grad

**alexandrabel90** · April 20th 2010, 10:41 AM

given that a space craft is approaching too close to the sun. the temperature of craft at (x,y,z) is T(x,y,z)=exp(-x^2-2y^2-3Z^2)where x,y,z are in metres. the craft is now at (1,1,1). if the craft travels at exp(8) metres pre sec, how fast will the temperature decrease if it proceeds in that direction?

my working,

the craft will decrease the fastest in the direction of neg grad T. hence the rate of decrease would be the magnitude of grad T...however i didnt manage to get the answer...

can someone help me?

thanks.

**Aryth** · April 20th 2010, 11:06 AM

Originally Posted by alexandrabel90

the craft will decrease the fastest in the direction of neg grad T. hence the rate of decrease would be the magnitude of grad T...however i didnt manage to get the answer...

As far as this question is concerned, I'm not sure if the gradient is all you need. But since you need help finding the gradient, I will help you find it, but I'm afraid I can't contribute much else to this problem. I assume that you need to plug in the point once you find the magnitude, which I did at the end of this problem. Hopefully this helps.

Remember that the gradient is just the partials with respect to x, y, and z, so that:

$\bold{\nabla T} = \left(\frac{\partial{T}}{\partial{x}}, \frac{\partial{T}}{\partial{y}}, \frac{\partial{T}}{\partial{z}}\right)$

Now we calculate the individual partial derivatives:

$\frac{\partial{T}}{\partial{x}} = -2xe^{-x^2 - 2y^2 - 3z^2}$

$\frac{\partial{T}}{\partial{y}} = -4ye^{-x^2 - 2y^2 - 3z^2}$

$\frac{\partial{T}}{\partial{z}} = -6ze^{-x^2 - 2y^2 - 3z^2}$

Which gives us a gradient of:

$\bold{\nabla T} = e^{-x^2 - 2y^2 - 3z^2}\left(-2x, -4y, -6z\right)$

Now we need to know the magnitude of the gradient, which would be given by:

$|\bold{\nabla T}| = e^{-x^2 - 2y^2 - 3z^2}\sqrt{4x^2 + 16y^2 + 36z^2}$

$= 2e^{-x^2 - 2y^2 - 3z^2}\sqrt{x^2 + 4y^2 + 9z^2}$

Now you plug in the point you got:

$= 2e^{-1 - 2 - 3}\sqrt{1 + 4 + 9}$

$= 2e^{-6}\sqrt{14}$

**alexandrabel90** · April 20th 2010, 11:29 AM

that was what i got for grad too..but apparently, finding the magnitude of grad is not the right answer for this question..hence im lost as to what i actually need to find to solve this question.

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