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