Directional derivative

**Ares_D1** · November 9th 2009, 06:10 PM

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?

**Jameson** · November 9th 2009, 07:16 PM

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.

**Ares_D1** · November 9th 2009, 08:27 PM

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.

**Ares_D1** · November 9th 2009, 08:57 PM

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,

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

That's the answer, thanks!

**HallsofIvy** · November 10th 2009, 03:26 AM

Originally Posted by Ares_D1

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

I understand the rules for partial differentiation, just wasn't sure how to interpret the $exp$ function.

Yes, "exp(x)" is defined as $exp(x)= e^x$ .

Thread: Directional derivative

LinkBack

Thread Tools

Display

Directional derivative

Similar Math Help Forum Discussions

directional derivative q

Directional Derivative

Directional derivative

Total Derivative vs. Directional Derivative

directional derivative

Search Tags