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Ali is hiking on the hill, whose height is given by f(u,v)=n^2 e^((u+n)/(v+n)). Currently, he is positioned at point (3, 5). Find the direction at which he moves down the hills quickly. Take n =12

Sagot :

Lublanaidn

Answer:

[tex]<-144e^{0.88},7.47e^{0.88}>[/tex]

Step-by-step explanation:

We are given that

[tex]f(u,v)=n^2e^{\frac{u+n}{v+n}}[/tex]

Point=(3,5)

n=12

We have to find the direction at which he moves down the hills quickly.

[tex]f(u,v)=144e^{\frac{u+12}{v+12}}[/tex]

[tex]f_u(u,v)=144e^{\frac{u+12}{v+12}}[/tex]

[tex]f_u(3,5)=144e^{\frac{3+12}{5+12}}[/tex]

[tex]f_u(3,5)=144e^{\frac{15}{17}}=144e^{0.88}[/tex]

[tex]f_v(u,v)=144e^{\frac{u+12}{v+12}}\times (-\frac{u+12}{(v+12)^2})[/tex]

[tex]f_v(3,5)=144e^{\frac{15}{17}}(-\frac{15}{(17)^2}[/tex]

[tex]f_v(3,5)=-\frac{2160}{289}e^{\frac{15}{17}}=-7.47e^{0.88}[/tex]

[tex]\Delta f(3,5)=<f_u(3,5),f_v(3,5)>[/tex]

[tex]\Delta f(3,5)=<144e^{0.88},-7.47e^{0.88}>[/tex]

The direction at which he moves down the hills quickly=-[tex]\Delta f(3,5)[/tex]

The direction at which he moves down the hills quickly=[tex]<-144e^{0.88},7.47e^{0.88}>[/tex]

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