Find the rate of change of temperature

[tangibleLime]

Homework Statement

The temperature at a point (x, y, z) is given by the following equation where T is measured in °C and x, y, z in meters.

T(x, y, z) = 200e^-x2-3y2-9z2

Find the rate of change of temperature at the point P(2, -1, 2) in the direction towards the point (3, -3, 3).

Homework Equations

The Attempt at a Solution

I calculated the partial derivatives of T(x,y,z) as...
$$T_{x} = -400xe^{-x^{2}-3y^{2}-9z^{2}}$$
$$T_{y} = -1200ye^{-x^{2}-3y^{2}-9z^{2}}$$
$$T_{z} = -3600ze^{-x^{2}-3y^{2}-9z^{2}}$$

Giving me the gradient...

$$\nabla T(x,y,z) = <-400xe^{-x^{2}-3y^{2}-9z^{2}}, -1200ye^{-x^{2}-3y^{2}-9z^{2}}, -3600ze^{-x^{2}-3y^{2}-9z^{2}}>$$

Plugging in the given values for the gradient, I get...

$$\nabla T(2,-1,2) = <\frac{-800}{e^{35}},\frac{1200}{e^{35}},\frac{-7200}{e^{35}}>$$

Meanwhile, I take the point (3,-3,3) and convert it into a direction unit vector (I think this is where I make a mistake).

$$v = <3,-3,3>$$

$$u = \frac{1}{\sqrt{27}} * <3,-3,3>$$

$$u = <\frac{1}{\sqrt{3}},-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}>$$

Combining all of this in D_uT...

$$D_{u}T = \frac{-800}{\sqrt{3}*e^{35}} + \frac{-1200}{\sqrt{3}*e^{35}} + \frac{-7200}{\sqrt{3}*e^{35}}$$

This gives me a wrong answer of...

$$\frac{-9200}{\sqrt{3}*e^{35}}$$

Any help that isn't a purposefully cryptic message that leaves me no closer to solving my problem is appreciated.

Thanks.

 

[LCKurtz]

Homework Statement

The temperature at a point (x, y, z) is given by the following equation where T is measured in °C and x, y, z in meters.

T(x, y, z) = 200e^-x2-3y2-9z2

Find the rate of change of temperature at the point P(2, -1, 2) in the direction towards the point (3, -3, 3).

...

Meanwhile, I take the point (3,-3,3) and convert it into a direction unit vector (I think this is where I make a mistake).

Yes, that is a mistake. You want the direction from (2,-1,2) to (3,-3,3).

 

[tangibleLime]

Thanks, I knew that was a little fishy. But unfortunately I still can't get the right answer.

To get a unit directional vector from those points, this is what I did:

$$P = (2, -1, 2)$$

$$Q = (3, -3, 3)$$

$$PQ = (3-2, -3-(-1), 3-2) = <1,-2,1> = v$$

$$\frac{<1,-2,1>}{\sqrt{1^2+(-2)^2+1^2}} = \frac{<1,-2,1>}{\sqrt{6}}$$

$$u = <\frac{1}{sqrt{6}},\frac{-2}{\sqrt{6}},\frac{1}{\sqrt{6}}>$$

I then use this directional unit vector u as I did originally, which I believe is the correct method, but it still results in an incorrect answer (all calculations verified by calculator). Am I doing something else wrong?

 

[LCKurtz]

I think the e³⁵ terms are wrong.

 

[tangibleLime]

Ah, yes, got it.

I was, for the first x in the exponent of e, squaring negative x instead of squaring x and taking the negative of that.

Thanks!

 

[tangibleLime]

Hmm, sorry but I have one more question.

This problem had two more parts.

Part B:
In which direction does the temperature increase fastest at P?

Part C:
Find the maximum rate of increase at P.

I was able to calculate part C by simply computing $$|\nabla T(2,-1,2)|$$. I thought I knew how to do part B, but apparently I am mistaken. It is my understanding that to maximize the directional derivative (which I believe is what I must do here), I simply take the unit vector in the same direction as my gradient vector, which turns out to be <1,-2,1> from my P-Q calculations a few posts above. However, the first element of the answer is -2... (it's an online system, with the answer set up as <-2, ____, ____> where I will in the second blanks). Shouldn't the answer be <1, -2, 1>? I thought it may just be a scalar of -2 at first, but multiplying that out gave me <-2,4,-2> which was incorrect.

 

[LCKurtz]

Hmm, sorry but I have one more question.

This problem had two more parts.

Part B:
In which direction does the temperature increase fastest at P?

Part C:
Find the maximum rate of increase at P.

I was able to calculate part C by simply computing $$|\nabla T(2,-1,2)|$$. I thought I knew how to do part B, but apparently I am mistaken. It is my understanding that to maximize the directional derivative (which I believe is what I must do here), I simply take the unit vector in the same direction as my gradient vector, which turns out to be <1,-2,1> from my P-Q calculations a few posts above.

But the point Q has nothing to do with the gradient vector at P. And <1,-2,1> is not a unit vector, not that that matters.

However, the first element of the answer is -2... (it's an online system, with the answer set up as <-2, ____, ____> where I will in the second blanks). .

Probably the reason they specified a -2 as the first component is that any positive multiple of the correct direction points in the same correct direction. So if you scale it properly to agree with the first component, the other two components will agree, making it easy for the computer to check your answer.

 

[HallsofIvy]

The gradient vector itself always points in the direction of fastest increase and it length is that rate of increase.