Question

Greatest rate of increase

Answer 1

So I have the function \[f(x,y,z) = (-3x-3xy+12y+4z)e ^{-xy}\] and the point with highest rate of increase is at (0,1,-9) and I'm supposed to find the value of that increase, which basically should be d/df, right?
I took the differentials as below
\[\frac{ df }{ dx } = (-3-3y)e ^{-xy} -y(-3x-3xy+12y+4z)e ^{-xy}\]
\[\frac{ df }{ dy } = (-3x+12)e ^{-xy} -x(-3x-3xy+12y+4z)e ^{-xy}\]
\[\frac{ df }{ dz } = 4e ^{-xy} - (-3x-3xy+12y+4z)e ^{-xy}\]

I then substituted the (0,1,-9) into the differentials and added then to get
\[\frac{ d }{ df } = \frac{ df }{ dx } + \frac{ df }{ dy } +\frac{ df }{ dz }\]
but the answer I'm getting is 58, which is not among the answers in the MC. My options are (A) 22 (B)19 (C)21 (D)25 (E)16

Could anybody please tell me where I went wrong?

Answer 2

firstly \(f_z = 4 e^{-xy}\) ?!

for the components, I get \[\left(\begin{matrix}f_x \\ f_y \\ f_z \end{matrix}\right) = \left(\begin{matrix} 18 \\ 12 \\ 4 \end{matrix}\right)\]

and the absolute value of that vector is option (A), you might wish to double check the arithmetic.


strictly speaking the " total differential" is \(df = f_x dx + f_y dy + f_z dz\), not sure what you'd do with it here

Answer 3

@IrishBoy123 okay, let me try that...

Answer 4

@IrishBoy123 oooh it works!! THANK YOU SO MUCH!!

Answer 5

cool, dramaqueen !

and "that vector" is actually known as the gradient ... in this case, of your scalar field \(f(x,y,z)\), ie :

\(grad (f) = \nabla f = \left(\begin{matrix}f_x \\ f_y \\ f_z \end{matrix}\right) = \left(\begin{matrix} \frac{ \partial f}{\partial x} \\ \frac{ \partial f}{\partial y} \\ \frac{ \partial f}{\partial z} \end{matrix}\right) \)