1. ## Differentials...2

Evaluate f(1, 2) and f(1.05, 2.1) and calculate Δz. Also, use the total differential dz to approximate Δz.

f(x, y) = x siny

z_x = siny

z_x = xcosy

dz = sinydx + xcosydy

To find f(1, 2) and f(1.05, 2.1), I evaluate the given function at the two points.

I am having trouble finding delta z and then using dz to approximate delta z.

The answer for delta z is given to be -0.00923 and the numerical answer for dz is given to be 0.00385.

How is this done?

2. ## Re: Differentials...2

The value for $\displaystyle \Delta z$ is $\displaystyle 1.05\sin{2.1} - 1\sin{2} = -0.00293$. This should be approximately $\displaystyle z_x\Delta x + z_y\Delta y = (\sin{2})(1.05-1) + (1\cos{2})(2.1-2) = 0.00385$.

This is an unusual example - usually the estimate is pretty good if $\displaystyle \Delta x$ and $\displaystyle \Delta y$ are "small".

- Hollywood

3. ## Re: Differentials...2

Originally Posted by hollywood
The value for $\displaystyle \Delta z$ is $\displaystyle 1.05\sin{2.1} - 1\sin{2} = -0.00293$. This should be approximately $\displaystyle z_x\Delta x + z_y\Delta y = (\sin{2})(1.05-1) + (1\cos{2})(2.1-2) = 0.00385$.

This is an unusual example - usually the estimate is pretty good if $\displaystyle \Delta x$ and $\displaystyle \Delta y$ are "small".

- Hollywood
Thanks for the input. I solved the problem after several tries.

4. ## Re: Differentials...2

f(x,y) = xsiny

Let z = f(x,y)

dz = sinydx + xcosydy

f(1,2) = 1sin(2) = 0.90930

f(1.05,2.1) = (1.05)sin(2.1) = 0.906370

Δx = 0.05

Δy = 0.1

Δz = (1.05)*sin(2.1) - sin(2)

Δz = -0.00293