Question

calcIII

Answer 1

i don't understand what i did wrong..

Answer 2

directional derivative , huh

Answer 3

mmhmm

Answer 4

for direction of most increase , just find gradient

Answer 5

according to my book, it's suppose to be a scalar

Answer 6

<2y, 2x + 8y>

plug in the point here

Answer 7

<-6,-20> for most increse

<6,20> for most decrease

same as you got

Answer 8

hmm from the text "directional derivative is a dot product del(f) dot u

Answer 9

derivative itself you have to find unit vector

<-6,-20>

magnitude

Sqrt[6^2+20^2]=\[2 \sqrt{109}\]

this is your unit vector

\[\frac{1}{2 \sqrt{109}}\{-6,-20\}=\left\{-\frac{3}{\sqrt{109}},-\frac{10}{\sqrt{109}}\right\}\]

Answer 10

o i used the direction for part a...

Answer 11

\[\left\{-\frac{3}{\sqrt{109}},-\frac{10}{\sqrt{109}}\right\}.\{-6,-20\}\]=Sqrt[209]

Answer 12

sorry sqrt[109]

Answer 13

got it, thx.must get sleep