题目
3.求下列曲面在所示点处的切平面与法线:-|||-(1) -(e)^2x=0, 在点(1,1,2);-|||-(2) dfrac ({x)^2}({a)^2}+dfrac ({y)^2}({b)^2}+dfrac ({z)^2}({c)^2}=1 在点 (dfrac (a)(sqrt {3)},dfrac (b)(sqrt {3)},dfrac (c)(sqrt {3)})()
题目解答
答案
解析
步骤 1：求解曲面 $y-{e}^{2x-2}=0$ 在点 (1,1,2) 处的切平面
令 $F(x,y,z)=y-{e}^{2x-z}$ , 则 ${F}_{x}(1,1,2)=-2,{F}_{y}(1,1,2)=1$ , ${F}_{z}(1,1,2)=1$ 。
切平面方程为 $-2(x-1)+(y-1)+(z-2)=0$ 。
步骤 2：求解曲面 $\dfrac {{x}^{2}}{{a}^{2}}+\dfrac {{y}^{2}}{{b}^{2}}+\dfrac {{z}^{2}}{{c}^{2}}=1$ 在点 $(\dfrac {a}{\sqrt {3}},\dfrac {b}{\sqrt {3}},\dfrac {c}{\sqrt {3}})$ 处的切平面
令 $F(x,y,z)=\dfrac {{x}^{2}}{{a}^{2}}+\dfrac {{y}^{2}}{{b}^{2}}+\dfrac {{z}^{2}}{{c}^{2}}-1$ , 则 ${F}_{x}(\dfrac {a}{\sqrt {3}},\dfrac {b}{\sqrt {3}},\dfrac {c}{\sqrt {3}})=\dfrac {2}{\sqrt {3}a}$ , ${F}_{y}(\dfrac {a}{\sqrt {3}},\dfrac {b}{\sqrt {3}},\dfrac {c}{\sqrt {3}})=\dfrac {2}{\sqrt {3}b}$ , ${F}_{z}(\dfrac {a}{\sqrt {3}},\dfrac {b}{\sqrt {3}},\dfrac {c}{\sqrt {3}})=\dfrac {2}{\sqrt {3}c}$ 。
切平面方程为 $\dfrac {2}{\sqrt {3}a}(x-\dfrac {a}{\sqrt {3}})+\dfrac {2}{\sqrt {3}b}(y-\dfrac {b}{\sqrt {3}})+\dfrac {2}{\sqrt {3}c}(z-\dfrac {c}{\sqrt {3}})=0$ 。
步骤 3：求解曲面 $y-{e}^{2x-2}=0$ 在点 (1,1,2) 处的法线
法线方程为 $\dfrac {x-1}{-2}=y-1=z-2$ 。
步骤 4：求解曲面 $\dfrac {{x}^{2}}{{a}^{2}}+\dfrac {{y}^{2}}{{b}^{2}}+\dfrac {{z}^{2}}{{c}^{2}}=1$ 在点 $(\dfrac {a}{\sqrt {3}},\dfrac {b}{\sqrt {3}},\dfrac {c}{\sqrt {3}})$ 处的法线
法线方程为 $\dfrac {x-\dfrac {a}{\sqrt {3}}}{\dfrac {2}{\sqrt {3}a}}=\dfrac {y-\dfrac {b}{\sqrt {3}}}{\dfrac {2}{\sqrt {3}b}}=\dfrac {z-\dfrac {c}{\sqrt {3}}}{\dfrac {2}{\sqrt {3}c}}$ 。
令 $F(x,y,z)=y-{e}^{2x-z}$ , 则 ${F}_{x}(1,1,2)=-2,{F}_{y}(1,1,2)=1$ , ${F}_{z}(1,1,2)=1$ 。
切平面方程为 $-2(x-1)+(y-1)+(z-2)=0$ 。
步骤 2：求解曲面 $\dfrac {{x}^{2}}{{a}^{2}}+\dfrac {{y}^{2}}{{b}^{2}}+\dfrac {{z}^{2}}{{c}^{2}}=1$ 在点 $(\dfrac {a}{\sqrt {3}},\dfrac {b}{\sqrt {3}},\dfrac {c}{\sqrt {3}})$ 处的切平面
令 $F(x,y,z)=\dfrac {{x}^{2}}{{a}^{2}}+\dfrac {{y}^{2}}{{b}^{2}}+\dfrac {{z}^{2}}{{c}^{2}}-1$ , 则 ${F}_{x}(\dfrac {a}{\sqrt {3}},\dfrac {b}{\sqrt {3}},\dfrac {c}{\sqrt {3}})=\dfrac {2}{\sqrt {3}a}$ , ${F}_{y}(\dfrac {a}{\sqrt {3}},\dfrac {b}{\sqrt {3}},\dfrac {c}{\sqrt {3}})=\dfrac {2}{\sqrt {3}b}$ , ${F}_{z}(\dfrac {a}{\sqrt {3}},\dfrac {b}{\sqrt {3}},\dfrac {c}{\sqrt {3}})=\dfrac {2}{\sqrt {3}c}$ 。
切平面方程为 $\dfrac {2}{\sqrt {3}a}(x-\dfrac {a}{\sqrt {3}})+\dfrac {2}{\sqrt {3}b}(y-\dfrac {b}{\sqrt {3}})+\dfrac {2}{\sqrt {3}c}(z-\dfrac {c}{\sqrt {3}})=0$ 。
步骤 3：求解曲面 $y-{e}^{2x-2}=0$ 在点 (1,1,2) 处的法线
法线方程为 $\dfrac {x-1}{-2}=y-1=z-2$ 。
步骤 4：求解曲面 $\dfrac {{x}^{2}}{{a}^{2}}+\dfrac {{y}^{2}}{{b}^{2}}+\dfrac {{z}^{2}}{{c}^{2}}=1$ 在点 $(\dfrac {a}{\sqrt {3}},\dfrac {b}{\sqrt {3}},\dfrac {c}{\sqrt {3}})$ 处的法线
法线方程为 $\dfrac {x-\dfrac {a}{\sqrt {3}}}{\dfrac {2}{\sqrt {3}a}}=\dfrac {y-\dfrac {b}{\sqrt {3}}}{\dfrac {2}{\sqrt {3}b}}=\dfrac {z-\dfrac {c}{\sqrt {3}}}{\dfrac {2}{\sqrt {3}c}}$ 。