Implicit differentiation help three variables

In summary: The evaluation of dF/dt is correct (24) if you use the correct value of dx/dt (which should be 2cost).In summary, the conversation discusses differentiating a composite function and finding the value of dF/dt at a specific point. The conversation explores various methods, including chain rules and Jacobians, but ultimately discovers that there was a typo in the original function given. The correct derivatives are calculated and the correct value of dF/dt is found to be 24.
  • #1
aluvieriti
3
0

Homework Statement



I have a question. How in general would one differentiate a composite function like F(x,y,z)=2x^2-yz+xz^2 where x=2sint , y=t^2-t+1 , and z = 3e^-1 ? I want to find the value of dF/dt evaluated at t=0 and I don't know how. Can someone please walk me through this?

Homework Equations


Mod note: Fixed the equation below to match the OP's change above.
F(x,y,z)=2x^2-yz+xz^2

dF/dx=4x-z^2 , dF/dy= -z , dF/dz = 2xz-y , dz/dt=0 , dx/dt=2cost, dy/dt=2t-1

dF/dx dx/dt + dF/dy dy/dt + dF/dz dz/dt= dF/dt

The Attempt at a Solution



I tried a couple of things, including chain rules and jacobians. I know that dF/dt should equal dF/dx dx/dt + dF/dy dy/dt + dF/dz dz/dt but for some reason this doesn't work, or I am doing something wrong. I start out by differentiating to get dF/dx=4x-z^2 , dF/dy= -z , dF/dz = 2xz-y , dz/dt=0 , dx/dt=2cost, dy/dt=2t-1 but this doesn't match the answer, which my book says is 24.

How do they get this, and where is my error? Thanks.
 
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  • #2
aluvieriti said:

Homework Statement



I have a question. How in general would one differentiate a composite function like F(x,y,z)=2x^2-yz+x^2 where x=2sint , y=t^2-t+1 , and z = 3e^-1 ? I want to find the value of dF/dt evaluated at t=0 and I don't know how. Can someone please walk me through this?
Do you have a typo in your definition for F(x, y, z)? As you wrote it, it could be simplified to 3x2 - yz.
aluvieriti said:

Homework Equations


F(x,y,z)=2x^2-yz+x^2

dF/dx=4x-z^2 , dF/dy= -z , dF/dz = 2xz-y
These three should be partial derivatives, and a couple of them are incorrect, at least based on what you wrote.

$$ \frac{\partial F}{\partial x} = 6x $$

aluvieriti said:
, dz/dt=0 , dx/dt=2cost, dy/dt=2t-1

dF/dx dx/dt + dF/dy dy/dt + dF/dz dz/dt= dF/dt

The Attempt at a Solution



I tried a couple of things, including chain rules and jacobians. I know that dF/dt should equal dF/dx dx/dt + dF/dy dy/dt + dF/dz dz/dt but for some reason this doesn't work, or I am doing something wrong. I start out by differentiating to get dF/dx=4x-z^2 , dF/dy= -z , dF/dz = 2xz-y , dz/dt=0 , dx/dt=2cost, dy/dt=2t-1 but this doesn't match the answer, which my book says is 24.

How do they get this, and where is my error? Thanks.
 
  • #3
Mark44 said:
Do you have a typo in your definition for F(x, y, z)? As you wrote it, it could be simplified to 3x2 - yz.These three should be partial derivatives, and a couple of them are incorrect, at least based on what you wrote.

$$ \frac{\partial F}{\partial x} = 6x $$

Sorry, yes there was a typo, I fixed it. Are the partials still wrong? Thanks for responding so quickly!
 
  • #4
aluvieriti said:
Sorry, yes there was a typo, I fixed it. Are the partials still wrong? Thanks for responding so quickly!

I want to rephrase the question. Since everyone else I have talked to thinks there was an error in the book, does everyone here agree?
 
  • #5
aluvieriti said:
I want to rephrase the question. Since everyone else I have talked to thinks there was an error in the book, does everyone here agree?
Here's your work from post #1.
aluvieriti said:
dF/dx=4x-z^2 , dF/dy= -z , dF/dz = 2xz-y
Your partial with respect to x has a sign error. The other partials are correct, and your three derivatives are correct.
 

1. What is implicit differentiation with three variables?

Implicit differentiation with three variables is a mathematical technique used to find the derivative of a function with three variables, where the dependent variable is not explicitly given in terms of the independent variables. It is used to find the rate of change of a function with respect to one of the variables, while keeping the other two variables constant.

2. When is implicit differentiation with three variables used?

Implicit differentiation with three variables is used in multivariable calculus, specifically in finding the partial derivatives of a function with three variables. It is also used in solving optimization problems, where a function is optimized with respect to three variables.

3. How is implicit differentiation with three variables different from implicit differentiation with two variables?

In implicit differentiation with two variables, the derivative is found with respect to one variable, while treating the other variable as a constant. In implicit differentiation with three variables, the derivative is found with respect to one variable, while treating the other two variables as constants.

4. What are the steps to perform implicit differentiation with three variables?

The steps to perform implicit differentiation with three variables are:1. Differentiate both sides of the equation with respect to the variable in question.2. Treat the other two variables as constants and use the basic rules of differentiation.3. Solve the resulting equation for the derivative.

5. Are there any common mistakes to watch out for when using implicit differentiation with three variables?

Yes, some common mistakes to watch out for include:- Forgetting to treat the other two variables as constants.- Not applying the chain rule correctly.- Making errors in algebraic manipulation while solving for the derivative.It is important to check your work carefully to avoid these mistakes.

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