Problem about rate of change (multivariable calculus)

In summary, the conversation discusses calculating dT/dt given the values of dx/dt, dy/dt, dz/dt, and the partial derivatives of T with respect to x, y, and z. The correct formula for dT/dt is given and the conversation also addresses a notation error made by the teacher.
  • #1
supermiedos
63
0

Homework Statement


Let T = f(x, y, z), where dx/dt = 4, dy/dt = 4 and dz/dt = -3
Calculate dT/dt if dT/dx = 4, dT/dy = 7 and dT/dz = 9


Homework Equations


dT/dt = ∂T/∂x (dx/dt) + ∂T/∂y (dy/dt) + ∂T/∂z (dz/dt)


The Attempt at a Solution


I tried to get an explicit formula for T. I thought that, if dT/dx = 4, then T = 4x + c. Similarly, T = 7y + c and T = 9z + c. If i add the above expressions, I get:

3T = 4x + 7y + 9z + C, so, T would be T = 4x/3 + 7y/3 + 3z + C, and I could use the chain rule and just substitute dx/dt and so on...

Is my reasoning fine?
 
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  • #2
Why do you need to do anything about T?

You have the equation for dT/dt, and you know the values of everything on its right-hand side.
 
  • #3
First, you can't get an explicit expression for T, there are infinitely many possibililties. But you don't need that, just the "chain rule"":
[tex]\frac{dT}{dt}= \frac{\partial T}{\partial x}\frac{dx}{dt}+ \frac{\partial T}{\partial y}\frac{dy}{dt}+ \frac{\partial T}{\partial z}\frac{dz}{dt}[/tex]
 
  • #4
But I need ∂T/∂x and I have dT/dx = 4 (same for the others). Don't tell me ∂T/∂x = dT/dx, because I'm going to cry
 
  • #5
Using d when you have a function with more than one variable is not correct notation. ∂T/∂x is the correct notation.
 
  • #6
Omg you are right. My teacher wrote it like dT/dx, and I never realized he wrote it wrong. Thank you for your help guys
 

Related to Problem about rate of change (multivariable calculus)

1. What is the rate of change in multivariable calculus?

The rate of change in multivariable calculus is a measure of how one variable changes with respect to another variable. It is calculated by finding the gradient or partial derivative of a function with multiple variables.

2. Why is the concept of rate of change important in multivariable calculus?

The concept of rate of change is important in multivariable calculus because it helps us understand how a function changes in multiple dimensions. This is crucial for solving real-world problems involving multiple variables and understanding the behavior of complex systems.

3. How do you calculate the rate of change in multivariable calculus?

To calculate the rate of change in multivariable calculus, we use the partial derivative. This involves taking the derivative of a function with respect to one variable while holding all other variables constant. The result is a slope or gradient that represents the rate of change at a specific point.

4. What is the difference between average rate of change and instantaneous rate of change?

The average rate of change is the average rate at which a function changes over a given interval. It is calculated by dividing the change in the output by the change in the input. On the other hand, the instantaneous rate of change is the rate of change at a specific point on a curve, which can be found using the derivative or slope of the tangent line at that point.

5. How is the rate of change related to optimization problems in multivariable calculus?

The rate of change plays a crucial role in optimization problems in multivariable calculus. In these problems, we are looking for the maximum or minimum value of a function, which can be found by setting the rate of change to zero. This allows us to find critical points and determine whether they correspond to a maximum, minimum, or saddle point.

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