Question

f(x,y)=(x^2+y^2) e^-x find the rate of change of f at (1,2) as you move in the direction <-2,3>

Answer 1

where are you stuck?

Answer 2

i found the derivatives and I know i have to multiply by <-2/sqrt15, 3/sqrt15> i am just not sure what i do with the (1,2) if i put it in to the function or?

Answer 3

make a unit vector

Answer 4

\[\left<f_x(a,b),f_y(a,b)\right>\cdot\frac{<a,b>}{\|<a,b>\|}\]

Answer 5

so is it (-2e^-x)/sqrt(13)

Answer 6

as a final answer?

Answer 7

there should be no x's in your answer

Answer 8

your answer should just be a number

Answer 9

so (-2sqrt(13))/(13e)

Answer 10

no

Answer 11

where are you getting the \(\sqrt{13}\)?

Answer 12

oops...just a sec...I was using the wrong number

Answer 13

i got \[\frac{6\sqrt{13}}{13e}\]

Answer 14

my formula is a little off too..

\[\left<f_x(a,b),f_y(a,b)\right>\cdot\frac{<\alpha,\beta>}{\|<\alpha,\beta>\|}\]


would be better

Answer 15

what did you get for the partial derivative of x

Answer 16

\[(-x^2+2x-y^2)e^{-x}\]

Answer 17

\[\frac{\partial}{\partial x}(x^2+y^2) e^{-x}=(x^2+y^2)e^{-x}(-1)+(2x)e^{-x}\]

\[=(-x^2-y^2)e^{-x}+(2x)e^{-x}=(-x^2+2x-y^2)e^{-x}\]

Answer 18

okay thanks i see what i did wrong

Answer 19

good :)

Answer 20

thank you so much

Answer 21

no problem