# Dy/dx of vector valued function F(x,y)

• shippo113
In summary: This equation is a first order differential equation and so the partial derivatives with respect to x are Fx(x,y) and Fy(x,y).
shippo113
F(x,y) = F1(x,y)i + F2(x,y)j, F is a vector valued function of x,y. How to find dy/dx?

The book gives the answer as

dy/dx = (F1(x,y))/(F2(x,y)). But I do not know how to get to this answer at all. Find the partial derivative is what I can do, but how to get to dy/dx when the main function is F(x,y)?

I would be tempted to look at the norm of the function and then differentiate.

excuse me but what does norm mean?

Think of it as dot product.
$$\mathbf{F}(x,y)=F_{1}(x,y)\mathbf{i}+F_{2}(x,y) \mathbf{j} \Rightarrow |\mathbf{F}(x,y)|^{2}=F_{1}(x,y)^{2}+F_{2}(x,y)^{2}$$

This appears to be just a vector function of two apparently independent variables x and y. Not a situation where you would normally even ask for dy/dx. What is the context of your question?

OK so this question is from the book called "Div, Grad, Curl, and all that: an informal text on vector calculus" by h. m. Schey. This is the the question 1-6 from the book.

The question asks to prove that the F(x,y) = F1(x,y)i + F2(x,y) is infact a solution to the differential equation dy/dx=(F2(x,y))/(F1(x,y)).

Sorry but I just realized from the small print in the ebook that the dy/dx=(F2(x,y))/(F1(x,y)) and not upside down.

shippo113 said:
F(x,y) = F1(x,y)i + F2(x,y)j, F is a vector valued function of x,y. How to find dy/dx?

The book gives the answer as

dy/dx = (F1(x,y))/(F2(x,y)). But I do not know how to get to this answer at all. Find the partial derivative is what I can do, but how to get to dy/dx when the main function is F(x,y)?
The question, as given, makes no sense. In order to find dy/dx, y must be a function of x. But just telling us that "F(x,y)= F1(x,y)i+ F2(x,y)j" does not imply that. It might well be that x and y are completely independent parameters.

Please check your book again and tell us what the problem really says.

It's always a challenge to figure out what the problem is when the poster doesn't understand it will enough to state it correctly. So here goes a wild guess. First I'm guessing that what the OP called F1(x,y) and F2(x,y) are really F1(x,y) and F2(x,y) and are alternate notations for the partials Fx(x,y) and Fy(x,y). Then I'm going to further guess that his question has something to do with the fact that if you differentiate F(x,y) = C implicitly with respect to x you get

Fx(x,y) + Fy(x,y)y' = 0

$$y' = -\frac{F_x(x,y)}{F_y(x,y)}$$

## 1. What does "dy/dx" represent in a vector valued function?

"dy/dx" represents the derivative of the vector valued function with respect to the variable x. In other words, it represents the rate of change of the function in the y-direction for a given change in the x-direction.

## 2. How do I calculate dy/dx for a vector valued function?

To calculate dy/dx for a vector valued function, you first need to find the individual derivatives of each component of the function with respect to x. Then, you can combine these derivatives to form the resulting vector representing dy/dx.

## 3. What does the dy/dx vector tell us about the function?

The dy/dx vector provides information about the direction and rate of change of the function at a specific point. The magnitude of the vector represents the slope of the tangent line to the curve at that point, while the direction of the vector represents the direction in which the function is changing.

## 4. Can I find the dy/dx vector at any point on the function?

Yes, you can calculate the dy/dx vector at any point on the function, as long as the function is differentiable at that point. This means that the function must have a well-defined tangent line at that point.

## 5. How is dy/dx of a vector valued function different from a regular derivative?

The dy/dx of a vector valued function is a vector itself, while a regular derivative is a single value. This means that the dy/dx vector contains information about both the direction and rate of change of the function, while a regular derivative only represents the slope of the function at a specific point.

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