Please help check my solution to gradient question.

  • Thread starter Thread starter ozone
  • Start date Start date
  • Tags Tags
    Gradient
Click For Summary

Homework Help Overview

The problem involves a wildebeest moving across a plain, with its position described in terms of distance east (x) and north (y) from a starting point, while also considering the rate of change of air temperature as a function of both location and time. The wildebeest's velocity and the temperature gradients in different directions are provided, leading to a question about the rate of change in air temperature experienced by the wildebeest.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to set up a function for the temperature change and calculates a gradient vector, seeking confirmation on the setup and efficiency of their approach. Some participants question the inclusion of a third dimension (k) in the context of the problem, while others express confusion about the relationship between the temperature changes per mile and the overall temperature change per hour.

Discussion Status

The discussion is ongoing, with various interpretations being explored. Some participants have provided alternative formulations of the problem, while others have raised questions about the assumptions made in the original setup. There is no explicit consensus yet, but the dialogue is contributing to a deeper understanding of the concepts involved.

Contextual Notes

Participants are grappling with the dimensionality of the problem and the implications of the temperature change rates provided. There is uncertainty regarding the relevance of the third dimension in the context of the wildebeest's movement and the overall temperature change rate.

ozone
Messages
121
Reaction score
0

Homework Statement



A wildebeest is charging across a plain. His path takes him to location (x,y) where x is his distance (in miles) east of his starting point and y is his distance in miles north of his starting point at time t. So x and y are functions of t. The air temperature is a function of both location and time.
Right now he is moving at a velocity of 12 miles per hour in the northward direction and 7 miles per hour westward. In the northward direction the temperature changes at a rate of -0.10 degrees per mile, and in the eastward direction the temperature changes at a rate of -0.05 degrees per mile. Also, overall the temperature (irrespective of location) is changing at a rate of -1.6 degrees per hour.
What is the rate of change in air temperature that the wildebeest is experiencing right now?



The Attempt at a Solution



I just want to make sure I set this up right. I first created a function called r(t). We then have

∇r = -7 i + 12 j + 1 k

Next I made a unit vector for the 3 respective changes in temperature
u = -.05/1.603)i - .1/1.603) j + 1.6/1.63 k

Next I just summed the dot product ∇r * u

I just wanted to make sure there are no errors in this solution, and that I set it up in an efficient manner. I have trouble when I have to start mixing (x,y) with t.

Thank you.
 
Physics news on Phys.org
I really don't understand at all. if the wildebeast can only move in a two dimensional plane (North-South and East-West), what does "k" represent? And how is the wildebeast moving in that direction at 1 mile per hour?

But then I don't understand the question itself! If the temperature is changing at "-0.10 degrees per mile" in the northern direction and at -0.05 degrees per mile" in the eastern direction, then how can it be changing at -1.6 degrees per hour irrespective of location?
 
Lol that is why I posted it here! A teacher at my college put this on a calc III test, and I want to make sure that I am prepared for an upcoming exam. I chose to set z = 1 in the gradient because I wanted to consider this problem as a 3 dimensional function where the z value would be a shift up and down a 3 dimensional axis perhaps.

However I'm not sure if this would yield a correct answer.

edit : the change in global temperature is what I considered to be our z value.
 
You could call it r(t)=(x,y,t). Then r'(t)=(-7,12,1)

∇T=(-.05,-.1,-1.6)

Then dT/dt=∇T\cdotr'(t)

Comments:

I'm not sure you need a unit vector.

Rate of change with respect to what, time?

Notice that dT/dt≠∂T/∂t=-1.6
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
Replies
11
Views
2K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 20 ·
Replies
20
Views
2K
Replies
1
Views
2K
  • · Replies 22 ·
Replies
22
Views
5K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K