Rate of change of temperature at (1,-1,1) towards (2,4,-4)

In summary, the conversation discusses finding the rate of change of temperature at a specific point (1, -1, 1) in the direction towards another point (2, 4, -4). The speaker plans to use the partial derivative of each component and dot it with a unit vector in the direction from (1, -1, 1) to (2, 4, -4). However, they later realize that (2, 4, -4) is a point and not a vector.
  • #1
Larrytsai
228
0

Homework Statement


The temperature at a point (x,y,z) is given by ,T(x,y,z)=200e^((-x^2-y^2)/(4-z^2)/9) where is measured in degrees Celsius and x,y, and z in meters. There are lots of places to make silly errors in this problem; just try to keep track of what needs to be a unit vector.
Find the rate of change of the temperature at the point (1, -1, 1) in the direction toward the point (2, 4, -4).

The Attempt at a Solution



What I have done so far, is taken the partial derivative of each component. What I plan to do is, plug in the points 1,-1, 1 and dot it with the unit vector of (2,4,-4)
can anyone confirm if this is the right path?

So for some reason I am still not getting the right answer, I have

Tx = [200[-2x]e^((-x^2-y^2)/(4-z^2)/9)] / 9(4 - z^2)
Ty = 200(-2y)e^((-x^2-y^2)/(4-z^2)/9) / 9(4 - z^2)
Tz = -200[-x^2-y^2 ](-18z)[[36-9z^2]^-1]e^((-x^2-y^2)/(4-z^2)/9)

then i dot it with this unit vector
(2,4,-4)- (1,-1,1) = (1,5,-5)
|u| = sqrt(51)

(400e^(-2/27))/(sqrt(51))[-1-5]+[7200(-5)e^(-2/27)]/((27^2)sqrt(51))
 
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  • #2
No, not "unit vector of (2, 4, -4)". That would be a vector from (0, 0, 0) to (2, 4, -4).
You want to use the unit vector in the same direction as a vector form (1, -1, 1) to (2, 4, -4).
 
  • #3
ohhh yeah i see now, I assumed that (2,4,-4) was a vector not a point. Okay thanks =)!
 

FAQ: Rate of change of temperature at (1,-1,1) towards (2,4,-4)

1. What is a directional derivative?

A directional derivative is a measure of the rate of change of a function in the direction of a given vector. It represents the slope of the function in a particular direction.

2. How is a directional derivative calculated?

The directional derivative is calculated by taking the dot product of the gradient of the function and the unit vector in the specified direction. This can also be written as the product of the magnitude of the gradient and the cosine of the angle between the gradient and the direction vector.

3. What is the significance of directional derivatives?

Directional derivatives are important in multivariable calculus and optimization problems, as they allow us to determine the direction in which a function has the steepest increase or decrease. They also help us understand the behavior of a function in different directions.

4. What are some real-world applications of directional derivatives?

Directional derivatives have various applications in fields such as physics, engineering, and economics. For example, in physics, they are used to calculate the direction of the maximum change in temperature or pressure in a given system. In economics, they can be used to find the direction in which a company can maximize its profits.

5. Can a directional derivative be negative?

Yes, a directional derivative can be negative. This indicates that the function is decreasing in the specified direction. A positive directional derivative indicates that the function is increasing in that direction, while a value of 0 indicates that there is no change in the function in that direction.

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