- #1

CraigH

- 222

- 1

Hi,

I am currently studying vector calculus, and to try and understand it I have created a question for myself, and it would be greatly appreciated if someone could verify that I have answered my question correctly, and I have done everything right.

Okay here it is; Point Q in the room is a heat source, how does the temperature vary as you move away from the origin (0,0,0)

http://img571.imageshack.us/img571/4250/gascube.png

http://imageshack.us/a/img841/4250/gascube.png

.The further away from point Q you are the less heat there is. This can be modeled by:

[itex] dt = \nabla t .dr[/itex]

.(dt) is a scalar, it is a change in temperature

.(dr) is a vector, it is a change in distance of x, y and z [itex](dr=dx\hat{x} +dy\hat{y} +dz\hat{z} )[/itex]

.(del t) is a vector, it is the amount t changes for x, y, and z, for a change in distance of x,y,and z

For the purposes of this example I am going to say that the heat at any point equals 1 divided by the square of the distance from q.

In the x direction

[itex] t = \frac{1}{(xq-x)^{2}}[/itex]

In the y direction

[itex] t = \frac{1}{(yq-y)^{2}}[/itex]

In the z direction

[itex] t = \frac{1}{(zq-z)^{2}}[/itex]

Where (xq,yq,zq)=(0,2.5,2.5) And (x,y,z) is the point you are at

So, [itex] \frac{dt}{dx} = 2(xq-x)^{-1}, \frac{dt}{dy} = 2(yq-y)^{-1}, \frac{dt}{dz} = 2(zq-z)^{-1}[/itex]

And the final answer is:

[itex] dt =(2(xq-x)^{-1}\hat{x} + 2(yq-y)^{-1}\hat{y} + 2(zq-z)^{-1}\hat{z}) . dr[/itex]

Is this correct?

Thank you

I am currently studying vector calculus, and to try and understand it I have created a question for myself, and it would be greatly appreciated if someone could verify that I have answered my question correctly, and I have done everything right.

Okay here it is; Point Q in the room is a heat source, how does the temperature vary as you move away from the origin (0,0,0)

http://img571.imageshack.us/img571/4250/gascube.png

http://imageshack.us/a/img841/4250/gascube.png

.The further away from point Q you are the less heat there is. This can be modeled by:

[itex] dt = \nabla t .dr[/itex]

.(dt) is a scalar, it is a change in temperature

.(dr) is a vector, it is a change in distance of x, y and z [itex](dr=dx\hat{x} +dy\hat{y} +dz\hat{z} )[/itex]

.(del t) is a vector, it is the amount t changes for x, y, and z, for a change in distance of x,y,and z

For the purposes of this example I am going to say that the heat at any point equals 1 divided by the square of the distance from q.

In the x direction

[itex] t = \frac{1}{(xq-x)^{2}}[/itex]

In the y direction

[itex] t = \frac{1}{(yq-y)^{2}}[/itex]

In the z direction

[itex] t = \frac{1}{(zq-z)^{2}}[/itex]

Where (xq,yq,zq)=(0,2.5,2.5) And (x,y,z) is the point you are at

So, [itex] \frac{dt}{dx} = 2(xq-x)^{-1}, \frac{dt}{dy} = 2(yq-y)^{-1}, \frac{dt}{dz} = 2(zq-z)^{-1}[/itex]

And the final answer is:

[itex] dt =(2(xq-x)^{-1}\hat{x} + 2(yq-y)^{-1}\hat{y} + 2(zq-z)^{-1}\hat{z}) . dr[/itex]

Is this correct?

Thank you

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