# Directional derivative

• November 9th 2009, 06:10 PM
Ares_D1
Directional derivative
Having a little trouble with this problem. I keep getting zero as an answer which is incorrect. I know where I'm making the mistake, just don't know how to fix it.

Problem:
Compute the directional derivative for $f(x,y,z) = zexp(-2xy)$ at point $(1,1,-1), t = (1i,-3j,2k)$.

Formula:
$\frac{df}{ds} = t \cdot (\frac{\partial f}{dx}i, \frac{\partial f}{dy}j, \frac{\partial f}{dz}k)$.

Basically I'm getting stuck with the partial derivatives. I seems to me that the partial of either x, y, or z will give 0. No?
• November 9th 2009, 07:16 PM
Jameson
Quote:

Originally Posted by Ares_D1
Having a little trouble with this problem. I keep getting zero as an answer which is incorrect. I know where I'm making the mistake, just don't know how to fix it.

Problem:
Compute the directional derivative for $f(x,y,z) = zexp(-2xy)$ at point $(1,1,-1), t = (1i,-3j,2k)$.

Formula:
$\frac{df}{ds} = t \cdot (\frac{\partial f}{dx}i, \frac{\partial f}{dy}j, \frac{\partial f}{dz}k)$.

Basically I'm getting stuck with the partial derivatives. I seems to me that the partial of either x, y, or z will give 0. No?

Fx means you consider x a variable and treat all other variables as constants. So if you have $f=ae^{-2bx} \Rightarrow f'=(ae^{-2bx})*(-2b)$ by the rule for differentiating e^x. So now just use z=a and y=b and that's Fx.

Repeat for Fy and Fz. Also, if this is supposed to be in the direction of vector t, you need to get the corresponding unit vector so you can just apply the direction.
• November 9th 2009, 08:27 PM
Ares_D1
Quote:

Originally Posted by Jameson
Fx means you consider x a variable and treat all other variables as constants. So if you have $f=ae^{-2bx} \Rightarrow f'=(ae^{-2bx})*(-2b)$ by the rule for differentiating e^x. So now just use z=a and y=b and that's Fx.

Repeat for Fy and Fz. Also, if this is supposed to be in the direction of vector t, you need to get the corresponding unit vector so you can just apply the direction.

Sorry, so $e^{-2xy} = exp(-2xy)$?

I understand the rules for partial differentiation, just wasn't sure how to interpret the $exp$ function. http://www.mathhelpforum.com/math-he...ons/icon11.gif
• November 9th 2009, 08:57 PM
Ares_D1
So with $e^{-2xy} = exp(-2xy)$ I have:

$\frac{df}{ds} = t \cdot (\frac{\partial f}{dx}i, \frac{\partial f}{dy}j, \frac{\partial f}{dz}k)$

$= (1,-3,-2) \cdot (-2yze^{-2xy}, -2xze^{-2xy}, e^{-2xy})$

$= -2yze^{-2xy} + 6xze^{-2xy} + 2e^{-2xy}$

$= 2e^{-2} - 6e^{/2} + 2e^{-2}$

$= -2exp(-2)$.

And with the corresponding unit vector,

$\frac{df}{ds} = \frac{-2exp(-2)}{\sqrt{14}}$

Sorry, so $e^{-2xy} = exp(-2xy)$?
I understand the rules for partial differentiation, just wasn't sure how to interpret the $exp$ function. http://www.mathhelpforum.com/math-he...ons/icon11.gif
Yes, "exp(x)" is defined as $exp(x)= e^x$.