Differentials — Question 9

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Question 9

A differential field is proposed as df=(y2+2x)dx+(2xy+ey)dy,f(0,0)=5.df=(y^2+2x)\,dx+(2xy+e^y)\,dy,\qquad f(0,0)=5. Tasks

  1. Reconstruct ff.

  2. Use dfdf to estimate f(0.02,−0.01)f(0.02,-0.01) from the origin.

  3. Compare the estimate with the exact value through second order.

Original worksheet page 1: question and worked solution for 2-5-009
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Question 9 – Solution

Strategy. Reconstruct the function first, then separate its linear and nonlinear increment terms.

Step 1: Reconstruct Integrating fx=y2+2xf_x=y^2+2x gives f=xy2+x2+C(y).f=xy^2+x^2+C(y). Since fy=2xy+C′(y)=2xy+eyf_y=2xy+C'(y)=2xy+e^y, C=ey+C0C=e^y+C_0. The initial value gives 1+C0=51+C_0=5, so f=xy2+x2+ey+4.\boxed{f=xy^2+x^2+e^y+4}.

Step 2: Differential estimate At (0,0)(0,0), fx=0,fy=1.f_x=0,\qquad f_y=1. For dx=0.02dx=0.02, dy=−0.01dy=-0.01, f(0.02,−0.01)≈5+df=5−0.01=4.99.f(0.02,-0.01)\approx 5+df=5-0.01=\boxed{4.99}.

Step 3: Exact comparison Exactly, f=0.02(0.01)2+(0.02)2+e−0.01+4≈4.99045183.f=0.02(0.01)^2+(0.02)^2+e^{-0.01}+4\approx 4.99045183. The estimate error is about 0.00045183\boxed{0.00045183}, while the second-order correction is dx2+dy2/2=0.00045dx^2+dy^2/2=0.00045. Thus the second-order estimate is 4.990454.99045, consistent with the exact value.

Original worksheet page 2: question and worked solution for 2-5-009

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