Moglby ktos sprawdzic czy dobrze policzone, bo niestety nie mam odpowiedzi, a zalezy mi zeby nie bylo błedow.
Oblicz pochodną kierunkową funkcji f w punkcie P w kierunku wektrora a.
a)
\(\displaystyle{ f(x,y)=x ^{3} +2xy^{2}}\)
\(\displaystyle{ P(0,1)}\) \(\displaystyle{ \vec{a} =[1,1]}\)
\(\displaystyle{ | \vec{a} | = \sqrt{2}}\)
\(\displaystyle{ \frac{ \vec{a} }{| \vec{a} |} = [ \frac{ \sqrt{2} }{2} ,\frac{ \sqrt{2} }{2} ]}\)
\(\displaystyle{ grad f =[3x^{2}+2y^{2}, 4xy]}\)
\(\displaystyle{ grad f (P)=[2,0]}\)
\(\displaystyle{ f'\vec{a} (P)=[ \frac{ \sqrt{2} }{2} ,\frac{ \sqrt{2} }{2} ] \cdot [2,0]=\sqrt{2}}\)
b) \(\displaystyle{ f(x,y,z)=x^{2}+y^{2}-z^{2}}\)
\(\displaystyle{ P(1,1,1)}\) \(\displaystyle{ \vec{a} =[4,4,0]}\)
\(\displaystyle{ | \vec{a} | = 4\sqrt{2}}\)
\(\displaystyle{ \frac{ \vec{a} }{| \vec{a} |} = [ \frac{ \sqrt{2} }{2} ,\frac{ \sqrt{2} }{2},0 ]}\)
\(\displaystyle{ grad f =[2x,2y-2z]}\)
\(\displaystyle{ grad f (P)=[2,2,-2]}\)
\(\displaystyle{ f'\vec{a} (P)= [ \frac{ \sqrt{2} }{2} ,\frac{ \sqrt{2} }{2},0 ]\cdot [2,2,-2]=2\sqrt{2}}\)
c)\(\displaystyle{ \frac{y+z}{x}}\)
\(\displaystyle{ P(2,1,1)}\) \(\displaystyle{ \vec{a} =[1,0,2]}\)
\(\displaystyle{ | \vec{a} | = \sqrt{5}}\)
\(\displaystyle{ \frac{ \vec{a} }{| \vec{a} |} = [ \frac{ \sqrt{5} }{5} ,0,\frac{ \sqrt{5} }{5} ]}\)
\(\displaystyle{ grad f =[ \frac{-(y+z)}{x^{2}}, \frac{1}{x} ,\frac{1}{x} ]}\)
\(\displaystyle{ grad f (P)= [ -\frac{3}{4}, \frac{1}{2} ,\frac{1}{2} ]}\)
\(\displaystyle{ f'vec{a} (P)=[ -frac{3}{4}, frac{1}{2} ,frac{1}{2} ]cdot [ frac{ sqrt{5} }{5} ,0,frac{ sqrt{5} }{5} = frac{sqrt{5} }{20}}\)