Help understanding equation involving a partial derivative

[Woolyabyss]

Mod note: Moved from a homework section
1. Homework Statement
N/A

Homework Equations

f(x + Δx,y) = f(x,y) + ∂f(x,y)/∂x*Δx

The Attempt at a Solution

Sorry this isn't really homework. We were given this equation today in order to derive the Taylor expansion formula in two variables and I'm not sure where it came from.It seems similar to the mean value theorem.

any help would be appreciated

 

[Mark44]

Mod note: Moved from a homework section
1. Homework Statement
N/A

Homework Equations

f(x + Δx,y) = f(x,y) + ∂f(x,y)/∂x*Δx

The Attempt at a Solution

Sorry this isn't really homework. We were given this equation today in order to derive the Taylor expansion formula in two variables and I'm not sure where it came from.It seems similar to the mean value theorem.

any help would be appreciated

It's helpful to sketch the surface z = f(x, y), showing the points $(x_0, y_0, f(x_0, y_0))$ and $(x_0 + \Delta x, y_0, f(x_0 + \Delta x, y_0))$. The formula above should really be $f(x_0 + \Delta a, y_0) \approx f(x_0, y_0) + \frac{\partial f(x, y)}{\partial x}|_{(x_0, y_0)} \Delta x$, since the right side is only an approximation to the left side.

What's happening here is that the expression on the right side gives the approximate function value using the line that is tangent to the surface z = f(x,y) at $(x_0, y_0, z_0)$ (with $z_0 = f(x_0, y_0)$), along a direction parallel to the x-axis. The value that is produced might be smaller than the actual function value on the surface, or it might be larger, depending on whether the surface is concave up or concave down, respectively, near the point $(x_0, y_0, z_0)$.

 

[Woolyabyss]

It's helpful to sketch the surface z = f(x, y), showing the points $(x_0, y_0, f(x_0, y_0))$ and $(x_0 + \Delta x, y_0, f(x_0 + \Delta x, y_0))$. The formula above should really be $f(x_0 + \Delta a, y_0) \approx f(x_0, y_0) + \frac{\partial f(x, y)}{\partial x}|_{(x_0, y_0)} \Delta x$, since the right side is only an approximation to the left side.

What's happening here is that the expression on the right side gives the approximate function value using the line that is tangent to the surface z = f(x,y) at $(x_0, y_0, z_0)$ (with $z_0 = f(x_0, y_0)$), along a direction parallel to the x-axis. The value that is produced might be smaller than the actual function value on the surface, or it might be larger, depending on whether the surface is concave up or concave down, respectively, near the point $(x_0, y_0, z_0)$.

Thanks for the reply. I'm still unsure about how the line to the tangent of the surface can be used to approximate the function in the example given.

 

[Mark44]

Thanks for the reply. I'm still unsure about how the line to the tangent of the surface can be used to approximate the function in the example given.

In pretty much the same way that $f(x_0 + \Delta x)$ can be approximated by $f(x_0) + f'(x_0) \Delta x$. It's really the same idea. Here's an example using a function of one variable.
Let $f(x) = \sqrt{x}$. Approximate $\sqrt{4.1}$
Here x₀ = 4, f(x₀) = 2, and $\Delta x = .1$
$\sqrt{4.1} = f(x_0 + \Delta x) \approx f(x_0) + f'(x_0) \Delta x = 2 + \frac{1}{2\sqrt{4}} .1 = 2.025$
So $\sqrt{4.1} \approx 2.025$.
Compare this answer with what I get from a calculator, approximately 2.02484567.

My answer overestimates the actual answer because the tangent line approximation gives me a number that is above the curve $y = \sqrt{x}$.

 

[epenguin]

Mod note: Moved from a homework section
1. Homework Statement
N/A

Homework Equations

f(x + Δx,y) = f(x,y) + ∂f(x,y)/∂x*Δx

The Attempt at a Solution

Sorry this isn't really homework. We were given this equation today in order to derive the Taylor expansion formula in two variables and I'm not sure where it came from.It seems similar to the mean value theorem.

any help would be appreciated

Last term is Δf (lim as → 0 bla bla). Take first RHS term to LHS and you see nothing but your first ever formula in beginning calculus.

They're assuming here only x varies and they must soon get to x and y both vary in which case they'd have to add a + ∂f/∂y*Δy