场地面积的误差分析
某场地长
l
l
l 的值为
l
∗
=
110
l^*=110
l∗=110m,宽
d
d
d 的值为
d
∗
=
80
d^*=80
d∗=80m,知
∣
l
−
l
∗
∣
⩽
0.2
|l-l^*|\leqslant0.2
∣l−l∗∣⩽0.2m
,
∣
d
−
d
∗
∣
⩽
0.1
,|d-d^*|\leqslant0.1
,∣d−d∗∣⩽0.1m,试求面积 S=
l
d
ld
ld 的绝对误差限与相对误差限。
解:
S= l d , ∂ S ∂ l = d , ∂ S ∂ d = l ld,\frac\partial S{\partial l}=d,\frac{\partial S}{\partial d}=l ld,S∂∂l=d,∂d∂S=l,
故
ε ( S ∗ ) ≈ ∣ ( ∂ S ∂ l ) ∗ ∣ ε ( l ∗ ) + ∣ ( ∂ S ∂ d ) ∗ ∣ ε ( d ∗ ) , \varepsilon(S^*)\approx\left|\left(\frac{\partial S}{\partial l}\right)^*\right|\varepsilon(l^*)+\left|\left(\frac{\partial S}{\partial d}\right)^*\right|\varepsilon(d^*), ε(S∗)≈ (∂l∂S)∗ ε(l∗)+ (∂d∂S)∗ ε(d∗),
其中
( ∂ S ∂ l ) ∗ = d ∗ = 80 m, ( ∂ S ∂ d ) ∗ = l ∗ = 110 r \left(\frac{\partial S}{\partial l}\right)^*=d^*=80\text{ m,}\quad\left(\frac{\partial S}{\partial d}\right)^*=l^*=110\text{ r} (∂l∂S)∗=d∗=80 m,(∂d∂S)∗=l∗=110 r
而
ε ( l ∗ ) = 0.2 m , ε ( d ∗ ) = 0.1 m \varepsilon(l^{*})=0.2m, \varepsilon(d^{*})=0.1m ε(l∗)=0.2m,ε(d∗)=0.1m
于是绝对误差限为
ε ( S ∗ ) ≈ ( 80 × 0.2 + 110 × 0.1 ) m 2 = 27 m 2 \varepsilon(S^{*})\approx(80\times0.2+110\times0.1)\mathrm{~m}^{2}=27\mathrm{~m}^{2} ε(S∗)≈(80×0.2+110×0.1) m2=27 m2
相对误差限为
ε r ( S ∗ ) = ε ( S ∗ ) ∣ S ∗ ∣ = ε ( S ∗ ) l ∗ d ∗ ≈ 27 8800 = 0.31 % \varepsilon_{r}(S^{*})=\frac{\varepsilon(S^{*})}{|S^{*}|}=\frac{\varepsilon(S^{*})}{l^{*}d^{*}}\approx\frac{27}{8800}=0.31\% εr(S∗)=∣S∗∣ε(S∗)=l∗d∗ε(S∗)≈880027=0.31%