Multi Calculus- Partial Derivatives

In summary, the conversation discusses finding the partial derivative Kt(0,0) for the given functions x(s,t), y(s,t), H(x,y), and K(s,t). The chain rule is suggested for solving Kt and the correct expressions for x_t and y_t are provided. The conversation ends with the correct answer being found.
  • #1
201
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Homework Statement



I am translating the question from another language so it might not be word to word with the original question.

assume x(s,t) and y(s,t) determined by these two functions:

sin(xt) +x+s=1
eyt+y(s+1)=1

The following function is defined H(x,y)=x2+y2-xy
such that
K(s,t)=H(x(s,t),y(s,t))

find the partial derivative Kt(0,0) (the derivative of K in terms of t)

Homework Equations





The Attempt at a Solution



first I wanted to solve the first two equation system in terms of x and y, but it didn't work.
tried also plugging them into H, no success.

Would appreciate any help.
Roni
 
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  • #2
Start by applying the chain rule to [itex]H[/itex]...
 
  • #3
gabbagabbahey said:
Start by applying the chain rule to [itex]H[/itex]...
what do you mean by the chain rule ?

should I first replace the x,y with s,t ?
I don't really see the lead ...
would appreciate if you could give any more hints but I'm going to try it also
 
  • #4
I tried the chain rule, but looks like I'm doing something wrong.

Kt=Kx*Xt+Ky*Yt

I am assuming K(x,y)=H(x,y), a correct assumption ?
I am getting 0 and that's not the answer.
Kx=2x-y
Ky=2y-x

no matter what I multiply these by, I must get a zero but the answer is not zero.
So I guess I am missing something
 
  • #5
Roni1985 said:
I tried the chain rule, but looks like I'm doing something wrong.

Kt=Kx*Xt+Ky*Yt

I am assuming K(x,y)=H(x,y), a correct assumption ?

Well, sort of. [itex]K(s,t)[/itex] is explicitly a function of [itex]s[/itex] and [itex]t[/itex] only. You know that [itex]x[/itex] and [itex] y[/itex] are also some function of [itex]s[/itex] and [itex]t[/itex] and that

[tex]K(s,t)=H(x(s,t),y(s,t))=[x(s,t)]^2+[y(s,t)]^2+x(s,t)y(s,t)[/itex]

So, it is more correct to say

[tex]K_t(s,t)=H_x x_t+H_y y_t[/itex]

I am getting 0 and that's not the answer.
Kx=2x-y
Ky=2y-x

Well, [itex]H_x=2x-y[/itex] and [itex]H_y=2y-x[/itex]...but what about [itex]x_t[/itex] and [itex]y_t[/itex]...what are you getting for those?
 
  • #6
gabbagabbahey said:
Well, sort of. [itex]K(s,t)[/itex] is explicitly a function of [itex]s[/itex] and [itex]t[/itex] only. You know that [itex]x[/itex] and [itex] y[/itex] are also some function of [itex]s[/itex] and [itex]t[/itex] and that

[tex]K(s,t)=H(x(s,t),y(s,t))=[x(s,t)]^2+[y(s,t)]^2+x(s,t)y(s,t)[/itex]

So, it is more correct to say

[tex]K_t(s,t)=H_x x_t+H_y y_t[/itex]



Well, [itex]H_x=2x-y[/itex] and [itex]H_y=2y-x[/itex]...but what about [itex]x_t[/itex] and [itex]y_t[/itex]...what are you getting for those?

Xt=-x*cos(xt)
Yt=(-yeyt)/(t*eyt+s+1)

But it doesn't matter what I'm getting for these because I multiply each one of them by a zero.
 
  • #7
ohh hold on, the s and t are zeros , so it doesn't mean x and y are zeros
wow

Edit:
got the answer,

thanks for the help
 
  • #8
Roni1985 said:
Xt=-x*cos(xt)
Yt=(-yeyt)/(t*eyt+s+1)

Your expression for [itex]x_t[/itex] doesn't look quite right.

But it doesn't matter what I'm getting for these because I multiply each one of them by a zero.

I'm not sure why you think this. [itex]H_x=2x-y\neq 0[/itex] and [itex]H_y=2y-x\neq 0[/itex]
 
  • #9
gabbagabbahey said:
Your expression for [itex]x_t[/itex] doesn't look quite right.

oh, it didn't influence my answer so I didn't see the mistake

I think it's got to be
-x*cos(xt)/(t*cos(xt)+1)

Thank you.
 

1. What is multi calculus and why is it important?

Multi calculus is a branch of mathematics that involves studying functions with multiple variables. It is important because it helps us understand and analyze complex systems that cannot be described using single variable calculus.

2. What are partial derivatives and how are they different from regular derivatives?

Partial derivatives are the derivatives of a multivariable function with respect to one of its variables, while holding the other variables constant. They are different from regular derivatives because they take into account the effect of all the other variables on the function.

3. How do you find partial derivatives?

To find the partial derivative of a multivariable function, you need to differentiate the function with respect to one variable while holding the other variables constant. This can be done using the standard rules of differentiation, such as the power rule and the chain rule.

4. What is the importance of partial derivatives in real-life applications?

Partial derivatives are used in a wide range of real-life applications, such as economics, physics, and engineering. They help us to understand the rate of change of a function with respect to one variable, while keeping all other variables constant. This allows us to make predictions and optimize systems in various fields.

5. What is the relationship between partial derivatives and the gradient vector?

The gradient vector is a vector that points in the direction of the steepest increase of a function. It is made up of the partial derivatives of the function with respect to each variable. In other words, the gradient vector is a generalization of partial derivatives to multivariable functions.

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