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]
