14.2-bo'lim : Gradient vektor, tangens tekisliklar va normal chiziqlar
1. \(\left( {3, - 3,2} \right)\) da \({x^2}y = 4z{{\bf{e}}^{x + y}} - 35\) ga teguvchi tekislik va normal chiziqni toping.
Yechim
Birinchidan, biz tenglamani tezda qayta yozishimiz kerak:
\[{x^2}y - 4z{{\bf{e}}^{x + y}} = - 35\]2-qadam
Endi bizga 1-bosqichdagi tenglamaning chap tomonidagi funksiyaning gradienti va uning \(\left( {3, - 3,2} \right)\)dagi qiymati kerak. Mana bu miqdorlar.
\[\nabla f = \left\langle {2xy - 4z{{\bf{e}}^{x + y}},{x^2} - 4z{{\bf{e}}^{x + y}}, - 4{{\bf{e}}^{x + y}}} \right\rangle \hspace{0.25in}\hspace{0.25in}\nabla f\left( {3, - 3,2} \right) = \left\langle { - 26,1, - 4} \right\rangle \]3-qadam
Tangens tekislik u holda,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{ - 26\left( {x - 3} \right) + \left( 1 \right)\left( {y + 3} \right) - 4\left( {z - 2} \right) = 0}}\hspace{0.25in} \Rightarrow \hspace{0.25in}\require{bbox} \bbox[2pt,border:1px solid black]{{ - 26x + y - 4z = - 89}}\]Oddiy chiziq,
\[\require{bbox} \bbox[2pt,border:1px solid black]{{\vec r\left( t \right) = \left\langle {3, - 3,2} \right\rangle + t\left\langle { - 26,1, - 4} \right\rangle = \left\langle {3 - 26t, - 3 + t,2 - 4t} \right\rangle }}\]