Differentials — Question 3

Question 3

For $f(x,y)=x^2+y^2$, start at $(3,4)$ and use $dx=0.1$, $dy=-0.2$.

Tasks

1. Compute $df$.

2. Compute the exact change $\Delta f$.

3. Derive the discrepancy $\Delta f-df$ and explain its sign.

Question 3 – Solution

Strategy. Expand the exact finite change so the first-order and second-order parts separate visibly.

Step 1: Differential $$df=2x\,dx+2y\,dy=6(0.1)+8(-0.2)=\boxed{-1.0}.$$

Step 2: Exact change $$\Delta f=(3.1)^2+(3.8)^2-(3^2+4^2)=9.61+14.44-25=\boxed{-0.95}.$$

Step 3: Discrepancy Algebraically, $$\Delta f=2x\,dx+2y\,dy+(dx)^2+(dy)^2 =df+(dx)^2+(dy)^2.$$ Therefore $$\boxed{\Delta f-df=0.01+0.04=0.05}.$$ It is nonnegative because the omitted terms are squares; the upward curvature makes the exact change exceed the linear prediction.