Respuesta :
Answer:
Maximum error for viscosity is 17.14%
Step-by-step explanation:
We know that everything is changing with respect to the time, "r" is changing with respect to the time, and also "p" just "v" will not change with the time according to the information given, so we can find the implicit derivative with respect to the time, and since
[tex]n = (\frac{\pi}{8}) (\frac{pr^4}{v})\\[/tex]
The implicit derivative with respect to the time would be
[tex]\frac{dn}{dt} = \frac{\pi}{8} ( \frac{r^4}{v} \frac{dp}{dt} + \frac{4pr^3}{v}\frac{dr}{dt} )[/tex]
If we multiply everything by dt we get
[tex]dn = \frac{\pi}{8} ( \frac{r^4}{v} dp + \frac{4pr^3}{v} dr})[/tex]
Remember that the error is given by [tex]\frac{dn}{n}[/tex] therefore doing some algebra we get that
[tex]\frac{dn}{n} = 4 \frac{dr}{r} + \frac{dp}{p}[/tex]
Since, r = 0.006 , dr = 0.00025 , p = 4*105 , dp = 2000 we get that
[tex]\frac{dn}{n} = 0.1714[/tex]
Which means that the maximum error for viscosity is 17.14%.
The estimate of the maximum error in the viscosity η using differentials is given by; 17.17%
We are given the viscosity formula as;
η = (π/8)(pr⁴/v)
Thus;
dη/dr = (π/8)(4pr³/v)
and dη/dp = (π/8)(r⁴/v)
By chain rule, the maximum error of viscocity η is approximated by;
(dη)/η = [(dη/dr)dr]/η + [(dη/dp)dp]/η
Plugging in the relevant values, we have;
(dη)/η = [(π/8)(4pr³/v)]*dr/[(π/8)(pr⁴/v)] + [(π/8)(r⁴/v)]*dp/[(π/8)(pr⁴/v)]
Simplifying gives;
(dη)/η = (4/r)dr + dp/p
From the question, we see that;
r = 0.006
and dr = 0.00025
Also; p = 4 × 10⁵
dp = 2000
Thus, plugging in the relevant values into (dη)/η = (4/r)dr + dp/p, we have;
(dη)/η = (4/0.006)*0.00025 + (2000/4 × 10⁵)
(dη)/η = 0.1667 + 0.005
(dη)/η = 0.1717 = 17.17%
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