We know that if y = f(x) then
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{dy\over
dx} = {\mathop{lim}}_{δx→0}{f(x + δx) − f(x)\over
δx} \quad .
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Provided that δx is small enough (but not infinitesimally small) {dy\over dx} ≈ {δy\over δx}, so
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\class{boxed}{ δy ≈ {dy\over
dx}δx\quad . }
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Example 4.23:
We can measure the volume of a sphere by measuring its radius r and then use the formula, V = (4∕3)π{r}^{3}. Suppose we measure r = 6.3 ± 0.02\text{ m}. Find the approximate error in V .
Solution:
If r = 6.3\text{ m} then V = {4\over 3}π6.3{0}^{3} = 1047.4{\text{ m}}^{3}. The small error δr = 0.02\text{ m} will cause an error in V given by δV ≈ {dV \over dr} δr = 4π{r}^{2}δr = 4π\kern 1.66702pt 6.3{0}^{2}\kern 1.66702pt 0.02 = 10.0{\text{ m}}^{3}. Hence
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\class{boxed}{V = 1047.4 ± 10.0{\text{ m}}^{3}. }
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Example 4.24:
We measure the height h of a tower at a distance d, by measuring d and the angle α with the horizontal. We then use the formula \mathop{tan}\nolimits α = (h∕d).
Solution:
Find error in h due to an error δα in α assuming d to be known exactly. We solve for h, h = d\mathop{tan}\nolimits α, dh∕dα = d{\mathop{sec}\nolimits }^{2}α. Therefore
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\class{boxed}{δh ≈ {dh\over
dα}δα = d{\mathop{sec}\nolimits }^{2}αδα\quad . }
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Example 4.25:
Given the relation between current, voltage and resistance, I = V∕R, with V = 250\text{ V}, R = 50\ Ω, find the change in the current I 1) if V increases by 1\text{ V}, and 2) if R increases by 1\ Ω.
Solution:
We use the rule for small changes for partial derivatives,
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δI ≈ {∂I\over
∂V }δV + {∂I\over
∂R}δR\kern 1.66702pt .
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We find
Using the numerical values, we find
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δI = {1\over
50} × 1 −{250\over
5{0}^{2}} × 1 = {1\over
50} − {1\over
10} = −{2\over
25} = −0.08\text{ A}
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