Question

If z = 8x² + y² and (x, y) changes from (1, 2) to (0.95, 1.9), compare the values of Δz and dz. (Round your answers to four decimal places.)

If z = x² − xy + 5y² and (x, y) changes from (1, −1) to (1.03, −1.05),
compare the values of Δz and dz. (Round your answers to four decimal places.)

Answer 1

To compare ∆z and dz, calculate the actual change in z for the specified changes in x and y, and use partial derivatives to estimate the change for dz. Perform calculations for both given functions and compare the numerical values, rounded to four decimal places.

Step-by-step explanation:

To compare the values of ∆z and dz, we first need to understand that ∆z represents the actual change in z when the variables x and y change, while dz represents the differential, which is an approximation of the change in z using the derivative.

• The first function given is z = 8x² + y². To find ∆z, we calculate the value of z at the new point and subtract the value of z at the original point. For dz, we take the partial derivatives with respect to x and y, evaluate them at the original point, and multiply by the respective changes in x and dx, and y and dy.
  ∆z for the first function: z(0.95, 1.9) - z(1, 2)
• dz for the first function: (dz/dx)(dx) + (dz/dy)(dy)

The second function is z = x² - xy + 5y². Similarly, to find ∆z, compute the difference in the function values at the new and original point, and for dz, use the partial derivatives like before.

• ∆z for the second function: z(1.03, -1.05) - z(1, -1)
• dz for the second function: (dz/dx)(dx) + (dz/dy)(dy)

Calculations will provide the numerical comparison between ∆z and dz for each case after rounding to four decimal places.