(harrow 0)dfrac (f(1.2+h)-f(1.2))(h)= __-|||-4、已知函数 (xy,x+y)=(x)^2+(y)^2+xy, 则 dfrac (partial f(x:y))(partial x)+dfrac (partial f(x,y))(partial y)= __-|||-5、设 =ln (2y-x) 则 dfrac ({partial )^2z}(partial xpartial y)(|)_(x=1)^x-1= __ _·-|||-6、设 =((x-y))^2 则 /(1,0,1)= __-|||-7、若 =f((x)^2-(y)^2,(e)^xv) 则 dfrac (du)(partial x)= __ dfrac (partial u)(partial y)= __ ---|||-____-|||-若 u=f(2x,xy,xyz) 则 (int )_(x)^1= __ _(y)^1= __ _(2)^1= __-|||-8、二元函数 z=f(x,y) 在点(x0,y0)处满足的关系为 () +-|||-A)可微→可导→连续 B)可微→可导,或可微→连续,-|||-但可导不一定连续-|||-C)可微→可导→连续 D)可导→连续,但可导不一定可微2、 lim _(xarrow y)+dfrac (ln (1-x-y))(sin (2x+2y)cdot cos xy)=-|||-__ .+-|||-3、设 (x,y)=(x)^2-4xy+3(y)^2, 则 lim _(harrow 0)dfrac (f(1.2+h)-f(1.2))(h)= __-|||-4、已知函数 (xy,x+y)=(x)^2+(y)^2+xy, 则 dfrac (partial f(x:y))(partial x)+dfrac (partial f(x,y))(partial y)= __-|||-5、设 =ln (2y-x) 则 dfrac ({partial )^2z}(partial xpartial y)(|)_(x=1)^x-1= __ _·-|||-6、设 =((x-y))^2 则 /(1,0,1)= __-|||-7、若 =f((x)^2-(y)^2,(e)^xv) 则 dfrac (du)(partial x)= __ dfrac (partial u)(partial y)= __ ---|||-____-|||-若 u=f(2x,xy,xyz) 则 (int )_(x)^1= __ _(y)^1= __ _(2)^1= __-|||-8、二元函数 z=f(x,y) 在点(x0,y0)处满足的关系为 () +-|||-A)可微→可导→连续 B)可微→可导,或可微→连续,-|||-但可导不一定连续-|||-C)可微→可导→连续 D)可导→连续,但可导不一定可微2、 lim _(xarrow y)+dfrac (ln (1-x-y))(sin (2x+2y)cdot cos xy)=-|||-__ .+-|||-3、设 (x,y)=(x)^2-4xy+3(y)^2, 则 lim _(harrow 0)dfrac (f(1.2+h)-f(1.2))(h)= __-|||-4、已知函数 (xy,x+y)=(x)^2+(y)^2+xy, 则 dfrac (partial f(x:y))(partial x)+dfrac (partial f(x,y))(partial y)= __-|||-5、设 =ln (2y-x) 则 dfrac ({partial )^2z}(partial xpartial y)(|)_(x=1)^x-1= __ _·-|||-6、设 =((x-y))^2 则 /(1,0,1)= __-|||-7、若 =f((x)^2-(y)^2,(e)^xv) 则 dfrac (du)(partial x)= __ dfrac (partial u)(partial y)= __ ---|||-____-|||-若 u=f(2x,xy,xyz) 则 (int )_(x)^1= __ _(y)^1= __ _(2)^1= __-|||-8、二元函数 z=f(x,y) 在点(x0,y0)处满足的关系为 () +-|||-A)可微→可导→连续 B)可微→可导,或可微→连续,-|||-但可导不一定连续-|||-C)可微→可导→连续 D)可导→连续,但可导不一定可微

求抛物线y=x2 与直线x+y+2=0 之间的最短距离.




题目解答

答案

解析

一、题目1：抛物线$y=x^2$与直线$x+y+2=0$之间的最短距离

考察知识：点到直线的距离公式、函数极值求法。
解题思路：
抛物线$y=x^2$上任意一点$(x,x^2)$到直线$x+y+2=0$的距离公式为：
$d(x)=\frac{|x+x^2+2|}{\sqrt{1^2+1^2}}=\frac{|x^2+x+2|}{\sqrt{2}}$
因$x^2+x+2=(x+\frac{1}{2})^2+\frac{7}{4}\geq\frac{7}{4}>0$，故$d(x)=\frac{x^2+x+2}{\sqrt{2}}$。
对$d(x)$求导：$d'(x)=\frac{2x+1}{\sqrt{2}}$，令$d'(x)=0$得$x=-\frac{1}{2}$。
代入$x=-\frac{1}{2}$得$y=(-\frac{1}{2})^2=\frac{1}{4}$，最短距离为：
$d\left(-\frac{1}{2}\right)=\frac{\left(-\frac{1}{2}\right)^2+\left(-\frac{1}{2}\right)+2}{\sqrt{2}}=\frac{\frac{7}{4}}{\sqrt{2}}=\frac{7\sqrt{2}}{8}$